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The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4, Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on, 20.06.2019 and it has been decided to implement it from the educational year 2019-20., , Mathematics and Statistics, (Arts and Science), Part - 2, STANDARD - XI, , Maharashtra State Bureau of Textbook Production and Curriculum Research,, Pune - 411 004, Download DIKSHA App on your smartphone. If you scan the Q.R.Code, on this page of your textbook, you will be able to access full text and the, audio-visual tudy material relevant to each lesson provided as teaching, and learning aid.

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PREFACE, Dear Students,, Welcome to the eleventh standard!, You have successfully completed your secondary education and have entered the, higher secondary level. You will now need to learn certain mathematical concepts and acquire, some statistical skills to add more applicability to your work. Maharashtra State Bureau of, Textbook Production and Curriculum Research has modified and restructured the curriculum in, Mathematics in accordance with changing needs., The curriculum of Mathematics is divided in two parts. Part 1 covers topics in Trignometry,, Algebra, Co-ordinate Geometry and Statistics. Part 2 covers Complex Numbers, Sets and, Relations, Calculus and Combinatorics. There is a special emphasis on applications. Activities, are added in chapters for creative thinking. Some material will be made available on E-balbharati, website (ebalbharati.in). It contains a list of specimen practical problems on each chapter. Students, should complete the practical exercises under the guidance of their teachers. Maintain a journal, and take teacher’s signature on every completed practical., You are encouraged to use modern technology in your studies. Explore the Internet for, more recent information on topics in the curriculum. Get more examples and practice problems, from the Internet. You will enjoy learning if you study the textbook thoroughly and manage to, solve problems., On the title page Q.R. code is given. It will help you to get more knowledge and clarity, about the contents., This textbook is prepared by Mathematics Subject Commitee and members of study, group. This book has also been reviewed by senior teachers and subject experts. The Bureau, is grateful to all of them and would like to thank for their contribution in the form of creative, writing, constructive criticism, and valuable suggestions in making this book useful to you and, helpful to your teachers., spirit., , The Bureau hopes that the textbook will be received well by all stakeholders in the right, You are now ready to study. Best wishes for a happy learning experience., , Pune, Date : 20 June 2019, Indian Solar Date : 30 Jyeshtha 1941, , (Dr. Sunil Magar), Director, Maharashtra State Bureau of Textbook, Production and Curriculum Research, Pune.

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Mathematics and Statistics Std XI (Part II), Arts and Science, Sr., No, , Area, , 1, , Complex, Numbers, , 2, , Sequences and, Series, , 3, , Permutations, and, combinations, , 4, , Method of, Induction and, Binomial, theorem, , 5, , Topic, , Complex, Numbers, , Sequences, and Series, , Permutations,, Combinations, , Method of, Induction, , Binomial, Theorem, , Sets, Sets and, relations, Relations, , Competency Statement, The students will be able to • understand set of complex numbers and, different ways of expressing complex, numbers., • perform algebraic operations on complex, numbers., • simplify algebraic expressions involving, complex numbers., • Revise AP, learn GP and HP., • Find the general term and the sum of the first n, terms of these sequences., • count the number of arrangements of given, objects satisfying specific conditions., • count the number of possible selections of, objects with certain conditions., • understand the method of induction and apply, it to verify mathematical statements., • expand binomial expressions and find its, general term., • simplify the binomial expression for negative, index or fractional power., • work with sets and operations on sets., • construct sets from given conditions., • solve problems on applications of set, theory., • identify the types of relations., • study equivalence relations.

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6, , Functions, , Functions, , • work with function defined on different, domains., • identify different types of functions., • carry out algebraic operations on functions., , 7, , Limits, , Limits, , • understand the concept of limit of a function., • determine the limits of functions if they exist., , 8, , Continuity, , Continuity, , • Define and study the continuity of a function, at a point and in an interval., , 9, , Differentiation, , Differentiation, , • understand and study the differentiability of a, function., • understand and study differentiation of various, functions.

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INDEX, Sr. No., , Chapter, , Page No., , 1, , Complex Numbers, , 1, , 2, , Sequences and Series, , 23, , 3, , Permutations and Combination, , 43, , 4, , Methods of Induction and Binomial Theorem, , 69, , 5, , Sets and Relations, , 87, , 6, , Functions, , 106, , 7, , Limits, , 133, , 8, , Continuity, , 160, , 9, , Differentiation, , 179, , Answers, , 196

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1, , COMPLEX NUMBERS, 1.1 A Complex number :, , Let's Study, , 1.1(a) Imaginary Number :, , •, , A complex number (C.N.), , •, , Algebra of C.N., , •, , Geometrical Representation of C.N., , •, , Polar & Exponential form of C.N., , Ex :, , •, , De Moivre's Theorem., , Note:, , A number of the form bi, where b∈ R, b ≠ 0,, i=, , −1 is called an imaginary number., −25 = 5i, 2i,, , 2, i, − 11i etc., 7, , The number i satisfies following properties,, Let's Recall, •, , Algebra of real numbers., , •, , Solution of linear and quadratic equations, , •, , Representation of a real number on the number, line, , •, , Representation of point in a plane, , •, , Trigonometric ratios, , i), , i×0=0, , ii), , If a∈R, then, , −a 2 =, , i 2 a 2 = ± ia, , iii) If a, b∈R, and ai = bi then a = b, 1.1 (b) Complex Number :, Definition : A number of the form a+ib, where, a, b∈ R and i = −1 with i2 = −1 is called a, complex number and is usually denoted by z., That is z = a+ib, a, b∈R and i = −1, , Introduction:, , Here 'a' is called the real part of z and is denoted, by Re(z) or R(z). 'b' is called imaginary part of z, and is denoted by Im(z) or I(z), , Consider, the equation x2 + 1 = 0. This, equation has no solution in the set of real numbers, because there is no real number whose square is, −1. We need to extend the set of real numbers to a, larger set, which would include solutions of such, equations., , The set of complex numbers is denoted by C, ∴ C = {a+ib / a, b∈ R, and i = −1 }, Ex :, , We introduce a symbol i (greek letter iota), such that i = −1 and i = −1. i is called as an, imaginary unit or an imaginary number., 2, , Swiss mathematician Leonard Euler, (1707-1783) was the first mathematician to, introduce the symbol i with i =, , −1 and i2 = −1., , 1, , z, 2+4i, 5i, 3−4i, , a+ib, 2+4i, 0+5i, 3−4i, , Re(z), , Im(z), 4, 5, −4, , 2, 0, 3, , 5+ −16, , 5+4i, , 5, , 4, , 2+ 5 i, , 2+ 5 i, , 2, , 5i, , 7+ 3, , (7+ 3 )+0i, , (7+ 3 ), , 0

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Note :, , 1.2.2, , 1), , A complex number whose real part is zero, is called a purely imaginary number. Such a, number is of the form z = 0 + ib = ib, , 2), , A complex number whose imaginary part is, zero is a real number., , Definition : The conjugate of a complex, number z = a + ib is defined as a − ib and is, denoted by z, Ex : 1), z, 3 + 4i, 7i −2, 3, 5i, , z = a + 0i = a, is a real number., 3), , A complex number whose both real and, imaginary parts are zero is the zero complex, number. 0 = 0 + 0i, , 4), , The set R of real numbers is a subset of the, set C of complex numbers., , 5), , Conjugate of a Complex Number:, , The real part and imaginary part cannot be, combined to form single term. e.g. 2 + 3i ≠ 5i, , z, 3 − 4i, −7i −2, 3, −5i, , 2+, , 3, , 2+, , 3, , 7+, , 5i, , 7−, , 5i, , 2), , Properties of z, , 1.2 Algebra of Complex Numbers :, , 1), 2), , ( z) = z, If z = z , then z is purely real., , 1.2.1 Equality of two Complex Numbers :, , 3), , If z = −z , then z is purely imaginary., , Now we define the four fundamental, operations of addition, subtraction, multiplication, and division of complex numbers., , Definition : Two complex numbers z1 = a+ib, and z2 = c + id are said to be equal if their, corresponding real and imaginary parts are, equal., i.e. a + ib = c + id, , if a = c, , 1.2.3 Addition of complex numbers :, Let z1 = a+ib, , and b = d, , and, , z2 = c+id, , then z1+ z2 = (a+ib) + (c+id), , Ex. : i) If x + iy = 4 + 3i then x = 4 and y = 3, , , , Ex. : ii) If 7a + i (3a − b) = 21 − 3i then find a, and b., , = (a+c) + (b+d) i, , In other words, Re(z1+ z2) = Re(z1) + Re(z2), and Im(z1+ z2) = Im(z1) + Im(z2), , Solution : 7a + (3a − b) i = 21 − 3i, By equality of complex numbers, , Ex. 1) (2 + 3i) + (4 + 3i) = (2+4) + (3+3)i, , 7a = 21, , , , ∴a=3, , = 6 + 6i, , 2) (−2 + 5i) + (7 + 3i) + (6 − 4i), , and 3a − b = −3 ∴ 3(3) + 3 = b, , = [(−2) +7+6] + [5 + 3 + (− 4)]i, , ∴ 12 = b, , = 11 + 4i, , Note : The order relation (inequality) of complex, number can not be defined. Hence, there does not, exist a smaller or greater complex number than, given complex number. We cannot say i < 4., , Properties of addition : If z1, z2, z3 are complex, numbers then, i) z1+ z2 = z2+ z1 (commutative), ii) z1 + (z2+ z3) = (z1+ z2) + z3 (associative), 2

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1.4 Fundamental Theorem of Algebra :, 'A polynomial equation with real coefficients, has at least one root' in C., or 'A polynomial equation with complex, coefficients and of degree n has n complex roots'., , Ex. 2 : Solve x2−(2 3 + 3i) x + 6 3 i = 0, Solution : Given equation is, x2−(2 3 + 3i) x + 6 3 i = 0, The method of finding the roots of, ax2 + bx + c = 0, is applicable even if a, b, c, , 1.4.1 Solution of a Quadratic Equation in, complex number system :, Let the given equation be ax2 + bx + c = 0, where a, b, c∈R and a ≠ 0, ∴, , −b ± b 2 − 4ac, are complex numbers. where x =, 2a, Here, a = 1, b = −(2 3 + 3i), c = 6 3 i, , The solution of this quadratic equation is, given by, , b2 − 4ac = [−(2 3 + 3i)]2 − 4 × 1 × 6 3 i, = 12 − 9 + 12 3 i − 24 3 i, , −b ± b 2 − 4ac, x =, 2a, Hence, the roots of the equation ax2 + bx + c = 0, , = 3 − 12 3 i, = 3 (1 − 4 3 i), , −b − b 2 − 4ac, −b + b 2 − 4ac, are, and, 2a, 2a, The expression (b2−4ac) = D is called the, discriminant., , So, the given equation has complex roots. These, roots are given by, x =, , If D < 0 then the roots of the given quadratic, equation are complex., , ∴ a2 − b2 + 2iab = 1− 4 3 i, ∴ a2 − b2 = 1 and 2ab = −4 3, a2 − b2 = 1 and ab = −2 3, , Ex. 1 : Solve x2 + x + 1 = 0, Solution : Given equation is x2 + x + 1 = 0, Comparing with ax2 + bx + c = 0 we get, , Consider (a2 + b2)2 = (a2 − b2)2 + 4 a2b2, , , a = 1, b = 1, c = 1, These roots are given by, , ∴ a2 + b2, , = 1 + 4 (12) = 49, = 7 ...(1), , and also a2 − b2 = 1 ...(2), , −b ± b − 4ac, 2a, 2, , Solving (1) and (2), 2a2 = 8, , −1 ± −3, =, 2, , ∴ a2 = 4, , ∴ a = ±2 and b = ± 3, ∴ We have four choices, , −1 ± 3 i, 2, , ∴ Roots are, , 2, , Let a + ib = 1 − 4 3 i, , Solved Examples :, , =, , ), , 3 + 3i ± 3(1 − 4 3 i ), , Now, we shall find 1 − 4 3 i, , Note : If p + iq is the root of equation ax2 + bx + c, = 0 where a, b, c∈R and a ≠ 0 then p − iq is also, a solution of the given equation. Thus, complex, roots occur in conjugate pairs., , x =, , (2, , −1 + 3 i, −1 − 3 i, and, 2, 2, , 8, , a = 2,, , b= 3, , ... (3), , a = 2,, , b=− 3, , ... (4)

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2), , Solve the following quadratic equations., i), , is called the real axis. Similarly, y = Im(z) is, represented on the Y-axis, so the Y-axis is called, the imaginary axis., , 8x2 + 2x + 1 = 0, , ii) 2x2 − 3 x + 1 = 0, iii) 3x2 − 7x + 5 = 0, iv) x2− 4x + 13 = 0 , 3), , Solve the following quadratic equations., i), , x2 + 3ix + 10 = 0, , ii) 2x2 + 3ix + 2 = 0, iii) x2 + 4ix − 4 = 0, iv) ix2− 4x − 4i = 0, 4), , Solve the following quadratic equations., i), , Fig. 1.2, , x2 − (2+i) x −(1−7i) = 0, , ii) x − (3 2 +2i) x + 6 2 i = 0, 2, , iii) x2 − (5−i) x + (18+i) = 0, iv) (2 + i)x2− (5−i) x +2 (1−i) = 0, 5), , e.g. (1) (1, 2) ≡ 1 + 2i, , (2) −4+3i ≡ (−4, 3), , (3) (0, 0) ≡ 0 + 0i, , (4) 5+0i ≡ (5, 0), , (5) (0, −1) ≡ 0 − i, , (6) −2−2i ≡ (−2,−2), , A diagram which represents complex, numbers by points in a plane with reference to, the real and imaginary axes is called Argand's, diagram on complex plane., , Find the value of, iii) x3 − x2+x+46, if x = 2+3i., , 25, iv) 2x3 − 11x2+44x+27, if x = 3−4i ., 5, v) x3 + x2−x+22, if x = 1−2i ., , 1.5.1 Modulus of z :, If z = a+ib is a complex number then the, modulus of z, denoted by | z | or r, is defined as, | z | = a 2 + b 2 . (From fig. 1.3), point P(a, b), represents the complex number z = a+ib., , vi) x4 +9 x3+35x2−x+4, if x = −5+ −4 ., vii) 2x4 + 5x3+7x2−x+41, if x = −2− 3 i., , ∴r=|z|=, , 1.5 Argand Diagram or Complex Plane :, A complex number z = x + iy, x, y∈R and, i = −1 is expressed as a point in the plane whose, co-ordinates are ordered pair (x, y). Jean Robert, Argand used the one to one correspondence, between a complex number and the points in a, the plane., , a 2 + b 2 = OP, , Hence, modulus of z is the distance of, point P from the origin where the point P, represents the complex number z in the plane., e.g. For, z = 4 + 3i,, Modulus of z = | z | = 16 + 9 = 25 = 5, 1.5.2 Argument of z :, , Let z = x + iy be a complex number., , OP makes an angle q with positive direction, of X-axis. q is called the argument or amplitude, of the complex number z = a+ib, denoted by, arg(z)., , Then the point P(x, y) represents the complex, number z = x + iy (fig.1.2) i.e. x + iy ≡ (x, y),, x = Re(z) is represented on the X-axis. So, X-axis, 10

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b, ∴ q = tan−1 =   = arg(z),, a, e.g. If z = 2+2i then, , 2, arg(z) = q = tan−1  , 2, p, ∴ tan−1(1) =, 4, Note : If tan x = y then its inverse funtion is given, by x = tan−1 y or x = arc tan y, eg:, p, 1,  1  p, 1) As tan, =, then tan−1 = , =, 3, 6,  3 6, , Fig. 1.3, b, a, ∴ sinq = r , cosq = r , r ≠ 0, , p,  π, 2) As tan  −  = − tan, = −1 then, 4,  4, , ∴ b = rsinq , a = rcosq, b, and tanq = a , if a ≠ 0, , tan−1 (−1) = −, , π, 4, , 1.5.3 Argument of z in different quadrants/axes :, , z = a + ib, , Example, , Quadrant/Axis, , θ = arg z, b, = tan−1   ,, a, , from Example, , (0 ≤ θ < 2π), a > 0, b = 0, , a > 0, b > 0, , a = 0, b < 0, , z=3, , z=1+i, , z = 5i, , On positive real, (X) axis, , θ=0, , θ=0, , In quadrant I, , b, θ = tan−1   ,, a, p, (0 < θ < ), 2, , 1 p, θ = tan−1   =, 4, 1, , On Positive, imaginary (Y) axis, , θ=, , p, 2, , a < 0, b > 0, , z=− 3 +i, , In quadrant II, , b, θ = tan−1   + π, a, p, ( < θ < π), 2, , a < 0, b = 0, , z = −6, , On negative real, (X) axis, , θ=π, , 11, ,  , , θ=, , p, 2, ,  1 , θ = tan−1 ,  +π, − 3, 5p, −π, =, +π=, ), 6, 6, θ=π

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We call the origin as pole. (figure 1.4), −z, , z, , −z, , z, , Fig.1.4, , Fig. 1.5, , As a = rcosq, b = rsinq, , Thus, q together with r gives the position of, the point A in the Argand's diagram., Hence, from the figure 1.5,, , ∴ z = a + ib becomes, z = rcosq + irsinq, ∴ z = r(cosq + isinq), , arg z = 45° =, , This is called polar form of complex number, z = a + ib, , p, 4, , , arg (− z ) = 135° =, , arg (−z) = 225° =, , 1.5.5 Exponential form :, , 5p, 4, , 3p, 4, , , arg z = 315° =, , 7p, 4, , Ex. 2 : Represent the following complex numbers, in the polar form and in the exponential form, , It is known and can be proved using special, series that eiq = cosq + isinq, , i) 4+4 3 i, , ∴ z = a + ib = r (cosq + isinq) = r eiq, , ii) −2, , iii) 3i, , iv) − 3 + i, , Solution :, , where r = | z | and q = arg z is called an, exponential form of complex number., , i), , Let, z = 4 + 4 3 i, a = 4, b = 4 3, , Solved Example:, Ex. 1 : Represent the complex numbers, , r =, , z = 1+i, z = 1−i, − z = −1+i, −z = −1−i in, Argand's diagram and hence find their arguments, from the figure., , As q lies in quadrant I, , 42 + (4 3 ) 2 = 16 + 48 = 64 = 8, , 4 3, b, , q = tan−1   = tan−1 , a,  4 , , Solution :, arg z is the angle made by the segment OA, with the positive direction of the X-axis. (Fig.1.5), , = tan−1( 3 ) =, , 13, , p, 3, , or 60°

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∴ The polar form of z = 4 + 4 3 i is, , π,  1 , = tan−1 ,  +π= −6 +π, − 3, 5π, =, 6, , , z = r (cosθ + i sinθ), z = 8 (cos 60° + i sin 60°), = 8 (cos, , π, π, + i sin ), 3, 3, , ∴ The exponential form of z = 8 e, , ∴ The polar form of, z = r (cosθ + i sinθ), 5π, 5π, z = 2 (cos 6 + i sin, ), 6, , π , i , 3, , ∴ The exponential form of, , ii) Let z = −2, ∴ a = −2, b = 0, Hence, r =, , z = re = 2 e, iθ, , (−2) 2 + 02 =, , 4=2, , Solution: z =, , 3π, 3π, + i sin, ), 4, 4, By using allied angles results in trigonometry, we, get, π, , 1, 3π, π, cos, = cos  π −  = − cos, =−, 4, 4, 4, , 2, π, , 1, 3π, π, sin, = sin  π −  = sin, =, 4, 4, 4, , 2, =, , 02 + 32 = 3, , As point (0, 3) lies on positive imaginary axis, πc, θ = 2 or 90°, ∴ The polar form of z = 3(cos90° + i sin90°), π, π, , = 3 (cos 2 + i sin 2 ), , As (−, , 2.e(cos, ,  −1  1  , +i, ∴ z = 2.e, ,  2  2 , , i π, 2, , = −1 + i, 5π i, , iv) Let, z = − 3 + i, , π i, , Ex. 4 : Express (i) 3.e 12 × 4.e 12, , b=1, , π, π , , 2  cos + i sin , 12, 12 , , ii), in a + ib form, 5π, 5π , , 2  cos, + i sin, , 6, 6 , , , (− 3 ) + 1 = 3 + 1 = 4 = 2, 2, , = r eiθ, , z = r (cosθ + i sinθ), , iii) Let z = 3i, , r =, , 3π i, , 2.e, .e 4, , in the a + ib form., , ∴r=, , ∴ The exponential form of z = 2eiπ, , ∴a=− 3,, , 2.e, .e 4, , 3π, 2.e, ,θ= 4, As the polar form of z is, , ∴ The polar form of z = 2(cos180°+i sin180°), , = 2 (cos π + isin π), , ∴ The exponential form of z = 3e, , 3π i, , Ex. 3 : Express z =, , As point (−2, 0) lies on negative real axis, θ = πc or 180°, , a=0, b=3, Hence, r =, ,  5π , i, ,  6 , , 2, , 3 , 1) lies in quadrant II, , b, q = tan−1   = π, a, 14

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5π i, , Solution: (i), , π i, , 3) If z = 3 + 5i then represent the z , z , − z , − z in, Argand's diagram., , 3.e 12 × 4.e 12, ,  5π π ,  + i, 12 , , = (3 × 4) e 12, = 12 e, , 6π i, 12, , 4) Express the following complex numbers in, polar form and exponential form., 1, i) −1+ 3 i ii) −i, iii) −1 iv) 1+i, 1+ 7i, 1+ 2i, v), vi), (2−i)2, 1−3i, , π i, 2, , = 12 e, π, π, , = 12  cos + i sin , 2, 2, , = 12 (0 + i) = 12i, , (ii), , 5) Express the following numbers in the form, x+iy, π, π, i), 3 (cos 6 + isin 6 ), , π, π , π, , i, 2  cos + i sin , 12, 2.e, e, 12, 12 , , =, 5π, i, 5π, 5π , , 2e 6, 2  cos, + i sin, , 6, 6 , , , ii), ,  2   12π − 56π  i  2   − 34π  i, =, =, e, e,  2 ,  2 ,  2    3π,  cos  −, = , 2,   4, , , 7π, 7π, 2.e(cos 4 + isin 4 ), , ,  5π ,  5π, iii) 7  cos  −,  + i sin  −,  6 ,  6, , , ,  3π  ,  + i sin  −  , ,  4 , , π i, , iv) e 3, ,  2  1,  1 , +i−,   −, = , , 2 ,  2  2 , 1 i, = − −, 2 2, , -4π, , v) e 3, , i, , , , , , 5π i, , vi) e 6, , 6) Find the modulus and argument of the, 1+ 2i, complex number, ., 1−3i, 7) Convert the complex number, z=, , EXERCISE 1.3, , ii), , iv) −3(1−i), , v) −4−4i , , vii) 3, , viii) 1 + i, , i) ( z ) = z , , i iii) −8 + 15i, 3 + 2.e, vi), , π, π, cos 3 + isin 3, , in the polar form., , 8) For z = 2+3i verify the following :, , 1) Find the modulus and amplitude for each of, the following complex numbers., i) 7 − 5i, , i−1, , ii) z z = |z|2, , iii) (z+ z ) is real iv) z − z = 6i, , 3 −i, , 9), , iv) 1 + i 3, , z1 = 1 + i, z2 = 2 − 3i. Verify the following :, z 1+ z 2, , = z1 + z2, , ii) z1− z2, , = z1 − z2, , iii) z1 . z2, , = z1 . z2, , i), , x) (1+2i)2 (1−i), 4 + 3i sin θ , 2) Find real values of q for which , ,  1 − 2i sin θ , is purely real., , iv), , 15, , z1, z2, , z1, = z, 2

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θ = tan−1, , , 1, [(−1)2+2×(−1)×i 3 +(i 3 )2], 4, 1, =, (1−2i 3 −3), 4, 1, =, (−2−2i 3 ), 4, , r = 11  2 ,, x, y, , = tan−1 1 =, , =, , π, 4, , ∴ z = r(cosθ + isinθ) =, ∴ z4 = (1 + i)4 =, , , , , 2 cos  i sin , 4, 4, , , , , , 2 cos  i sin , 4, 4, , , =, , 4, 4 , , 2 cos,  i sin, 4, 4 , , , =, , 2  cos   i sin  , , =, , 2  1  i(0) , , =, , 2  1   2 = − 2 + 0i, , =, , 4, ,  1  i 3 , Similarly it can be verified that , , 2, , , 1  i 3, =, 2, , 1  i 3  1  i 3 , , , , 2, 2, , , ,  1  i 3 ,  1  i 3 , ,  and w2 = , where w = , 2, 2, , , , , , x3 = 1, , 2p i, , 4p i, , Also note that 1 = e2pi, w = e 3 , w2 = e 3, , ∴ (x −1)(x2 + x +1) = 0, , Properties of 1, w, w2, i) w is complex cube root of 1., ∴ w3 = 1, , ∴ x −1 = 0 or x2 + x +1 = 0, 1  (1) 2  4 11, ∴ x = 1 or x =, 2 1, , ii) w3 −1 = 0, i.e. (w −1) (w2 +w+1) = 0, ∴ w =1 or w2 +w+1 = 0, but w ≠1, ∴ w2 +w+1 = 0, , 1  3, x = 1 or x =, 2, , 1  i 3, 2, ∴ Cube roots of unity are, x = 1 or x =, , 1, 1, and w2 = w, w, 3, iv) w = 1 so w3n = 1, , 1  i 3 −1 − i 3, ,, 2, 2, Among the three cube roots of unity, one is, real and other two roots are complex conjugates, of each other., , iii) w2 =, , 1,, ,  1  i 3 , = , , 2, , , , 2, ,  1  i 3 ,  = w2, , 2, , , , Hence, cube roots of unity are 1, w, w2, , ∴ x3 −1 = 0, , ∴, , 2, , 1  i 3, Let, = w, then, 2, , Number 1 is often called unity. Let x be the, cube root of unity i.e. 1, , ∴, , 2, , Thus cube roots of unity are 1,, , 1.7 Cube roots of unity :, , ∴, , −1−i 3, 2, , v), , w4 = w3 .w = w so w3n+1 = w, , vi) w5 = w2 .w3 = w2.1 = w2 So w3n+2 = w2, , 2, , vii) w = w2, viii) w 2 = w, 17, , 2

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1.8 Set of points in complex plane, , Illustration:, , If z = x + iy represents the variable point, P(x,y) and z1 = x1 + iy1, represents the fixed point, A (x1,y1) then (i) z−z1 represents the length of, AP, , For z1 = 2 +3i , z2 = 1 +i and z = x + iy, (i), , z − z1 = ( x + iy ) − (2 + 3i ) = x + iy − 2 − 3i, 2, 2, = ( x − 2) + i ( y − 3) = ( x − 2) + ( y − 3), , represents the distance between (x,y) and, (2,3)., (ii) If z − z1 = 5 , then, , Fig. 1.6, , ( x + iy ) − (−1 + i ) = ( x + 1) + i ( y − 1) = 5, , (2) z−z1= a represents the circle with centre, A(x1,y1) and radius a., , ∴ ( x + 1) 2 + ( y − 1) 2 = 5, ∴ ( x + 1) 2 + ( y − 1) 2 = 52, represents the circle with centre (−1,1) and, radius 5., , = P(z), , (iii) If z − z1 = z − z2 then, , A(z1), , ( x + iy ) − (2 + 3i ) = ( x + iy ) − (−1 + i ), ∴ ( x − 2) + i ( y − 3) = ( x + 1) + i ( y − 1), ∴ ( x − 2) 2 + ( y − 3) 2 = ( x + 1) 2 + ( y − 1) 2, ∴ x2 − 4 x + 4 + y 2 − 6 y + 9 = x2 + 2x + 1 + y 2 − 2 y + 1, , Fig. 1.7, , x2 − 4 x + 4 + y 2 − 6 y + 9 = x2 + 2x + 1 + y 2 − 2 y + 1, (3) z−z1=z−z2represents the perpendicular, bisector of the line joining the points A and, ∴ −6 x + 4 y + 11 = 0 i.e. 6 x + 4 y − 11 = 0, B., represents the perpendicular bisector of line, joining points (2,3) and (−1,1)., SOLVED EXAMPLES, , (z2), A(z1), , Ex. 1 : If w is a complex cube root of unity, then, prove that, 1, 1, i), + w2 = −1, w, , (z), , ii) (1+ w2)3 = −1, iii) (1−w+w2)3 = −8, , Fig. 1.8, , Solution : Given, w is a complex cube root of, unity., ∴ w3 = 1 Also w2 +w+1 = 0, 18

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z1 − z2 = (a−c) + (b−d)i, z1. z2 = (ac−bd) + (ad+bc)i, , ∙, , 4), , A) −1, , z1  ac + bd   bc − ad , =, +, i, z2  c 2 + d 2   c 2 + d 2 , , 5), , For any non-zero complex number z = a+ib, 1, a, −b, =, +i 2 2, z a2+b2, a +b, , 6), , ∙, , For any positive integer k,, i4k = 1, i4k+1 = i, i4k+2 = −1, i4k+3 = − i, , ∙, , The conjugate of z = a+ib is z , is given by z, = a−ib, , ∙, , The polar form of the complex number, , 7), , ∙, , 1), , x 2 + y 2 is called modulus and, y, cosq = x , sinq =, (q is called argument, r, r,  y, of z) arg (z) = tan−1   ., x, w is complex cute root of unity then, 1 + w + w2 = 0 , w3 = 1., =, , B) 1,1, , C) 1, 0, , 8π, B) 256 and 3, 4π, D) 64 and 3, , If arg(z) = q, then arg (z) =, A) − q, , 9), , B) q, , C) π−q, , 2π, A) − 3, , π, B) 3, , π, C) − 3, , −3, , C) 0, , 1), , D) 4, , Simplify the following and express in the, form a+ib., i) 3+ −64, , ii) (2i3)2, , iii) (2+3i)(1−4i), , 5, 4+3i, iv) 2 i(−4−3i) v) (1+3i)2(3+i) vi), 1−i, , D) −1, , 2, 4, vii) (1+ i ) (3+ i ) (5+i)−1 viii), , −6 is equal to, , A) −3 2, , B) z lies on y-asis, , D) z lies on a rectangle C) z lies on a circle, , i592+i590+i588+i586+i584, The value of is 582 580 578 576 574 is equal, i +i +i +i +i, to :, B) 1, , 2π, D) 3, , 10) If z = x+iy and |z−zi| = 1 then, , II) Answer the following., , C) 4i, , D) π+q, , If −1+ 3 i =reiq , then q = ................. ., , If n is an odd positive integer then the value of, 1 + (i)2n + (i)4n + (i)6n is :, B) 0, , D) −1,1, , The modulus and argument of (1+i 3 )8 are, respectively, , A) z lies on x-asis, , A) −2, 3), , If w(≠1) is a cube root of unity and (1+w)7, = A + Bw, then A and B are respectively the, numbers, , 2π, C) 256 and 3, 8), , D) 3, , z z, +, z z, A) cos2q B) 2cos2q C) 2cosq D) 2sinq, , Select the correct answer from the given, alternatives., , A) −4i, 2), , C) 0, , 2π, A) 2 and 3, , MISCELLANEOUS EXERCISE - 1, I), , B) 1, , If z=r(cosq+isinq), then the value of, , A) 0, 1, , z = x+iy is r (cosq + i sinq) = r eiθ where, r, , If w is a complex cube root of unity, then the, value of w99+w100+w101 is :, , 5, +2i7+i9, ix) 3i, i6+2i8+3i18, , B) 3 2 C) 3 2 i D)− 3 2 i, 21, , x), , 5 + 3i, 5 − 3i, , 5+7i, 5+7i, +, 4+3i, 4−3i

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2), , Solve the following equations for x, y∈R, , 8, ,  1+ i   1− i , 11) Show that ,  +,  = 2.,  2  2, , i), , (4−5i)x + (2+3i)y = 10−7i, x+iy, ii), = 7−i, 2+3i, , 12) Convert the complex numbers in polar form, and also in exponential form., , iii) (x+iy) (5+6i) = 2+3i, , 2 + 6 3i, 5 + 3i, , iv) 2x+i9y (2+i) = xi7+10i16, , i), , 3), , Evaluate i) (1−i+i2)−15, , ii) z = −6+ 2 i, , 4), , Find the value of, i), , ii) (i131+i49), , iii), , x3+2x2−3x+21, if x = 1+2i., , ii) x4+9x3+35x2−x+164, if x = −5+4i., 5), , 7), 8), 9), , iii), , v) 2i, , vi) −3i, , vii), , 1 + 3i, 2, , iv), , 16) Simplify i), , −1 − i, 2, , iii), , 1, 1, +, i, 2, 2, , 3, 1, +, 1−2i 1+i, , 3+4i, 2−4i, , 18) If α and β are complex cube roots of unity,, prove that (1−α)(1−β) (1−α2)(1−β2) = 9, 19) If w is a complex cube root of unity, prove, that (1−w+w2)6 +(1+w−w2)6 = 128, , Find the real numbers x and y such that, 5+6i, y, x, + 3+2i = −1+8i, 1+2i, , 20) If w is the cube root of unity then find the, value of, 18, , 18, ,  −1 − i 3 ,  −1 + i 3 , ,  + , , 2, 2, , , , , , 1, i 10, 1, i 10, +, ) +(, −, ) =0, 2, 2, 2, 2, , v, ,  65 1 , ii)  i + 145 , i , , , i29+i39+i49, i30+i40+i50, , i238+i236+i234+i232+i230, i228+i226+i224+i222+i220, , 17) Simplify, , Represent 1+2i, 2−i, −3−2i, −2+3i by points, in Argand's diagram., 5, Show that z = (1−i)(2−i)(3−i) is purely, imaginary number., , 10) Show that (, , 1−2i, 1+2i, +, is real., 3−4i, 3+4i, , 15) Show that, , Find the modulus and amplitude of each, complex number and express it in the polar, form., ii) 6−i, , a+ib, , prove that x2+y2 =1., a−ib, 3, , v) 3−4i vi) 6+8i, , i) 8+15i, , −3 3 3 i, +, 2, 2, ,  −1 + −3 , 14) Show that z = ,  is a rational, 2, , , number., , i) −16+30i ii) 15−8i iii) 2+2 3 i, , 6), , z=, , 13) If x+iy =, , Find the square roots of, iv) 18i, , 8, , v, , 22, , v

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2, , SEQUENCES AND SERIES, 2.2 Arithmetic Progression- ( A.P.), , Let's Study, , In a sequence if the difference between, any term and its preceding term (tn+1−tn) is, constant,then the sequence is called an Arithmetic, Progression ( A.P.), , •, , A.P. and G.P., , •, , Sum of n terms of a G.P., , •, , Sum of infinite terms of a G.P., , •, , H.P. and A.G.P., , Consider the following sequences, 1) 2, 5, 8, 11, 14, …, , •, , A.M., G.M, H.M., , 2), , 4, 10, 16, 22, 28, …, , 3), , 4, 16, 64, 256, …, 1, 1 1, ,, ,, ,…, 25, 125, 5, , Let's Recall, , 4), , 2.1 Sequence :, , 5), , A set of numbers where the numbers are, arranged in a definite order,like the natural, numbers, is called a sequence., , The sequences 1), 2) and 5) are A. P. but the, terms in sequences 3) and 4) are not in A. P. as, the difference between their consecutive terms is, not constant., , Examples : Natural numbers, Even integers, between 10 and 100, Squares of integers., , -3, 2, 7, 12, 17, …, , If t1, t2, t3....., tn are in A.P. then tn+1 − tn = d, is constant, for all n., , In general, a sequence is written as t1, t2, t3,, t4....,tn .... where t1 - first term, t4 - fourth term, ...,, tn - nth term., , Hence the sequence can also be written as, a, a+d, a+2d, .... . Its nth term is tn = a + (n-1)d,, t1 = a and sum of n terms i.e., , Finite sequence – A sequence containing finite, number of terms is called a finite sequence., , Sn = t1 + t2 + ..... + tn =, , n, [2a + (n-1)d]., 2, , If t1, t2, t3....., tn are in A.P. then, , It is written as {t1, t2, t3....., tn} for some positive, integer n., , i) t1+ k, t2+ k, t3+ k, ..... tn + k are also in A.P., ii) k t1, kt2, kt3, .... , ktn are also in A.P. (k ≠ 0), , Infinite sequence – A sequence is said to be, infinite if it is not a finite sequence., , Let's Learn, , It is written as {t1, t2, t3.....} or {tn} n ≥ 1, , 2.3 Geometric progression :, , Sequences that follow specific patterns are, called progressions., , A sequence t1, t2, t3....., tn ..... is G.P. (Geometric, t n +1, progression) if the common ratio, = r is, tn, constant for all n., , In the previous class, we have studied, Arithmetic Progression (A.P.)., 23

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Hence a G.P. can also be written as a, ar2, ar3, ...., where a is first term and r is common ratio., , Solution : Here t1 = 1 , t2=, , -3, t2, −3, 2, Consider = t =, =, 2, 1, 1, , Examples :, i), , 2, 4, 8, 16, .... [here a = 2, r = 2], , 1, ii) 1, 1 , 1 , 1 , .... [a = 1, r =, ], 3, 3, 9 27, , Here the ratio of any two consecutive terms, is constant hence the given sequence is a G.P., , 2.3.1 The General term or the nth term of a G.P., For G.P., t1 = a, t2 = ar, t3 = ar2, t4 = ar3, ..., , Now t9 = ar, , 9-1, , If a and r are the first term and common ratio, of a G.P. respectively. Then its nth term is given, by tn= arn-1. (Verify), ,  −3 , =ar =1  ,  2 , , 3 , 3, 3 3 , .... is 243?, , 3, -6, 12, -24, …, , Solution : Here a =, , Here a = 3 , r = -2, , and tn = a rn−1, , Since tn= arn-1 = 3 (-2)n-1, , ∴ a rn-1 = 243, , 3,, , tn= 243, , n, , ∴ ( 3 ) n = 3 2 = 35, , 1, r=5, , 1, Since tn = arn-1 = 5  , 5, , 3, r=, , 3 .( 3 )n-1 = 243 = 35, , 1 1, ii) 5, 1, ,, ,…, 5 25, Here a = 5 ,, , 8, , 8, , Ex. 2) Which term of the sequence, , Ex. Find nth term of the following G.P., , n −1, , 1, = , 5, , ∴, , n−2, , ., , n, =5, 2, , ∴ n = 10, ∴ Tenth term of the G.P. is 243., , Properties of Geometric Progression., If t1, t2, t3, .....tn are in G.P. then, (i), , t3,  9   2  −3, =,   −  = 2, t2,  4  3 , , and, , iii) 1, -1, 1, -1, 1, -1, .... [a = 1, r = −1], , i), , 9, −3, , t3= ,, 4, 2, , Ex. 3) For a G.P. If a = 3 and t7 = 192 find r and t11., Solution : Given a = , t7 = ar = 192, 192, ∴ ( r) = 192 , r =, = 64, 3, ∴r = 2 ,, ∴ r= 2 ., also t11= a r = 3 (2) = ., , 1 1 1, , , , ..... 1 are also in G.P., t1 t2 t3, tn, , ii) (k = 0) k t1, kt2, kt3, .... ktn are also in G.P., iii) t1n, t2n, t3n, ... are also in G.P., SOLVED EXAMPLES, , 1, Ex 4) In a G.P. ,if the third term is, and sixth, 5, 1, term is, , find its nth term ., 625, 1, 1, Solution : Here t3= , t6 =, 5, 625, , −3 9 −27, , ,, , … is a, 2 4 4, G.P., if it is a G.P. Find its ninth term., Ex. 1) Verify whether 1,, , 24

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1, 5, 1, t6 = ar5 =, 625, t3 = ar2 =, , Let’s Note :, i) Three numbers in G.P. can be conveniently, a, assumed as, , a, ar, r, , …(1),, …(2), , Divide equation (2) by equation (1), we get, , iii) Assume 4 numbers in a G.P. as, a a, , , ar, ar3 (Here ratio is r2), r3 r, iv) Assume 5 numbers in a G.P. as, a a, , , a, ar, ar2, r2 r, , ar 5 (1 / 625), =, ar 2, (1 / 5), r3=, , 1, 1, = 3, 125, 5, , 1, ∴r= 5 ., , Ex 6) Find three numbers in G.P. such that their, sum is 42 and their product is 1728., , Substitute in equation (2), we get , , Solution : Let the three numbers in G.P. be, a, , a, ar ., r, , 2, 1, 1, a  = 5, 5, , ∴ a = 5., 1, tn= arn-1 = 5  , 5, , As their product is 1728, a .a.ar = 1728, r, ∴ a3 = 1728 = (12)3, , n −1, , = 5 × (5)1-n = (5)2-n., , 5n-2, Ex 5) If for a sequence {tn}, tn = n-3 show that, 4, the sequence is a G.P. Find its first term and the, common ratio., 5n-2, Solution: tn, = 4n-3, , ∴ a = 12., According to first condition, their sum is 42 , a, ∴ r + a + ar = 42, 12, ∴ r + 12 + 12r = 42, 12, ∴ r + 12r = 30, , 5n-1, ∴ tn+1 = 4n-2, 5n-1, 4n-3, 5n-1, t +1, 4n-2, Consider n, = n−2 = 4n-2 × 5n-2, tn, 5, 4n−3, , Multiply by r, ∴12 + 12r2 = 30r, , 5, 5n −1− n + 2, n − 2 − n + 3 = 4 = constant, ∀ n ∈ N., 4, 5, The given sequence is a G.P. with r = ., 4, 16, and t1 = 5 ., , ∴12r2 − 30r + 12 = 0, Dividing by 6, we get , ∴ 2r2 - 5r + 2 = 0, ∴ (2r-1) ( r-2 ) = 0, ∴ 2r =1 or r = 2, 1, ∴r= 2 r=2, 25

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Ex 8) If p, q, r, s are in G.P. then show that, ( q-r)2 + ( r-p ) 2 + (s-q)2 = (p-s)2, , 1, If a=12, r = 2 then the required numbers are, 24, 12, 6., , r, s, q, Solution : Since, p, q, r, s are in G.P. p = q = r, , If a = 12, r =2 then the required numbers are, 6, 12, 24., , ∴ q2 = pr , r2 = qs , qr = ps, , ∴ 24, 12, 6 or 6, 12, 24 are the three required, numbers in G.P., , consider L.H.S. = (q − r)2 + (r − p)2 + (s − q)2, = q2-2qr + r2 + r2-2rp + p2 + s2-2sq + q2, , Ex 7) Find four numbers in G. P. such that their, product is 64 and sum of the second and third, number is 6., a a, Solution : Let the four numbers be 3 , , ar , ar3, r r, (common ratio is r2), According to the first condition, a, a, 3, r3 × r × ar × ar = 64, ∴ a4 = 64, , = pr -2qr +qs + qs -2rp + p2 +s2 -2sq + pr, = -2qr +p2+s2 = -2ps +p2+s2 (⸪ qr = ps ), = (p-s)2 = R.H.S., Ex 9) Shraddha deposited Rs. 8000 in a bank which, pays annual interest rate of 8%.She kept it with the, bank for 10 years with compound interest. Find, the total amount she will receive after 10 years. , [given (1.08 )­­10 = 2.1575)], , ∴ a = (26)1/4 = 23/2, ∴a= 2 2 ., , Solution:, The Amount deposited in a bank is Rs 8000 with, 8% compound interest., , a, + ar = 6, r, , Now using second condition, 2 2 + 2 2r = 6., r, Multiplying by r,, , Each year , the ratio of the amount to the, 108, principal to that year is constant =, 100, Hence we get a G.P. of successive amounts., , 2 2 + 2 2 r2 = 6r, Dividing by 2 , 2+, , 2 r2 = 3r, , 2 r2 – 3r +, 2 r (r -, , r=, , amount, 108, The ratio of principal =, ., 100, , 2 =0,, , 2 r 2 -2r –r+, (r -, , We consider the amount at the end of each year, for Rs 100 , the amount is 108., , 2 = 0,, , 2 ) - 1 (r −, , For P = 8000 ,, 2 ) = 0., , the amount after 1 year is, , 2 ) ( 2 r - 1) = 0., 2 or r =, , 1, 2, , 8000 + 8000 ×, , ., , 8, = 8000, 100, , 8 , , 1 +, ,  100 , , = 8000 ×, , If a = 2 2 , r = 2 then 1, 2, 4, 8 are the four, required numbers in G.P., 1, If a = 2 2 , r =, then 8, 4, 2, 1 are the four, 2, required numbers in G.P., , 108, 100, , the amount after 2 years is 8000 ×, = 8000 ×, 26, , 108, 108, × 100, 100, 108, 100, , 2

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108, 108, ×, 100, 100, 108, × 100, , 7, , r = 3 find t6., 243, iii ) If r = − 3 and t6 = 1701, find a., 2, iv) If a = , t6 = 162, find r ., 3, , the amount after 3 years is 8000 ×, , 108, 100, , = 8000 ×, , 3, , ii), , ., , Therefore after 10 years the amount is, 108, = 8000, 100, , 10, , 3), , Which term of the G.P. 5, 25, 125, 625, … is, 510?, , 4), , For what values of x, the terms, 4, , x , 4 are in G.P?, 3, 27, , = 8000 (1.08)10, = 8000 × 2.1575 = 17260., , 5), , Thus Shraddha will get Rs 17260 after 10 years., The formula to find amount by compound interest, is, N, R , , 1, +, A=P , ,  100 , A , R , = 1 +, , P,  100 , , ∴, , 6), 7), , N, , R , , Note that, ( 1 +,  is a G.P.],  100 , , 9), , EXERCISE 2.1, , 2, 6, 18, 54, …, , 1, -5, 25, -125 …, 1 , 1 , 1 , …, iii), 5,, 5 5 5 25 5, iv) 3, 4, 5, 6, …, , 2), , The fifth term of a G.P. is x , eighth term of, a G.P. is y and eleventh term of a G.P. is z, verify whether y2 = x z ., , 11) The number of bacteria in a culture doubles, every hour. If there were 50 bacteria, originally in the culture, how many bacteria, will be there at the end of 5th hour ?, , ii), , v), , Find four numbers in G.P. such that sum of, the middle two numbers is 10/3 and their, product is 1., , 10) If p,q,r,s are in G.P. show that p+q , q+r , r+s, are also in G.P., , Check whether the following sequences are, G.P. If so, write tn., i), , 5n-3, If for a sequence, tn = 2n-3 , show that the, sequence is a G.P. Find its first term and the, common ratio., Find three numbers in G.P. such that their, sum is 21 and sum of their squares is 189., , 8 ) Find five numbers in G. P. such that their, product is 1024 and fifth term is square of the, third term., , N, , 1), , If a=, , 12) A ball is dropped from a height 0f 80 ft., th, 3, The ball is such that it rebounds   of the, 4, height it has fallen. How high does the ball, rebound on 6th bounce? How high does the, ball rebound on nth bounce?, , 7, 14, 21, 28, …, , For the G.P., 1, i), If r = , a = 9 find t7, 3, 27

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13) The numbers 3, x and x + 6 form are in G.P., Find (i) x, (ii) 20th term (iii) nth term., 14) Mosquitoes are growing at a rate of 10%, a year. If there were 200 mosquitoes in, the begining. Write down the number of, mosquitoes after (i) 3 years (ii) 10 years, (iii) n years., , 2), , If r is positive and r > 1, it is convenient to, a(rn–1), write Sn = r – 1, , 3), , If r = 1 then G.P. is a, a, a ...a (n times),, So, Sn = a. n, , 4), , Sn – Sn-1 = tn, , 15) The numbers x − 6, 2x and x2 are in G.P. Find, (i) x (ii) 1st term (iii) nth term., , SOLVED EXAMPLES, Ex 1) If a = 1, r = 2 find Sn for the G.P., , Let's Learn, , Solution : a = 1 , r = 2, , 2.3.2 Sum of the first n terms of a G.P. (Sn), ,  r n −1 ,  2n − 1 , Sn = a , =, 1, ×, , ,  = 2n – 1 ., r, −, 1, 2, −, 1, , , , , , Consider the G.P. t1, t2, t3, ..... tn .... . we write, the sum of first n terms, t1+ t2+ t3 + ..... + tn as, , n, , ∑t, r =1, , r, , Ex 2) For a G.P. 0.02, 0.04, 0.08, 0.016, …,, find Sn., 0.04, Solution : Here a = 0.02, r = 0.02 = 2, , = Sn, , Note : ∑ is the notation of summation, the sum is, of all tr (1 < r < n), ,  r n −1 ,  2n − 1 , Sn = a ,  = 0.02 , ,  r −1 ,  2 −1 , , n, , In ∑ the variable is r., r=1, , Theorem : If a, ar, ar2, ...., arn-1 (r ≠ 1) is a G.P. then, Sn = a + ar2 + .... + arn−1 = t1+ t2+ t3 + ..... + tn, n, , = ∑ tr =, r =i, , = 0.02 . (2n -1), Ex 3) For the following G.P. 3, -3, 3, -3, …,, find Sn., , a (1 − r n ), ... ( r ≠ 1), (1 − r ), , Solution :, Case (i), , Proof : Consider Sn = a + ar + ar2 + .... + arn-1, Sn = a(1+r+r2+r3+ ... + rn-1), ... (1), , If n is even , n = 2k, , Multiplying both sides by r we get, r Sn = a(r+r2+r3+ ... + rn) , , ... (2), , S2k = (3-3) + (3-3) + (3-3) + ( 3-3) + ... 2k terms, , Subtract (2) from (1) we get Sn – Sn r = a (1-rn), , = (3+3+ ... k terms) + (-3, -3, -3, ... k terms), , ∴ Sn (1-r) = a (1-rn), ∴ Sn =, , = 3k - 3k = 0, , a(1– r ), 1 – r , r ≠ 1., n, , Case (ii), If n is odd , n = 2k +1 , , Let’s note :, 1) If r is positive and r < 1, it is convenient to, a(1– rn), write Sn = 1 – r, , S2k+1 = (3 + 3 + 3 .... k terms), + (-3, -3, -3, ... k terms) + 3 = 3k - 3k + 3 = 3, 28

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Ex 4) For a G.P. if a=6 , r =2 , find S10., , Dividing (2) by (1) we get,, s6, 144, r 6− 1, =, = 16, 3, s3, r −1, ,  r n −1 , Solution: Sn = a , ,,  r −1 , , (r3− 1)(r3+1), = 9,, (r3−1), ,  210 − 1 ,  1023 , S10 = 6 ,  =6 ,  = 6 (1023) = 6138 .,  1 ,  2 −1 , , , r3 +1 = 9,, , r3 = 8 = 23,, , r=2, Substitute r = 2 in (1) We get, , Ex 5) How many terms of G.P., 2 ,22,23,,24.,... are needed to give the sum 2046., ,  23 − 1 , a,  = 16,,  2 −1 , , Solution : Here a = 2 , r = 2 , let Sn = 2046.,  2n − 1 ,  r n −1 , ∴ 2046 = a , =, 2, ,  = 2 (2n-1), , −, 2, 1, r, −, 1, , , , , ,  8 −1 , a,  = 16,,  2 −1 , , 1023 = 2n-1 , 2n = 1024 = 2 10 ∴ n = 10, , a (7) = 16 ,, a = 16 / 7, , Ex 6) If for a G.P. r=2 , S10 =1023 , find a ., , Ex 9) Find 5+55+555+5555+ ... upto n terms., ,  210 − 1 , Solution : S 10 = a , ,  2 −1 , , Solution:, Let Sn,, , ∴ 1023 = a (1023), ∴ a =1., , = 5+55+555+5555+... upto n terms., = 5 (1+11+111+... upto n terms), , Ex 7) For a G.P. a = 3 , r = 2 , Sn = 765 , find n ., , 5, (9+99+999+... upto n terms), 9, 5, =, [(10-1) + (100-1) + (1000-1) + ... to n, 9, brackets], 5, =, [(10 + 100 + 1000 +... upto n terms), 9, − (1+1+1+ …..upto n terms)], =, ,  2n − 1 , Solution :Sn= 765 = 3 ,  = 3 (2n −1) ,, −, 2, 1, , , ∴ 255 = 2n – 1,, 2n = 256 = 28,, n = 8., Ex 8) For a G.P. if S3= 16, S6=144, find the first, term and the common ratio of the G.P., , [(a = 10, r = 10) and (a = 1, r = 1)], , Solution : Given,  r3 −1 , S3 = a ,  = 16,  r −1 ,  r 6 −1 , S6 = a ,  = 144,  r −1 , , =, , n, 5   10 − 1  , −n, 10, 9   10 − 1  , , =, , 5 10 n, (10 − 1) − n , 9  9, , … (1), ... (2), , 29

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Ex 10) Find the sum to n terms 0.3+0.33+0.333+, ... n terms, , Ex 12) For a sequence , if Sn = 5(4n-1), find the, nth term, hence verify that it is a G.P., Also find r., , Solution :, , Solution :Sn = 5(4n-1), Sn-1= 5 (4n-1-1), , Sn, , = 0.3+0.33+0.333+ upto n terms, , We know that tn = Sn – Sn-1, , = 3 [0.1+0.11+ 0.111+ ... n terms], , = 5(4n −1) − 5 (4n-1−1), , Multiply and divide by 9, 3, = 9 [0.9+0.99+ 0.999+ ... n terms], , , , = 5(4n − 4n-1), , 3, = 9 [(1-0.1)+(1-0.01)+(1-0.001)+ ... to n, terms], 3, = 9 [(1+1+1 ...... n terms), , 3, = 9, ,  1, n ,  n − 9 (1 − 0.1 ) , , = 5 (4n-1) (4 – 1), , , , = 5 (4n-1)(3), = 15 (4n-1), , Then, tn+1 = 15 (4n), , [(a = 1, r = 1) and (a = 0.1, r = 0.1)], , (1 − 0.1n) , n, −, 0, ., 1, , (1 − 0.1) , , , , , ∴ tn, , , - (0.1+0.01+ 0.001+ ...... n terms)], , 3, = 9, , = 5 (4n) – 5 – 5(4n−1)+5, , 15 (4n), tn+1, and t = 15 (4n-1), n, = 4n−n+1 = 4, = constant ∀ n ∈ N., r = 4., The sequence is a G.P. with tn=15(4n-1)., , Ex 11) Find the nth term of the sequence, 0.4 , 0.44, 0.444,…, , Ex 13) A teacher wanted to reward a student by, giving some chocolates. He gave the student two, choices. He could either have 60 chocolates at, once or he could get 1 chocolate on the first day,, 2 on the second day, 4 on the third day and so, on for 6 days.Which option should the student, choose to get more chocolates?, , Solution :, Here t1=0.4, t2=0.44 = 0.4 + 0.04, t3= 0.444 = 0.4+0.04+0.004, tn= 0.4 + 0.04 + 0.004 + 0.0004 + ... upto n terms, , Ans : We need to find sum  of chocolates in 6, days., , here tn is the sum of first n terms of a G.P., , According to second option teacher gives, 1 chocolate on the first day , 2 on the second day,, 4 on the third day, and so on.Hence it is a G. P ., with a = 1, r = 2 ., , with a = 0.4 and r = 0.1,  1 − 0.1n  0.4, tn= 0.4 , [1- 0.1)n],  =,  1 − 0.1  0.9, 4, = 9 [1- (0.1)n]., ,  r n −1 , By using Sn = a , ,  r −1 , 30

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S6, , Mr. Pritesh will get Rs. 5,85,640 in the fifth year, and his total earnings through salary in 10 years, will be Rs 63,74,960., ,  26 − 1 , = 1, ,  2 −1 , = 64 -1 = 63, , Hence the student should choose the second way, to get more chocolates., , EXERCISE 2.2, 1), , Ex 14) Mr. Pritesh got a job with an annual, salary package of Rs. 4,00,000 with 10% annual, increment. Find his salary in the 5th year and also, find his total earnings through salary in 10 years., , For the following G.P.s , find Sn, i), 3, 6, 12, 24, ..., 3, q2, ii) p, q, , q2 , …, p p, iii) 0.7, 0.07, 0.007, ...., , [Given (1.1)4 = 1.4641, (1.1)10 = 2.59374], , iv), , Solution : In the year he will get a salary of, Rs. 4,00,000., , 2), , He gets an increment of 10% so in the second year, his salary will be, , For a G.P., i), ii), ,  110 , 4, 00, 000 × ,  = 4, 40, 000,  100 , , 3), , ii), , , 2, ,  110  and so on ..., 4, 00, 000 × , ,  100 , hence it is a G.P. with a = 4,00,000 & r = 1.1., , 4), , So, his salary in the fifth year will be, 4, ,  110 , t5 = ar = 4, 00, 000 × ,  = 585640.,  100 , 4, , 5), ,  r −1 , ,  r −1 , , will be S10 = a , , [⸪ (1.1)10 = 2.59374], , 6), ,  2.59374 − 1 , = 4,00,000 × , , 0.1, , ,  1.59374 , = 4,00,000 , ,  0.1 , , 7), , = 4,00,000 [15.9374] = 63,74,960., 31, , For a G.P. sum of first 3 terms is 125, and sum of next 3 terms is 27,, find the value of r ., , If t4 =16 , t9 =512 , find S10, , Find the sum to n terms, i), ii), , 10, , If a =2, r = 3, Sn = 242 find n., , For a G.P., i), If t3 = 20 , t6= 160 , find S7, ii), , His total income through salary in 10 years, , 2, a = 2, r = − 3 , find S6, If S5 = 1023 , r = 4, Find a, , For a G.P., i), , In the third year his salary will be, , 5 , -5, 5 5 , -25…, , 3 + 33 + 333 + 3333 + ..., 8 +88 + 888 + 8888 + ..., , Find the sum to n terms, i), , 0.4 + 0.44 + 0.444 + ..., , ii), , 0.7 + 0.77+ 0.777 + ..., , Find the sum to n terms of the sequence, i), , 0.5, 0.05, 0.005, ..., , ii), , 0.2, 0.02, 0.002, ...

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8), , 9), , For a sequence, if Sn = 2 ( 3n-1) , find the, nth term, hence show that the sequence is, a G.P., , 1, 1, 1, 1, Example : 1 + 2 + 4 + 8 + 16 + ..., Solution :, 1, here a = 1, r = 2 , i.e. r < 1, , If S,P,R are the sum , product and sum of the, reciprocals of n terms of a G. P. respectively,, , ∴ Sum to infinity is given by, , n, , S , then verify that   = P2., R, , a, 1−r =, , 10) If Sn ,S2n ,S3n are the sum of n,2n,3n terms of, a G.P. respectively , then verify that, , Visual proof : This can be visualised taking a, rectangle of 2×1 containing smaller rectangles of, 1, 1, area 1, 2 , 4 , ... square units., , Sn (S3n - S2n) = ( S2n - Sn ) 2., 10, , 11) Find (i), , ∑ (3 × 2 ), r, , r =1, , 10, , (ii), , 1, 1, =, =2, 1 1, 1−    , 2 2, , ∑ 5× 3, , r, , r =1, , 12) The value of a house appreciates 5% per year., How much is the house worth after 6 years, if its current worth is Rs. 15 Lac. [Given :, (1.05)5 = 1.28, (1.05)6 = 1.34], 13) If one invests Rs. 10,000 in a bank at a rate of, interest 8% per annum, how long does it take, to double the money by compound interest?, [(1.08)5 = 1.47], , Fig. 2.1, These rectangles are seen to be completely in the, big rectangle and slowly fills up the big rectangle, of area two square units., , 2.4 Sum of infinite terms of G. P., , 1, 1, That is 1 + 2 + 4 + .... = 2., , Consider a G.P. of the positive terms. The sum of, a (rn − 1) a (1 − rn), first n terms is , =, (1 − r), (r − 1), , SOLVED EXAMPLES, , If r > 1, r −1 is constant but rn approaches ∞, as n approaches ∞, so the infinite terms cannot be, summed up., , Ex. 1) Determine whether the sum of all the terms, in the series is finite. In case it is finite find it., , If r < 1, rn approaches 0, as n approaches ∞ and, , i), , 1, 1 , 1 ,…, 3 32 33, , ii), , 3 , −9 27 , −81 …, 5 25 125 625, , the sum Sn =, , a, a (1 − rn), approaches, 1−r, (1 − r), ∞, , Here the infinite sum ∑ tr is said to be, r =1, , a, 1- r, , iii) 1, -3, 9, -27, 81, …, , 32

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Solution:, i), , iii) 2. 5 = 2 + 0.5 + 0. 05+ 0.005 + 0.0005 + ..., After the first term,the terms are in G.P. with, a = 0.5 , |r| = |0.1| < 1, , 1, 1, 1, , |r| =, =, <1, 3, 3, 3, ∴ Sum to infinite terms is finite., Here a =, , ∴ Sum to infinite terms is finite and is , 5, 0.5, a, 0.5, =, =, = 9, 0.9, 1−r, 1− 0.1, , 1, 1,  ,   1, 3, 3 =, a, S=, =, =, 1 − r 1−  1 , 2 2,  , , , 3, 3, ii), , 5, 23, 2. 7 = 2 + 9 =, 9, , -3, 3, 3, 3, Here a = , r = - , |r| =, = <1, 5, 5, 5, 5, , EXERCISE 2.3, , ∴ Sum to infinite terms is finite., , 3, 3,  , , , 3, a, 5, S=, = 5 =   = 8, 1−r, 3, 8, 1 − (− ),  , 5, 5, , 1), , i), , iii) Here a=1 , r = -3 |r| = |−3| = 3 <1, , 1, 1, 1, 1 ,…, 2 4 8 16, , ii) 2, 4 , 8 , 16 , …, 3 9 27, , ∴ sum to infinity does not exist, , iii) -3, 1,, , 2.4.1 Expressing recurring decimals as rational, numbers :, , −1 1, ,, ,…, 3 9, , −2, iv) 1 , , 4 , −8 , 16 , …, 5 5 5 5 5, , We know that recurring decimal fraction can, be written as rational numbers e.g. 0.666 ... 0.6, 2, = . This can also be checked using G.P., 3, , v) 9, 8.1, 7.29, ..., 2), , Ex i) 0.66666…, , Express the following recurring decimals as, a rational number., 0. 7, , = 0.6 + 0.06 + 0.006 +…, , i), , the terms are in G.P. with a=0.6, |r| = |0.1| < 1, , ii) 2. 4, , ∴ Sum to infinite terms is finite and is , 0.6, 6, 2, a, 0.6, = 1−0.1 =, = 9 = 3, 1−r, 0.9, ii), , Determine whether the sum to infinity of the, following G.P.s exist, if exists find them, , iii) 2.3 5, iv) 51.0 2, 3), , 0. 46 = 0.46+0.0046+0.000046 + …, the terms are in G.P. with a = 0.46 ,, |r| = |0.01| < 1., , 4), , ∴ Sum to infinite terms is finite and is , 0.46, 46, a, 0.46, = 1 − (0.01) =, = 99 ., 1−r, 0.99, 33, , 2, and sum, 3, to infinity is 12. Find the first term., If the common ratio of a G.P. is, , If the first term of the G.P. is 16 and its sum, 96, to infinity is, find the common ratio., 17

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5), , 6), , The sum of an infinite G.P. is 5 and the sum, of the squares of these terms is 15 find the, G.P., Find (i), , ∞, , ∑ 4(0.5), , (ii), , r, , r =1, ,  1, , 1, 1, = 2+ 2 n− 2, , r, , 1, (iii) ∑ ( −8 (iv), )  − ,  2, r =0, , ∞, , ∑ 0.4, , n, , n=1, , =, , Ex. 2) Find the nth term of H.P., , with a= 5 and d = -4, Hence tn= a + (n−1) d, = 5 + (n−1) (−4), , 2.5 Harmonic Progression ( H. P. ), , = 5 -4n +4 = 9-4n., , Definition : A sequence t1, t2, t3, t4 , …, tn, ..., , 1, For H.P. tn= 9−4n, , (tn≠ 0, n ∈ N) is called a harmonic progression if, 1 1 1, 1, t1 , t2 , t3 …, t , … are in A.P., , 2.6 Types of Means:, , n, , e.g., i) 1 , 1 , 1 … are in H. P. as, 7 11 15, 1, , 1 , 1 i.e. 7, 11, 15 … are in A.P., ii), , 1, −1 −1, , 1,, ,, ,..., 5, 3 7, , Solution: Since 5, 1, -3, -7, … are in A.P., , 8) A ball is dropped from a height of 10m. It, bounces to a height of 6m, then 3.6m and so, on. Find the total distance travelled by the, ball., , 1, 11, , 3+n, 2 ., , 2, Therefore in H.P. tn = 3+n, , 7) The mid points of the sides of a square of, side 1 are joined to form a new square. This, procedure is repeated indefinitely. Find the, sum of (i) the areas of all the squares (ii) the, perimeters of all the square., , 1, 7, , 1, 2, , 3, n, = 2 + 2, , r =1, , r, , ∞, , ∞, , ∑  − 3 , , For A.P. tn = a + (n-1 ) d = 2 + (n-1 ), , 2.6.1 Arithmetic mean (A. M.):, If x and y are two numbers, their A.M. is given by , x+y, A= 2 ., , 1, 15, , 1 3 3, ,, ,, … is H.P. as 4, 14 , 16 , ... are, 4 14 16, 3 3, , We observe that x, A , y form an A.P., i.e. A - x = y - A ∴ 2A = x + y ∴ A =, , in A.P., , x+y, 2, , 2.6.2 Geometric mean (G. M.) :, , SOLVED EXAMPLES, , If x and y are two numbers having same sign, (positive or negative), their G.M. is given by, G = xy ., , Ex. 1) Find the nth term of the H.P., 1 , 2 , 1 , 2…, 2 5 3 7, 5, 7, Solution : Here 2,, , 3,, … are in A.P. with, 2, 2, 1, 1 2 1 2, a= 2 and d = 2 hence , , ,, , … are in, 2 5 3 7, H.P., , We observe that x, G, y form a G.P., y, G, i.e., =, ∴ G2 = x y, G, x, ∴ G=, 34, , xy

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2.6.3 Harmonic mean (H.M.) :, If x and y are two numbers, their H.M. is given, 2xy, by H = x+y ., , ∴, , 1 1 1, , ,, is in A.P., x H y, 1, =, H, , 1, x, , +, 2, , 1, y, , x+y, 2xy, = 2xy ∴ H = x+y, , G A, =, ≥1, , H G, G, ∴ H ≥1∴G≥H, , n, 1 1 1, 1, + + + .... +, x1 x2 x3, xn, , Ex. 1 : Find A.M.,G.M.,H.M. of the numbers, 4 and 16, Solution : Here x = 4 and y = 16, x+y, 4+16 20, A= 2 = 2 = 2 = 10, , Theorem : If A, G and H are A.M., G.M., H.M., of two positive numbers x and y respectively, then, ii) A ≥ G ≥ H, , G=, , Proof : Let A, G and H be A.M., G.M and H.M., of two positive numbers x and y, , i), , x+y, 2, , RHS = AH =, , 2, A1, A2, A3, A4, 22 are in AP with, , 1, =, (x+y -2, 2, =, , 1, 2, , 64 = 8, , Solution: Let A1, A2, A3, A4 be the four terms, between 2 and 22 so that, , 2xy, x+y, , x+y, 2 −, , ConsiderA-G =, , 4 ×16 =, , Ex. 2 : Insert 4 terms between 2 and 22 so that the, new sequence is in AP., , = xy = G2 = L.H.S., ii), , xy =, , 2×4×16 128 32, 2xy, H = x+y = 4+16 = 20 = 5, , 2xy, xy , H = x+y, , x+y, 2 , G=, , (III), , SOLVED EXAMPLES, , 2) If x = y then A = G = H, , A=, , (From II), , A≥G≥H, , G = x1.x2 .x3 ... xn, n, , Then, , (II), , From (I) and (III), , x1+ x2+ x3+... + xn, A=, n, , i) G2 = AH, , (I), , Now since G2 = AH, , Note 1) : These results can be extended to n, numbers as follows, , H=, , ), , ∴A−G≥0 ∴A≥G, A, ∴, ≥1, G, , We observe that x, H, y form an HP ., i.e., , (, , 2, 1, x − y <0, 2, Since squares are always, non-negative., , =, , 2, , x +, , a = 2, t6 = 22, n = 6 ., , xy, , ∴, xy ), 2, , y -2, , 22 = 2 + (6-1)d = 2 + 5d, 20 = 5d, d = 4, , x, , A1 = a+d = 2+4 = 6,, , y, , A2 = a+2d = 2+8 = 10,, 35

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A3 = a+3d = 2+3×4 = 2+12 = 14, , t3= G2 = ar2 = 1(3)2 = 9, , A4 = a+4d = 2 + 4×4 = 2+16 = 18., , ∴ 3 and 9 are the two required numbers between, 1 and 27., , ∴ the 4 terms between 2 and 22 are 6, 10, 14, 18., , Ex: 5 The A.M. of two numbers exceeds their, 18, G.M. by 2 and their H.M. by 5 . Find the, Numbers., , 2, 1, Ex: 3 Insert two numbers between, and, so, 9, 12, that the resulting sequence is a HP., 1, Solution : Let the required numbers be, and, H1, 1, H2, , Solution : Given A= G + 2, ∴, , G = A − 2 .... (I), 18, Also A= H + 5, , ∴ 2 , 1 , 1 , 1 are in HP., 9 H1 H2 12, , ∴H=A−, , 18, ..... (II), 5, , We know that G 2 = A H, 18, (A−2) 2= A (A − 5 ), .... from (I) and (II), 18, A2− 4A + 22 = A2 –, A, 5, 18, A – 4A = − 4, 5, , 9, ∴ , H1 , H2, 12 are in A.P., 2, 9, 9, t1 = a = , t4 = 12 = a+3d ∴ 2 + 3d = 12, 2, 24−9, 15, 9, ∴ 3d = 12 –, = 2, = 2, 2, 5, ∴d= 2, 9, 5 14, t2 = H1 = a+d =, +, = 2 =7., 2, 2, , −2A = − 4 × 5,, ∴, , A = 10, Also G =A−2 = 10 − 2 = 8, Now, let the two numbers be x and y., , 9, 5 19, t3 = H2= a+2d =, +2×, = 2, 2, 2, 2, 1, and, are to be, 19, 7, , As A =, ∴, , 2, 1, inserted between, and, 9, 12, Ex: 4 Insert two numbers between 1 and 27 so, that the resulting sequence is a G.P., , x+y, 2 = 10, x+y = 20,, , y = 20 - x ... (III), Now G =, , xy = 8, , ∴, , xy = 64, , ... (IV), , ∴, , x(20 − x) = 64, , ... from (III) and (IV), , 20x – x2 = 64, , Solution: Let the required numbers be G1 and G2, , x2-20x + 64 = 0, , ∴, , 1, G1, G2, 27 are in G.P., , (x-16) (x-4) = 0, , ∴, , t1 =1 , t2 = G1 , t3 = G2 ,t4 = 27, , x =16 or x = 4, , ∴, , a = 1 , t4 = ar3 = 27 ∴ r3 = 27, , If x = 16 then y = 4, , ∴, , r3 = 27 = 33 ∴ r = 3, , If x = 4 then y = 16., , t2 = G1 = ar = 1 × 3 = 3., , The required numbers are 4 and 16., 36

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9), , EXERCISE 2.4, 1) Verify whether the following sequences are , H.P., i), , 1 , 1 , 1 , 1 ,…, 3 5 7 9, , ii), , 1 1 1 1, ,, ,, ,, ,…, 3 6 12 24, , Let's Learn, 2.7 ARITHMETICO-GEOMETRIC, PROGRESSION (A.G.P.) :, Definition : A sequence in which each term is, the product of the corresponding terms of an A.P., and G.P. is called and arithmetico - geometric, progression (AGP)., , iii) 5, 10 , 10 , 10 , …, 17 32 47, 2) Find the nth term and hence find the 8 thterm, of the following HPs, , 3), , 4), , i), , 1 1 1 1, ,, ,, , ,…, 2 5 8 11, , ii), , 1 1 1 1, ,, ,, ,, ,…, 4 6 8 10, , iii), , 1, 1 1 1, ,, ,, ,, …, 5 10 15 20 ,, , e.g. Consider the sequence, A.P., G.P., , 6), , Insert two numbers between, , Insert two numbers between 1 and −27 so, that the resulting sequence is a G.P., , 8), , If the A.M. of two numbers exceeds their, 18, G.M. by 2 and their H.M. by, , find the, 5, numbers., , (a+2d), 8, 9, (r2), , (a+3d), 11, 27, (r3), , Here the first factors of the terms are in AP and the, second factors are in G.P., Therefore the given sequence forms an A.G.P., nth term of A.G.P., t1 = a×1, t2 = (a+d)r, t3 = (a+2d)r2, ...., ∴ tn = [a + (n − 1)d] rn-1, 2.7.1 Sum of n terms of A.G.P.:, , 1, 1, and, so, 4, 3, that the resulting sequence is a HP., , 7), , (a+d), 5, 3, (r), , i.e. 2×1, 5×3, 8×9, 11×27,..., , Find H.M. of two positive numbers A.M. and, 15, G.M. are, and 6, 2, Find GM of two positive numbers whose, A.M. and H.M. are 75 and 48, , (a), 2, 1, (1), , A.G.P. is a, (a+d)r, (a+2d) r2 , (a+3d) r3 , …, , Find A.M. of two positive numbers whose, 16, G.M. and H. M. are 4 and 5 respectively., , 5), , Find two numbers whose A.M. exceeds their, 25, 1, GM by, and their HM by 26, 2, , Let Sn = a + (a + d)r + (a + 2d)r2 + ..., + [a + (n − 1)d]rn−1, ∴ Sn = a + ar + dr + ar2 + 2dr2 + ...., + arn−1 + (n−1)drn−1, , ...(i), , ∴ rSn = ar + ar2 + dr2 + ar3 +2dr3 + ..., + arn + (n −1)drn, Subtracting (ii) from (i) we get, 37, , ...(ii)

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Sn − rSn = a + dr + dr2 + ... drn−1 − arn − (n −1)drn, , EXERCISE 2.5, , ∴ Sn (1− r) = a + dr + dr + ...., 2, , + drn−1 − [a + n(n−1)d]rn,  1 r, Sn (1−r) = a + dr ,  (1  r ), , n 1, , 1), , , n,  − [a + (n−1)d]r, , , Find Sn of the following arithmetico geometric sequences, i), , a, dr(1−r n−1) [a+(n−1)d]rn, Sn = 1−r + (1−r)2 −, 1− r, , 2, 4x, 6x2, 8x3, 10x4, …, , ii) 1, 4x, 7x2, 10x3, 13x4, …, iii) 1, 2×3, 3×9, 4×27, 5×81, …, , (r ≠1), Note that, sum to infinity of A.G.P. is, a, dr, +, 1−r, (1−r)2, , iv)) 3, 12, 36, 96, 240, …, 2), , SOLVED EXAMPLE, , Find the sum to infinity of the following, arithmetico - geometric sequence, 1, 2 , 3 , 4 , …, 4 16 64, , i), , Ex 1) Find tn and the sum of n terms of, , ii) 3, 6 , 9 , 12 , 15 , …, 5 25 125 625, , 1, 4, 12, 32, 80, 192, ..., Solution: Given sequence can be written as, , iii) 1,, , 1×1, 2×2, 3×4, 4×8, 5×16, …, , −4 7 −10, ,, ,, …, 3 9 27, , Properties of Summation, , We observe that first factors in each term 1, 2, 3,, 4, 5, ... are in A.P. with a = 1,d = 1, , i), , and the second factors in each term 1, 2, 4, 8, 16,, … are in G.P with r = 2, , n, , n, , r 1, , r 1, ,  ktr = k  tr , ... k ≠ 0, where k is a nonzero constant., , ∴ tn= [a + (n − 1)d]rn−1, = [1 + (n − 1)]2n−1 = n.2n−1, We know that Sn of A.G.P. is given by, (1−rn−1) [a+(n−1)d]rn, a, Sn = 1−r + dr (1−r)2 −, ( r ≠1), 1− r, Substituting the values of a,d,r we get ,, , n, , ii), , , , iii), , ∑, , iv), , (ar + br) =, , r 1, n, , , r 1, , ar +, , n, , , r 1, , br, , 1= n, , 1, , n, , n, , k = k, , 1  kn ,, , 1, , 1, (1−2n−1), [1+(n−1)1]2n, Sn = 1− 2 + 1 × 2 (1− 2)2 –, 1− 2, 1, Sn =, + 2 (1 − 2n-1) + n. 2n, −1, , n, , ... k ≠ 0, , 1, , Result: 1) The Sum of the first n natural numbers, =, , Sn = −1 + 2− 2n + n . 2n, , n, , r =, r 1, , n (n+1), 2, , Result 2) The Sum of squares of first n natural, numbers, , Sn = 1 – 2n + n 2n, Sn = 1− 2n(1−n) = 1+ (n−1) 2n ., , =, 38, , n, ,  rr =, r 1, , 2, , n(n+1)(2n+1), 6

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Result 3) The sum of the cubes of the first n, n, ,  n(n + 1) , natural numbers = ∑ rr = , , r =1,  2 , , 50, , 2, , The above results can be proved using, Mathematical Induction. (Chapter No. 4), , n, , n, , r =1, , r =1, , r =1, , 2, , 2, , 2, , 2, , 2, , tr = 5 + (r-1) 2 = 5+ 2r -2 = 2r+3, , = ( 12+ 22 + 32 + ... + 292 ) – (12 + 22 ), 29, , =∑ r 2, , r =1, , 2, , ∑, r =1, , r, , r =1, , Also the second factors 5,7,9,11, ... are also in, A.P. with a= 5 , d = 2 ., , Solution: 3 +4 + 5 + ... + 29, 2, , r =1, , 50(50 + 1), - 50, 2, , tr = a + ( r-1)d = 1 + (r-1) 2 = 2r-1., , Ex 2) Find 3 +4 + 5 + ... + 29 ., 2, , 50, , Solution : Note that first factors of each term, 1, 3, 5, 7, ... are in A.P. with a =1, d=2., , = 4 ( n2+n) -7n = 4n2 + 4n -7n = 4n2-3n., 2, , r =1, , 50, , ... upto n terms., , n(n + 1) ,  - 7n,  2 , , r =1, , r =1, , Ex 4) Find 1 × 5 + 3 × 7 + 5 × 9 + 7 × 11, , ∑ r1 = 8 , , r =1, , 50, , = 5050., , r =1, , n, , n, , 50, , = 50 (101), , ∑ r(8r − 7), , = 8 ∑ rr - 7, , ∑ 4r - ∑ 1, , = 50 ( 2 × 51-1), , n, , n, , =, , (4r − 1), , r =1, , = 2 × 50 × 51 – 50, , r − ∑ r7, ∑ r(8r − 7) = ∑ 8r, , Solution :, , ∑, , = 4∑ r -∑ 1 = 4 ×, , SOLVED EXAMPLES, Ex 1) Evaluate, , =, , 3, , n, , ∑ r(2r − 1) (2r + 3), , ∴ Sn =, , 2, , 29(29+1)(58+1) 2(2+1)(4+1), –, 6, 6, = (29 × 30 × 59) / 6 – (2 × 3 × 5) / 6, , =, , =, , r =1, , n, , ∑ r(4r, r =1, , 2, , + 4r − 3), , n, , n, , n, , r =1, , r =1, , r =1, , = 4 ∑ rr2 + 4 ∑ rr − ∑ r3, , = (29 × 5 × 59) – 5, = 5 (29 × 59 – 1) = 5 (1711-1), = 5 (1710) = 8550, , =4, , Ex 3) Find 1002 – 992 + 982 – 972 + ... + 22 -12, , n(n+1)(2n+1), n(n+1), +4, - 3n, 6, 2, , Solution : 100 – 99 + 98 – 97 + ... + 2 -1, ,  2(n + 1)(2n + 1) , =n ,  + 2 n (n+1) - 3n, 3, , , = (1002 + 982 + 962 + ... + 22 ) – (992 + 972 +, 952+ .... + 12), , n, = 3 [2(2n2 + n + 2n +1) + 6(n + 1) - 9], , 2, , 50, , =, , ∑, , =, , ∑, , r =1, , 2, , 2, , 2, , 2, , 2, , =, , n, [(4n2 + 6n + 2) + 6n − 3], 3, , =, , n, (4n2 + 12n − 1), 3, , 50, , (2 r) 2, , ∑, r =1, , (2r − 1)2, , 50, , r =1, , (4 r2 - 4r2 + 4r -1), 39

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2.8 Power Series, Some functions can be expressed as infinite sums, of powers of x. They are called powers series., Some examlpes of Power Series are,, ex, , 2), , x3, x5, x7, sin x = x − 3! + 5! − 7! + ..., , 3), , cos x = 1 − , , 4), , e-x, , If, , 1×2 + 2×3 + 3×4 + 4×5 + ... upto n terms, 1 + 2 + 3 + 4 + ... upto n terms, , , , =, , 100, , find n., 3, , 10) If S1 , S2 and S3 are the sums of first n natural, numbers, their squares and their cubes, respectively then show that -, , x3, x2, x4, x, = 1+, +, +, +, + ..., 1! 2! 3! 4!, , 1), , 9), , 9 S22 = S3 (1+ 8 S1 )., , x4, x6, x2, + 4! −, + ..., 6!, 2!, , Let's Remember, , x4, x, x2 x3, +, −, + 4! ..., 1! 2! 3!, , 1), , For A.P. tn = a + (n-1) d, , 2), , For G.P. tn= a r, , 3), , A.M. of two numbers A=, , x+y, 2, , The proofs of the equations given above are, obtained at more advanced stage in mathematics., , 4), , G.M. of two numbers G =, , xy, , EXERCISE 2.6, , 6), , 2xy, H.M. of two numbers H= x+y, G2 =AH, , 7), , If x = y then A = G = H, , 8), , If x ≠ y then H < G < A., , 5), , = 1− , , If x  < 1 then, log (1+x) = x −, , 1), , Find the sum, n, , 2), , Find, , ∑ r(3r, , 3), , Find, , ∑ r, , r =1, n, , r =1, n, , ∑, , 2, , x4, x3, x2, + 3 − 4, 2, , n, , ∑ r(r + 1) (2r − 1), , a (r n - 1), , Sn =, (r - 1), , r =1, , − 2r + 1), , MISCELLANEOUS EXERCISE - 2, ,  1 + 2 + 3..... + r , , r, , , (I) Select the correct answer from the given, alternative., ,  13 + 23 + ..... + r 3 , , , r (r + 1), , , , 4), , Find, , 5), , Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ..., upto n terms., , 6), , Find the sum 22+42 + 62 + 82 + ... upto n terms., , 7), , Find (702 – 692) + (682-672)+(662-652) + ..., + ( 22-12 ), , 8), , Find the sum 1×3×5 + 3×5×7 + 5×7×9 + ... +, (2n-1) (2n+1) (2n+3), , r =1, , 5), , n-1., , 1), , The common ratio for the G.P. 0.12,0.24,0.48,, is –, A) 0.12, , 2), , 40, , B) 0.2, , C) 0.02, , D) 2 ., , The tenth term of the geometric sequence, 1 −1, , , 1, -2, ... is –, 4 2, −1, 1, A) 1024 B), C) -128 D), 128, 1024

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3), , A) 3, 4), , C) 12th, , B) 5, , In a G.P., the fourth term is 48 and the eighth, term is 768. Find the tenth term., , 2), , Find the sum of the first 5 terms of the G.P. , 2, whose first term is 1 and common ratio is, 3, , 3), , 4, 243, For a G.P. a =, and t7 =, , find the, 3, 1024, value of r ., , D) 13th, , C) 15, , D) -5, , 4), , B) 2, , C) 4, , D) 8, , 5), , B) -, , 1, 2, , C), , 10, 3, , D), , B), , 9, 2, , C), , 5, 2, , Find three numbers in G.P. such that their, , 6), , Find five numbers in G.P. such that their, product is 243 and sum of second and fourth, number is 10., , 7), , For a sequence Sn = 4 ( 7n-1) verify that the, sequence is a G.P., , 8), , Find 2+22+222+2222+ ... upto n terms., , 9), , Find, the nth term of the sequence , 0.6,0.66,0.666,0.6666, ..., , 3, 10, , The tenth term of H.P. 2 , 1 , 2 , 1 , ... is 9 7 19 12, 1, 27, , 5n - 2, , verify whether, 7 n -3, the sequence is a G.P. If it is a G.P. , find its, first term and the common ratio., For a sequence , if tn=, , sum is 35 and their product is 1000., , 5 5, 5, Sum to infinity of a G.P.  5 ,- , , - ,, 2 4, 8, 5, , ... is 16, , A), 9), , B) 11th, , 1), , 21, and their, 4, product is 1 then the common ratio is –, , A) 5, 8), , D) -1., , The sum of 3 terms of a G.P. is, A) 1, , 7), , C) 1, , If common ratio of the G.P is 5 , 5th term is, 1875 , the first term is A) 3, , 6), , B) 2, , (II) Answer the following., , then r = ?, , Which term of the geometric progression, 1, 2, 4, 8, ... is 2048., A)10th, , 5), , t6, 1458, t3 = 54, , If for a G.P., , D) 27., , Which of the following is not true, where , A, G, H are the AM, GM, HM of a and b, respectively. (a, b > 0), , 10) Find, , a+b, B) G = ab, 2, 2ab, C) H =, D) A = GH ., a+b,  , A) A =, , 11) Find, , A) 6, 15/2  , , B) 15, 25, , C) 3, 12, , D) 6/5, 3/2 ., , ∑ r(5r, r =1, , 2, , + 4r -3 ), , n, , ∑ rr(r -3 )(r-2), r =1, n, , 12) Find, , 12 + 22 + 32 + ... + r 2, r, ∑, 2r + 1, r =1, , 13) Find, , 13 + 23 + 33 + ...r 3, r, ∑, 2, r =1, ( r + 1), , 10) The G.M.of two numbers exceeds their H.M., 6, by , the A.M. exceeds G.M. by 3 the two, 5, 2, numbers are ..., , n, , n, , 14) Find 2 × 6 + 4 × 9 + 6 × 12 + ... upto n terms., 15) Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ..., upto n terms., 41

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25) If for a G.P. first term is (27)2 and seventh, term is (8)2 , find S8., , 12 + 22, 12, 12 + 22 + 32, 16) Find, +, +, + ... upto n, 2, 1, 3, terms., , 26) If pth, q th and rth terms of a G.P. are x,y,z, respectively .Find the value of xq-r .y r-p .zp-q, , 17) Find 12 + 13 + 14 + 15 + ... 20, 2, , 18) If, , 2, , 2, , 2, , 2, , 3, 1 + 2 + 3 + 4 + 5 + ... upto n terms, =, 1× 2 + 2 × 3 + 3 × 4 + 4 × 5 + ... upto n terms 22, , 27) Which 2 terms are inserted between 5 and 40, so that the resulting sequence is G.P., , Find the value of n ., , 28) If p,q,r are in G.P. and p1/x = q 1/y = r 1/z ,verify, whether x,y,z are in A.P. or G.P. or neither., , 19) Find ( 50 –49 ) + (48 -47 )+(46 -45 ) + ..., + ( 22-12 )., 2, , 20) If, , 2, , 2, , 2, , 2, , 2, , 29) If a,b,c are in G.P. and ax2+2bx+c=0 and, px2+2qx+r=0 have common roots then verify, that p b2 – 2 q b a + r a2 = 0, , 5, 1× 3 + 2 × 5 + 3 × 7 + ... upto n terms, = , find, 3, 3, 3, 1 + 2 + 3 + ... upto n terms, 9, , 30) If p,q,r,s are in G.P., show that (p2+ q2 + r2), (q2 +r2+ s2 ) = (pq+qr+rs)2, , the value of n., 21) For a G.P. if t2 = 7 , t4 = 1575 find a, , 31) If p,q,r,s are in G.P. , show that (pn + qn),, (qn + rn) , (rn+sn) are also in G.P., , 22) If for a G.P. t3= 1/3 , t6 = 1/81 find r, n, , 2, 23) Find ∑ r , r =1  3 , , r, , 32) Find the coefficientof x6 in the expansion of, e2x using series expansion ., , ., , 33) Find the sum of infinite terms of, , 24) Find k so that k-1,k , k+2 are consecutive, terms of a G.P., , v, , 1+, , v, , 42, , v, , 4, 7, 10, 13, +, +, +, + ..., 5, 25 125 625

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3 PERMUTATIONS AND COMBINATIONS, The theory of permutations and combinations, is central in problems of counting a large number, of objects that are impossible to count manually., The theory of permutations and combinations, enables us to count objects without listing or, enumerating them., , Let's Study, •, , Fundamental principles of counting, , •, , Factorial notation, , •, , Permutations, , •, , Permutations of distinct objects, , •, , Permutations, identical, , •, , Circular permutations, , •, , Combinations, , when, , some, , objects, , Let us begin with the following simple, example. Every smartphone requires a passcode, to unlock it. A passcode is formed by four of the, ten digits on the screen. The order of these four, digits cannot be changed for passcode to work., How many distinct passcodes are possible? Note, that a passcode consists of four digits. The first, digit of a passcode can be any of the ten digits,, the second digit can be any of the ten digits, and, similarly for the third and fourth digits. This, gives a total of 10 × 10 × 10 × 10 = 10,000 as the, number of distinct possible passcodes., , are, , Let's Recall, •, •, , The number system., The four basic mathematical operations:, addition,, subtraction,, multiplication,, division., , Consider one more example which is not, as easy as the last example. The school cricket, team has eleven players. The school wants a, photograph of these players, along with the, principal and the two vice principals of the, school, for school magazine. Seven chairs are, arranged in a row for the photograph. Three chairs, in the middle are reserved for the principal and, the two vice principals. Four players will occupy, the remaining four chairs and seven players will, stand behind the chairs. The question then is: “In, how many different ways can the eleven players, take positions for the photograph?” This example, will be considered later in the chapter. Till then,, we can try, on our own, to find the number of, different ways in which the eleven players can sit, or stand for the photo., , 3.1 INTRODUCTION, Counting is a fundamental activity in, Mathematics. Learning to count was our first, step in learning Mathematics. After learning to, count objects one by one, we used operations, of addition and multiplication to make counting, easy. We shall now learn two more methods of, counting to make complicated counting easier., These two methods are known as permutations, and combinations., Permutations refer to the number of different, arrangements of given objects, when the order of, objects is important. Combinations are related to, the number of different selections from a given, set of objects, when the order of objects in the, selections is immaterial., 43

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3.2.1 Addition Principle:, , Let us understand three principles of counting, that are fundamental to all methods of counting,, including permutations and combinations., , Consider the situation where a boy wants, to go for movie. He has three T-shirts of three, different colours: white, green, and blue. He also, has four shirts of four different colours: red, green,, yellow, and orange. How many choices does he, have to wear? This situation can be represented, using a tree diagram as follows. (see fig 3.2), , 3.2 Fundamental principles of counting, Tree Diagram, We have learnt in set theory that subsets of, a set can be represented in the form of a Venn, diagram. An alternative method is to draw a tree, diagram if the subsets are disjoint. For example,, The games students play at school are of two types, 1) Indoor games, 2) Outdoor games. Available, indoor games in school are chess, carrom and, table tennis. While outdoor, games in school are, cricket, volleyball, basketball and badminton., Such information is presented with the help of, tree-diagram as follows: (see fig 3.1), , What to wear?, Shirt, , Chess, , Table Tenniss, , Cricket, Basket, ball, , Green, , Yellow, , Orange, , Green, , White, Blue, , The tree diagram shows that the boy has 4 choices, in shirt and 3 choices in T-shirt, i.e. seven choices, in all., , Outdoor, , Carrom, , Red, , Fig. 3.2, , School Games, Indoor, , T-Shirt, , The boy can choose from 4 Shirts OR from, 3 T-shirts. Hence, the number of ways of choosing, are 3 + 4 = 7 in all., , Volley ball, Badminton, , This example shows that the total number, of outcomes is obtained by adding the number of, outcomes of each characteristic when only one, characteristic is to be chosen., , Fig. 3.1, Diagrams of this nature are called tree diagrams., A tree diagram shows the division of a set into, disjoint subsets., , Statement of The Addition Principle., Addition Principle : Suppose one operation, can be done in m ways and another operation can, be done in n ways, with no common way among, them. If one of these operations is to be performed, then there are m + n ways to do it., , The fundamental principles of counting, provide an efficient way of finding the number of, different ways to carry out two or more activities,, either simultaneously or successively (that is, one, after another)., , 44

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Ex. 1) A restaurant offers five types of fruit, juices and three types of milk shakes. If a customer, wants to order a drink, how many choices does, the customer have?, , So, the tree diagram shows that there are six, possible outcomes, i.e. 2 × 3 = 6 ways to serve, Ice cream., In the example of Ice cream, the final count, of six is obtained by multiplying 3 (number, of flavours) by 2 (number of serving options),, in either order., , Solution : Since the restaurant offers five fruit, juices, the customer has five ways of selecting a, fruit juices. Similarly, since the restaurant offers, three types of milk shakes, the customer has three, ways of selecting a milk shake. Finally, since the, customer wants to select only one drink, there are, 5 + 3 = 8 choices for the customer., , Definition of The Multiplication Principle, If one operation can be carried out in m, ways, followed by the second operation that can, be carried out in n ways, and these two operations, are independent then the two operations can be, carried out in m × n ways., , Ex. 2) Consider an experiment of drawing, a card from a pack of 52 playing cards. What is, the number of ways in which the drawn card is a, spade or a club?, , Ex. 1) : Samadhan Bhojanalay offers a thali, that has four items: roti, rice, vegetable and dal., Following choices are available and one option is, to be selected for each item., , Solution : Since there are 13 spade cards and, 13 club cards hence the number of ways in which, the drawn card is spade or club is 13 + 13 = 26, ways., , Roti: chapati, tandoor roti, Rice: plain rice, jeera rice, dal khichadi, , Note : The word ‘OR’ in the statement, suggests addition, i.e. ‘+’., , Vegetable: dum aloo, paneer masala, mixed, veg, , 3.2.2 Multiplication principle, , Dal: dal fry, dal tadka, , Now, consider the following situation. An, Ice cream is served either in a cup or in a cone. Also,, Ice cream is available in three flavours: vanilla,, chocolate, and strawberry. This information can, be represented in the form of a tree diagram as, follows. (see Fig 3.3), , How many different menus are possible?, Solution : Since one option is to be selected, for each item, the number of different possible, thali choices are identified as follows., Roti 2, Rice 3, Vegetable 3, Dal 2 and hence, the total number of different possible thali menus, is 2 × 3 × 3 × 2 = 36., , Ice Cream, cup, Vanila, , Chocolate, , Cone, Strawberry, , Vanila, , Chocolate, , Strawberry, , Ex. 2) A company decides to label each of, its different products with a code that consists of, two letters followed by three digits. How many, different products can be labeled in this way?, , Fig. 3.3, There are 2 ways of choosing mode of serving, and 3 ways of choosing flavours., 45

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3.3 Invariance Principle, , Solution : Since the first two characters, in a label are letters, they can be formed in, 26 × 26 = 676 ways. The next three characters are, digits and can be formed in 10 × 10 × 10 = 1000, ways. The total number of distinct labels is, by, the multiplication principle, given by 676 × 1000, = 6,76,000., , The result of counting objects in a set does, not depend on the order in which these objects are, counted or on the method used for counting these, objects., For example, 1) In earlier example of, addition principle of choosing first from, 4 Shirts, then from 3 T-Shirts is same as, choosing first from 3 T-Shirts and then from, 4 Shirts., , Activity 1) Suresh has 4 pencils and 2 erasers., He wants to take one pencil and one eraser for the, examination. Can we find the number of ways in, which he can select a pencil and an eraser?, , 2) In earlier example of multiplication, principle choosing first from 2 modes of servings, and then from 3 flavours is same as choosing first, from 3 flavours and then from the 2 modes of, servings., , Activity 2) Sunil has 4 ballpens of one, company and 3 ballpens of another company. In, how many ways can he select a ball pen?, In the above activities, can we decide when, to use the addition principle and when to use the, multiplication principle? Can we give reasons?, What are answers in the above examples?, , SOLVED EXAMPLES, Ex. 1) : From the figure below, find the total, number of routes from A to B. e.g. one upper route, is A→D→N→E→B. (See fig 3.4), , Remark: The addition and multiplication, principles can be extended from two to any finite, number of activities, experiments, events, or, operations., Extended Addition Principle : Suppose, there are three possible choices with no common, outcome for any two, the first choice can be made, in m ways, second in n ways and third in r ways., If only one of the choices is to be made, it has, m + n + r possible ways., Extended, Multiplication, Principle:, Suppose an experiment consists of three, independent activities, where first activity has, m possible outcomes, second has n possible, outcomes, third has r possible outcomes. Then the, total number of different possible outcomes of the, experiment is m × n × r., , Fig 3.4, Solution: A D, → N E, → B is one route, C, F, N, B is another route, through N. A, through N. Thus there are in all 2 × 3 = 6 routes, through N. There are 2 routes through S. Hence, the total number of routes is 6 + 2 = 8., , 46

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Ex. 2) : Suppose 5 chocoloates of different, type are to be distributed among 4 children and, there is no condition on how many chocolates a, child can get (including zero.) How many different, ways are possible for doing so?, , The addition principle finally gives the total, number of ways of forming an even number less, than 1000 using digits 2,3,7,8 is 2+8+32=42., EXERCISE 3.1, , Solution: The first chocolate can be given, to any of the four children. Therefore, there are, four different ways of giving the first chocolate., Similarly, the second chocolate can be given in, four different ways, and similarly for each of, the remaining chocolates. The multiplication, principle then gives the total number of different, ways as 4 × 4 × 4 × 4 × 4 = 45 = 1024., Ex. 3) : How many even numbers can, be formed using the digits 2, 3, 7, 8 so that the, number formed is less than 1000?, Solution: The condition that the number, formed from the given digits is less that 1000, means that this number can have upto 3 digits. Let, us therefore consider the three cases separately,, case i) One digit numbers, Since the number must be even, the onedigit number can be only 2 or 8. Hence, there are, 2 ways of forming a one digit even number., , 1., , A teacher wants to select the class monitor, in a class of 30 boys and 20 girls. In how, many ways can the monitor be selected if the, monitor must be either a girl or a boy?, , 2., , A Signal is generated from 2 flags by putting, one flag above the other. If 4 flags of different, colours are available, how many different, signals can be generated?, , 3., , How many two letter words can be formed, using letters from the word SPACE, when, repetition of letters (i) is allowed, (ii) is not, allowed?, , 4., , How many three-digit numbers can be, formed from the digits 0, 1, 3, 5, 6 if, repetitions of digits (i) are allowed, (ii) are, not allowed?, , 5., , How many three-digit numbers can be, formed using the digits 2, 3,4,5,6 if digits, can be repeated?, , 6., , A letter lock contains 3 rings and each, ring containing 5 letters. Determine the, maximum number of false trials that can be, before the lock is open., , 7., , In a test, 5 questions are of the form 'state,, true or false'. No student has got all answers, correct. Also, the answer of every student, is different. Find the number of students, appeared for the test., , 8., , How many numbers between 100 and 1000, have 4 in the units place?, , 9., , How many numbers between 100 and 1000, have the digit 7 exactly once?, , case ii) Two digit numbers, Since the required number is even, the units, place of a two-digit number must be either 2 or, 8 i.e. 2 ways. The ten’s place can be filled with, any of the four given digits. Therefore, there are, 4×2=8 ways of forming a two-digit even number., case iii) Three digit numbers, Finally,since the requried number is even,, the units place can be filled in two ways, ten’s, place can be filled in four ways, and hundred’s, place can also be filled in four ways. The number, of ways of forming a three digit even number is, 4×4×2=32, , 10. How many four digit numbers will not, exceed 7432 if they are formed using the, digits 2,3,4,7 without repetition?, 47

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Illustrations:, , 11. If numbers are formed using digits 2, 3, 4, 5,, 6 without repetition, how many of them will, exceed 400?, , Factorial 1 = 1! = 1, Factorial 2 = 2! = 2 × 1 = 2, , 12. How many numbers formed with the digits, 0, 1, 2, 5, 7, 8 will fall between 13 and 1000, if digits can be repeated?, , Factorial 3 = 3! = 3 × 2 × 1 = 6, Factorial 4 = 4! = 4 × 3 × 2 × 1 = 24, and so on., Note: Though 0 is not a natural number, we, define 0! = 1., , 13. A school has three gates and four staircases, from the first floor to the second floor. How, many ways does a student have to go from, outside the school to his classroom on the, second floor?, , Properties of the factorial notaction., For any positive integers m,n.,, , 14. How many five-digit numbers formed using, the digit 0, 1, 2, 3, 4, 5 are divisible by 5 if, digits are not repeated?, 3.4 Factorial Notation., The theory of permutations and combinations, uses a mathematical notation known as the, factorial notation. Let us first understand the, factorial notation, which is defined for natural, number., Definition: For a natural number n, the, factorial of n, written as n! or n and read as “n, factorial”, is the product of n natural numbers, from 1 to n., , 1), , n! = n × (n - 1)!, , 2), , n > 1, n! = n × (n - 1) × (n - 2)!, , 3), , n > 2, n! = n × (n - 1) × (n - 2) × (n - 3)!, , 4), , (m + n)! is always divisible by m! as well as by, n! e.g. (3 + 4)! is divisible by 3! as well as 4!, , 5), , (m × n)! ≠ m! × n!, , 6), , (m + n)! ≠m! + n!, , 7), , m > n, (m - n)! ≠ m!-n! but m! is divisible, by n!, , 8), , (m ÷ n)! ≠m! ÷ n!, , Ex. 1) : Find the value of 6!, Solution : 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720, Ex. 2) : Show that (7-3)! ≠ 7! – 3!, , That is, n! or n is expressed as, , Solution: (7-3)! = 4! = 4 × 3 × 2 × 1 = 24., , 1 × 2 × 3 ×....× (n−2) × (n−1) × n., , , , 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040, , Also, 3! = 3 × 2 × 1= 6., , Note: The factorial nonation can also be, defined as the product of the natural numbers, from n to 1., , 7! - 3! = 5040 − 6 = 5034, Therefore, (7-3)! ≠7!-3!, , That is, n! = n × (n − 1) ×....× 3 × 2 × 1, For example, 5! = 5 × 4 × 3 × 2 × 1 = 120, It is read as “5 factorial is equal to 120.”, , 48

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Ex. 3): Find n if (n+6)! = 56 (n+4)!, Solution: (n+6)! = 56 (n+4)!, ∴ (n+6) (n+5) (n+4)! = 56 (n+4)!, , 2., , Compute:, 12!, 6!, , (i), , ∴ (n+6) (n+5) = 56, , (iii) (3 × 2)!, , Here instead of solving quadratic in n, we write, 56 as product of 2 consecutive numbers as, 56 = 8×7, ∴ (n+6) (n+5) = 8×7, 3., , (vi), , 6!-4!, 4!, , (vii), , 8!, 6!-4!, , (viii), , 8!, (6-4)!, , 5 × 6 × 7 × 8 × 9 × 10, , (iv) 5 × 10 × 15 × 20, , 12 ! 12 !, +, 5 !7 ! 6 !6 !, 4., , , , 1, 1, 12 ! , +, ,  5! × 7 × 6 ! 5! × 6 × 6 !, , Evaluate :, (i), , n!, for, r !(n - r )!, , n = 8, r = 6, , (iii) n = 15, r = 10, , (ii) n = 12, r = 12,, (iv) n = 15, r = 8, , =, , 12 !  1 1 , +, 5!6 !  7 6 , , =, , 12 !  13 , 5!6 !  6 × 7 , , =, , 12 ! ×13, (5! × 6) × (6 ! × 7), , (iii), , =, , 13!, 6! × 7 !, , (iv) (n + 1)! = 42 × (n − 1)!, , 5., , 6., , Evaluate:, (iii) 10! - 6!, , n 3 1!, = +, 8! 6 ! 4 !, , (ii), , n 4 3, = +, 6 ! 8! 6 !, , 1! 1! 4, = −, n ! 4 ! 5!, , (v) (n + 3)! = 110 × (n + 1)!, , EXERCISE 3.2, , 8!, , Find n, if, (i), , = R. H. S., , (i), , Write in terms of factorials, , (iii) 6 × 7 × 8 × 9, , ( 7! = 7 × 6! 6! = 6 × 5!), , 1., , 3! × 2!, , (ii) 3 × 6 × 9 × 12 × 15, , 12 ! 12 ! 13!, +, =, 5 !7 ! 6 !6 ! 6 !7 !, Solution: Consider,, , =, , (iv), , 9!, 3! 6!, , (i), , Ex. 4) : Show that, , L. H. S. =, ,  12 ,  !,  6, , (v), , Equating bigger factors from either side, ∴ (n+6) = 8, , n=8−6=2, , (ii), , (ii) 10!, (iv) (10 - 6)!, 49, , Find n, if:, (i), , (17 − n)!, = 5!, (14 − n)!, , (iii), , n!, n!, :, = 5:3, 3!(n − 3)! 5!(n − 5)!, , (ii), , (15 − n)!, = 12, (13 − n)!

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(iv), , n!, n!, :, = 1: 6, 3!(n − 3)! 5!(n − 7)!, , Suppose the 3 seats are numbered 1, 2 and 3, and the 3 persons are named A, B and C., , (v), , (2n)!, n!, :, = 24:1, 7 !(2n - 7)! 4 !(n - 4)!, , We can fill the 1st chair in 3 ways. Having, done that, we can fill the 2nd chair with any of, the remaining two people, hence in 2 ways. The, 3rd chair is then filled in a unique way as there is, only one person left. Thus the total number of the, seating arrangements are, , 7., , Show that, n!, n!, (n + 1)!, +, =, r !(n − r )! (r − 1)!(n − r + 1)! r !(n − r + 1)!, , 8., , Show that, 9!, 9!, 10 !, +, =, 3!6! 4 !5! 4 !6 !, , 9., , 10., , 3 × 2 × 1 = 3! = 6, This can also be checked by listing them as, follows, , Show that, (2n)!, = 2n (2n − 1)(2n − 3)...5.3.1, n!, , ABC BAC CAB, ACB BCA CBA., , Simplify, (2n + 2)!, (i), (2n)!, , Extending this result for n persons to be, placed in n chairs in a row, we get n! ways of, arrangement., , (n + 3)!, (ii) 2, (n − 4)(n + 1)!, , 1, 1, 1, n ! (n - 1)! (n - 2)!, , (iii), , Note that all persons are distinct and chairs, have their own ordinal numbers (1st, 2nd, 3rd etc.), , (iv) n[n !+ (n − 1)!] + n 2 (n − 1)!+ (n + 1)!, , Now we see a different problem. There are, 4 persons and 2 chairs in a row to be filled. The, 1st chair can be filled in 4 ways. Having done this,, the 2nd chair can be filled in 3 ways. So the number, of different arrangements is 4 × 3 = 12., , (v), , n + 2 3n + 1, −, n ! (n + 1)!, , (vi), , 1, 1− n, +, (n − 1)! (n + 1)!, , (vii), , 1, 3, n2 − 4, −, −, n ! (n + 1)! (n + 2)!, , See the enumeration, AB, , BA CA DA, , (viii), , n2 − 9, 6, 1, +, −, (n + 3)! (n + 2)! (n + 1)!, , AC, , BC, , AD, , BD CD DC, , 3.5 Permutations: (When all objects are distinct), , CB, , DB, , Extending this argument if 7 different, objects are available and 3 boxes are in a row. We, want to place one object in each box. There are, 7 × (7 - 1) × (7 - 2) = 7 × 6 × 5 different ways to, do it., , If we want to place 3 persons on 3 chairs in a, row, in how many ways can we obtain the seating, arrangement?, 50

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A permutation is formally defined as follows., , Now, using the multiplication principle of, counting, the total number of ways of filling r, , Permutation : A permutation is an, arrangement, in a definite order of a number of, objects, taken some or all at a time., , places using n distinct objects is denoted by nPr, and given by the product, n ×(n-1)×(n-2)×… ×([n-(r-2)]×[n-(r-1)], , The number of distinct permutations of r, distinct objects chosen from a given collection of, n distinct objects is denoted by nPr, nPr, or P(n,r)., , = n ×(n-1)×(n-2)×… ×(n-r+2)×(n-r+1), Hence nPr = n ×(n-1)×(n-2)×… ×(n-r+2)×(n-r+1), , 3.5.1 Permutations when all objects are, distinct [r ≤ n]:, , If we multiply and divide this product by, , Theorem 1. The number of permutations of, n distinct objects taken r (r ≤ n) at a time, without, repetitions, is, , (n-r)×(n-r-1)×… ×3 ×2 ×1,, we find that, n, , n × (n - 1) × (n - 2) × ... × (n - r + 1)., Proof. The number of ways of arranging n, distinct objects taken r at a time without repetitions, is same as the number of ways r places can be, filled using n objets (r ≤ n)., , =, , 1st 2nd, , 3rd, , (n − r ) × (n − r − 1) × ... × 3 × 2 ×1, (n − r ) × (n − r − 1) × ... × 3 × 2 ×1, , n × (n − 1) × (n − 2) × ... × (n − r + 1) × (n − r ) × (n − r − 1) × ... × 3 × 2 ×1, (n − r ) × (n − r − 1) × ... × 3 × 2 ×1, , =, , For this, consider the following table., Place, , Pr = n × (n − 1) × ... × (n − r + 1) ×, , n!, (n - r )!, , We have thus found that, , ..., , (r-2)th, , (r-1)th, , rth, , Number n n-1 n-2 ..., of ways, , [n-(r3)], , [n-(r2)], , [n-(r1)], , n, , Pr =, , n!, (n − r )!, , (for r ≤ n), , Note:, , The table shows that the first place can be, filled with any of the n objects. As the result,, there are n ways of filling the the first place. After, putting one object in the first place, only n-1, objects are available because repetitions are not, allowed. Therefore, the second place can be filled, in n-1 ways. After putting two distinct objects in, the first two places, only n-2 objects are available, for the third place, so that the third place can filled, in n-2 ways., , When r = n i.e. all n objects are placed in a, row., n, , Pn = n × (n − 1) × (n − 2) × ... × [n − (n − 1)], , = n × (n − 1) × (n − 2) × ... ×1 = n!, Alternatively, from the above formula, we, obtain, n, , Pn =, , n!, n!, =, (n − n)! 0 ! = n!, , (since 0!=1, by definition), , Continuing in this way, after putting r-1, objects in the first r-1 places, the number of, available objects is [n – (r-1)] for the rth place., Hence, the rth place can be filled in [n–(r-1)] ways., 51

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place. The other four digits can be arranged in, 4, P4 = 4! = 24 ways to form numbers that have 3, in the unit’s place. This shows that 24 of the 120, numbers have 3 in the unit’s place., , SOLVED EXAMPLES, Ex. 1) Find the value of 5P2., Solution: 5 P2 =, , , 5!, 5! 5 × 4 × 3!, = =, (5 − 2)! 3!, 3!, , Similarly, each of the other four digits is in, the unit’s place in 24 of the 120 numbers., , = 5 × 4 = 20., , The sum of the digits in the units place among, all 120 numbers is 24 (3+4+6+7+8) = 24×28 =, 672., , Ex. 2) How many different ways are there to, arrange letters of the word 'WORLD'? How many, of these arrangements begin with the letter R?, How many arrangements can be made taking, three letters at a time?, , Similary, the sum of the digits in the ten’s, place among all 120 numbers is 672. The same is, also the sum of the digits in each place among all, 120 numbers., , Solution: The word WORLD has 5 letters, W, O, R, L, D. These can be arranged among, themselves in 5P5=5!=120! different ways., , The required sum is 672 units + 672 tens, + 672 hundreds + 672 thousands + 672 ten, thousands., , If an arrangement begins with R, the, remaining four letters can be arranged among, themselves in 4P4=4!=24 ways., , = 672 × (1+10+100+1000+10000), = 672×11111 = 74,66,592, , The Number of arrangements of 5 letters,, 5!, taken 3 at a time, is 5P3= = 60, 2!, , Ex 5) A teacher has 2 different books on, English, 3 different books on Physics, and, 4 different books on Mathematics. These books, are to be placed in a shelf so that all books on any, one subjects are together. How many different, ways are there to do this?, , Ex. 3) How many three digit numbers can be, formed from the digits 2,4,5,6,7 if no digit is, repeated?, Solution: Every arrangement of digits gives, a different number. Therefore, the problem is to, find the number of arrangements of five digits, 5!, taken two at a time This is given by 5 P=, = 60, 3, 2!, , Solution : The five digits can be arranged in, 5, P5 = 5! = 120 ways., , Solution : First, let us consider all books on, each subjects to be one set, so that there are three, sets, say E,P,M. These three can be arranged in, 3, P3 = 3! = 6 different ways. Now, in each of these, ways, the 2 books on English can be arranged in, 2, P2 = 2! = 2 different ways the 3 books on physics, can be arragned in 3P3 =3! = 6 different ways, and, the four books on Mathematics can be arranged, in 4P4 = 4! = 24 different ways., , Now, consider any one of the five given, digits, say 3. Suppose the digit 3 is in the unit’s, , The required number of arrangments is then, given by 6×2×6×24 = 1728, , Ex 4) How many numbers can be formed with the, digits 3, 4, 6, 7, 8 taken all at a time? Find the sum, of all such numbers., , 52

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Ex 6) Find n if nP5 = 42 × nP3, , SOLVED EXAMPLES, , Solution: We are given that nP5 = 42 × nP3, That is,, , ∴, , Ex 1: It is required to arrange 8 books on a, shelf. Find the number of ways to do this if two, specified books are, , n!, n!, = 42, (n − 5)!, (n − 3)!, , n!, n!, = 42, (n − 5)!, (n − 3)(n − 4)(n − 5)!, , (i) always together, Solution:, , (i) Consider 2 books as one single set. These, 2 books can be arrannged among themselves, in 2! ways. The total 7 books can be arranged, among themselves in 7! ways., , ∴ (n − 3)(n − 4) = 42, ∴ (n − 3) (n − 4) = 7 × 6, , ∴n − 3 = 7, , The total number of ways = 2! 7! = 2 × 5040, = 10080., , ∴ n = 3 + 7 = 10, 3.5.2 Permutations when repetitions are, allowed:, , (ii) We take one of the two books and remaining, 6 books and arrange these in 7! ways. Then, we can place the remaining book in such, a way to avoid 2 places adjacent to the, 1st book, so there are 6 places available hence, 6 × 7! = 6 × 5040 = 30240., , We now consider problems of arranging, n objects taken r at a time when repetitions are, allowed., Theorem 2. The number of arrangements of n, distinct objects taken r at a time, when repetitions, are allowed, is same as the number of ways of, filling r places using n distinct objects when, repetitions are allowed. Consider the following, table., Place, Number, of ways, , 1st, n, , 2nd, n, , (ii) never together., , 3rd … (r-2)th (r-1)th, n …, n, n, , Alternative Method :, The number is also obtained by subtracting, from the total arrangements those when these, books are together i.e., 8! − 2 × 7! = (8 − 2) × 7! = 6 × 7! = 6 × 5040, = 30240., , rth, n, , Ex 2: In how many ways can 7, examination papers be arranged so that papers, 6 and 7 are never together?, , Because repetitions are allowed, each place, can be filled in n different ways., , Solution : By the same argument as in, example above, the number of ways in which any, two papers are never together is., , Using multiplication principle, it can be, concluded that the number of permutations of n, distinct objects taken r at a time, when repetitions, are allowed, is given by, , (7-2) (7-1)! = 5×6! = 3600, Ex 3: A family of 3 brothers and 5 sisters is to, be arranged for a photograph in such a ways that,, (i) all brothers sit together. (ii) no two brothers sit, together., , n×n×....×n(r times) = nr, , 53

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Solution :, , number of permutations of n-m+1 objects taken all, at a time is (n-m+1)!. The m specified objects are, together, but can be rearranged among themselves., The number of permutations of these m objects, taken all at a time is m!. Since this argument, holds for each of the (n-m+1)! permutations,, the required number of permutations of n objects, taken all together, when m specified objects are, always together is m! (n-m+1)!, , (i) Since all 3 brothers are together, treat them as, one person, so that there are 5+1=6 persons., the number of arranging them is, 6, , P6 = 6! = 720., , Once this is done, the three brothers can be, arranged among themselves in 3P3 = 3! = 6, ways., The total number of arrangements is then, given by 6!×3! = 720 × 6 = 4320., , Remarks:, , (ii) 5 sisters can be arranged among themselves, in 5P5 = 5! = 120 ways., , 1., , The number of permutations of n distinct, objects taken all at a time, when 2 specified, objects are always together, is 2×(n-1)!., , 2., , The number of permutations of n distinct, objects taken r at a time, when a specified, object is always to be included, is., r × (n-1)P (r-1)., , Consider the following arrangement., * S1 * S2 * S3 * S4 * S5 *, Where * indicates a position where one, brother can be placed so that no two brothers, are together. Since there are 6 such positions and, 3 brothers, the number of arrangements is, , P=, 3, , 6, , First keep the specified object aside. Arrange, (r−1) from the remaining (n−1) objects in, (n−1), P(r−1) ways. Then place the specified object, in r possible ways. Hence the total number of, arrangements is r × (n−1)P(r−1)., , 6!, = 120, 3!, , The required number of arrangements is, then given by, 5, , 3., , P5 × 6P3 = 120 × 120, , = 14400., If some of the n objects are always kept, together in a permutation problem, then the, following theorem is useful for such cases., , The number of permutations of n distinct, objects taken r at a time, when a specified, object is not be included in any permutation,, is (n-1)Pr., EXERCISE 3.3, , Theorem. The number of permutations taken, all at a time, when m specified objects among n, always come together, is m!(n-m+1)!, Proof. Since the specified m objects always, come together, let us consider them as a single, object. This makes the number of distinct objects, n-m+1 for the purpose of permutations. The, , 1., , Find n, if nP6 : nP3 = 120:1, , 2., , Find m and n, if (m+n)P2 = 56 and, (m-n), , 54, , P2 = 12, , 3., , Find r, if 12Pr-2 : 11Pr-1 = 3:14, , 4., , Show that (n+1) (nPr) = (n-r+1) [(n+1)Pr]

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5., , 6., , How many 4 letter words can be formed, using letters in the word MADHURI if, (a) letters can be repeated (b) letters cannot, be repeated., , 12. Find the number of 6-digit numbers using, the digits 3,4,5,6,7,8 without repetition., How many of these numbers are, , Determine the number of arrangements of, letters of the word ALGORITHM if., , 13. A code word is formed by two different, English letters followed by two non-zero, distinct digits. Find the number of such code, words. Also, find the number of such code, words that end with an even digit., , (a) divisible by 5, (b) not divisible by 5., , (a) vowels are always together., (b) no two vowels are together., (c) consonants are at even positions., , 14. Find the number of ways in which 5 letters, can be posted in 3 post boxes if any number, of letters can be posted in a post box., , (d) O is the first and T is the last letter., 7., , 8., , In a group photograph, 6 teachers are in the, first row and 18 students are in the second, row. There are 12 boys and 6 girls among the, students. If the middle position is reserved, for the principal and if no two girls are, together, find the number of arrangements., , 15. Find the number of arranging 11 distinct, objects taken 4 at a time so that a specified, object., (a) always occurs, , 16. In how many ways can 5 different books be, arranged on a shelf if, , Find the number of ways so that letters of, the word HISTORY can be arranged as,, , (i) there are no restrictions, , (a) Y and T are together, , (ii) 2 books are always together, , (b) Y is next to T., , (iii) 2 books are never together, , (c) there is no restriction, , 17. 3 boys and 3 girls are to sit in a row. How, many ways can this be done if, , (d) begin and end with vowel, (e) end in ST, , (i) there are no restrictions, , (f) begin with S and end with T, 9., , (b) never occurs., , (ii) there is a girl at each end, (iii) boys and girls are at alternate places, , Find the number of arrangements of the letters, in the word SOLAPUR so that consonents, and vowels are placed alternately., , (iv) all boys sit together, 3.5.3 Permutations when some objects are, identical, , 10. Find the number of 4-digit numbers that can, be formed using the digits 1, 2, 4, 5, 6, 8 if, , Consider the problem of arranging letters, in the words like GOOD, INDIA, GEOLOGY,, MATHEMATICS, or PHILOSOPHY, where, some letters occur more than once and hence all, letters are not distinct., , (a) digits can be repeated, (b) digits cannot be repeated, 11. How many numbers can be formed using the, digits 0, 1, 2, 3, 4, 5 without repetition so, that resulting numbers are between 100 and, 1000?, , Consider the word ODD. First we consider, the 2 D’s as distinct objects D1 and D2., 55

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The total number of words is OD1D2, OD2D1,, D1OD2, D2OD1, D1D2O, D2D1O = 6 = 3!, , SOLVED EXAMPLES, , But D1 and D2 are identical so, OD1D2 =, OD2D1, D1D2O = D2D1O, D2OD1 = D1OD2 and as, D1 and D2 can be arranged among themselves in, 2! ways, (D1D2, D2D1), , Ex 1 : Find he number of permutations of the, letters of the word UBUNTU., , 6, 3!, =, = 3 different words, 2, 2!, formed (ODD,DOD,DDO), , Therefore, number of permutations of the, 6!, letters of the word UBUNTU =, = 120., 3!, , Theorem : Consider a set of n objects where, n1 objects are identical and the remaining n - n1, are distinct. The number of permutations of these, n!, n objects is, . Note that n1 < n., n1 !, , Ex 2 : How many arrangements can be made,, with the letters of the word CALCULATOR? In, how many of these arrangements, vowels occur, together?, , Solution : The word UBUNTU consists of, 6 letters, in which letter ‘U’ is repeated 3 times., , Thus there are, , Solution : The word CALCULATOR, consists of 10 letters, in which ‘C’ is repeated two, times, ‘A’ is repeated two times, ‘L’ is repreated, two times and rest all are different., , Proof : Let m be the total number of, arrangements, where n1 out of n objects are, identical. The number of permutations of n, different objects is n!. Now n1 objects can be, rearranged in n1! ways among themselves if they, were all different. Thus each arrangement where, these n1 objects are identical corresponds to n1!, different arrangements if they were all different., n!, Hence m × n1! = n!. Therefore m =, n1 !, , Therefore, number of permutations of the, 10 !, letters of the word CALCULATOR =, ., 2! 2! 2!, The word CALCULATOR consists of, 4 vowels A, U, A, O. Let us consider them as a, single letter say P., , Remarks., 1., , Therefore, now we have 7 letters P, C, L, C,, L, T, R in which ‘C’ is repeated two times, ‘L’ is, repeated two times. The number of arrangement, 7!, of these 7 letters is given, = 1260. After, 2! 2!, this is done, 4 vowels (in which ‘A’ is repeated, 4!, 2 times) can be arranged in, = 12 ways, 2!, , The number of permutations of n objects, not, all distinct, where n1 objects are of one type, and n2 objects are of a second type, taken all, n!, at a time is, and note that n1 + n2 ≤ n., n1 !n2 !, The proof is similar to the proof of the above, theorem., , 2., , The number of permutations of n objects, not, all distinct, where ni objects are of type i, i=1,, n!, 2, …, k, taken all at a time is, ,, n1 !n2 !...nk !, n1+ n2+...nk ≤ n., , Therefore, number of arrangements of the, letters of the word CALCULATOR in which, vowels are together = 1260 × 12 = 15120., , 56

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10. Find the number of different ways of, arranging letters in the word ARRANGE., How many of these arrangement do not have, the two R’s and two A’s together?, , EXERCISE 3.4, 1., , Find the number of permutations of letters in, each of the following words., , 11. How many distinct 5 digit numbers can be, formed using the digits 3, 2, 3, 2, 4, 5., , (i) DIVYA, (ii) SHANTARAM, , 12. Find the number of distinct numbers formed, using the digits 3, 4, 5, 6, 7, 8, 9, so that odd, positions are occupied by odd digits., , (iii) REPRESENT, (iv) COMBINE, (v) BALBHARATI, 2., , 3., , 14. Find the number of distinct words formed, from letters in the word INDIAN. How, many of them have the two N’s togethers?, , A coin is tossed 8 times. In how many ways, can we obtain. (a) 4 heads and 4 tails? (b) at, least 6 heads?, , 4., , A bag has 5 red, 4 blue, and 4 green marbles., If all are drawn one by one and their, colours are recorded, how many different, arrangements can be found?, , 5., , Find the number of ways of arranging letters, of the word MATHEMATICAL How many, of these arrangements have all vowels, together?, , 6., , 13. How many different 6-digit numbers can be, formed using digits in the number 659942?, How many of them are divisible by 4?, , You have 2 identical books on English,, 3 identical books on Hindi, and 4 identical, books on Mathematics. Find the number of, distinct ways of arranging them on a shelf ., , 15. Find the number of different ways of, arranging letters in the word PLATOON, if. (a) the two O’s are never together. (b), consonants and vowels occupy alternate, positions., 3.5.4 Circular permutation:, We can imagine a circular arrangement, of n different objects to be transformed into an, arrangement in a line by cutting the circle in a, place. This cut can be made at n different places., , Find the number of different arrangements, of letters in the word MAHARASHTRA., How many of these arrangements have, (a) letters R and H never together? (b) all, vowels together?, , For example, consider n = 4, , 7., , How many different words are formed if the, letters R is used thrice and letters S and T are, used twice each?, , 8., , Find the number of arrangements of letters, in the word MUMBAI so that the letter B is, always next to A., , Fig. 3.5, , Find the number of arrangements of letters, in the word CONSTITUTION that begin, and end with N., , Thus given a single circular arrangement, of 4 objects, it corresponds to 4 different linear, arrangements (arrangements in a row)., , 9., , 57

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Let the circular arrangements be m., We know that the total number of linear, arrangement of 4 different objects is 4!, Each arrangement in row corresponds to, some circular arrangement., Therefore number of circular arrangements, 4!, is m =, = 3!, 4, , Fig. 3.5 (b), , Similarly, each circular arrangement of n, objects corresponds to n different arrangements, in a row., , Consider the 4 directions, E, N, W, S on, Mariner's compass. If 4 stickers of different, colours are to be placed at the different letters., , Therefore number of circular arrangements, n!, is, = (n − 1)!, n, , The number of arrangements is 4!. Here, rotation of W, S, E, N is not allowed. Hence, arrangement is like a linear arrangement. Such, , Note:, , arrangement is called as an arrangement with, respect to round table., , 1), , A circular arrangement does not have a, fixed starting point and any rotation of, it is considered to be the same circular, arrangement. This arrangement will be the, same with respect to each other if a rotation, is made but clockwise and anticlockwise, arrangements are different., , Now consider the situation where m out of, the n objects are alike and cannot be distinguished, from one another. The following result gives the, formula for the number of circular arrangements, of these n objects., Theorem. The number of circular, arrangements of n objects, of which m objects are, (n-1)!, alike (identical), is given by, m!, Proof. As argued earlier, m like, (indistinguishable) objects can be rearranged, among themselves in m! ways without affecting, the arrangements of the n objects. The number, of circular arrangements of n objects is, (n-1)! and the number of arrangements of m like, (indistinguishable) objects among themselves is, m!, the required number of arrangements is given, (n -1)!, by, m!, , Fig. 3.5 (a), In circular arrangement, (with respect to each, other) all these count as one arrangement., 2), , If clockwise and anticlockwise circular, arrangements are considered to be the same, then each circular arrangement corresponds, to 2n different linear arrangements. Thus, the numbers of such circular arrangement, is, , Does this argument remind us of a similar, argument that we came across earlier?, , n ! (n -1)!, =, ., 2n, 2, 58

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Remark: The number of circular permutations, of r objects taken from n distinct objects can be, found under two different conditions as follows., , p2 = 2! ways. Hence, the required number of ways, in which two particular students come together, = 7! × 2! = 10080., , (a) When, clockwise, and, anticlockwise, arrangements are considered to be different,, then the required number of circular, , Ex 2 : In how many ways 6 men and, 3 women can be seated at a round table so that, every man has woman by his side., , arrangements is given by, , n, , 2, , pr, r, , (b) When, clockwise, and, anticlockwise, arrangements are not to be considered, different, then the required number of circular, n, , pr, 2r, Verify these two statements for n = 6, r = 3., arrangements is given by, , Fig. 3.6, , SOLVED EXAMPLES, , Solution : 3 women have (3 − 1)! = 2 ways, of circular seatings., , Ex 1 : In how many ways can 8 students, be arranged at a round table so that 2 particular, students are together, if, , In each seating a women has one place on, each side for a man. Thus there are 6 different, places for 6 men which can be filled in 6! ways., , (i) students are arranged with respect to, each other? (That is the seats are not numbered.), , Hence, the total number of required, arrangements = 6! × 2 = 720 × 2 = 1440., , (ii) students are arranged with respect to the, table? (That is the seats are numbered serially.), , Ex 3 : Find the number of ways in which, 12 different flowers can be arranged in a garland, so that 4 particular flowers are always together., , Solution : Considering those 2 particular, students as one student, we have 7 students., , Solution : Considering 4 particular flowers, as a single flower, we have 9 flowers which, can be arranged to form a garland in 8! ways., But 4 particular flowers can be arranged in 4!, ways. Hence, the required number of ways, 1, = (8! × 4!) = 483840., 2, , (i) In circular arrangement 7 students can, be arranged at a round table, in 6! ways and, 2 students can be arranged among themselves in, 2, p2 = 2! ways. Hence, the required number of ways, in which two particular students come together =, 6! × 2! = 1440., (ii) Here the arrangement is like an, arrangement in a row., , Ex 4 : How many necklaces of 12 bead each,, can be made from 18 beads of different colours?, , So 7 students can be arranged in 7! ways and, 2 students can be arranged amongst themselves in, 59

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Solution : Here, clockwise and anticlockwise, arrangements are same. Hence, total number of, , When 6 gentlemen are arranged around a, table, there are 6 positions, each being between, two gentlemen, for 6 ladies when no two ladies sit, side by side. Now, the number of ways in which, 6 gentlemen can be seated around a table, = (6 − 1)! = 5!, , 18, P12, P12, circular permutations =, =, ., 24, 2 ×12, 18, , Ex 5 : Three boys and three girls are to be, seated around a table in a circle. Among them, the, boy X does not want any girl as neighbour and, girl Y does not want any boy as neighbour. How, many such distinct arrangements are possible?, , Now, corresponding to each seating, arrangement for the gentlemen, the 6 ladies can, be seated in 6! ways., Thus, the required number of arrangements, = (5!) (6!) = 86400., Ex 7 : In how many different ways can, 4 married couples occupy seats around a circular, table if (i) Spouses sit opposite to each other?, (ii) Men and women alternate?, , Fig. 3.7, Solution : The arrangement is as shown in, the figure (fig. 3.7). The boy X will have B2 and, B3 as neighbours. The girl Y will have G2, G3 as, neighbours. The two boys B2, B3 can be arranged, in two ways. The two girls G2, G3 can be arranged, in two ways., , Fig. 3.8, , Hence, the total number of arrangements, = 2 × 2 = 4., , Solution : (i) Spouses sit opposite to each, other. Since there is no fixed starting point let one, woman occupy one seat. Let, W1H1, W2H2, W3H3, and W4H4 be the four couples. Since there is no, fixed starting point of the circle, Let W1 occupy, one seat then H1 has to sit exactly opposite. There, are 3 seats on either side of W1. Then W2 can, occupy any of the 6 seats which confirms the seat, of her husband H2. Now there are 4 seats left. W3, can occupy anyone of them so that her husband's, seat is fixed. Now W4 has 2 choices, fixing the, seat of her husband. In all there are 6 × 4 × 2 =, 48 ways., , Ex 6 : In how many different arrangements, can 6 gentlemen and 6 ladies sit around a table if, (i) there is no restriction. (ii) no two ladies sit side, by side?, Solution : (i) There is no restriction., Here, the total number = 6 + 6 = 12., 12 persons can be arranged in circular, permutation in (12−1)! = 11! ways., (ii) No two ladies sit side by side., 60

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ii) Let 4 women sit on alternate seats around, a table starting at any place. This is done in 3!, ways. It leaves 4 alternate seats empty. They are, fixd by 4 men in 4! ways. Hence, the total number, of ways is 4! × 3! = 144., , 10. Fifteen persons sit around a table. Find, the number of arrangements that have two, specified persons not sitting side by side., Properties of Permutations, (i), , (ii) nP0 = 1, , EXERCISE 3.5, 1., 2., , 3., , (iii) nP1 = n, , In how many different ways can 8 friends sit, around a table?, , (iv) nPr = n × (n-1)P(r-1), , A party has 20 participants. Find the number, of distinct ways for the host to sit with them, around a circular table. How many of these, ways have two specified persons on either, side of the host?, , Find the number of ways for 15 people to sit, around the table so that no two arrangements, have the same neighbours., , 5., , A committee of 10 members sits around a, table. Find the number of arrangements that, have the president and the vice president, together., , 6., , Five men, two women, and a child sit around, a table. Find the number of arrangements, where the child is seated (a) between the two, women. (b) between two men., , 7., , Eight men and six women sit around a table., How many of sitting arrangements will have, no two women together?, , 8., , Find the number of seating arrangements for, 3 men and 3 women to sit around a table so, that exactly two women are together., , 9., , Four objects in a set of ten objects are alike., Find the number of ways of arranging them, in a circular order., , , , = n(n-1) × (n-2)P(r-2), , , , = n(n-1)(n-2) × (n-3)P(r-3) and so on., n, , (v), , Delegates from 24 countries participate in, a round table discussion. Find the number, of seating arrangments where two specified, delegates are. (a) always together, (b) never, together., , 4., , Pn = n!, , n, , n, , pr, , p( r −1), , = n − r +1, , 3.6 Combinations, Permutations involve ordered arrangements, of objects. We shall now consider situations, where the order of objects in an arrangement is, immaterial, but only selection of a set of objects, in the arrangement is considered. A selection of, objects without any consideration of the order is, called a combination., Consider the earlier example from, permutations where 2 chairs were filled from, a group of 4 persons. We make a little change, in the problem. We want to select a group of, 2 people and not consider the order. So the, arrangements AB and BA correspond to the same, group. Similarly BC and CB are given in the same, group. The list is given as follows.,  AB , , ,  BA , ,  BC , , ,  CB , ,  CA , , ,  AC , ,  AD , , ,  DA , ,  BD , , ,  DB , ,  CD , , ,  DC , , P2 4 × 3, =6, =, 2 ×1, 2, different groups selected. This is called the, combination number of selecting a group of 2, from 4 persons denoted by 4C2., 4, , Thus there are altogether, , 61

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Definition Combination. A combination of, a set of n distinct objects taken r at a time without, repetition is an r-element subset of the n objects., , as has been stated earlier., 3., , If nCr = nCs, then either s = r or s = n-r., , Note: The order of arrangement of the, elements is immaterial in a combination., , 4., , n, , If we want to choose a team of 3 players from, a set of 8 different players, we first get the number, 8, P3, i.e. different ordered sets of 3 players and, since any set of 3 gives 3! ordered sets, we divided, 8, P3 by 3!. Thus the number of combinations of, , 5., , n, , 6., , n, , 7., , C0 + nC2 + nC4 + … = nC1 + nC3 + nC5 + …, = 2(n-1), , 8., , n, , 9., , n, , P3, 8!, =, 3! (8 − 3)!3!, , Combination (nCr) : From n different, objects, the number of ways of selecting a group, or a set of r objects (without considering order) is, denoted by nCr or C(n,r) or nCr. It is the number, of combinations of r nCr objects from n distinct, objects., n!, n, Theorem : Cr =, (n − r )!r !, , C0 + nC1 + … nCn = 2n, , n, , (i), , Pr, n!, =, r ! (n − r )!r !, , 3.6.1 Properties of combinations., =, , n!, (n - r )![n - (n - r )]!, , , , =, , n!, (n - r )!r !, , , , =, , n, , n, , C0 =, , =, , n!, r !(n - r )!, , C3 =, , 7!, 3!(7 - 3)!, , 7, , =, , 7!, 3!(4)!, , =, , 7 × 6 × 5 × 4!, 3 × 2 × 1× 4 !, , =, , 7×6×5, = 35, 3 × 2 ×1, , (ii), , 10, , C7 =, , Cr., , Thus, nCn−r = nCr for 0 ≤ r ≤ n., 2., ,  n   n −1 , C(r-1) =   , ,  r   r −1 , , (n-2), , C(r-2) =..., , n, Cr has maximum value if (a) r =, when n, 2, n -1, n +1, is even (b) r =, or, when n is odd., 2, 2, , nCr, , n, , Consider nCn - r, , (n-1), , Ex 1 : Find the value of (i) 7C3 (ii) 10C7 (iii) 52C3, Solution : We know that, , corresponds to r! different ordered arrangements., So, we count the number of different sets or r, objects, without considering the order among, , 1., , n, Cr=  , r, , SOLVED EXAMPLES, , Proof : First we find the number of ‘ordered’, sets of r objects from n distinct objects., n!, ∴ n Pr =, . Now, each set of r objects, (n − r )!, , them, is =, , Pr, r!, , Cr + nCr-1 = n+1Cr, , 8, , 3 players from 8 players is, , n, , Cr =, , n!, n!, = = 1, because 0 ! = 1, 0 !(n − 0)! n !, 62, , 10 !, 7 !(10 - 7)!, , =, , 10 !, 7 !(3)!, , =, , 10 × 9 × 8, = 120, 3× 2

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(iii), , C3 =, , 52, , =, , collinear, therefore they form only one line and no, triangle i.e. we have counted (pC2 − 1) extra lines, and pC3 extra triangles., , 52 !, 52 !, =, 3!(52 - 3)! 3!(49)!, 52 × 51× 50, = 22100, 3 × 2 ×1, , ∴ If p-points (p≥3) are collinear, then number, of straight lines = nC2 − (pC2 − 1) and number of, triangles = nC3 − pC3, , Ex 2 : Find n and r, if nCr-1 : nCr : nCr+1 = 14:8:3, n, , Ex 4 : Four cards are drawn from a pack of, 52 playing cards. In how many different ways can, this be done? How many selections will contain, , n, , Cr, 8, C, 14 7, =, Solution : n r −1 = = and n, Cr +1 3, 8 4, Cr, ∴, , r +1 8, r, 7, =, = and, n−r 3, n − r +1 4, , (i) exactly one card of each suit?, (ii) all cards of the same suit?, (iii) all club cards?, , 3, 4 , ∴ (n−r) =  r  −1 and (n−r) = (r + 1), 8, 7 , 3, 4 , ∴  r  −1 = (r + 1), 8, 7 , , (iv) at least one club card?, (v) three kings and one queen?, (vi) three black and one red cards?, , Solving this we get r = 7 and n = 10, , Solution : A pack of 52 cards contains, 4 different suits, viz. Club, Spade, Diamond and, Heart. Each suit contains 13 cards. Club and Spade, are black coloured cards. Diamond and Heart are, red colour cards. i.e. pack of 52 cards contains, 26 black and 26 red colour cards., , Ex 3 : There are n points in a plane. Find the, number of straight lines and triangles that can be, obtained by joining points on a plane if, (i) no three points are collinear, (ii) p-points are collinear (p≥3), , From a pack of 52 cards, any 4 cards can be, 52 !, drawn in 52C4 =, ways., 4 ! × 48!, , Solution : (i) Straight line can be drawn by, joining any two points and traingle can be drawn, by joining any three non-collinear points., , (i) Exactly one card of each suit., , From n-points, any two points can be selected, n, in C2 ways., , One club card can be selected in 13C1 = 13, ways. One heart card can be selected in 13C1 = 13, ways. One spade card can be selected in 13C1 = 13, ways. One diamond card can be selected in 13C1 =, 13 ways., , ∴ Number of straight lines = nC2, Since no three points are collinear, any three, non-collinear points can be selected in nC3 ways., ∴ Number of triangles = nC3, , ∴ Using fundamental principle, exactly one, card of each suit can be selected in 13 × 13 × 13 ×, 13 = (13)4 ways., , (ii) If P points are non-collinear then, we can, obtain PC2 straight lines and PC3 triangles from, those p points. But we are given that p points are, 63

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Ex 5 : Find n, if nC8 = nC6, , (ii) All cards are of the same suit., From 4 suits, any one suit can be selected in, 4, C1 = 4 ways. After this is done, any four cards, from selected suit can be drawn in 13C4 = 715, ways., , Solution : If nCx = nCy , then either x = y or x = n−y, , ∴ Using fundamental principle, all 4 cards, of the same suit can be selected in 4 × 715 = 2860, ways., , Ex 6 : Find r, if 16C4 + 16C5 + 17C6 + 18C7 = 19Cr, , ∴ 8 = n−6, , (⸪ x ≠ y), , ∴ n = 14, , Solution : (16C4 + 16C5) + 17C6 + 18C7 = 19Cr, ∴ (17C5 + 17C6) + 18C7 = 19Cr, , (iii) All club cards, , ∴ (18C6 + 18C7) = 19Cr, , From 13 club cards, any 4 club cards can be, drawn in 13C4 = 715 ways., , ∴, , C7 = 19Cr, , 19, , ∴ r = 7 or r = 19−7 = 12, , (iv) At least one club card, From a pack of 52 cards, any 4 cards can be, drawn in 52C4 ways if there is no condition., , Ex 7 : Find the difference between the maximum, values of 8Cr and 11Cr, , From 52C4 selections, remove those selections, that do not contain any club card, so that in the, remaining selection we have at least one club, card., , Solution :, Maximum value of 8Cr occurs at r =, , 8, =4, 2, , ∴ maximum value of 8Cr = 8C4 = 70, , If the selection does not contain any club card, i.e. 4 cards are drawn from remaining 39 non-club, cards, then this can be done in 39C4 ways., , Maximum value of 11Cr occurs at r =, or at, , ∴ Number of selections which contain at, least one club card = 52C4 − 39C4 ., , 12, =6, 2, , ∴ Maximum value of, = 462, , (v) Three king cards and one queen card, , 10, =5, 2, , Cr =, , 11, , C5 =, , 11, , C6, , 11, , From 4 kings, any 3 king cards can be, selected in 4C3 = 4 ways and from 4 queen cards, 1 queen card can be selected in 4C1 = 4 ways., , ∴ difference between the maximum values, of 8Cr and 11Cr = 11C5 − 8C4 = 462 − 70 = 392, , ∴ Using fundamental principle, three king, cards and one queen card can be selected in 4 × 4, = 16 ways., , EXERCISE 3.6, 1., , (vi) Three black cards and one red card, , Find the value of (a) 15C4 (b) 80C2, (c) 15C4 +15C5 (d) 20C16 - 19C16, , From 26 black cards, any 3 black cards can, be selected in 26C3 ways and from 26 red cards,, any one red card can be selected in 26C1 ways., , 2., , Find n if, (a) 6P2 = n 6C2, (b) 2nC3: nC2 = 52 : 3, , Using fundamental principle, three black and, one red card can be selected in 26C3 × 26C1 ways., , (c) nCn-3 = 84, 64

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3., , Find r if 14 C2r : 10C2r-4 = 143:10, , 4., , Find n and r if., , 14. Find the number of triangles formed by, joining 12 points if, (a) no three points are collinear, , (a) nPr = 720 and nCn-r = 120, , (b) four points are collineas., , (b) nCr-1 : nCr : nCr+1 = 20: 35:42, 5., , If nPr = 1814400 and nCr = 45, find n+4 Cr+3, , 6., , If nCr-1 = 6435, nCr = 5005, nCr+1 = 3003,, , 15. A word has 8 consonants and 3 vowels., How many distinct words can be formed if, 4 consonants and 2 vowels are chosen?, , find rC5., , 16. Find n if,, , 7., , 8., , Find the number of ways of drawing 9 balls, from a bag that has 6 red balls, 8 green balls,, and 7 blue balls so that 3 balls of every colour, are drawn., , (i) nC8 = nC12, , Find the number of ways of selecting a team, , (v) nCn-2=15, , (ii), , (iv) 2nCr-1 = 2nCr+1, , 17. Find x if nPr = x nCr, , After a meeting, every participant shakes, hands with every other participants. If the, number of handshakes is 66, find the number, of participants in the meeting., , 18. Find r if 11C4+ 11C5 + 12C6 + 13C7 = 14Cr, 4, , 19., , 10. If 20 points are marked on a circle, how many, chords can be drawn?, , (b) n = 15, , (c) n = 12 , , (d) n = 8, , Find the value of ∑, , ( 21− r ), , C4, , r =1, , 20. Find the differences between the greatest, values in the following:, (a) 14Cr and 12Cr ,, , 11. Find the number of diagonals of an n-sided, polygon. In particular, find the number of, diagonals when., (a) n = 10 , , C3n = 23C2n+3, , (iii) 21C6n = 21C(n2+5), , of 3 boys and 2 girls from 6 boys and 4 girls., 9., , 23, , (b) 13Cr and 8Cr ,, , (c) 15Cr and 11Cr,, 21. In how many ways can a boy invite his, 5 friends to a party so that at least three join, the party?, , 12. There are 20 straight lines in a plane so that, no two lines are parallel and no three lines, are concurrent. Determine the number of, points of intersection., , 22. A group consists of 9 men and 6 women., A team of 6 is to be selected. How many, of possible selections will have at least, 3 women?, , 13. Ten points are plotted on a plane. Find the, number of straight lines obtained by joining, these points if, , 23. A committee of 10 persons is to be formed, from a group of 10 women and 8 men. How, many possible committees will have at least, 5 women? How many possible committees, will have men in majority?, , (a) no three points are collinear., (b) four points are collinear., 65

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(1) Linear Permutation :, , 24. A question paper has two sections. section I, has 5 questions and section II has 6 questions., A student must answer at least two question, from each section among 6 questions he, answers. How many different choices does, the student have in choosing questions?, , (a) The number of permutation of n different, objects taken r at a time when repetition, of r objects in the permutation is not, allowed is given by, , 25. There are 3 wicketkeepers and 5 bowlers, among 22 cricket players. A team of 11, players is to be selected so that there is exactly, one wicketkeeper and at least 4 bowlers in, the team. How many different teams can be, formed?, , n, , Pr =, , (b) The number of permutations of n different, objects, taken r objects at a time, when, repetition of r objects in the permutation, is allowed, is given by nr., , 26. Five students are selected from 11. How, many ways can these students be selected if., (a) two specified students are selected?, (b) two specified students are not selected?, , (c) The number of permutations of n objects,, when p objects are of one kind, q objects, are of second kind, r objects are of third, kind and the rest, (if any), are of different, , Let's Remember, •, , kind is, , Factorial notation :, n! or n = 1,2,3.... (n−2) (n−1) n,, , The arrangements in a circle are called, circular permutations., , Principle of Addition :, If an event can occur either in m or n, mutually exclusive alternate ways, then the, total number of ways in which the event can, occur is m + n., , •, , (a) The number of circular permutations of, n different objects = (n−1)!, (b) The number of ways in which n things of, which p are alike, can be arranged in a, , Principle of Multiplication :, , circular order is, , If an event has m possible outcomes, and, another independent event has n possible, outcomes, then there are m.n possible, outcomes for the two events together., •, , n!, p!q!r!, , (ii) Circular Permutation:, , (0! = 1), •, , n!, where r ≤ n, (n - r )!, , •, , (n - 1)!, p!, , Combination :, A combination is a selection. Total number, of selections of 'n' different objects, taken 'r', at a time is denoted by nCr or nCr or C(n,r),, n, or or   and is given by, r , n, Pr, n!, n, Cr =, =, r!, r !(n - r )!, , Permutation :, A permutation is an arrangement, in a, definite order, of a number of objects, taken, some or all at a time., , 66

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•, , Properties of nCr:, , (a), , n, , 4), , Cr = nCn−r, , (b) If nCx = nCy, then either x = y or x + y = n, (c), , n, , Cr + nCr−1 = n+1Cr, , A) 9×8!, , 1 ≤ r ≤ n (Pascal's rule), , 5), , (d) The number of ways of selecting one or, more things from n different things is given, by 2n−1., (e), , n, , 6), , n, when n is even, 2, , (ii) r =, , n -1 n +1, or, when n is odd., 2, 2, , MISCELLANEOUS EXERCISE - 3, (I) Select the correct answer from the given, alternatives., 1), , A) 5, 2), , B) 3, , C) 8, , D) 15, , B) 4, , C) 42, , 9), , D) 10, , C) 144, , D) 256, , B) 56, , C) 24, , D) 8, , B) 750, , C) 40, , D) 11340, , There are 10 persons among whom two are, brothers. The total number of ways in which, these persons can be seated around a round, table so that exactly one person sits between, the brothers, is equal to:, D) 3!×8!, , The number of arrangements of the letters of, the word BANANA in which two N's do not, appear adjacently., A) 80, , In how many ways can 8 Indians and,, 4 American and 4 Englishmen can be seated, in a row so that all person of the same, nationality sit together?, A) 3! 8! , C) 4! 4! , , B) 288, , A) 2!×7! B) 2!×8! C) 3!×7!, , A college has 7 courses in the morning and, 3 in the evening. The possible number of, choices with the student if he wants to study, one course in the morning and one in the, evening is-., A) 21, , 3), , 8), , D) 8×9!, , A question paper has two parts, A and B,, each containing 10 questions. If a student has, to choose 8 from part A and 5 from part B, In, how many ways can he choose the questions?, A) 320, , A college offers 5 courses in the morning, and 3 in the evening. The number of ways a, student can select exactly one course, either, in the morning or in the evening., , C) 9×9!, , Find the number of triangles which can be, formed by joining the angular points of a, polygon of 8 sides as vertices., A) 16, , 7), , B) 8×8!, , In how many ways 4 boys and 3 girls can be, seated in a row so that they are alternate., A) 12, , Cr has maximum value, if, , (i) r =, , In how many ways can 10 examination, papers be arranged so that the best and the, worst papers never come together?, , B) 60, , C) 40, , D) 100, , 10) The number of ways in which 5 male and, 2 female members of a committee can be, seated around a round table so that the two, females are not seated together is, , B) 3! 4! 8! 4!, D) 8! 4! 4!, , A) 840, , 67, , B) 600, , C) 720, , D) 480

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(II) Answer the following., 1), , Find the value of r if 56Pr+2: 54Pr-1= 30800:1, , 2), , How many words can be formed by writing, letters in the word CROWN in different, order?, , 3), , 4), , 11) A hall has 12 lamps and every lamp can be, switched on independently. Find the number, of ways of illuminating the hall., , Find the number of words that can be formed, by using all the letters in the word REMAIN, If these words are written in dictionary order,, what will be the 40th word?, , 12) How many quadratic equations can be, formed using numbers from 0,2,4,5 as, coefficients if a coefficient can be repeated, in an equation., , Capital English alphabet has 11 symmetric, letters that appear same when looked at in a, mirror. These letters are A, H, I, M, O, T, U,, V, W, X, and Y. How many symmentric three, letter passwords can be formed using these, letters?, , 5), , How many numbers formed using the digits, 3,2,0,4,3,2,3 exceed one million?, , 6), , Ten students are to be selected for a project, from a class of 30 stdudents. There are, 4 students who want to be together either, in the project or not in the project. Find the, number of possible selections., , 7), , 10) Nine friends decide to go for a picnic in two, groups. One group decides to go by car and, the other group decides to go by train. Find, the number of different ways of doing so if, there must be at least 3 friends in each group., , 13) How many six-digit telephone numbers can, be formed if the first two digits are 45 and no, digit can appear more than once?, 14) A question paper has 6 questions. How many, ways does a student have to answer if he, wants to solve at least one question?, 15) Find the number of ways of dividing 20, objects in three groups of sizes 8,7,and 5., 16) There are 4 doctors and 8 lawyers in a panel., Find the number of ways for selecting a team, of 6 if at least one doctor must be in the team., , A student finds 7 books of his interest, but, can borrow only three books. He wants, to borrow Chemistry part II book only if, Chemistry Part I can also be borrowed. Find, the number of ways he can choose three, books that he wants to borrow., , 8), , 30 objects are to be divided in three groups, containing 7,10,13 objects. Find the number, of distinct ways for doing so., , 9), , A student passes an examination if he, secures a minimum in each of the 7 subjects., Find the number of ways a student can fail., , v, , 17) Four parallel lines intersect another set of, five parallel lines. Find the number of distinct, parallelograms formed., 18) There are 12 distinct points A,B,C,.....,L, in, order, on a circle. Lines are drawn passing, through each pair of points, i) How many lines are there in total., ii) How many lines pass through D., iii) How many triangles are determined by, lines., iv) How many triangles have on vertex C., , v, , 68, , v

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4 METHOD OF INDUCTION AND BINOMIAL THEOREM, Step 1 : (Foundation) The 1st domino falls, down., , Let's Study, •, •, •, •, •, , Mathematical Induction, Binomial Theorem, General term of expansion, Expansion for negative and fractional index, Binomial coefficients, , (followed by it 2nd also falls down. Then 3rd,, 4th and so on.), Step 2 : (Assumption) Assume if kth domino, falls down., , Let's Learn, Introduction :, The earliest implicit proof by induction was, given by Al Karaji around 100 AD. The first, explicit formulation of the principle was given by, Pascal in 1665. The Mathematical Induction is a, powerful method, easy to use for proving many, theorems., , Fig. 4.1, Step 3 : (Succession) Followed by kth domino,, (k + 1)th domino will also fall down., , 4.1 Principle of Mathematical Induction :, , Step 4 : (Induction) It is true that all the dominos, will fall down., , Principle of Mathematical Induction consists of, the following four 4 steps:, , We will see how to use the principle of, mathematical induction to prove statements., , Step 1 : (Foundation) To prove P(n) is true for, n=1, , Stepwise Explanation :, , (It is advisable to check if P(n) is true for n = 2, 3, also if P(1) is trivial)., , Step 1. (Foundation) Formulate the statement of, the theorem as P(n) say, for any positive integer, n and verify it for integer n = 1. In fact, it is, often instructive, though not necessary, to verify, the statement for n = 2 and 3. This gives better, insight into the theorem., , Step 2 : (Assumption) To assume P(n) is true, for n = k., Step 3 : (Succession) To prove that P(n) is true, for n = k + 1., Step 4 : (Induction) To conclude that P(n) is true, for all n ∈ N, , Step 2. (Assumption) Assume that the statement, P(n) is true for a positive integer k., , Row of dominos standing close to each other gives, us the idea of how the Principle of Mathematical, Induction works., , Step 3. (Succession) Prove the statement for, n = k + 1., 69

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Step 4. (Induction) Now invoke the principle, of Mathematical induction. Conclude that the, theorem is true for any positive integer n., , Step 4 : (Induction) Now by the Principle of, Mathematical induction, the statement P(n) is, proved for all positive integers n., , Illustration :, , SOLVED EXAMPLES, , Let us prove a theorem with this method., The theorem gives the sum of the first n positive, integers., , Ex.1 By method of induction, prove that., n, 1.3 + 2.5 + 3.7 + .... + n (2n+1) = (n+1)(4n+5), 6, for all n ∈ N, , It is stated as P(n) : 1 + 2 + 3 . . . + n = n(n+1)/2., Step 1 : (Foundation), , Solution :, , To prove P(n) is true for n = 1, , Let P(n) ≡ 1.3 + 2.5 + 3.7 + .... + n (2n+1),, , 1(1 + 1), L.H.S = 1 R.H.S =, = 1 which is trivially, 2, true., , for all n ∈ N, n, = (n+1)(4n+5), 6, , Check that 1 + 2 =, 1+2+3=, , 2 × (2 + 1), and, 2, , Step (I) : (Foundation) To prove P(1) is true, Let n = 1, , 3 × (3 + 1), , so P(2) and P(3) are also, 2, , L. H. S. = 1.3 = 3, R. H. S. =, , true., , =, , Step 2 : (Assumption) Assume that P(n) is true, for n = k and in particular,, , 1, (2) (9) = 3, 6, , ∴ L. H. S. = R. H. S., , k (k +1), 1+2+3+...+k=, ., 2, , ∴ P(1) is true, Step (II) : (Assumption) Assume that let P(k), is true, ∴ 1.3 + 2.5 + 3.7 + .... + k (2k+1), k, =, (k+1)(4k+5) , ...(i), 6, , Step 3 : (Succession) To prove P(n) is true for, n = k +1 that is, 1 + 2 + 3 + . . . + k + (k+1) =, , 1, (1 + 1)(4.1 + 5), 6, , (k + 1)(k + 2), 2, , Here L.H.S. = 1 + 2 + 3 + . . . + k + (k+1), , Step (III) : (Succession) To prove that P(k+1), is true., , k (k +1), + (k+1) (by step 2), 2, k, = (k + 1) ( + 1), 2, =, , i.e. 1.3 + 2.5 + 3.7 + .... + (k+1) [2(k+1) +1], (k +1), (k+1+1) [4(k+1) +5], 6, i.e. 1.3 + 2.5 + 3.7 + .... + (k+1) (2k+3), , =, , (k + 1)(k + 2), = R.H.S., 2, Thus, P(k + 1) is proved., =, , =, 70, , (k +1), (k+2) (4k+9), 6

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Step (II) : Assume that P(k) is true., , Now, L.H.S. = 1.3 + 2.5 + 3.7 + .... + (k+1) (2k+3), , =1.3+2.5+...+k(2k+1)+(k+1)(2k+3), , ∑ ax, , k, (k+1)(4k+5)+(k+1) (2k+3), 6, , ... from (i), , 1 − x k , =a ,  ....(i),  1− x , , k, , = a + ax + ax2 + .... + axk−1, , r =1, , =, , Step (III) : To prove that P(k+1) is true, , = (k+1)  k (4k + 5) + 2k + 3, , , 6, , , , r −1, , 1 − x k +1 , i.e. a + ax + ax2 + .... + axk = a , ,  1− x , ,  4k 2 + 5k + 12k + 18, = (k+1) , 6, (k + 1)(k + 2)(4k + 9), =, 6, , , , , = R.H.S., ∴ P (k+1) is true., , 1 − x k , k, =a ,  + ax [by (i)], 1, −, x, , , , , , =, , Step (IV) : (Induction) From all steps above by, the principle of mathematical induction,, P(n) is true for all n ∈ N., , , , a[1 − x k + x k − x k +1 ], =, (1 − x), , ∴ 1.3 + 2.5 + 3.7 + .... + n (2n+1), n, = (n+1)(4n+5), for all n ∈ N., 6, , , , 1 − x k +1 , =a , ,  1− x , , , , = R. H. S., , , , Now, L.H.S. = a + ax + ax2 + ... + axk−1 + axk, , Ex.2 By method of induction, prove that., n, , ∑ ax, r =1, , ∴ P(k+1) is true., ,  1− x , =a ,  , for all n ϵ N, x ≠ 1.,  1− x , n, , r −1, , n, , Solution : Let P(n) ≡, , ∑ ax, , a (1 − x k ) + ax k (1 − x), (1 − x), , Step (IV) : From all steps above by the principle, of mathematical induction, P(n) is true for, all n ∈ N., , r −1, , r =1, ,  1 − xn , r −1, ax, ∴∑, =a ,  , for all n ∈ N, x ≠ 1., r =1,  1− x , n, ,  1 − xn , 2, n−1, = a + ax + ax + .... + ax = a , ,  1− x , Step (I) : To prove that P(1) is true, , Ex.3 By method of induction, prove that., , Let n = 1, , 52n − 1 is divisible by 6, for all n ∈ N., , ∴ L. H. S. = a, , Solution : 52n − 1 is divisible by 6, if and only if, ,  1− x , R. H. S. = a ,  =a,  1− x , ∴ L. H. S. = R. H. S., , 52n − 1 is a multiple of 6, , , Let P(n) be 52n − 1 = 6m, m ∈ N., , Step (I) : To prove that P(1) is true, Let n = 1, , ∴ P(1) is true, , ∴ 52n − 1 = 25 − 1 = 24 = 6.4, 71

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∴ 52n − 1 is a multiple of 6, ∴ P(1) is true., , Since k ≥ 4, k + 1 > 4+1, i.e. k + 1 ≥ 5,, also k +1 ≥ 2 (why?), and from Step II, k ! ≥ 2k ; k ≥ 4., Therefore, L.H.S. = (k+1)k! ≥ 2.2k = 2k+1 =R.H.S., , Step (II) : Assume that P(k) is true., ∴ 52k − 1 = 6a, , ∴ 52k, , where a ∈ N, , = 6a + 1 , , i.e. (k+1)! ≥ 2k+1, k + 1 ≥ 4, , ...(i), , Therefore P(k+1) is true., , Step (III) : To prove that P(k+1) is true, i.e. to prove that 52(k+1) − 1 is a multiple of 6, i.e. 52k+2 − 1 = 6b,, b∈N, Now 52k+2 − 1 = 52k.52 − 1, , = (6a + 1) 25 − 1, by (i), , = 150a + 24 = 6(25a + 4), , = 6b, , Step (IV) : (Induction) From all steps above,, P(n) is true for ∀ n ∈ N, n ≥ 4., Ex., , Solution : The statement P(n) has L.H.S. a, recurrence relation tn+1 = 3tn + 4, t1 = 1 and, R.H.S. a general statement tn = 3n − 2., , Step (IV) : From all the steps above, P(n) = 52n −1 is divisible by 6,, for all n ∈ N, , Step I : (Foundation) To prove P(1) is true., L.H.S. = 1, R.H.S = 31 − 2 = 3 − 2 = 1, So P(1) is true., , Note :, 1), , 5<5 is not a true statement, whereas 5 ≤ 5,, 5 ≥ 5 are true statements., , 2), , 2 = 3, 2 > 3, 2 ≥ 3 are not true statements,, whereas 2 < 3, 2 ≤ 3are true statements,, , 5) Given that (recurrence relation), tn+1 = 3tn + 4, t1 = 1, prove by induction that, (general statement) tn = 3n − 2., , =7, , For n = 2, L.H.S. = t2 = 3t1 + 4 = 3(1) + 4, , Now R.H.S. = t2 = 32 − 2 = 9 − 2 = 7. P(2), is also true., Step II : (Assumption) Assume that P(k) is true., , Ex. 4) By method of induction, prove that, n ! ≥ 2n ; ∀ n ∈ N, n ≥ 4., , i.e. for tk+1 = 3tk + 4, t1 = 1, then tk = 3k − 2, Step III : (Sucession) To prove that P(k + 1) is, ture., , Solution : Step I : (Foundation) Since P(n) is, stated for n ≥ 4. Take n = 4, , i.e. to prove tk+1 = 3k+1 − 2, , L.H.S. = 4! = 24, R.H.S. = 2 = 16., 4, , Since tk+1 = 3tk + 4, and tk = 3k − 2 (From, Step II) tk+1 = 3(3k − 2) + 4 = 3k+1 − 6 + 4, , Since 24≥16, P(n) is true for n = 4, [P(n) is not true for n = 1, 2, 3 (Verify!)], , = 3k+1 − 2., Therefore P(k + 1) is true., , Step (II) : (Assumption) Assume that let P(k), is true., , Step IV: (Induction) From all the steps, above P(n), tn = 3n − 2 is true for ∀ n ∈ N, where, tn+1 = 3tn + 4, t1 = 1., , i.e. k ! ≥ 2 ; k ∈ N, k ≥ 4., k, , Step (III) : (Succession) To prove that P(k+1), is true., i.e. to prove that (k + 1)! ≥ 2k+1, k + 1 ≥ 4., L.H.S. = (k +1)! = (k + 1)k!, 72

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1.6 + 2.9+ 3.12 +....+ (k + 1) (3k + 6) =, , (k+1)3+3(k+1)2+2(k+1)+3, L.H.S. = 1.6 + 2.9+ 3.12 +....+ (k + 1) (3k + 6), = 1.6+2.9+3.12 +....+ k(3k+3) + (k+1), (3k+6), = k3 + 3k2 + 2k + 3 + (k + 1) (3k + 6)by(i), = k3 + 3k2 + 2k + 3 + 3k2 + 6k+ 3k + 6, = k3 + 3k2 + 3k + 1 + 3k2+6k+3+2k+2+3, = (k + 1)3 + 3(k2 + 2k + 1) + 2(k + 1) + 3, = (k + 1)3 + 3(k + 1)2 + 2(k + 1) + 3, = R. H. S., , Ex.6 By method of induction, prove that., 2n > n, for all n ∈ N., Solution : Let P(n) = 2n > n, Step (I) : To prove that P(1) is true, Let n = 1, L.H.S. = 21 = 2, R.H.S. = 1, 2 > 1 Which is true, ∴ P(1) is true, Step (II) : Assume that P(k) is true, k ∈ N, ∴ 2k > k , , ∴ P(k + 1) is true., , ...(i), , If P(k) is true then P (k+1) is true., , Step (III) : To prove that P(k+1) is true, i.e. 2k+1 > k + 1, Now 2k+1 = 2k.21 > k.2, ...by (i), ∴ 2k+1 > 2k, ∴ 2k+1 > k + k, , Now we examine the result for n = 1, L.H.S. = 1.6 = 6, R. H. S. = 13 + 3(1)2 + 2 (1) + 3, , , =9, , ∴ L. H. S. ≠ R. H. S., ∴ P(1) is not true, ∴ P (n) is not true for all n ∈ N., , ∴ 2k+1 > k + 1 (∴ k≥1), ∴ P(k + 1) is true., Step (IV) : From all steps above and by the, principle of mathematical induction, P(n), is true for all n ∈ N., ∴ 2n > n, for all n ∈ N., , EXERCISE 4.1, Prove by method of induction, for all n ∈ N., , Remarks : (1) In the proof of P(n) by method, of induction, both the conditions viz., (i) P(1) is true and (ii) P(k+1) is true when, P(k) is true, must be satisfied. (2) In some, problems, second step is satisfied but the, first step is not satisfied. Hence the result is, not valid for all n ∈ N., for example,, let P(n) ≡ 1.6+2.9+3.12+....+n(3n + 3) =, n3 + 3n2 + 2n + 3, Let us assume that P(k) is true., ∴1.6+2.9+3.12+....+k(3k + 3) = k3 + 3k2 + 2k + 3, ...(i), We have to prove that P(k+1) is true,, i.e. to prove that, , (1), , 2 + 4 + 6 + ..... + 2n = n (n+1), , (2), , 3 + 7 + 11 + ..... + to n terms = n(2n+1), , (3), , n(n+1)(2n+1), 6, n, 12+32 + 52 + .... + (2n−1)2= (2n−1)(2n+1), 3, 3, 3, 3, 1 +3 + 5 + .... to n terms = n2 (2n2−1), n, 1.2+2.3+3.4+.....+n(n+1) =, (n+1)(n+2), 3, n, 1.3+3.5+5.7+ ..... to n terms = (4n2+6n−1), 3, 1, 1, 1, 1, +, +, +....+, (2n − 1)(2n + 1), 1.3 3.5, 5.7, n, =, 2n + 1, , (4), (5), (6), (7), (8), , 73, , 12 + 22 + 32 + .... + n2 =

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Similarly,, , y, , ∴  2x − 2  = 32x5 − 40x4y + 20x3y2 − 5x2y3, , , 5, , (, , 5, , , , 5 4 y, xy −, 8, 32, , +, , Ex.3 : Expand, , (, , 5+ 3, , ), , (, , 4, , 3 and n = 4, , = C0, , 5, , 4, , 2, , 4, , 1, , 1, , 3, , + C1, , 2, , 0, , 3, , 4, , 3, , 2, , 4, , , , 0, , ∴, , (, , ∴, , 4, , ), , ∴, , (, (, , 4, , (, , )( ), , (, , ), , 2 +1, , 5, , 5, , −, , (, , ), , 2 −1, , 5, , −, , (, , ), , 2 −1, , 5, , 3, , 5, , 2, , 5, 5, , 5, , 5, , − 1)4, , − 4C 3 ( ) 1 + 4C 4 ( ) 0, , 5, , 0, , 2, , 3, , 1, , , , +, , 4, , 5, , ) ( ), + 5 ( 2) + 1, ( 2 + 1) = ( 4 2 ) +, + (5 2 ) + 1, , 5, , 5, , ), , ( 20 2 ), , −, , +, , =, , 5, , C4 (2)1(0.02)4 + 5C5 (2)0(0.02)5, , Now 5C0 = 5C5 = 1, 5C1 = 5C4 = 5, 5C2 = 5C3 = 10, ∴, , 20 +, , +, , 5, , C2 (2)3(0.02)2 + 5C3 (2)2(0.02)3 +, , 2 + 1 = 1 4 2 + 5(4) + 10 2 2 + 10(2), , 5, , −, , = 5C0 (2)5(0.02)0 + 5C1 (2)4(0.02)1 +, , 5, , (, , 5, , =, , Solution : (2.02)5 = [2+0.02]5, , 0, , 5.4, Now C0 = C5 = 1, C4 = C1 = 5, C2 = C3 =, 2.1, = 10, 5, , − 4(10) + 1(1), , Ex. 6 : Find the value of (2.02)5 correct upto 4, decimal places., , 1, , 5, , 4.3, =6, 2.1, , (99)4 = 1(10) − 4(10) + 6(10), , ∴, , , , 4, , ), , = 82, , Now 4C0 = 4C4 = 1, 4C1 = 4C3 = 4, 4C2 =, , 5, , ( 2 + 1) = C ( 2 ) + C ( 2 ), , C ( 2) + C ( 2) + C ( 2), + C ( 2), , ∴, , ), , , , 4, , 5+, , Solution :, , (, , (, , 2 +1, , 4, , Ex.4 : Evaluate, , ∴, , ), , (99)4 = 4C0 ( )4 − 4C1 ( )3 + 4C2 ( )2, , ∴, , ( 5 )(3 3 ) + 1(1)(9), 3 ) =25+ ( 20 15 ) +90+ (12 15 ) +9, 3 ) = 124 + ( 32 15 ), , 5+, , 5, , 5, , Solution : We have (99)4 = (, , +4, , 5, , ), , 2 −1, , Ex. 5 (Activity) : Using binomial theorem, find, the value of (99)4, , 4, , 4, , (, , (, , = 82, , 3, , 5 + 3 =1(25)(1)+4 5 5 3 3 +6(5)(3), , , ∴, , 4, , −, , = 2(20 + 20 + 1), , 4, , 4, , (, , 5, , − 4 2 − 20 + 20 2 − 20 + 5 2 − 1, , 4.3, Now C0 = C4 = 1, C1 = C3 = 4, C2 =, = 6,, 2.1, 4, , ), , 2 +1, , = 4 2 + 20 + 20 2 + 20 + 5 2 + 1, , ( 5) ( 3), (, ), ( )( ), + C ( 5) ( 3) + C ( 5) ( 3), + C ( 5 ) ( 3 ), 5+ 3, , 4, , 4, , 5, , ... (ii), Subtracting (ii) from (i) we get,, , Solution : Here a = 5 , b =, Using binomial theorem,, 4, , ) = ( 4 2 ) −20+ ( 20 2 ) − 20+ (5 2 ) −1, , 2 −1, , + 20, , ... (i), 76, , (2.02)5 = 1(32)(1) + 5(16)(0.02), , , , + 10(8)(0.0004) + 10(4)(0.000008), , , , + 5(2)(0.00000016), , , , + 1(0.0000000032)

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Ignore last two terms for four decimal places, ∴, , (2.02)5 = 32 + 1.60 + 0.0320 + 0.0003, , ∴, , (2.02)5 = 33.6323., , (5), , Using binomial theorem, find the value of, (i) (102)4, , (6), , Ex. 7 : Without expanding, find the value of, , (ii) (1.1)5, , Using binomial theorem, find the value of, (i) (9.9)3 , , (ii) (0.9)4, , (2x−1)5 + 5 (2x−1)4 (1−x) + 10 (2x−1)3(1−x)2+, 10 (2x−1)2(1−x)3+5(2x−1)(1−x)4+ (1−x)5, , (7), , Solution : We notice that 1, 5, 10, 10, 5, 1 are the, values of 5C0 , 5C1 , 5C2 , 5C3 , 5C4 and 5C5, respectively., , − 4(x+1) (x−1)3 + (x−1)4, , (i) (x+1)4 − 4(x+1)3 (x−1) + 6 (x+1)2 (x−1)2, , (ii) (2x−1)4 + 4(2x−1)3 (3−2x) +, 6 (2x−1)2 (3−2x)2 + 4(2x−1)1 (3−2x)3, + (3−2x)4, , Hence, given expression can be written as, 5, , C0(2x−1)5 + 5C1(2x−1)4 (1−x), , + 5C2(2x−1)3(1−x)2 + 5C3(2x−1)2(1−x)3, , (8), , Find the value of (1.02)6, correct upto four, places of decimals., (9) Find the value of (1.01)5, correct upto three, places of decimals., (10) Find the value of (0.9)6, correct upto four, places of decimals., , + 5C4(2x−1)(1−x)4+ 5C5 (1−x)5, =[(2x−1) + (1−x)]5, = (2x − 1 + 1 − x)5, ∴, , = x5, (2x−1)5+5(2x−1)4 (1−x) +10 (2x−1)3(1−x)2, + 10 (2x−1)2(1−x)3 + 5(2x−1)(1−x)4 + (1−x)5, = x5, , 4.3, , (, , Expand (i), , (2), , Expand (i) (2x + 3), , (3), , Find the value of, , 3+ 2, 2, , ( ), (ii) ( 2 + 5 ), (i), , 3 +1, , 4, , −, , 4, , (ii), , t2 = nC1 an−1 b1, , (, , 5− 2, , 1, , (ii)  2 x − , x, , , 4, , ), , 3 −1, , (, , 5, , (4), , (, , ), , General term in expansion of (a+b)n, In the expansion of (a+b)n, we denote the, terms by t1, t2, t3, ...., tr, tr+1, .... tn .... then, t 1 = nC 0 a n b 0, , EXERCISE 4.2, (1), , Without expanding, find the value of, , ), , t3 = nC2 an−2 b2, . . ., . . ., . . ., tr = nCr−1 an−r+1 br−1, , 5, , 6, , tr+1 = nCr an−r br, tr+1 is called a general term for all r ∈ N and, , 4, , + 2− 5, , ), , 0 ≤ r ≤ n. Using this formula, we can find, any term of the expansion., , 5, , 4.3 Middle term (s) in the expansion of (a+b)n:, , Prove that, , (i) In (a+b)n if n is even then the number of, terms in the expansion is odd. So the only, , ) = 970, ( 3 + 2) + (, (ii) ( 5 + 1) − ( 5 − 1) = 352, 6, , (i), , 5, , 3− 2, , 6, , n+2, middle term is , ,  2 , , 5, , 77, , th, , term.

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(ii) In (a+b)n if n is odd then the number of, terms in the expansion is even. So the two, middle terms are  n + 1 ,  2 , th, n, +, 3, ,  term., , ,  2 , , , , term and, , t5 = 8C4 (2x2)8−4, , , ,  81 , = 70(16x ) , 4 ,  16 x , , , , = 5670x4, , ∴, , = 1120x4, , 16, x4, , Fifth and sixth terms are the middle terms., We have tr+1 = nCr an−r br,, For t5 , r = 4, , ∴, , t5 = C4 (2x), 9, , 9−4, ,  1 , − ,  4x , , 4, , 4, 9.8.7.6, 1 , 5 , = . . . (32x)  − , 4321,  4x , , , ,  1 , = 126 (2x5) , 4 ,  256 x ,  63 x , = , ,  4 , For t6 , r = 5, 5,  −1 , 9, 9−5, ∴ t6 = C5 (2x)  ,  4x , , , The fifth term in the expansion of, 8, ,  2 3 , 4,  2 x +  is 5670x, 2x , , Ex. 2 : Find the middle term(s) in the expansion, 8, , 5,  −1 , 9.8.7.6, 4, = . . . (2x)  , 4321,  4x , , 2, ,n=8, x, 8+ 2, n+2, Now n is even, hence ,  =  2  = 5,  2 , Solution : Here a = x2, b =, , ∴, , , , 4, , 8, ,  2 2, of  x + , x, , , = 70 (x8), , ∴, , 4, , 8.7.6.5,  3 , =, (2x2)4  , 4.3.2.1,  2x , , , , 4, ,  n +1 ,  n +3, Now n is odd ,  =5  2  =6, , ,  2 , , For t5, r = 4,  3 ,  ,  2x , , , , 8.7.6.5 2 4  2 , =, (x )  , 4.3.2.1, x, , 9, , 3, ,n=8, 2x, , tr+1 = nCr an−r br,, , 4, , 1 , ,  2x − , 4x , , 1, Solution : Here a = 2x, b = −, ,n=9, 4x, , 8, , Solution : Here a = 2x2, b =, , 2,  , x, , Ex. 3 : Find the middle terms in the expansion of, , Ex. 1 : Find the fifth term in the expansion of, , Since,, , 2 8−4, , th, , SOLVED EXAMPLES, ,  2 3 ,  2x + , 2x , , , t5 = C4 (x ), 8, ,  −1 , = 126 (16x4) , 5 ,  1024 x , 63, = −, 32x,  63 x , 63, ∴ The middle terms are ,  and −,  4 , 32x, , Fifth term is the only middle term., For t5, r = 4, We have tr+1 = nCr an−r br,, 78

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Ex. 4 : Find the coefficient of x7 in the expansion, , ∴ r, =4, ∴ coefficient of x−2, , 11, , 1, , of  x 2 + , x, , , 4, 4, 10.9.8.7, = 10C4 (2)10−4  −1  =, (2)6  −1 , , , , , 4.3.2.1,  3,  3, , 1, Solution : a = x2, b = , n = 11, x, We have tr+1 = nCr an−r br,, , , tr+1 =, , 11, , 1,  ,  x, , Cr (x ), , 2 11−r, , , , = 11Cr x22−2r x−r, , , , = Cr x, 11, , 22−3r, , ∴, , 22 − 3r = 7, , ∴, , r, , ∴, , 11, , =, , 4480, 3, , ∴ Coefficient of x−2 is, , 4480, 3, , 10, , 2, expansion of  x − 2 , x , , , =5, 11.10.9.8.7, = . . . ., = 462, 54321, , C5, , , , Ex. 6 : Find the term independent of x, in the, , = x7, , x22−3r, , 1, = 210 (64)  , 9, , r, , To get coefficient of x7, we must have, , , , , Solution : Here a =, , We have tr+1 = nCr an−r br, , ∴ Coefficient of x7 is 462., , 10, , Solution : Here a = 2x, b =, , −1, , n = 10, 3x 2, , We have tr+1 = nCr an−r br,, , = Cr, , , , = Cr x, , , , = 10Cr (−2)r x, , =, , , , = Cr (2), , 10, , r, , 10−r, , x, , 10−r, , ∴, ,  −1  −2r, ,  x,  3, , (−2)r x−2r, 10 −5 r, 2, , 10 − 5r, 2, , = x0, =0, , ∴ 10 − 5r = 0, , r, , , ,  10− r , , ,  2 , , r, , 10 −5 r, , x 2, , , r, , 10, , 10, ,  −2 ,  2, x , , To get the term independent of x, we must have, ,  −1 , Cr (2x2)10−r , 2 ,  3x , , , , ( x), , 10 − r, , , , 10, , Ex. 5 : Find the coefficient of x−2 in the expansion, 1 , , of  2 x −, , 3x 2 , , , −2, , n = 10, x2, , x,b=, ,  −1  10−3r, = 10Cr (2)10−r ,  x,  3, , ∴r, , =2, , ∴ the term independent of x is, , To get coefficient of x , we must have, −2, , 10, , x10−3r, = x−2, ∴ 10 − 3r = −2, ∴ −3r, = −12, , C2 (−2)2 =, , 10.9, (−2)2 = 45(4) = 180, 2.1, , ∴ the term independent of x is 180., 79

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(4), , EXERCISE 4.3, (1), , (i), , In the following expansions, find the, indicated term., ,  2 3  , 3rd term,  2x + , 2x , , , , , , (ii)  x 2 −, , 8, ,  x a, (iv)  − , a x, , 7, , 10, , 11, , 4, , 5th term, 3 , x , , (5), , In the expansion of (k+x)8, the coefficient, of x5 is 10 times the coefficient of x6. Find, the value of k., , 12, , (6), , Find the term containing x6 in the expansion, of (2−x) (3x+1)9, , 4, (v) In  3a+  , 10th term, a, , , 13, , (7), , The coefficient of x2 in the expansion of, (1+2x)m is 112. Find m., , In the following expansions, find the, indicated coefficients., , 4.4, , Binomial Theorem for Negative Index or, Fraction., , (iii)  4x − 5  , 7th term, , ,  5 2x , 1, (iv) In  + a 2  , 9th term, 3, , , (i), , , 3 2, x in  x 2 +, , x , , , 9, ,  5 5, (ii) x in  2x − 3 , x , , , 3, , 18, , (iii) x in  1 +x 2 , , , x, , 9, , If n is negative then n! is not defined., , 8, , 8, , 1 , , (iv) x in  x − , 2x , , , We state binomial theorem in another form., n n−1, n(n − 1) n−2 2, (a+b)n = an +, a b+, a b, 2!, 1!, , 5, , −3, , +, , 15, , , , , 1 , , , (v) x−20 in  x3 − 2 , 2x, (3), , 1, (ii)  x 2 + , x, , , 1, , (v)  x 4 − 3 , x , , , 11, , 9, , (2), , 12, ,  x y,  + , y x, , 2, , (iii)  x 2 − , x, , , 8, , (i), , Find the middle terms in the expansion of, , n(n − 1)....(n − r+1) an−r br + .....+ bn, r!, , Find the constant term (term independent, of x) in the expansion of, (i), , 1 , ,  2 x+ 2 , 3x , , , 9, , 10, , 3, , (iii)  x − 2 , x , , 5, (v)  2x 2 − , x, , , , , , Here tr+1 =, , 15, , 2, , , (ii)  x − 2 , x, 1, , (iv)  x 2 − , x, , , n(n − 1)(n − 2), an−3 b3 + ....., 3!, , n(n − 1)(n − 2)....(n − r+1) n−r r, a b, r!, , Consider the binomial theorem, (1+x)n = 1 + nC1x + nC2 x2 + .... + xn, , 9, , = 1 + nx +, , 9, , n(n − 1) 2, x + ....+ xn, 2!, , This is a finite sum., , 80

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The theorem has an extension to the case, where 'n' is negative or faction. We state it, here without proof., , (3), (4), , For |x|<1, (1+x)n = 1 + n x +, , n(n − 1), x2 +, 2!, , (5), , , , n(n − 1)(n − 2), 3!, , , , n(n − 1)....(n − r+1) r, x + ...., r!, , x3 + ..., , (6), , n(n − 1)(n − 2)...(n − r+1)x r, ,r≥0, r!, , Remarks : (1) If |x|<1 and n is any real number,, not a positive integer, then, , −4, , 2, 3, , ,  b  (−4)(−5)  b  (−4)(−5)(−6)  b , = a 1+(-4)  −  +,  −  + ..., −  +, 2, !, 3, !, a, a, a, , , , , , , , , , −4, , −4, , The general term is given by, , , b 20 b 2 120 b3, −4 , a, 1, +, 4, +, +, + ..., =, , 2, 3, a 2 a, 6 a, , , , ( − 1) r n(n − 1)(n − 2)...(n − r+1) r, tr+1 =, x, r!, ,  4b 10b 2 20b3, , + 2 + 3 + ..., = a 1 +, a, a, a, , , −4, , If n is any real number and |b|<|a|, then, n, , x x 2 x3, = 1 − − − − ..., 2 8 16, ,   b , = a 1 −  ,   a , , , , n(n − 1)(n − 2), n(n − 1), (1−x)n = 1− nx + 2!, x2 −, x3 +..., 3!, ,   b , b, (a + b) = a 1+   = an 1+ ,   a ,  a, , 2, , Ex.1 : State first four terms in the expansion of, 1, where |b|<|a|, (a − b ) 4, 1, Solution : We have, = (a−b)−4, 4, (a − b ), , Here there are infinite number of terms in the, expansion. The general term is given by, , (ii), , 1 − x = (1 − x), , 1, , SOLVED EXAMPLES, , Here n is not an integer and the terms on, the RHS are infinite, the series does not, terminate., , tr+1 =, , 1, = (1+x)−2 =1−2x+3x2 − 4x3+..., (1 + x) 2, 1, = (1−x)−2 =1+2x+3x2 + 4x3+..., (1 − x) 2, 1, x x 2 x3, 1 + x = (1 + x) 2 = 1 + − + + ..., 2 8 16, , n, , n, , Ex. 2 : State first four terms in the expansion of, 1, , |b|<|a|, (a+b), 1, Solution : =, = (a+b)−1, (a+b), , Note : While expanding (a+b)n where n is a, negative integer or a fraction, reduce the binomial, to the form in which the first term is unity and the, second term is numerically less than unity., , b, = a −1 1 + ,  a, , Particular expansion of the binomials for, negative index, fraction. |x|<1, (1), , 1, =(1+x)−1 = 1− x+x2−x3+x4−x5+..., 1+ x, , (2), , 1, =(1−x)−1 = 1+ x+x2+x3+x4+x5+..., 1− x, , −1, , 2, 3, , ,  b  (−1)(−2)  b  (−1)(−2)(−3)  b , a −1 1 + (−1)   +, +,  ,   + ..., 2!  a , 3!, a, a, , , ,  b b 2 b3, , = a 1 − + 2 − 3 + ..., a,  a a, , −1, , 81

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Ex. 3 : State first four terms in the expansion of, 2, (2−3x)−1/2 if |x|<, 3, 2, Solution : |x|<, 3, , ,  1  1 , −, , 1  2   2   1 , = 5 1 + +,  , 2,  25 ,  10, , ,  1   1   −3 ,   −   1, , ,  2  2  2  , , +, +, ..., , , , 6,  125 , , , 1, 1, 1, , = 5 1 + −, +, − ...,  10 200 2000, , = 5 [ 1 + 0.1 − 0.005 + 0.0005], (upto 4 decimal places), = 5[1.0955], = 5.4775, , ∴ 3x < 1, 2, We have (2−3x)−1/2, =2, , −1/2, ,  3x , 1 − , 2 , , , −1, , 2, , ,  1  1 ,   1   −3 x   − 2   − 2 − 1  −3 x  2, , , , 2, = 2 1 +  −  , +, , , 2!,  2 ,   2  2 , , −1, , ,  1  1  1, ,  −   − − 1  − − 2  −3 x 3, , 2  2  2, , , , +, ..., +, , , 3!,  2 , , , , EXERCISE 4.4, (1), , ,  1  3 , −, −, , 3 x  2   2   9 x 2 , −1, 2 , 1+ +, = 2, , , 4, 2!, ,  4 , , − 1 ) ( − 3 ) ( − 5 )  −27 x, (, 2, 2, 2, +, , , , 6, , = 2, , −1, , 2, , 8, , 3, , (2), , , , ,  + ...., , , , (3), , Solution:, , (iii) (1−x2)−3, , (iv) (1+x)−1/5, , State, by writing first four terms, the, expansion of the following, where |b|<|a|, (i) (a−b)−3, , (ii) (a+b)−4, , (iii) (a+b)1/4, , (iv) (a−b)−1/4, , Simplify first three terms in the expansion, of the following, (i) (1+2x)−4, , (ii) (1+3x)−1/2, , (iii) (2−3x)1/3, , (iv) (5+4x)−1/2, , (v) (5−3x)−1/3, , 30 = (25 + 5)1/2, ,  1, = 5 1 + ,  5, , (ii) (1−x)1/3, , (v) (a+b)−1/3, , Ex. 4 : Find the value of 30 upto 4 decimal, places., , = (25)1/2, , (i) (1+x)−4, (v) (1+x2)−1, ,  3 x 27 x 2 135 x 3, , 1 + 4 + 32 + 128 + ..., , , , 5 , , 1 + ,  25 , , State, by writing first four terms, the, expansion of the following, where |x|<1, , (4), , 1, 2, , Use binomial theorem to evaluate the, following upto four places of decimals., (i), , 1, , 99 , , (iii) 4 16.08, , 2, , (v) (0.98)−3, 82, , (ii) 3 126, (iv) (1.02)−5

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(II) Answer the following., , MISCELLANEOUS EXERCISE - 4, (I), , Select the correct anwsers from the given, alternatives., , (i), , Prove, by method of induction, for all, n∈N, n, 8 + 17 + 26 + .... + (9n−1) =, (9n+7), 2, , (1), , The total number of terms in the expression, of (x+y)100 + (x−y)100 after simplification is :, , (ii), , 12 + 42 + 72 + ... + (3n−2)2 =, , A) 50, , (iii) 2 + 3.2 + 4.22 + ... + (n+1)2n-1 = n. 2n, , (2), , (3), , (4), , (5), , (6), , B) 51, , C) 100, , (1), , D) 202, , The middle term in the expansion of (1+x)2n, will be :, A) (n−1)th B) nth C) (n+1)th D) (n+2)th, , (iv) 1 + 2 + 3 + ... +, 3.4.5, , 5.6.7, , , , In the expansion of (x2−2x)10, the coefficient, of x16 is, A) −1680 B) 1680 C) 3360 D) 6720, The term not containing x in expansion of, 10, 1, , 2, x, +, (1−x) ,  is, x, , A) 11C5 B) 10C5 C) 10C4 D) 10C7, , =, , n, (n + 2)(n + 3)(n + 4), , n(n + 1), 6(n + 3)(n + 4), , (2), , Given that tn+1 = 5tn − 8, t1 = 3, prove by, method of induction that tn = 5n−1 +2, , (3), , Prove by method of induction, n, ,  3 −4   2n + 1 −4n  , ∀ n ∈ N, ,  =, , −2n + 1,  1 −1   n, , The number of terms in expansion of, (4y+x)8−(4y−x)8, , (4), , Expand (3x2 + 2y)5, , A) 4, , (5), ,  2x 3 , − , Expand ,  3 2x , , (6), , Find third term in the expansion of, , B) 5, , C) 8, , D) 9, , The value 14C1 + 14C3+ 14C5+ ..... + 14C11 is, A) 214−1 B) 214−14 C) 212 D) 213−14, , (7), , 4.5.6, , n, (6n2−3n −1), 2, ,  2 y3 ,  9x − , 6 , , , The value 11C2 + 11C4+ 11C6+ 11C8 is equal to, A) 210−1 B) 210−11 C) 210+12 (D) 210−12, , (8), , In the expansion of (3x+2) , the coefficient, of middle term is, A) 36, B) 54 C) 81, D) 216, , (9), , The coefficient of the 8th term in the, expansion of (1+x)10 is :, A) 7, B) 120 C) 10C8 D) 210, , (7), , 4, , 4, , 4, , Find tenth term in the expansion of, 12, ,  2 1,  2x + , x, , (8), , Find the middle term (s) in the expansion of, (i)  2a − 3 ,  3 2a , , (10) If the coefficient of x2 and x3 in the expansion, of (3+ax)9 are the same, then the value of a, is, 9, 9, 7, 7, A) −, B) −, C), D), 7, 7, 9, 9, , (iii) (x2+2y2)7, , 85, , 6, , 10, , , 1 , (ii)  x − , 2y , , , 2, (iv)  3x − 1 ,  2 3x , , 9

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(9), , Find the coefficients of, , (16) Show that there is no term containing x6 in, 11, , (i), , 1, x6 in the expansion of  3x 2 − , 3x , , , (ii), ,  1, , x in the expansion of  2 + x 4 , x, , , 3, , the expansion of  x 2 −  ., x, , (17) Show that there is no constant term in the, , 9, , 18, , , x2 , expansion of  2 x − , 4, , , 60, , (18) State, first four terms in the expansion of, , (10) Find the constant term in the expansion of, ,  4x, 3 , + , (i) ,  3 2x , 2, , 9, ,  2x , 1 − , 3 , , , 12, , 1, , (ii)  2x 2 − , x, , , loga xn = n loga x, x > 0. n ∈ N, , (ii), , 152n-1+1 is divisible by 16, for all n ∈ N., , −1, , 2, , (19) State, first four terms in the expansion of, (1−x)−1/4, , (11) Prove by method of induction, (i), , 9, , (20) State, first three terms in the expansion of, (5 + 4x)−1/2, (21) Using binomial theorem, find the value of, , (iii) 52n − 22n is divisible by 3, for all n ∈ N., , 995 upto four places of decimals., 1, (22) Find approximate value of, upto four, 4.08, places of decimals., 3, , (12) If the coefficient of x16 in the expansion of, (x2 + ax)10 is 3360, find a., (13) If the middle term in the expansion of, , (23) Find the term independent of x in the in, , 6, , b, ,  x +  is 160, find b., x, , , 2, , expansion of (1 −x )  x + , x, , , 6, , 2, , (14) If the coefficient of x2 and x3 in the expansion, of (3 + kx)9 are equal, find k., , (24) (a + bx) (1 − x)6 = 3 −20x + cx2 + ..... then, find a, b, c., , (15) If the constant term in the expansion of, , (25) The 3rd term of (1+x)n is 36x2. Find 5th term., , 11, ,  3 k  is 1320, find k., x + 8 , x , , , (26) Suppose (1+kx)n = 1−12x + 60x2 − .... find, k and n., , v, , v, , 86, , v

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5, , SETS AND RELATIONS, ‘Happy people’, ‘Clever student’ are all relative, terms. Here, the objects are not well-defined. In, the last two collections. We can determine the, objects clearly. Thus, we can say that objects are, well-defined., , Let's Study, •, , Representation of a set, , •, , Types of sets, , •, , Intervals, , •, , Operations on sets, , •, , Ordered pair, , •, , Types of relations, , 5.1.1 Set : Definition:, A collection of well-defined objects is, called a set., The object in a set is called its element or, member., , Let's Recall, , We denote sets by capital letters A,B,C. etc., The elements of a set are represented by small, letters a, b, c, x, y, z etc. If x is an element of a set, A we write x∈A, and read as ‘x belongs to A’. If, x is not an element of a set A, we write x∉A, and, read as ‘x does not belong to A.’, , 5.1 Introduction:, The concept of a set was developed by, German mathematician George Cantor (18451918), You have already learnt about sets and some, basic operations involving them in the earlier, standards., , e.g. zero is a whole number but not a natural, number., ∴ 0∈W (Where W is the set of whole, numbers) and 0∉N (Where N is the set of natural, numbers), , We often talk about group or collection of, objects. Surely you must have used the words, such as team, bouquet, bunch, flock, family for, collection of different objects., , 5.1.2 Representation of a set:, There are two methods of representing a set., 1) Roster or Tabular method or List method, 2) Set-Builder or Rule Method, 3) Venn Diagram, , It is very important to determine whether a, given object belongs to a given collection or not., Consider the following collections:, i), ii), iii), iv), v), , Successful persons in your city., Happy people in your town, Clever students in your class., Days in a week., First five natural numbers., , 1., , Roster Method:, , In the Roster method, we list all the elements, of the set within braces {,} and separate the, elements by commas., Ex : State the sets using Roster method., , First three collections are not examples of, sets, but last two collections represent sets. This is, because in first three collections, we are not sure, of the objects. The terms ‘successful persons,’, , i), , B is the set of all days in a week., B = {Monday, Tuesday, Wednesday, , Thursday, Friday, Saturday, Sunday}, , 87

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ii), , 5.1.4 Types of Sets:, , C is the set of all vowels in English alphabets., C = {a, e, i, o, u}, , 1) Empty Set:, , Let’s Note:, , A set containing no element is called an, empty or a null set and is denoted by the, symbol φ or { } or void set., , 1) If the elements are repeated, they are written, only once., 2) , While listing the elements of a set, the, order in which the elements are listed is, immaterial., , e.g. A = {x/x∈N, 1<x<2} = { }, , 2., , Set-Builder Method:, In the set builder method, we describe the, elements of the set by specifying the property, which determines the elements of the set uniquely., , 2) Singleton set:, , Ex : State the sets using set-Builder method., , e.g. Let A be a set of all integers which are, neither positive nor negative., , i), , Y = {Jan, Feb, Mar, Apr, ...., Dec}, Y = {x/x is a month of a year}, , ii), , B = {1, 4, 9, 16, 25, .....}, , Here n(A) = 0, A Set containing only one element is called, a singleton set., , ∴A = {0} Here n (A) = 1, 3) Finite set:, , B = {x/x∈N and x is a square}, , The empty set or set which contains finite, number of objects is called a finite set., , 3), , Venn Diagram:, The pictorial representation of a set is, called Venn diagram. English Logician John Venn, introduced such diagrams. We can use triangles,, circles, rectangles or any closed figure to represent, a set., In a Venn diagram the elements of the sets, are shown as points, , e.g set of letters in the word 'BEAUTIFUL', A = {B, E,A ,U, T, I, F, L}, n(A) = 7, A is a finite set, 4) Infinite set:, A set which is not finite, is called an infinite, set., , A = {1,2,3} B = {a,b,c,d,e,f} C = {4,5,6}, , e.g. set of natural numbers, set of rational, numbers., Note :, 1) An empty set is a finite set., , A, , B, Fig. 5.1, , 2) N, Z, set of all points on a circle, are, infinite sets., , C, , 5) Subset:, , 5.1.3 Number of elements of a set:, , A set A is said to be a subset of set B if every, element of A is also an element of B and we, write A ⊆ B., , The number of distinct elements contained in, a finite set A is denoted by n(A)., Thus, if A = {5, 2, 3, 4}, then n(A) = 4, n(A) is also called the cardinality of set A., , Note: 1) φ is subset of every set., 2) A ⊆ A, Every set is subset of itself., 88

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6), 7), , Superset: If A ⊆ B, then B is called a, superset of A and we write, B ⊇ A., , 10) Equivalent sets:, Two finite sets A and B are said to be, equivalent if n (A) = n (B), , Proper Subset: A nonempty set A is said to, be a proper subset of the set B, if all elements, of set A are in set B and at least one element, of B is not in A., , e.g. A = {d, o, m, e}, B = {r, a, c, k}, , , , n (A) = n (B) = 4, , , ∴ A and B are equivalent sets., , i.e. If A ⊆ B and A ≠ B then A is called a, proper subset of B and we write A ⊂ B., , 11) Universal set: If in a particular discussion, all sets under consideration are subsets of a, set, say U, then U is called the universal set, for that discussion., , e.g. Let A = {1, 3, 5} and B = {1, 3, 5, 7}., Then, evey element of A is an element of B, but A ≠ B., ∴ A⊂B, i.e. A is a proper subset of B., , e.g. The set of natural numbers N, the set, of integers Z are subsets of real numbers, R. Thus, for this discussion R is a universal, set., , Remark: If there exists even a single, element in A which is not in B then A is not, a subset of B and we wrtie, A ⊄ B., 8), , , , In general universal set is denoted by U or, X., , Power Set:, The set of all subsets of a given set A is, called the power set of A and is denoted by, P(A), Thus, every element of power set A is, a set., , 5.1.5 Operations on sets:, 1) Complement of a set:, The complement of the set A is denoted by, , e.g. consider the set A={a,b}, let us write, all subsets of the set A. We know that φ is, a subset of every set, so φ is a subset of A., Also {a}, {b}, {a,b} are also subsets of A., Thus, the set A has in all four subsets viz., φ, {a}, {b}, {a,b}, , A' or Ac . It is defined as, A' = {x/x∈U, x∉A} = set of all elements in, U which are not in A., Ex. Let X = {0, 1, 2, 3, 4, 5, 6, 7, 8} be the, Universal Set and A = {2, 4, 6, 8}, , ∴ P(A) = {φ, {a}, {b}, {a,b}}, 9) Equal sets:, Two sets are said to be equal if they contain, the same elements i.e. if A⊆ B and B ⊆ A., e.g. Let X be the set of letters in the word, 'ABBA' and Y be the set of letters in the, word 'BABA'., , Fig. 5.2, , ∴ X = {A, B}, Y = {B, A}, , ∴ The complement of the set A is, , Thus the sets X and Y are equal sets and we, denote it by X = Y, , A' = {0, 1, 3, 5, 7}, 89

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Properties:, , 3) Intersection of sets:, , i) (A')' = A ii) φ' = U, , iii) U' = φ, , The intersection of two sets A and B is the, set of all elements which are both in A and, B is denoted by A∩B, , 2) Union of Sets:, The union of two sets A and B is the set of, all elements which are in A or in B, (here, 'or' is taken in the inclusive sense) and is, denoted by A ∪ B, , Thus, A ∩ B = {x/x∈A and x∈B}, , Thus, A ∪ B = {x/x∈A or x∈B}, The Union of two sets A and B i.e., A ∪ B is represented by a shaded part in, Venn-diagram as in fig. 5.3 and fig. 5.4., , A∩B, Fig. 5.5, The shaded portion in Fig. 5.5 represents, the intersection of A and B i.e. A∩B, , , Fig. 5.3, , Ex. A={1,3,5,7,9}, , Fig. 5.4, , B = {1,2,3,4,5,6,7,8}, , Ex. A = {x/x is a prime number less than 10}, , Find A∩B., , B= {x/x∈N, x is a factor of 8}, , Solution: A∩B = {1,3,5,7}, , find A∪B., , Ex. (Activity) : If A = {x/x∈N, x is a factor of 12}, , Solution : We have A = {2,3,5,7}, , B = {x/x∈N, x is a factor of 18}, , B = {1, 2, 4, 8}, , Find A∩B, , ∴ A∪B = {1, 2, 3, 4, 5, 7, 8}, , Solution:, , Properties:, i) A ∪ B = B ∪ A , , A={ , , , , , }, , (Commutativity), , B = { , , , , , }, , ii) (A ∪ B) ∪ C = A ∪ (B ∪ C), , , (Associativity), , iii) A ∪ φ = A , , Identity for union, , iv) A ∪ A = A , , Idempotent law, , ∴ A∩B = { , , , }, = common factors of 12 & 18, Ex. : A = {1,3,5,7,9}, B = {2,4,6,8, 10}, A∩B = ?, , v) If A∪ A' = U, , Solution : ∴ A∩B = { } = φ, , vi) If A⊂B then A∪B = B, , If A∩B = φ, A and B are disjoint sets., , vii) U∪A = U, viii) A⊂(A∪B), B⊂(A∪B), 90

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Properties:, , Note:, , i) A∩B = B ∩A (Commutativity), , i) A-B ⊆ A and B-A ⊆ B., , ii) (A ∩ B) ∩ C = A ∩ (B ∩ C), (Associativity), , ii) The sets A-B, A∩B and B-A are mutually, disjoint sets, i.e. the intersection of any of, these two sets is the null (empty) set., , Idempotent law, , iii), , A-B = A∩B' , B-A = A'∩B, A∪B = (A-B)∪(A∩B) ∪ (B-A), , , iii), , φ∩ A = φ , , iv), , A ∩ A = A , , v), , A∩ A = ∅, , iv), , vi), , if A⊂B then A∩B =A, , Shaded portion in fig. 5.7 represents A∪B, , vii) U∩A = A, , (Identity for intersection), , viii) (A∩B) ⊂A, (A∩B) ⊂B, ix) , , a) A∩(B∪C) = (A∩B)∪ (A∩C), , b) A∪(B∩C) = (A∪B) ∩ (A∪C), ..... Distributivity, De Morgan's Laws:, , Fig. 5.7, , For any two sets A and B, , v) (A-B) ∪ (B-A) = A∆B is called symmetric, difference of sets A and B, , 1) (A∪B)' = (A'∩B'), 2) (A∩B)' = (A'∪B'), Verify the above laws by taking, , Illustration :, , U = {1,2,3,4,5} A = {1,3,4}, , If A = {4,5,6,7,8} and B = {3,5,6,8,9} then, A−B = {4,7}, B−A = {3,9}, , B = {4,5}, , and A∆B = (A−B)∪(B−A) = {4,7}∪{3,9}, = {3,4,7,9}, , 4) Difference of Sets:, Difference of set A and set B is the set of, elements which are in A but not in B and is, denoted by A-B, or A∩B', , Properties:, i) A∆B = (A∪B)−(A∩B), , The shaded portion in fig. 5.6 represents, A-B. Thus, A-B = {x/x∈A, x∉B}, , ii), , A∆A = φ, , iii) A∆φ = A, , iv), , If A∆B = A∆C then B=C, , v), , A∆B = B∆A, , vi), , A∩(B∆C) = (A∩B) ∆ (A∩B), , Properties of Cardinalityof Sets:, For give sets A, B, 1) n (A∪B) = n(A) + n(B) − n(A∩B), 2) When A and B are disjoint sets, then, , Fig. 5.6, , n(A∪B) = n (A) + n(B), as A∩B = ∅,, , Similary B-A = {y/y∈B, y∉A}, , n(A∩B) = 0, 91

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3) n(A∩B')+ n(A∩B) = n(A), , 4) Semi-open Interval:, , 4) n(A'∩B)+ n(A∩B) = n(B), , (a, b] = {x/x∈R, a < x ≤ b}, , 5) n(A∩B')+ n(A∩B) + n(A'∩B) = n(A∪B), 6) For any sets A, B, C., , Fig. 5.11, , n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B), , (a, b] excludes a but includes b., , -n(B∩C) - n(A∩C) + n(A∩B∩C), , i.e. a∉(a, b] but b∈(a, b], , 7) If n(A) = m n[P(A)] = 2m Where p(A) is, power set of A, , 5) i) The set of all real numbers greater than, a i.e. (a,∞) = {x/x∈R, x > a}, , 8) n(A∆B) = n(A) + n(B) − 2n (A∩B), 5.1.6 Intervals:, 1) Open Interval: Let a, b ∈ R and a < b then, the set {x/x∈R, a < x < b} is called open, interval and is denoted by (a,b). All the, numbers between a and b belong to the, open interval (a,b) but a, b themselves do, not belonging to this interval., , Fig. 5.12, ii) The set of all real numbers greater than, or equal to a, , , [a,∞) = {x/x∈R, x ≥ a}, , Fig. 5.13, Fig. 5.8, , 6) i) The set of all real numbers less than b., ie. (-∞, b), ∴(-∞, b) = {x/x∈R, x < b}, , ∴ (a,b) = {x/x∈R, a < x < b}, 2) Closed Interval: Let a, b∈R and a < b, then the set {x/x∈R, a ≤ x ≤ b} is called, closed interval and is denoted by [a,b]. All, the numbers between a and b belong to the, closed interval [a, b]. Also a and b belong to, this interval., , Fig. 5.14, ii) The set of all real numbers less than or, equal to b i.e. (-∞, b], , Fig. 5.15, , Fig. 5.9, [a, b] = {x/x∈R, a ≤ x ≤ b}, , , , 3) Semi-closed Interval:, , 7) The set of all real numbers R is (-∞, ∞), , [a, b) = {x/x∈R, a ≤ x < b}, , ∴(-∞, b] = {x/x∈R, x ≤ b}, , Fig. 5.16, , Fig. 5.10, , R = (-∞,∞) = {x/x∈R, -∞ < x < ∞}, , Note that a∈[a, b) but b∉[a, b), 92

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Rules of inequality of real numbers., 1), 2), , c), , If 0<a<b and k>0 then a±k<b±k, ka<kb,, ak <bk, , subtracting, 4, we get, [Note that, as we do not know value of x, we, do not multiply by (x − 3)], , If 0<a<b and k<0 then ka>kb, ak>bk, , Ex. : Solve the following inequalities and write, the solution set using interval notation., a), , x, 40, x 3, , −7 < 2x + 5 ≤ 9, , x  4( x  3), 0, x 3, , : Subtracting 5, we get, , x  4 x  12, 0, x3, , −7−5 < 2x ≤ 9−5, −12 < 2x ≤ 4, , 3 x  12, 0, x 3, , Dividing by 2, we get, −6 < x ≤ 2, , 3( x  4), 0, x3, , So, x ∈ (−6,2]., b), , Dividing by -3, we get, , x2 + 2x <15, Subtracting 15, we get, , Say p(x) =, , x2 + 2x − 15 < 0, , Say p(x) = (x − 3) (x + 5) < 0, , Regions, , As x−3 and x + 5 is 0, for x = 3, −5, consider, them as critical points. They divide number, line into 3 regions as follow., x<−5, , (−5<x<3), , (x>3), , −6, , 0, , 4, , (0−3), (0+5), = (−3)(5), = −15, , (4−3), (4+5), = (1)(9), =9, , p(x)<0, , p(x)>0, , Test Points, , (−6−3), Value (x−3), (−6 + 5), (x+5) of at, = (−9) (−1), test points, =9, Sign of, p(x), , p(x)>0, , x4, 0, x 3, , Take x = 3, 4 as critical points. It divides, number line into 3 regions as follows., , Factor x2 + 2x − 15, we get, , Regions, , x, 4, x 3, , (x ≤ 3), , (3 ≤ x ≤ 4), , (x ≥ 4), , Test Points, , 2, , 3.5, , 5, , Sign of x−4, , −ve, , −ve, , +ve, , Sign of x−3, , −ve, , +ve, , +ve, , p(x)>0, Except at, x=3, , p(x)<0, , p(x)>0, , Sign of, p(x), , The solution set is (x < 3) and (x ≥ 4)., So x ∈ (−∞, 3) ∪ [4, ∞)., Here x = 3 is not in the solution set, as it, makes denominator 0., Ex. 8 : If A = [−5,3], B = (3,7), and (5,8]. Find, (a) A∪B (b) B∪C (c) A∪C (d) A∩B (e) B∩C, (f) A∩C (g) A∪C' (h) B − C (i) C − B, , Therefore the solution set is (−5 < x < 3), So, x (−5, 3)., 93

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Solution :, , Ex. 3 : If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, , (a), , A∪B = [−5, 3] ∪ (3, 7) = [−5, 7), , C = {5, 6, 7, 8}, D = {7, 8, 9, 10}, , (b), , B∪C = (3, 7) ∪ (5, 8] = (3, 8], , find i) A∪B ii) A∪B∪C iii) B∪C∪D, , (c), , A∪C = [−5, 3] ∪ (5, 8], , Are the sets A, B, C, D equivalent?, , (d), , A∩B = [−5, 3] ∩ (3, 7) = φ, , Solution: We have, , (e), , B∩C = (3, 7) ∩ (5, 8] = (5, 7), , (f), , A∩C = [−5, 3] ∩ (5, 8] = φ, , (g), , A∪C' = [−5, 3] ∪ (5, 8]' = [−5, 3] ∪, {(−∞, 5] ∪ (8, ∞)} = (−∞, 5] ∪ (8, ∞), , (h), , B − C = (3, 7) − (5, 8] = (3, 5], , (i), , C − B = (5, 8] − (3, 7) = [7, 8], , i) A∪B = {1, 2, 3, 4, 5, 6}, ii) A∪B∪C = {1, 2, 3, 4, 5, 6, 7, 8}, iii), , B∪C∪D = {3, 4, 5, 6, 7, 8, 9, 10}, , As n(A) = n(B) = n(C) = n(D) = 4, the sets A, B,, C, D are equivalent., Ex. 4 : Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}be the, universal set, A = {1, 3, 5, 7, 9}, B = {2, 3, 4, 6, 8, 10}, C = {6, 7, 8, 9}, , SOLVED EXAMPLES, , find i) A' ii) (A∩C)' iii) (A') iv) (B-C)', , Ex. 1: If A= {x/x∈N, x is a factor of 6}, , Solution: We have, , B = {x/x∈N, x is a factor of 8}, , i) A' = {2, 4, 6, 8, 10}, , find the A-B and B-A, , ii) (A∩C) = {7, 9}, , Solution : A = {1, 2, 3, 6}, , ∴ (A∩C)' = {1, 2, 3, 4, 5, 6, 8, 10}, , B = {1, 2, 4, 8}, ∴A-B = {3, 6}, B-A = {4, 8}, , iii), , (A')' = {2, 4, 6, 8, 10} = {1, 3, 5, 7, 9} = A, , iv) , , B-C = {2, 3, 4, 10}, , ∴ (B-C)' = {1, 5, 6, 7, 8, 9}, , 1, , Ex.2: A   / x  N , x  8, x, , , Ex. 5 : Let X be the universal set, for the nonempty sets A and B, verify the De Morgan's laws, , 1, , B =  / x  N , x  8 Find A-B and B-A,  2x, , , i) (A∪B)' = A'∩B', ,  1 1 1 1 1 1, Solution: A = 1, , , , , , ,  2 3 4 5 6 7, , ii) (A∩B)' = A'∪B', , 1 1 1 1 1 1 1 1 , B =  , , , , , , , ,  2 4 6 8 10 12 14 16 , , A = {1, 2, 3, 4, 5} B = {1, 2, 5, 6, 7}, ,  1 1 1, ∴A-B = 1, , , ,  3 5 7, , (A∩B) = {1, 2, 5}, , 1 1 1 1 1 , and B-A =  , , , , ,  8 10 12 14 16 , , B' = {3, 4, 8, 9, 10}, , Where X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, Solution: A∪B = {1, 2, 3, 4, 5, 6, 7}, A' = {6, 7, 8, 9, 10}, , 94

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i) (A∪B)' = {8, 9, 10}, , ...(1), , ii) n(A∩B) = n(A) + n(B) - n(A∪B), , A'∩B' = {8, 9, 10}, , ...(2), , n(A∩B) = 35 + 22 - 47, , from (1) and (2),, , , , (A∪B)' = A'∩B', , iii), , ii) (A∩B)' = {3,4,6,7,8,9,10} ...(3), A'∪B' = {3,4,6,7,8,9,10}, , ...(4), , n(A'∩B) = n(B) - n(A∩B), , , , = 22 - 10, , , , = 12, , iv) , , from (3) and (4), , = 10, , n(A∪B') = n(X) - n[(A∪B)']', , (A∩B)' = A'∪B', , , , = n(X) - n(A'∩B), , Ex.6: If P = {x/x + 14x + 40 = 0}, , , , = 50 - 12, , , , = 38, , 2, , Q = {x/x2 - 5x + 6 = 0}, , Similarly Q = {3, 2}, R = {-20, 3} and, , Ex. 8 : In a board examination, 40 students, failed in Physics, 40 in Chemistry and 35., In Maths, 20 failed in Maths and Physics,, 17 in Physics and Chemistry, 15 in Maths, and Chemistry and 5 in all the three, subjects. If 350 students appeared in the, examination, how many of them did not fail, in any subject?, , X = {-20, -10, -4, 2, 3, 4}, , Solution:, , i) P∪Q = {-10, -4, 3, 2}, , P = set of students failed in Physics, , R = {x/x2 + 17x - 60 = 0} and the universal, set X = {-20, -10 - 4, 2, 3, 4}, find, i) P∪Q, , ii) Q∩R, , iii) P∪(Q∩R), , iv) P∩(Q∪R), , v) (P∪Q)' vi) Q'∩R', , Solution: P = {x/x2 + 14x + 40 = 0}, ∴ P = {-10, -4}, , ii) , , Q∩R = {3}, , iii), , P∪(Q∩R) = {-10, -4, 3}, , iv), , P∩(Q∪R) = φ, , v) , , (P∪Q)' = {-20, 4}, , C = set of students failed in Chemistry, M = set of students failed in Maths, , vi) Q' = {-20, -10, -4, 4}, R' = {-10, -4, 2, 4}, Q'∩R' = {-10, -4, 4}, Ex. 7: If A and B are the subsets of the universal, set X and n(x) = 50, n(A) = 35, n(B) = 22 and, n(A'∩B') = 3 , find, , Fig. 5.17, , i) n (A∪B) ii) n(A∩B), iii) n(A'∩B) iv) n (A∪B'), , From figure 5.17, we have, n(X) = 350, n(P) = 40, n(C) = 40, n(M) = 35, , Solution:, , n(M∩P) = 20, n(P∩C) = 17, n(M∩C) = 15, , i) n(A∪B) = n(X) - n[(A∪B)'], = n(X) - n(A'∩B') (De Morgan's Law), = 50 - 3, , and n(M∩P∩C) = 5, The number of students who failed in at least one, subject = n(M∪P∪C), , = 47., 95

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we have,, , i) , n(A∪B∪C) = n(X) - n[(A∪B∪C)]', = n(X) - n (A'∩B'∩C') = 1600 - 30 = 1570, , n (M∪P∪C) =n(M) + n(P) + n(C)-n (M∩P) n(P∩C) - n (M∩C)+ n (M∩ P ∩ C ) , = 35 + 40 + 40 - 20 - 17 -15 + 5 = 68, , Since, (A∪B∪C) = n(A) + n(B) + n(C) , , n(A∩B) - n(B∩C) - n(C∩A), , The number of students who did not fail in any, subject = n(X) - n (M∪P∪C) = 350 - 68 = 282, , , , + n(A∩ B∩C), , Ex. 9 : A company produces three kinds of products, A, B and C. The company studied the perference of, 1600 consumers for these 3 products. It was found, that the product A was liked by 1250, the product, B was liked by 930 and product C was liked by, 1000. The proudcts A and B were liked by 650,, the products B and C were liked by 610 and the, products C and A were liked by 700 consumers., None of the products were liked by 30 consumers., , , , Find number of consumers who liked., , = 650 - 350 = 300, , i) All the three products, , Similarly n(A'∩B∩C) = 610 - 350 = 260, , ii) Only two of these products., , and n(A∩B'∩C) = 700 - 350 = 350, , Solution: Given that totally 1600 consumers, were studied. ∴n (X) = 1600, Let A be the set of, all consumers who liked product A. Let B the set, of all consumers who liked product B and C be, the set of all consumers who liked product C., , ∴ The number of consumers who liked, only two of the three products is, , n(A) = 1250,, , Maximum & Minimum of Sets:, , n(B) = 930, n(C) = 1000., , (a) , , min {n(A∪B)} = max{n(A), n(B)}, , (b), , max {n(A∪B)} = n(A) + n(B), , (c) , , min {n(A∩B)} = 0, , ∴ 1570 = 1250 + 930 + 1000 - 650 - 610, - 700 + n(A∩B∩C), , ∴ n(A∩B∩C) = 1570 + 1960 - 3180, , , = 350, , ∴ The number of consumers who liked all, the three products is 350., ii) n[(A∩B)∩C'] = n (A∩B) - n [(A∩B)∩C],  n (P∩Q') = n(P) - n [P∩Q], , = n(A∩B∩C') + n(A∩B'∩C) + n(A'∩B∩C), = 300 + 350 + 260 = 910, , n(A∩B) = 650, n(B∩C) = 610, n(A∩C) = 700, n(A'∩B'∩C') = 30, , (d) max {n(A∩B)} = min{n(A), n(B)}, Illustration:, e.g. If n(A) = 10, n(B) = 20 then, min {n(A∪B) = max{10,20} = 20 and, max {n(A∪B)} = 10 + 20 = 30, so 20≤n(A∪B)≤30., Also min {n(A∩B)} = 0 and, max {n(A∩B)} = min {10,20} = 10, so, 0≤n(A∩B)≤10., , Fig. 5.18, 96

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Ex. 10) In a survey of 100 consumers 72, like product A and 45 like product B. Find the, least and the most number that must have liked, both products A and B., , 4) If A, B, C are the sets for the letters in the, words 'college', 'marriage' and 'luggage', respective, then verify that [A-(B∪C)] =, [(A-B)∩(A-C)], , Soln.: Let n(A) and n(B) be number of, consumers who like proudct A and B respectively,, i.e. n(A) = 72, n(B) = 45., , 5) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6},, C = {4, 5, 6, 7, 8} and universal set, X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify, the following:, , Since, n(A∪B) = n(A) + n(B) − n(A∩B), = 72 + 45 − n(A∩B) = 117 − n(A∩B)., , i) A∪(B∩C) = (A∪B) ∩ (A∪C), ii) A∩(B∪C) = (A∩B) ∪ (A∪C), , But as only 100 consumers are surveyed., Therefore n(A∪B)≤ 100 i.e. 117 − n(A∩B)≤100., ∴ 117 − 100≤n(A∩B), 17≤n(A∩B)., , (A∪B)' = (A'∩B)', , iv) , , (A∩B)' = A'∪B', , v) A = (A∩B)∪ (A∩B'), , So the least number of consumers who like, both the products is 17., , vi) B = (A∩B)∪ (A'∩B), vii) (A∪B) = (A−B) ∪ (A∩B) ∪ (B−A), , Since n(A∩B)≤n(A), n(A∩B)≤n(B), then, , viii) A ∩ (B∆C) = (A∩B) ∆ (A∩C), , n(A∩B)≤min{n(A), n(B)} = min {72, 45}, = 45. So the most number of consumers who like, both products is 45., , ix) , , n (A∪B) = n(A) + n(B) - n(A∩B), , x) n (B) = (A'∩B) + n(A∩B), 6) If A and B are subsets of the universal set X, and n(X) = 50, n(A) = 35, n(B) = 20,, , EXERCISE 5.1, 1) Describe the following sets in Roster form, , n(A'∩B') = 5, find, , i) A= {x/x is a letter of the word 'MOVEMENT'}, , i) n (A∪B) ii) n(A∩B), iv) n(A∩B'), , ii) B = {x/x is an integer,  3  x  9 }, 2, 2, , iii) n(A'∩B), , 7) In a class of 200 students who appeared, certain examinations, 35 students failed, in CET, 40 in NEET and 40 in JEE,, 20 failed in CET and NEET, 17 in NEET, and JEE, 15 in CET and JEE and 5 failed, in all three examinations. Find how many, students,, , iii) C = {x/x = 2n + 1, n∈N}, 2) Describe the following sets in Set-Builder, form, i) {0}, , iii), , ii) (0, ±1, ±2, ±3}, , 1 2 3 4 5 6 7 , iii)  , , , , , , ,  2 5 10 17 26 37 50 , , i) did not fail in any examination., ii) failed in NEET or JEE entrance., , iv) {0, -1, 2, -3, 4, -5, 6, ...}, , C = {x/2x2-x-3 = 0} then, , 8) From amongst 2000 literate individuals of a, town, 70% read Marathi newspapers, 50%, read English newspapers and 32.5% read, both Marathi and English newspapers. Find, the number of individuals who read., , find i) (A∪B∪C) ii) (A∩B∩C), , i), , 3) If A = {x/6x2+x-15 = 0}, B = {x/2x2-5x-3 = 0}, , 97, , at least one of the newspapers.

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ii) neither Marathi and English newspaper., , 5.2, , iii) Only one of the newspapers., , 5.2.1 Ordered Pair:, A pair (a, b) of numbers, such that the order,, in which the numbers appear is important, is called, an ordered pair. Ordered pairs (a,b) and (b, a) are, different. In ordered pair (a,b), 'a' is called first, component and 'b' is called second component., , 9) In a hostel, 25 students take tea, 20 students, take coffee, 15 students take milk, 10, student take bot tea and coffee, 8 students, take both milk and coffee. None of them, take tea and milk both and everyone takes, atleast one beverage, find the total number, of students in the hostel., 10), , Relations:, , Note : i) (a,b)=(c, d), if and only if a = c and b = d., , , There are 260 persons with a skin disorder., If 150 had been exposed to the chemical, A, 74 to the chemical B, and 36 to both, chemicals A and B, find the number of, persons exposed to, , ii) (a, b) = (b, a) if and only if a = b, , Ex. : Find x and y when (x + 3, 2) = (4, y - 3), Solution: Using the definition of equality of two, ordered pairs, we have, (x +3, 2) = (4, y - 3), , i) Chemical A but not Chemical B, , ⇒ x + 3 = 4 and 2 = y -3, , ii) Chemical B but not Chemical A, , ⇒ x = 1 and y = 5, , iii) Chemical A or Chemical B, 11), , Write down the power set of A = {1,2,3), , 12), , Write the following intervals in Set-Builder, form, , i), , (-3, 0), , , ii) [6,12],, , iii) (6, ∞), , , iv) (-∞, 5], , v) (2, 5], , vi) [-3, 4), , , , 5.2.2 Carstesian Product of two sets:, Let A and B be two non-empty sets then,, the cartensian product of A and B is denoted by, A×B and is defined as set of all ordered pairs, (a, b) such that a ∈A and b∈B, Thus, A×B = {(a, b) / a∈A, b∈B}, e.g. 1) If A = {1, 2} and B = {a, b, c}, , , 13) A college awarded 38 medals in volley ball,, 15 in football and 20 in basket ball. The, medals awarded to a total of 58 players and, only 3 players got medals in all three sports., How many received medals in exactly two, of the three sports?, , , , 2) If A = φ or B = φ, we define, , ∴ A×B = φ, Note:, Let A and B be any two finite sets with, n(A) = m and n(B) = n then the number of elements, in the Cartesian product of two finite sets A and B, is given by n(A×B) = mn, , 14) Solve the following inequalities and write, the solution set using interval notation., i) −9 < 2x + 7 ≤ 19, , then A×B = {(1, a) (1, b), (1, c),, (2, a), (2, b), (2, c)}, , ii) x2 − x > 20, , Ex. 2 : Let A = {1. 3} ; B = {2, 3, 4} Find the, number of elements in the Cartesian product of, A and B., , 2x, iii), ≤ 5 , iv) 6x2 + 1 ≤ 5x, x-4, 15) If A = (-7, 3], B = [2, 6] and C = [4, 9] then, find (i) A∪B (ii) B∪C (iii) A∪C (iv) A∩B, (v) B∩C (vi) A∩C (vii) A'∩B (viii) B'∩C', (ix) B-C (x) A-B, , Solution: Given A = {1, 3} and B = {2, 3, 4}, ∴ n(A) = 2 and n (B) = 3, ∴ n (A×B) = 2×3 = 6, 98

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5.2.3 Cartesian product of a set with itself:, , e.g. if A = {2, 3, 4, 5, 6) and B = {6, 7, 8,, 10} then the relation aRb if a is factor of b gives, the subset {(2, 6), (2, 8), (2, 10), (3, 6), (4, 8),, (5, 10)} of A×B., , A × A = A2 = {(a, b) / a, b ∈ A}, Where (a, b) is an ordered pair., A3 = A ×A×A = {(a, b, c) /a, b, c∈A}, , On the other hand, if we consider the subset, S' = {(2, 7), (3, 10), (4, 7)} of A×B, then there is, a unique relation R' given by this subset. i.e. 2R'7,, 3R'10, 4R'7. Thus S' has defined R'., , Where (a, b, c) is an ordered triplet., e.g. If A = {4, 5} then we have, A2 = A ×A = {(4, 4), (4, 5), (5, 4), (5, 5)}, , For every relation between A and B,, there is a unique subset defined in A×B and for, every subset of A×B, there is a unique relation, associated with it. Hence we can say that every, relation between A and B is a subset of A×B and, all relationships are ordered pairs in A×B., , A3 = A ×A×A= {(4, 4, 4), (4, 4, 5), (4, 5, 4), (5, 5, 5), (5, 4, 4) (5, 4, 5), (5, 5, 4), (4, 5, 5)}, Note: If n(A) = m then n (A × A × A) = m3. i.e., n(A3) = [n(A)]3, In general, n (A × A × ........ r times) = n(A)r, , When A = B, that is when we have relation, between the elements of A, that will be subset of, A×A. If a is nonempty, any relation in A×A is, called binary relation on A., , 5.2.4 Definitions, of, relation,, Domain,, Co- domain and Range of a Relation:, Relation:, , Ex. : Let A = {1, 2, 3, 4, 5} and B = {1, 4, 5}, , The concept of the term Relation is drawn, from the meaning of relation in English language,, according to which two objects or quantities are, related if there is recognizable link between them., , Let R be a relation such that (x, y) ∈ R, implies x < y. We list the elements of R., Solution: Here A = {1,2, 3, 4, 5} and B = {1, 4, 5}, ∴ R = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4),, (3, 5), (4, 5)}, , Members in a family are often related to each, other. We describe relation as A is friend of B, F is, father of S, P is sister of Q etc. In Mathematics we, can have different relations. e.g. among integers, we can define the relation as m is factor of n. If, this relation is denoted by R, then we write 2R4,, 3R6, 5R10., , Arrow diagram for this relation R is given by, , , A well defined relation between elements of, A and B is given by aRb, a ∈ A and b ∈ B. Thus, the relation gives ordered pair (a,b) and defines a, subset of A×B., , A, , B, , Fig. 5.19, , 99

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Domain:, , Ex. 3.: If R' is defined in Z×Z as aR'b if (a+b) is, even. Let R be defined Z×Z as a Rb if (a−b), is even., , The set of all first components of the, ordered pairs in a relation R is called the domain, of the relation R., , Note that R and R' define the same relation, and subset related to them is {a, b)/a and b, both even or a and b both odd}., , i.e. domain (R) = {a/(a, b) ∈ R}, Range:, , Note : i) Since φ ⊂A×A, φ is a relation on A, and is called the empty or void relation, on A., , The set of all second components of all, ordered pairs in a relation R is called the range of, the relation., , ii) Since A×A ⊂ A×A, A×A is a relation, on A called the universal relation on, A. i.e. R = A×A, , i.e. range (R) = {b/(a, b) ∈ R}, Co-domain:, , Ex. 4 : If A = {2, 4, 6}, , If R is a relation from A to B then set B is, called the co-domain of the relation R., , then R = A×A = {(2, 2), (2, 4), (2, 6),, (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)} and, R = A×A is the universal relation on A., , Note : Range is a subset of co-domain, (check!)., , Note : The total number of relation that, can be defined from a set A to a set B is the, number of possible subsets of A×B., if n(A) = m1 and n(B) = m2, then n(A×B) = m1m2, and the total number of relations is 2m1m2., , 5.2.5 Binary relation on a set:, Let A be nonempty set then every subset of, A×A is binary relation on A. , Illustrative Examples:, Ex. 1.: Let A = {1, 2, 3} and, , 5.2.6 Identity Relation:, , R = {(1, 2), (2, 2), (3, 1), (3, 2)}, , If the relation in A is given by aRa for every, a ∈ A, then the subset given by this relation is the, diagonal subset {a,a)/a∈A}. Hence, aRb if b = a., This is called identity relation., , A×A = {(1,1), (1,2), (1,3), (2,1), (2,2,),, (2,3), (3,1), (3,2), (3,3)}, R⊂A×A and therefore, R is a binary relation, on A., , 5.2.7 Types of relations:, Let A be a non-empty set. Then a binary, relation R on A is said to be, , Ex. 2.: Let N be the set of all natural numbers and, R = {(a, b) / a, b ∈ N and 2a + b = 10}, , (i) Reflexive, if (a, a) ∈ R for every a∈A i.e., aRa for every a∈A, , Since R⊂N×N, R is a binary relation on N., R = {(1, 8), (2, 6), (3, 4), (4, 2)}, , (ii) Symmetric, if (a, b)∈R, ⇒ (b, a) ∈R for all a, b∈A, i.e. aRb ⇒ bRa for all a, b∈A, , We can state domain, range and co-domain, of the relation., They are as follows:, , (iii) Transitive, if (a, b)∈R and (b, c) ∈ R, ⇒ (a, c)∈R for all a, b, c∈A., , Domain (R) = {1, 2, 3, 4}, Domain (R) = {2, 4, 6, 8}, , Note: Read the symbol ' ⇒' as 'implies'., , Co-domain (R) = N, 100

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Equivalence relation:, , iii), , A relation which is reflexive, symmetric, and transitive is called an equivalence relation., , Transitive, since ∆1≅ ∆2 and ∆2≅ ∆3, then, ∆1≅ ∆3 for all ∆1 , ∆2 , ∆3∈S. Hence, the given, relation is an equivalence relation., , Congruence Modulo:, , Illustrative examples:, , Let n be any positive integer. a = b (mod m) ⇔, (a − b) is divisible by m. read as 'a is congruent, to b modulo m' e.g: 14 ≡ 4 (mod 5) since 14 − 4 =, 10 is divisible by 5., , Ex. 1: Let R be a relation on Q, defined by, R = {(a, b)/a, b∈Q and a-b∈Z}, Show that R is an equivalence relation., Solution: Given R = {(a, b)/a, b∈Q and a-b∈Z}, , Ex. R = {(a,b) : a,b ∈ Z, a ≡ b (mod m)} is, an equivalence relation., , i) Let a ∈Q then a - a = 0∈Z, ∴ (a, a)∈R, So, R is reflexive., , Solution: Since a − a = 0 is divisible by m, ∴ a ≡ a (mod m) ∴ R is reflexive., If a ≡ b (mod m) i.e. a − b is divisible by, m then b − a is also divisible by m, , ii) (a, b)∈R⇒ (a-b)∈Z, i.e. (a-b) is an integer, ⇒ -(a-b) is an integer, ⇒ (b-a) is an integer, ⇒ (b, a) ∈ R, Thus (a, b) ∈R ⇒ (b, a) ∈ R, ∴ R is symmetric., , i.e. b ≡ a (mod m). ∴ R is symmetric., If a ≡ b (mod m) and b ≡ c (mod m), i.e. a − b is divisible by m and b − c is, divisible by m., Then a − c = a − b + b − c = (a − b) + (b − c), is also divisible by m. i.e. a ≡ c (mod m)., , iii) (a, b) ∈ R and (b, c) ∈ R, ⇒ (a-b) is an integer and (b-c) is an integer, ⇒ {(a-b) + (b-c)} is an integer, ⇒ (a-c) is an integer, ⇒ (a, c) ∈ R, Thus (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∴ R is transitive., Thus, R is reflexive, symmetric and, transitive., ∴ R is an equivalence relation., , ∴ R is transitivie., So, R is an equivalence relation., Note:, Since each relation defines a unique subset, of A×B., 1) , , If n(A) = m and n(B) = n then number of, relations S from A to B is 2mn., , 2) If R is relation on A and n(A) = m then, 2, number of relations on A is 2n, , Ex. 2 : Show that the relation "is congruent to" on, the set of all triangles in a plane is an equivalence, relation., , SOLVED EXAMPLES, , Solution: Let S be the set of all triangles in a, plane. Then, the congruence relation on S is, i) Reflexive, since ∆ ≅ ∆ for every ∆ ∈ S, , Ex. 1: If (x+1, y-2) = (3, 1) find the value of x, and y., , ii) Symmetric, since ∆1≅ ∆2 ⇒ ∆2 ≅ ∆1 for all, ∆1, ∆2 ∈ S., , Solution: Since the order pairs are equal, the, coresponding elements are equal., 101

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Ex. 6: Express {(x, y)/x2 +y2 = 25 where x,y∈ W}, as a set of ordered pairs., , ∴ x + 1 = 3 and y-2 = 1, ∴ x = 2 and y = 3, Ex. 2: If A = {1, 2}, find A×A, , Solution: We have x2 +y2 = 25, , Solution: We have A = {1, 2}, , ∴ x = 0, y = 5 ⇒ x2 +y2 = (0)2 +(5)2 = 25, x = 4, y = 3 ⇒ x2 + y2 = (4)2 +(3)2 = 25, , ∴ A×A = {(1, 1), (1, 2), (2, 1), (2, 2)}, , x = 5, y = 0 ⇒ x2 + y2 = (5)2 + (0)2 = 25, , Ex. 3: If A = {1, 3, 5} and B = {2, 3} find A×B, and B×A show that A×B ≠ B×A, , x = 3, y = 4 ⇒ x2 + y2 = (3)2 + (4)2 = 25, ∴ The set = {(0, 5), (3, 4), (4, 3), (5, 0)}, , Solution: We have A×B = {(1, 2), (1, 3), (3, 2),, (3, 3), (5, 2), (5, 3)} and B×A = {(2, 1), (2, 3),, (2, 5), (3, 1), (3, 3), (3, 5)}, , Ex. 7: Let A = {1, 2, 3} and B = {2, 4, 6} Show, that R = {(1, 2), (1, 4), (3, 2), (3, 4)} is a relation, from A to B. Find i) domain (R) ii) Co-domain, (R) iii) Range (R)., , All elements in A×B, B×A except (3, 3) are, different., ∴ A×B ≠ B×A, , Solution : Here A = {1, 2, 3} , B = {2, 4, 6}, , Ex. 4 : (Activity) If A×B = {(3, 2), (3, 4), (5, 2),, (5, 4)} then find A and B, , and R = {(1, 2), (1, 4), (3, 2), (3, 4)}, , Solution: Clearly, we have, A = Set of all first components of A×B, , i) Domain (R) = Set of first components of, R = {1, 3}, , ∴ A = {, , ii) Co-domain (R) = B = {2, 4, 6}, , ,, , Since R⊂A×B, R is a relation from A to B, , }, , iii) Range (R) = Set of second components, of R = {2, 4}, , B = Set of all second components of A×B, ∴ B = {, , ,, , }, , Ex. 5: A and B are two sets given in such a way, that A×B contains 6 elements. If three elements of, A×B are (1, 3), (2, 5) and (3, 3), find its remaining, elements., , Ex. 8: Let A = {1, 2, 3, 4, 5} and B = {1, 4, 5}., Let R be a relation from A to B. Such that (x, y)∈R, if 2x < y., , Solution: Since (1, 3), (2, 5) and (3, 3) are in A×B, it follows that 1, 2, 3 are elements of A and 3, 5, are elements of B. As A×B contains 6 elements, and if we let., , i), , List the elements of R., , ii), , Find the domain, co-domain and range of R., , Solution : A = {1, 2, 3, 4, 5} and B = {1, 4, 5}, i) The elements of R are as follows:, , , A = {1, 2, 3} and B = {3, 5}, then, A×B = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3) (3,, 5)} giving n (A×B) = 6., , R = {(1, 4), (1, 5), (2, 5),}, , ii) Domian (R) = {1, 2}, , Hence, the remaining elements of A×B are, (1, 5), (2, 3) and (3, 5)., 102, , , , Range (R) = {4, 5}, , , , Co-domain (R) = {1, 4, 5} = B

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Ex. 9: Let A = {1, 2, 3, 4, 5, 6} Define a relation R, from A to A by R = {(x, y) / y = x + 1}., Write down the domain, co-domain and range of R., Solution : A relation R from A to A is given by, R = (x, y) / y = x + 1} is given by, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), Domian (R) = {1, 2, 3, 4, 5}, Range (R) = {2, 3, 4, 5, 6}, , Verify, i) A × (B ∩ C) = (A × B) ∩ (A × C), , , ii) A × (B ∪ C) = (A × B) ∪ (A × C), , 6) Express {(x, y) / x2 + y2 = 100,, where x, y ∈ W} as a set of ordered pairs., 7) Let A = {6, 8} and B = {1, 3, 5}, Show that R1 = {(a, b)/a∈ A, b∈B, a−b, is an even number} is a null relation., R2 = {(a, b)/a∈ A, b∈B, a+b is odd number}, is an universal relation., , Co-domain (R) = {1, 2, 3, 4, 5, 6}, Note:, 1) R represents set of real numbers on number, line., 2) R × R represents set of all ordered pairs on, cartesian plane., 3) R × R × R represents set of all ordered, triplets in three dimensional space., , 8), , Write the relation in the Roster form. State, its domain and range., , i) R1 = {(a, a2) / a is prime number less, than 15}, ii) R2 = {(a,, , EXERCISE 5.2, 1) , , 1, ) / 0<a ≤ 5, a∈N}, a, , iii) R3 = {(x, y) / y = 3x, y∈ {3, 6, 9, 12},, x∈ {1, 2, 3}, , If (x−1, y+4) = (1, 2) find the values of x and, y., , iv) R4 = {(x, y) / y > x+1, x = 1, 2 and y = 2,, 4, 6}, , 1 3, 2) If (x + 1 , y − 1) = ( , ), find x and y,, 2 2, 3 3, , v) R5 = {(x, y) / x+y = 3, x, y∈ {0, 1, 2, 3}, , 3) If A = {a, b, c}, B = (x , y} find, A×B, B×A, A×A, B×B., , vi) R6 = {(a, b) / a ∈N, a < 6 and b = 4}, vii) R7 = {(a, b) / a, b ∈N, a + b = 6}, , 4) If P = {1, 2, 3) and Q = {14},, find sets P × Q and Q × P, , viii) R8 = {(a, b) / b = a + 2, a ∈z, 0 < a < 5}, 9) Identify which of if the following relations, are reflexive, symmetric, and transitive., [Activity], , 5) Let A = {1, 2, 3, 4), B = {4, 5, 6},, C = {5, 6}., , Relation, R = {(a,b) : a,b ∈ Z, a−b is an integer}, R = {(a,b) : a,b ∈ N, a+b is even}, R = {(a,b) : a,b ∈ N, a divides b}, R = {(a,b) : a,b ∈ N, a2 − 4ab + 3b2 = 0}, R = {(a,b) : a is sister of b and a,b ∈ G = Set of girls}, R = {(a,b) : Line a is perpendicular to line b in a place}, R = {(a,b) : a,b ∈ R, a < b}, R = {(a,b) : a,b ∈ R, a ≤ b3}, 103, , Reflexive, , Symmetrice, , Transitive, , , , , , 

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2) If aN = {ax : x ∈N}, then set 6N∩8N =, Let's Remember, , A) 8N, , B) 48N, , C) 12N, , D) 24N, , •, , A∪B = {x x ∈ A OR x ∈ B}, , 3) If set A is empty set then n[P [P [P (A)]]] is, , •, , A∩B = {x x ∈ A AND x ∈ B}, , A) 6 , , •, , A' = {x x ∈ U, x ∉ A} where U is universal, set., , •, , A−B = {x x ∈ A, x ∉ B}, , •, , A∆B = (A−B) ∪ (B−A), , 4) In a city 20% of the population travels by, car, 50% travels by bus and 10% travels by, both car and bus. Then, persons travelling, by car or bus are, , •, , n(A∪B) = n(A) + n(B) − n(A∩B), , A) 80%, , •, , n(A∪B∪C) = n(A) + n(B) + n(C) − n(A∩B), − n(B∩C) − n(A∩C) + n(A∩B∩C), , •, , A×B = {(x, y) x ∈ A, y ∈ B}, , •, , Relation R from set A to set B is a subset of, A×B., , 5) If the two sets A and B are having 43, elements in common, then the number of, elements common to each of the sets A × B, and B × A is, , •, , Domain of R = {x (x, y)∈ R, x ∈ A, y ∈ B}, ⊆ A., , •, , Range of R = {y (x, y)∈ R, x ∈ A, y ∈ B} ⊆, B (Co-domain of R)., , •, , R is a relation on set A and if (x, x)∈ R for all, x ∈ A then R is reflexive., , •, •, , •, , A) 432, , B) 16, , B) 40%, , B) 2 43, , C) 2 , , C) 60%, , C) 43 43, , D) 4, , D) 70%, , D) 2 86, , 6) Let R be a relation on the set N be defined, by {(x, y) / x, y∈ N, 2x + y = 41} Then R is, A) Reflexive, C) Transitive, , B) Symmetric, D) None of these, , 7) The relation ">" in the set of N (Natural, number) is, , R is a relation on set A and if (x, y)∈ R then, (y, x) ∈ R for all x, y ∈ A then R is symmetric., R is a relation on set A and if (x, y)∈ R and, (y, z)∈ R then (x, z)∈ R for all x, y, z ∈ A, then R is transitive., , A) Symmetric, , B) Reflexive, , C) Transitive, , D) Equivalent relation, , 8) A relation between A and B is, A) only A × B, , If R is reflexive, symmetric and transitive, then R is an equivalence relation., , B) An Universal set of A × B , C) An equivalent set of A × B, D) A subset of A × B, , MISCELLANEOUS EXERCISE 5, , 9) If (x,y) ∈ R × R, then xy = x2 is a relation, which is, , (I) Select the correct answer from given, alternative (Q:1 to Q:10)., 1) For the set A = {a, b, c, d, e} the correct, statement is, A) {a, b} ∈ A, , B) {a}∈ A, , C) a∈ A , , D) a∉ A, , A) Symmetric, , B) Reflexive, , C) Transitive, , D) Equivalence, , 10), , If A = {a, b, c} The total no. of distinct, relations in A × A is, , A) 3 , 104, , B) 9 , , C) 8 , , D) 29

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(II) Answer the following., , 6) If A = {1, 2, 3}, B = {4, 5, 6} check if the, following are relations from A to B. Also, write its domain and range., , 1) Write down the following sets in set builder, form, , i) R1 = {(1, 4), (1, 5), (1, 6)}, , i) {10, 20, 30, 40, 50},, ii) {a, e, i, o, u), , ii) R2 = {(1, 5), (2, 4), (3, 6)}, , ii) {Sunday, Monday, Tuesday, Wednesday,, Thursday, Friday, Saturday}, , ii) R4 = {(4, 2), (2, 6), (5, 1), (2, 4)}, , iii) R3 = {(1, 4 ), (1, 5), (3, 6), (2, 6), (3, 4)}, 7), , Determine the domain and range of the, following relations., i) R = {(a, b) / a ∈ N, a < 5, b = 4}, ii) R = {(a, b) / b = a−1 a ∈ z, a < 3}, , 2) If U = {x/x∈N, 1 ≤ x ≤ 12}, A = {1, 4, 7, 10} B = {2, 4, 6, 7, 11}, C = {3, 5, 8 9, 12}, Write down the sets, i) A∪ B, , ii) B∩C , , iii) A-B, , iv) B∩C', , v) A∪B∪C , , vi) A ∩(B∪C), , 8), Find R : A gA when A = {1,2,3,4} such that, i) R = (a, b) / a−b = 10}, ii) R = {(a, b) / a − b ) ≥ 0}, , 3) In a survey of 425 students in a school, it, was found that 115 drink apple juice, 160, drink orange juice and 80 drink both apple, as well as orange juice. How many drink, neither apple juice nor orange juice?, , 9), , R = {1,2,3} g { 1,2,3} given by R = {(1,1), (2,2), (3,3), (1,2) (2,3)} Check if R is, a) relflexive, b) symmentric, c) transitive, 10) Check if R : Z g Z, R = {(a, b) 2 divides, a-b} is equivalence relation., , 4) In a school there are 20 teachers who teach, Mathematics or Physics. Of these, 12 teach, Mathematics and 4 teach both Physics and, Mathematics. How many teachers teach, Physics?, 5) i) If A = {1, 2, 3} and B = {2, 4}, state the, elements of A × A, A × B, B × A, B × B,, (A × B) ∩ (B × A), , 11), , Show that the relation R in the set, A = {1, 2, 3, 4, 5} Given by R = {(a, b) /, a−b is even} is an equivalence relation, , 12), , Show that following are equivalence, relation, , a) R in A is set of all books. given by, , , ii) If A = {−1, 1} , find A × A × A, , R = { (x, y) / x and y have same number, of pages}, , b) R in A = {x∈Z 0 < x < 12} given by, , , R = {(a, b) / a−b  is a multiple of 4}, , c) R in A = {x∈ N/x ≤10} given by, , , v, , v, 105, , v, , R = {(a, b) a=b}

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6, , FUNCTIONS, Since, every element from A is associated, to exactly one element in B, R1 is a well defined, function., , Let's Study, •, , Function, Domain, Co-domain, Range, , •, , Types of functions, , •, , Representation of function, , •, , Basic types of functions, , •, , Piece-wise defined and special functions, Let's : Learn, , 6.1 Function, , Fig. 6.2, , Definition : A function (or mapping) f from a set, A to set B (f : A→B) is a relation which associates, for each element x in A, a unique (exactly one), element y in B., , R2 is not a function because element 'd' in A, is not associated to any element in B., , Then the element y is expressed as y = f(x) ., y is the image of x under f., f is also called a map or transformation., If such a function exists, then A is called the, domain of f and B is called the co-domain of f., Illustration:, Examine the following relations which are given, by arrows of line segments joining elements in A, and elements in B., , Fig. 6.3, R3 is not a function because element a in A, is associated to two elements in B., The relation which defines a function f from, domain A to co-domain B is often given by an, algebraic rule., For example, A = Z, the set of integers and, B = Q the set of rational numbers and the function, n, f is given by f(n) =, here n ∈ Z, f(n) ∈ Q., 7, , Fig. 6.1, 106

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6.1.1 Types of function, One-one or One to one or Injective function, Definition : A function f : A → B is said to be, one-one if different elements in A have different, images in B. The condition is also expressed as, f (a) = f(b) ⇒ a = b [As a ≠ b ⇒ f (a) ≠ f(b)], Onto or Surjective function, , Fig. 6.6, , Definition: A function f : A → B is said to be onto, if every element y in B is an image of some x in A, (or y in B has preimage x in A), , f3 is onto but not one-one as f (a) = f (b) = 1, but a ≠ b., , The image of A can be denoted by f (A)., f(A) = {y ∈ B | y = f (x) for some x ∈ A}, f (A) is also called the range of f., Note that f : A → B is onto if f(A) = f ., Also range of f = f (A) ⊂ co-domain of f., Illustration:, , Fig. 6.7, f4 is neither one-one, nor onto, 6.1.2 Representation of Function, Verbal, form, , Output exceeds twice the input by 1, Domain : Set of inputs, Range : Set of outputs, , Arrow, form on, Venn, Diagram, , Fig. 6.4, f1 is one-one, but not onto as element 5 is in, B has no pre image in A, , Fig. 6.8, , Ordered, Pair, (x, y), Fig. 6.5, f2 is one-one, and onto, 107, , Domain : Set of pre-images, Range: Set of images, f = {(2,5), (3,7), (4,9), (5,11)}, Domain : Set of 1st components from, each ordered pair = {2, 3, 4, 5}, Range : Set of 2nd components from, each ordered pair = {5, 7, 9, 11}

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Rule /, Formula, , Tabular, Form, , y = f(x) = 2x + 1, Where x ∈ N, 1 < x < 6, f(x) read as ‘f of x’ or ‘function of x’, Domain : Set of values of x for, which f (x) is defined, Range : Set of values of y for which, f (x) is defined, x, 2, 3, 4, 5, , y, 5, 7, 9, 11, , Fig. 6.10, Since every x has a unique associated value of y., It is a function., , Domain : x values, Range: y values, Graphical, form, , Fig. 6.11, This graph does not represent a function as vertical, line intersects at more than one point some x has, more than one values of y., Fig. 6.9, Domain: Projection of graph on, x-axis., Range: Projection of graph on y-axis., , Horizontal Line Test:, If no horizontal line intersects the graph of a, function in more than one point, then the function, is one-one function., Illustration:, , 6.1.3 Graph of a function:, If the domain of function is in R, we can show, the function by a graph in xy plane. The graph, consists of points (x,y), where y = f(x)., Vertical Line Test, Given a graph, let us find if the graph represents a, function of x i.e. f(x), A graph represents function of x, only if no, vertical line intersects the curve in more than one , point., 108, , Fig. 6.12

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Solution : From graph when x = −4, y = 0, so g (−4) = 0, From graph when x = 3, y = −5 so g (3) = −5, , The graph is a one-one function as a horizontal, line intersects the graph at only one point., , Function Solution:, Ex. 3) If t (m) = 3m2 − m and t (m) = 4, then find m, Solution : As, t (m) = 4, 3m2 − m = 4, 3m2 − m − 4 = 0, 3m2 − 4m + 3m − 4 = 0, m (3m − 4) + 1 (3m − 4) = 0, (3m − 4) (m + 1) = 0, 4, Therefore , m = 3 or m = −1, , Fig. 6.13, The graph is a one-one function, 6.1.4 Value of funcation : f(a) is called the value, of funcation f(x) at x = a, Evaluation of function:, , Ex. 4) From the graph below find x for which, f (x) = 4, , Ex. 1) Evaluate f(x) = 2x2 − 3x + 4 at, , , x = 7 & x = −2t, , Solution : f (x) at x = 7 is f (7), f (7) = 2(7)2 - 3(7) + 4, = 2(49) − 21 + 4, = 98 − 21 + 4, = 81, f (−2t) = 2(−2t)2 − 3(−2t) + 4, = 2(4t2) + 6t + 4, = 8t2 + 6t + 4, , Fig. 6.15, Solution : To solve f (x) = 4 i.e. y = 4, , Ex. 2) Using the graph of y = g (x) , find g (−4), and g (3), , Find the values of x where graph intersects, line y = 4, , Fig. 6.14, , Fig. 6.16, 109

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2) Identity function, If f : R → R then identity function is defined by, f (x) = x, for every x ∈ R., , Therefore , x = −1 and x = 3., Function from equation:, Ex. 5) (Activity) From the equation 4x + 7y = 1, express, i), , y as a function of x, , ii), , x as a function of y, , Identity function is given in the graph below., , Solution : Given equation is 4x + 7y = 1, i), , From the given equation, 7y =, y=, , = function of x, , So y = f (x) =, ii), , From the given equation, 4x =, x=, , = function of y, , Fig. 6.18, , So x = g (y) =, , Domain : R or (−∞, ∞) and Range : R or (−∞, ∞), [Note : Identity function is also given by, I (x) = x]., , 6.1.5 Some Basic Functions, (Here f : R → R Unless stated otherwise), 1), , 3), , Constant Function, , Power Functions : f (x) = axn , n ∈ N, (Note that this function is a multiple of nth, power of x), , Form : f (x) = k , k ∈ R, i), , Example : Graph of f (x) = 2, , Fig. 6.17, , Square Function, Example : f (x) = x2, , Fig. 6.19, , Domain : R or (−∞, ∞) and Range : {2}, , Domain : R or (−∞, ∞) and Range : [0, ∞), 110

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Properties:, 1), , Graph of f (x) = x2 is a parabola opening, upwards and with vertex at origin., , 2), , Graph is symmetric about y - axis ., , 3), , The graph of even powers of x looks similar, to square function. (verify !) e.g. x4, x6., , 4), , (y − k) = (x − h)2 represents parabola with, vertex at (h, k), , 5), , If −2 ≤ x ≤ 2 then 0 ≤ x2 ≤ 4 (see fig.) and, if −3 ≤ x ≤ 2 then 0 ≤ x2 ≤ 9 (see fig)., , Fig. 6.21, , ii) Cube Function, Example : f (x) = x3, , Domain : R or (−∞, ∞) and Range : R or (−∞, ∞), Properties :, 1), , Graph of f (x) = ax + b is a line with slope,  b, ‘a’, y-intercept ‘b’ and x-intercept  −  .,  a, , 2) Function : is increasing when slope is positive, and deceasing when slope is negative., ii), , Quadratic Function, , Form : f (x) = ax2 + bx + c (a ≠ 0), , Fig. 6.20, Domain : R or (−∞, ∞) and Range : R or (−∞, ∞), Properties:, 1) The graph of odd powers of x (more than 1), looks similar to cube function. e.g. x5, x7., 4), , Polynomial Function, f (x) = a0 xn + a1 xn-1 + ... + an-1 x + an, , Fig. 6.22, , is polynomial function of degree n , if a0 ≠ 0, and, ai s are real., , Domain : R or (−∞, ∞) and Range : [k, ∞), , i), , Properties :, , Linear Function, , Form : f (x) = ax + b (a ≠ 0), , 1), , Example : f (x) = − 2x + 3, x ∈ R, 111, , Graph of f (x) = ax2 + bx + c and where, a > 0 is a parabola.

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iii), , If D = b2 − 4ac < 0, the parabola lies above, x-axis and y ≠ 0 for any x. Here y is positive, for all values of x. e.g. f(x) = x2 + 4x + 5, , iii) Cubic Function, Example : f (x) = ax3 + bx2 + cx + d (a ≠ 0), Domain : R or (−∞, ∞) and, Range : R or (−∞, ∞) , , Fig. 6.23, Consider, y = ax2 + bx + c, b2, b2, b, = a x2 +, +, c, −, x, +, 4a, 4a2, a, b 2 b2−4ac, = a x + 2a − 4a, b2−4ac, b 2, x, +, =, a, y + 4a, 2a, , Fig. 6.24, , With change of variable, , Property:, , b 2 − 4ac, b, y, +, x, +, X=, ,Y=, 4a, 2a, , 1), , f (x) = (x − 1) (x2 + x + 1) cuts x-axis at only, one point (1,0) , which means f (x) has one real, root & two complex roots., , this is a parabola Y = aX2, This is a parabola with vertex, b, b2−4ac, − 2a ,, 4a, ,  −b − D , or  ,,  where,  2a 4a , , Note that, any polynomial of odd degree must, have at least one real root, because the complex, roots appear in conjugate pairs., , D = b2 − 4ac and the parabola is opening upwards., There are three possibilities., , 5) Radical Function, , For a>0,, i), , If D = b − 4ac = 0, the parabola touches, x-axis and y ≥ 0 for all x. e.g. g(x), = x2 − 2x + 1, , ii), , If D = b − 4ac > 0, then parabola intersects, x-axis at 2 distinct points. Here y is negative, for values of x between the 2 roots and, positive for large or small x., , Graph of f (x) = x3 − 1, , Ex: f(x) =, , 2, , 1., , x,n∈N, , Square root function, f(x) =, , 2, , n, , x,x≥0, , (Since square root of negative number is not a, real number, so the domain of x is restricted to, positive values of x)., 112

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i.e. 9 ≥ 9 − x2 ≥ 0, i.e. 3 ≥, , 9 − x2 ≥ 0, , ∴ 3≥y≥0, , ...(II), , , From (I) and (II), y ∈ [0,3] is range of, f (x)., 2., , Cube root function, f(x) =, , 3, , x,, , Fig. 6.25, Domain : [0, ∞) and Range : [0, ∞), Note :, 1), , If x is positive, there are two square roots of, x. By convention x is positive root and, − x is the negattive root., , 2), , If −4 < x < 9, as x is only deifned for, x ≥ 0, so 0 ≤ x < 3., , Ex. 6 : Find the domain and range of f (x) =, , Fig. 6.26, , 9− x ., 2, , Soln. : f (x) =, , Domain : R and Range : R, , 9 − x is defined for, 2, , Note : If −8 ≤ x ≤ 1 then −2 ≤, , 9 − x ≥ 0, i.e. x − 9 ≤ 0 i.e. (x − 3)(x + 3), ≤0, 2, , 2, , Ex. 7 : Find the domain f (x) =, , Therefore [−3, 3] is domain of f (x)., (Verify !), To find range, let, , 3, , x ≤ 1., , x3 − 8 ., , Soln. : f (x) is defined for x3 − 8 ≥ 0, i.e. x3 − 23 ≥ 0, (x − 2) (x2 + 2x + 4) ≥ 0, , 9 − x2 = y, , In x2 + 2x + 4, a = 1 > 0 and D = b2 − 4ac, = 22 − 4 × 1 × 4 = −12 < 0, , Since square root is always positive, so, y≥0, ...(I), , Therefore, x2 + 2x + 4 is a positive, quadratic., , Also, on squaring we get 9 − x2 = y2, Since, 3 ≤ x ≤ 3, , i. e. x2 + 2x + 4 > 0 for all x, , i.e. 0 ≤ x2 ≤ 9, i.e. 0 ≥ − x2 ≥ − 9, , Therefore x − 2 ≥ 0, x ≥ 2 is the domain., , i.e. 9 ≥ 9 − x2 ≥ 9 − 9, , i.e. Domain is x ∈ [2, ∞), 113

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6) Rational Function, , Solution : f (x) is defined for all x ∈ R exccept, when denominators is 0., 5, Since, 4x + 5 = 0 ⇒ x = − ., 4, , Definition: Given polynomials, p(x), p(x), q(x) f (x) = q(x) is defined for x if q(x) ≠0., ,  5, So Domain of f (x) is R − −  .,  4, , 1, Example : f (x) = x , x ≠ 0, , 6 − 4x2, To find the range, let y =, 4x + 5, i.e. y(4x + 5) = 6 − 4x2, i.e. 4x2 + (4y)x + 5y − 6 = 0., This is a quadratic equation in x with y as, constant., Since x ∈ R − {−5/4}, i.e. x is real, we get, Solution if, D = b2 − 4ac ≥ 0, i.e. (4y)2 − 4(4)(5y − 6) ≥ 0, 16y2 − 16(5y − 6) ≥ 0, y2 − 5y + 6 ≥ 0, (y − 2) (y − 3) ≥ 0, , Fig. 6.27, , Therefore y ≤ 2 or y ≥ 3 (Verify!), , Domain : R-{0} and Range : R −{0} , , Range of f(x) is (−∞, 2] ∪ [3,∞), , Properties:, 1), , 7), , As x → 0 i.e. (As x approaches 0) f (x) →, ∞ or f (x) → −∞ , so the line x = 0 i.e y-axis, is called vertical asymptote.( A straight line, which does not intersect the curve but as x, approaches to ∞ or −∞ the distance between, the line and the curve tends to 0, is called an, asymptote of the curve.), , Exponential Function, , Form : f (x) = ax is an exponential function with, base a and exponent (or index) x, a ≠ 0,, a > 0 and x ∈ R., Example : f (x) = 2x and f(x) = 2−x, , 2), , As As x → ∞ or x → −∞, f (x) → 0, y = 0 the, line i.e y-axis is called horizontal asymptote., p(x), 3) The domain of rational function f (x) = q(x), is all the real values of except the zeroes of, q(x)., , Ex. 8 : Find domain and range of the function, , Fig. 6.28, , 6 − 4x2, f (x) =, 4x + 5, , Domain: R and Range : (0, ∞), 114

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Properties:, , i.e. 6.2x − (2x)2 − 8 ≥ 0, , 1), , As x → −∞ , then f (x) = 2x → 0, so the graph, has horizontal asymptote (y = 0), , i.e. (2x)2 − 6.2x + 8 ≤ 0, , 2), , By taking the natural base e (≈ 2.718), , graph of f (x) = ex is similar to that of 2x in, appearance, , 2x ≥ 2 and 2x ≤ 4 (Verify !), , i.e. (2x − 4)(2x − 2) ≤ 0, 2x ≥ 21 and 2x ≤ 22, x ≥ 1 and x ≤ 2 or 1 ≤ x ≤ 2, Doamin is [1,2], 8), , Logarithmic Function:, , Let, a > 0, a ≠ 1, Then logarithmic function loga x,, y = loga x if x = ay., for x > 0, is defined as, y = log x ⇔, , a, log arithmicform, , ay = x, , exponentialform, , Properties:, 1), , so logaa = 1, , Fig. 6.29, 3), , 4), , For a > 0, a ≠ 1, if ax = ay then x = y. So ax is, one-one function. (check graph for horizontal, line test)., , 2), , As ax = ay ⇔ x = y so loga x = loga y ⇔, x= y, , 3), , Product rule of logarithms., For a, b, c > 0 and a ≠ 1,, , r > 1, m > n ⇒ rm > rn and, r < 1, m > n ⇒ rm < rn, , loga bc = loga b + loga c (Verify !), , Ex. 9 : Solve 52x+7 = 125., , 4), , Solution : As 52x+7 = 125, , loga, , 3−7 −4, 2 = 2 = − 2, , Ex. 10 : Find the domain of f(x) =, Solution : Since, , Quotient rule of logarithms., For a, b, c > 0 and a ≠ 1,, , i.e : 52x+7 = 53, ∴ 2x + 7 = 3, and x =, , As a0 = 1, so loga1 = 0 and as a1 = a,, , 5), x, , 6−2 −2, , 3− x, , b, = loga b − loga c (Verify !), c, , Power/Exponent rule of logarithms., For a, b, c > 0 and a ≠ 1,, loga bc = c loga b, , x is defined for x ≥ 0, , f (x) is defined for 6 − 2x − 23−x ≥ 0, 23, i.e. 6 − 2 − x ≥ 0, 2, x, , 115, , (Verify !)

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Ex. 12 : Evaluate lne9 − lne4., , 6) For natural base e , loge x = ln x as Natural, Logarithm Function., , Solution : lne9 − lne4 = logee9 − logee4, = 9 logee − 4 logee, = 9(1) − 4(1) (∴ lne = 1), = 5,  x 3 ( x + 3) , Ex. 13 : Expand log , 2,  2 ( x − 4 ) , Solution : Using Quotient rule, = log [x3(x+3)] − log [2 (x−4)2 ], Using Product rule, = [log x3 + log (x+3)] − [log2 + log (x−4)2], Using Power rule, = [3log x + log (x+3)] − [log2 − 2log (x−4)], , Fig. 6.30, , = 3log x + log (x+3) − log2 + 2log (x−4), , Here domain of ln x is (0, ∞) and range is (−∞, ∞)., 8), , Ex. 14 : Combine, , Logarithmic inequalities:, , 1, 3ln (p + 1) −, ln r + 5ln(2q + 3) into single, 2, logarithm., , (i) If a>1, 0<m<n then loga m < loga n, e.g. log10 20 < log10 30, , Solution : Using Power rule,, , (ii) If 0>a<1, 0<m<n then loga m > loga n, e.g. log0.1 20 > log0.1 30, , Using Quotient rule, , (iii) For a, m>0 if a and m lies on the same side of, unity (i.e. 1) then loga m>0., , = ln, , e.g. log2 3>0, log0.3 0.5>0, , ( p + 1), r, , 3, , + ln(2q +3)5, , Using Product rule, , (iv) For a, m>0 if a and m lies on the different, sides of unity (i.e. 1) then loga m>0., ,  ( p + 1) 3, , = ln , (2q + 3)5 , r, , , , e.g. log0.2 3<0, log3 0.5>0, Ex. 11 : Write log72 in terms of log2 and log3., , Ex. 15 : Find the domain of ln(x − 5)., , Solution : log 72 = log(2 .3 ), 3, , 1, 2, , = ln (p + 1) − ln r + ln(2q +3)5, 3, , 2, , Solution : As ln(x − 5) is defined for (x − 5) > 0, , = log 23 + log 32 (Power rule), , that is x > 5 so domain is (5, ∞)., , = 3 log 2 + 2 log 3(Power rule), 116

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Let’s note:, 1), , log (x + y) ≠ log x + log y, , 2), , log x log y ≠ log (xy), , 3), 4), , log x, x, ≠ log, log y, y, , Fig. 6.31, , (log x) ≠ n log, n, , n, , 9), , Trigonometric function, , The graphs of trigonometric functions are, discusse in chapter 2 of Mathematics Book I., , 9) Change of base formula:, log x, For a, x, b > 0 and a, b ≠ 1, loga x = log b a, b, logx x, 1, Note: loga x = log a = log a (Verify !), x, x, , f (x), sin x, cos x, tan x, , log 81, Ex. 16 : Evaluate log4 9, 4, , Domain, R, R,  π 3π , R − ± 2 , ± 2 ..., , Range, [−1,1], [−1,1], R, , Solution : By Change of base law, as the base is, same (that is 4), EXERCISE 6.1, , log4 81, log4 9 = log9 81 = 2, 1), , Ex. 17 : Prove that, 2logb a4. logc b3. logac5 = 120, , Check if the following relations are functions., , (a) , , Solution : L.H.S. = 2logb a4. logc b3. logac5, = 4 × 2logb a × 3logc b × 5logac, Using change of base law,, log b, log c, log a, = 4 × 2 log b × 3 log c × 5 log a, = 120, , Fig. 6.32, (b), , Ex. 18 : Find the domain of f(x) = logx +5 (x2 − 4), Solution : SInce loga x is definedd for a, x > 0 and, a ≠ 1 f(x) is defined for (x2 − 4) > 0, x + 5 > 0,, x + 5 ≠ 1., i.e. (x − 2)(x + 2) > 0, x >−5, x ≠ −4, i.e. x < −2 or x > 2 and x > −5 and x ≠ −4, , Fig. 6.33, 117

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(c), , 7), , a−x, If f (x) = b−x , f (2) is undefined, and, f (3) = 5, find a and b., , 8), , Find the domain and range of the follwoing, functions., (a) f (x) = 7x2 + 4x − 1, x+4, (b) g (x) = x−2, , Fig. 6.34, 2), , Which sets of ordered pairs represent, functions from A = {1, 2, 3, 4} to B = {−1, 0,, 1, 2, 3}? Justify., (a) {(1,0), (3,3), (2,−1), (4,1), (2,2)} , , (d) f (x) =, , (b) {(1,2), (2,−1), (3,1), (4,3)} , (c) {(1,3), (4,1), (2,2)}, (d) {(1,1), (2,1), (3,1), (4,1)}, 3), , 4), , Check if the relation given by the equation, represents y as function of x., (a) 2x + 3y = 12, (b) x + y2 = 9, (c) x2 − y = 25, (d) 2y + 10 = 0, (e) 3x − 6 = 21, , (c) f, , 9), , (d) f (x + 1), , (e) f (−x),  f (2 + h) − f (2) , (f) ,  ,h≠0., h, , , 5), , Find x, if g (x) = 0 where, 5x−6, (a) g (x) = 7, , (b)g (x) =, , 18−2x, 7, , (e) f (x) =, , ( x − 2)(5 − x), , (f) f (x) =, , x −3, 7−x, , Express the area A of a square as a function, of its (a) side s (b) perimeter P., , Let f be a subset of Z × Z defined by, , 2, , 12) f = {(ab,a+b) : a,b ∈ Z} . Is f a function from, Z to Z? Justify., , (c) g (x) = 6x2 + x −2, , 14) Check the injectivity and surjectivity of the, following functions., , (d) g (x) = x3 − 2x2 − 5x + 6, 6), , x +1, , 10) Express the area A of circle as a function of its, (a) radius r (b) diameter d (c) circumference, C., 11) An open box is made from a square of, cardboard of 30 cms side, by cutting squares, of length x centimeters from each corner and, folding the sides up. Express the volume, of the box as a function of x. Also find its , domain., , (b) f (−3), , 1, 2, , 3, , (g) f (x) = 16 − x 2, , If f (m) = m2 − 3m + 1, find, (a) f (0), , x+5, 5+ x, , (c) h (x) =, , Find x, if f (x) = g (x) where, , (a) f : N → N given by f (x) = x2, , (a) f (x) = x4 + 2x2, g (x) = 11x2, , (b) f : Z → Z given by f (x) = x2, , (b) f (x) =, , (c) f : R → R given by f (x) = x2, , x −3, g (x) = 5 − x, 118

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21) Write the following expressions as a single, logarithm., , (d) f : N → N given by f (x) = x3, (e) f : R → R given by f (x) = x3, , (a) 5log x + 7log y − log z, , 14) Show that if f : A → B and g : B → C are, one-one, then g ◦ f is also one-one., , (b), , (c) ln (x+2) + ln (x−2) − 3ln (x+5), , 15) Show that if f : A → B and g : B → C are, onto, then g ◦ f is also onto., , 22) Given that log 2 = a and log 3 = b,, , 16) If f (x) = 3(4x+1) find f (−3) ., , write log, , 17) Express the following exponential equations, in logarithmic form, (a)25 = 32, , (b) 540 = 1, , (c) 23 = 23, 1, (e) 3−4 = 81, (g) e2 = 7.3890, , (d) 9 = 27, , 1, , (a) b logb a = a, , 3/2, , (b) log bm a =, , 1, logba, m, , (c) a logc b = b logc a, , (f) 10−2 = 0.01, (h) e1/2 = 1.6487, , 24) If f (x) = ax2 − bx + 6 and f (2) = 3 and, f (4) = 30, find a and b, , 18) Express the following logarithmic equations, in exponential form, 1, (a) log264 = 6, (b) log5 25 = −2, (c) log100.001=−3 , , (d) log1/2 (−8) = 3 , , (e) ln 1 = 0 , 1, (g) ln, = − 0.693, 2, , (f) ln e = 1, , 25) Solve for x., (a) log2 + log(x+3) − log(3x−5) = log3, 11 , , (b) 2log10x = 1 + log10  x + , 10 , , 21, (c) log2x + log4x + log16x = 4, (d) x + log10(1+2x) = x log105 + log106, , 19) Find the domain of, , 1, 1, x+ y, 26) If log ,  = 2 log x + 2 log y,,  3 , y, x, show that y + x = 7., , (a) f (x) = ln (x −5) , (b) f (x) = log10 (x2−5x +6), 20) Write the following expressions as sum or, difference of logarithms, (b) log, , (, , x3 y, ,  x− y, 27) If log ,  = log x + log y ,,  4 , show that (x+ y)2 = 20 xy, , ), ,  a 3 ( a − 2 )2 , , (c) ln ,  b2 + 5 , , ,  3 x − 2 ( 2 x + 1)4 , (d) ln , ,  ( x + 4 ) 2 x + 4 , , 96 in terms of a and b., , 23) Prove that, , (i) e−x = 6, , pq , (a) log ,  ,  rs , , 1, 1, log (x−1) +, log (x), 3, 2, , 28) If x = logabc, y = logbca, z = logcab, then prove that, , 2, , 119, , 1, 1, 1, +, +, =1, 1+ x 1+ y 1+ z

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6.2 Algebra of functions:, , iii) As (fg) (x) = f (x) g (x), , Let f and g be functions with domains A and B., f, Then the functions f + g, f − g, fg, g are defined, on A∩ B as follows., , (f o g ) (3m) = f (3m)g (3m), = [(3m)2 + 2] [5(3m) − 8], = [9m2 + 2] [15m−8], , Operations, , = 135m3 − 72m2 + 30m −16, , (f + g) (x) = f (x) + g (x), ,  f , f(x), iv) As   (x) = g(x) , g (x) ≠ 0, g, , (f − g) (x) = f (x) − g (x), (f . g) (x) = f (x) . g (x), , 02 + 2, f(0),  f ,   (0) = g(0) = 5(0) − 8, g, 2, 1, = −8 = −, 4, ,  f , f(x),   (x) = g(x) where g (x) ≠ 0, g, Ex. 1 : If f (x) = x2 + 2 and g (x) = 5x − 8, then find, , Ex. 2 : Given the function f (x) = 5x2 and, , i), , (f + g) (1), , g (x) =, , ii), , (f − g) (−2), , i), , ii), , (f o g) (3m), , iv), , f, (0), g, , 4 − x find the domain of, f, , (f + g) (x) ii) (f o g) (x) iii) g (x), , Solution : i) Domain of f (x) = 5x2 is (−∞, ∞) ., To find domain of g (x) =, 4−x≥0, , Solution : i) As (f + g) (x) = f (x) + g (x), , , (f + g) (1) = f (1) + g (1), , , , = [(1) + 2] + [5(1) − 8], , , , = 3 + (− 3), , , , =0, , 4− x, , x−4≤0, Let x ≤ 4, So domain is (−∞, 4]., , 2, , Therefore, domain of (f + g) (x) is, (−∞, ∞) ∩(−∞, 4], that is (−∞, 4], ii) Similarly, domain of (fog) (x) = 5x2 4 − x, is (−∞, 4], , ii) As (f − g) (x) = f (x) − g (x), ,  f , 5x2, iii) And domain of   (x) =, is (−∞, 4), 4− x, g, As , at x = 4 the denominator g (x) = 0 ., , (f − g) (−2) = f (−2) − g (−2), = [(−2)2 + 2] − [5(−2) − 8], = [4 + 2] − [−10−8], , 6.2.1 Composition of Functions:, , = 6 + 18, , A method of combining the function f : A→B with, g : B→C is composition of functions, defined as , (f ◦ g) (x) = f [g (x)] an read as ‘f composed with g’, , = 24, , 120

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ii), , As (g ◦ f) (x) = g [f (x)] and g (x) = x2 − 1, Replace x by f (x), to get, (g ◦ f) (x) = [f (x)]2 − 1, 2, ,  2 , , =,  −1,  x+5, Now let x = 3, 2, ,  2 , (g ◦ f) (3) = ,  −1,  3+5, Fig. 6.35, , 2, , 2, =   −1, 8, , Note:, 1), , 2), , 2, , 1, =   −1, 4, , The domain of g ◦ f is the set of all x in A such, that f (x) is in the B. The range of g ◦ f is set, of all g [f(x)] in C such that f(x) is in B., , =, , Domain of g ◦ f ⊆ Domain of f and Range, of g ◦ f ⊆ Range of g., , = −, , Illustration:, , 15, 16, , 1, Ex 4 : If f (x) = x 2, g (x) = x + 5, and h (x) = x ,, x ≠ 0, find (g ◦ f ◦ h) (x), Solution : (g ◦ f ◦ h) (x), = g {f [h (x)], , A cow produces 4 liters of milk in a day., Then x number of cows produce 4x liters of milk, in a day. This is given by function f (x) = 4x = 'y'., Price of one liter milk is Rs. 50. Then the price, of y liters of the milk is Rs. 50y. This is given by, another function g(y) = 50y. Now a function h(x), gives the money earned from x number of cows in, a day as a composite function of f and g as h(x) =, (g ◦ f )(x) = g[f (x)] = g(4x) = 50(4x) = 200x., , 1, = g f x, = g f 1, x, , 2, Ex. 3 : If f (x) = x+5 and g (x) = x2 − 1 , then find, , =, , i) (f ◦ g) (x) ii) (g ◦ f) (3), , 1, x, , 2, , 2, , +5, , 1, = x2 + 5, , Solution :, i), , 1 − 16, 16, , 2, As (f ◦ g) (x) = f [g (x)] and f (x) = x+5, Replace x from f (x) by g (x) , to get, , Ex. 5 : If h (x) = (x − 5)2, find the functions f and, g, such that h = f ◦ g ., → In h (x), 5 is subtracted from x first and then, squared. Let g (x) = x − 5 and f (x) = x2, (verify), , 2, (f ◦ g) (x) = g(x)+5, 2, , = x2−1+5, 2, , = x2+4, , 1, Ex. 6 : Express m (x)= x3+7 in the form of f ◦ g ◦ h, → In m (x), x is cubed first then 7 is added and , then its reciprocal taken. So,, 121

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1, h(x) = x3, g(x) = x + 7 and f (x) = x , (verify), 6.2.2 Inverse functions:, Let f : A → B be one-one and onto function and , f (x) = y for x ∈ A. The inverse function, f -1 : B → A is defined as f -1 (y) = x if f(x) = y, , , Ex. 8 : Verify that f (x) =, , x−5, 8 and g (x) = 8x + 5, , are inverse functions of each other., Solution : As f (x) =, g (x), , x−5, 8 , replace x in f (x) with, , g(x)−5, 8x+5−5, 8x, =, =, 8, 8, 8 =x, and g (x) = 8x + 5, replace x in g (x) with f (x), f [g (x)] =, ,  x − 5, , g [f (x)] = 8 f (x) + 5 = 8 , +5=x−5+5,  8 , =x, As f [g (x)] = x and g [f (x)] = x , f and g are inverse, functions of each other., , Fig. 6.36, Note:, 1) As f is one-one and onto every element, y ∈ B has a unique element x ∈ A such that, y = f (x) ., , Ex. 9 : Determine whether the function, 2x+1, f (x) = x−3 has inverse, if it exists find it., , 2) If f and g are one-one and onto functions, such that f [g(x)] = x for every x ∈ Domain, of g and g [f (x)] = x for every x ∈ Domain, of f, then g is called inverse of function f., Function g is denoted by f -1 (read as f inverse)., i.e. f [g(x)] = g [f(x)] = x then g = f-1 which, Moreover this means f [f -1 (x)] = f -1[f (x)] = x, 3), , Solution : f -1 exists only if f is one-one and onto., Consider f (x1) = f (x2),, Therefore,, 2x2+1, 2x1+1, =, x2−3, x1−3, , 1, f -1(x) ≠ [f (x)]−1, because [f (x)]−1 = f (x), [f (x)]−1 is reciprocal of function f (x) where as, , (2x1+1) (x2−3) = (2x2+1) (x1−3), 2x1x2 − 6x1 + x2 − 3 = 2x1x2 − 6x2 + x1 − 3, , f (x) is the inverse function of f (x)., -1, , − 6x1 + x2 = − 6x2 + x1, , e.g. If f is one-one onto function with f (3) =, 7 then f -1 (7) = 3 ., , 6x1 + x2 = 6x2 + x1, 7x2 = 7x1, , Ex. 7 : If f is one-one onto function with, f (x) = 9 − 5x , find f -1 (−1) ., , x2 = x1, Hence, f is one-one function., , Soln. : → Let f -1 (−1) = m, then − 1 = f (m), Therefore,, −1 = 9 − 5m, 5m = 9 + 1, 5m = 10, m=2, That is f (2) = −1, so f -1 (−1) = 2 ., , 2x+1, Let f (x) = y, so y = x−3, Express x as function of y , as follows, 2x+1, y = x−3, y (x −3) = 2x + 1, 122

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xy −3y = 2x + 1, xy −2x = 3y + 1, x(y −2) = 3y + 1, ∴, , 3y+1, x = y−2 for y ≠ 2., , Thus for any y ≠ 2,, we have x such that f (x) = y, f−1 is well defined on R - {2}, If f(x) = 2 i.e. 2x + 1 = 2 (x − 3), i.e. 2x + 1 = 2x − 6 i.e. 1 = −6, Fig. 6.37, , Which is contradiction., 2 ∉ Range of f., , As (1,3) lies on line y = 4 − x , so it is, shown by small black disc on that line., (1,2) is shown by small white disc on the line y=, x +1, because it is not on the line., , Here the range of f (x) is R − {2}., x is defined for all y in the range., Therefore f (x) is onto function., , 1) Signum function :, , As f is one-one and onto, so f exists., -1, , Definition: f (x) = sgn (x) is a piecewise function, , As f (x) = y , so f (y) = x, -1, , −1 if x < 0, , f (x) =  0 if x = 0,  1 if x > 0, , , , 3y+1, Therefore, f -1 (y) = y−2, Replace x by y , to get, 3x+1, f -1 (x) = x−2 ., 6.2.3 Piecewise Defined Functions:, A function defined by two or more equations, on different parts of the given domain is called, piecewise defind function., ,  x + 1 if x < 1, e.g.: If f (x) = ,  4 − x if x ≥ 1, Here f (3) = 4 − 3 = 1 as 3 > 1 ,, Fig. 6.38, , whereas f (−2) = −2 + 1 = −1 as −2 < 1 and, , Domain: R and Range: {−1, 0, 1}, , f (1) = 4 − 1 = 3., , 123

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Properties:, 1), , For x > 0, the graph is line y = 1 and for x <, 0 , the graph is line y = −1 ., , 2), , For f (0) = 0 , so point (0,0) is shown by, black disc, whereas points (0,−1) and (0,1), are shown by white discs., , Fig. 6.40, 6), , If |x| < m , then it represents every x whose, distance from origin is less than m, 0 ≤ x < m, and 0 ≥ x > −m That is −m < x < m. In interval, notation x ∈ (− m, m), , Absolute value function (Modulus function):, Definition: f (x) = |x|, is a piece wise function, , Fig. 6.41, 7), ,  x if x ≥ 0, f (x) = , − x if x < 0, , If |x| ≥ m , then it represents every x whose, distance from origin is greater than or equal to, m, so, x ≥ m and x ≤ −m . In interval notation, x ∈ (− ∞, m] ∪ [m, ∞), , Fig. 6.42, 8), , If m < |x| < n, then it represents all x whose, distance from origin is greater than m but less, than n . That is x ∈ (− n,−m) ∪ (m, n) ., , Fig. 6.43, Fig. 6.39, , 9), , Domain : R or (−∞,∞) and Range : [0,∞), , Verify by taking different values for x and y, (positive or negative)., , Properties:, 1), , Triangle inequality |x + y| ≤ |x| + |y| ., , Graph of f (x) = |x| is union of line y = x from, quadrant I with the line y = −x from quadrant, II. As origin marks the change of directions, of the two lines, we call it a critical point., , 10) |x| can also be defined as |x| =, = max{x, −x}., , 2), , Graph is symmetric about y-axis ., , Solution : If |x| ≤ m , then − m ≤ x ≤ m, , 3), , Graph of f (x) = |x−3| is the graph of |x| shifted, 3 units right and the critical point is (3,0)., , 4), , f (x) = |x| , represents the distance of x from, origin., , 5), , x2, , Ex. 10 : Solve |4x −5| ≤ 3., Therefore, −3 ≤ 4x −5 ≤ 3, −3 + 5 ≤ 4x ≤ 3 + 5, 2 ≤ 4x ≤ 8, 8, 2, ≤x≤ 4, 4, 1 ≤x≤2, 2, , If |x| = m, then it represents every x whose, distance from origin is m, that is x = + m or, x = −m ., 124

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Ex. 11 : Find the domain of , , 3) Greatest Integer Function (Step Function):, Definition: For every real x , f (x) = [x] = The, greatest integer less than or equal to x . [x] is also, called as floor function and represented by  x  ., , 1, ||x|-1|-3, , Solution : As function is defined for ||x|−1|−3>0, Therefore ||x|−1|>3 , , Illustrations:, , So |x|−1>3 or |x|−1<−3 , That is, , 1) f (5.7)= [5.7] = greatest integer less than or, equal to 5.7, , |x|>3 + 1or |x|<−3+1 , , Integers less than or equal to 5.7 are 5, 4, 3, 2 of, which 5 is the greatest., , |x|>4 or |x|<−2 , But |x|<−2 is not possible as |x|>0 always, So −4 < x < 4 , x ∈ (−4, 4) ., , 2) f (−6.3)= [−6.3] = greatest integer less than or, equal to −6.3., , Ex. 12 : Solve |x −1| + |x +2| = 8., , Integers less than or equal to −6.3 are −10, −9, −8,, −7 of which −7 is the greatest., , Solution : Let f (x) = |x −1| + |x +2|, , ∴ [−6.3] = −7, , Here the critical points are at x =1 and x =−2 ., , 3) f (2) = [2] = greatest integer less than or equal, to 2 = 2., , They divide number line into 3 parts, as follows., , 4) [π] = 3 5) [e] = 2, The function can be defined piece-wise as follows, f (x) = n, if n ≤ x < n + 1 or x ∈ [n, n +1), n ∈ I, , Fig. 6.44, Region, , Test, Value, , I, x < −2, , −3, , (x −1) < 0, − (x −1) − (x +2), , II, , 0, , (x +2) < 0 = −2 x −1, (x −1) < 0, − (x −1) + (x +2), , 2, , (x +2) > 0 = 3, (x −1) > 0 (x −1) + (x +2), , −2≤ x ≤1, III, x >1, , Sign, , −2 if − 2 ≤ x < −1, −1 if − 1 ≤ x < 0, , if 0 ≤ x < 1, f (x) = 0, 1, if 1 ≤ x < 2, , if 2 ≤ x < 3, 2, , f (x), , or x ∈ [ −2, −1), or x ∈ [ −1, 0 ), or x ∈ [ 0,1), or x ∈ [1, 2 ), or x ∈ [ 2, 3), , (x +2) > 0 = 2 x +1, , As f (x) = 8, 9, From I, −2x −1= 8 ∴ −2 x = 9 ∴ x = − 2 ., From II, 3 = 8 , which is impossible, hence there, is no solution in this region., 7, From III, 2x +1= 8 ∴ 2x = 7 ∴ x = 2 ., 7, 9, Solutions are x = −, and x = 2 ., 2, , Fig. 6.45, Domain = R and Range = I (Set of integers), 125

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Properties:, , Properties:, , 1), , If x ∈ [2,3), f (x) = 2 shown by horizontal, line. At exactly x = 2 , f (2) = 2 , 2 ∈ [2,3), hence shown by black disc, whereas 3 ∉, [2,3) hence shown by white disc., , 1), , If x ∈ [0,1], f (x) = {x} ∈ [0,1) shown by, slant line y = x . At x = 0, f (0) = 0, 0 ∈ [0,1), hence shown by black disc, whereas at, x = 1, f (1) = 1, 1 ∉ [0,1) hence shown by, white disc., , 2), , Graph of y = [x] lies in the region bounded, by lines y = x and y = x−1. So x−1 ≤ [x] < x, , 2), , Graph of y = {x} lies in the region bounded, by y = 0 and y = 1. So 0 ≤ {x} < 1, ,  0 if x ∈ I, [x] + [−x] = ,  −1 if x ∉ I, , 3), , 3), , Ex. 13:{5.2}+{−5.2} = 0.2+0.8 = 1where 5.2 ∈ 1, , Ex. [3.4] + [−3.4] = 3 + (−4) = −1where 3.4 ∉ I, , {7} +{−7} = 0 + (0) = 0 where 7 ∈ I, , [5] + [−5] = 5 + (−5) = 0 where 5 ∈ I, 4), , 0 if x ∈ I, {x} + {−x} = , 1 if x ∉ I, , 4), , [x+n] = [x] + n, where n ∈ I, , {x ± n} = {x}, where n ∈ I, , Ex. 14 : {2.8+5}= {7.8} = 0.8 and {2.8} = 0.8, {2.8 − 5} = {−2.2} = −2.2 − (−2.2) = −2.2 − (−3), = 0.8 ( {x} = x − [x]), , Ex. [4.5 + 7] = [11.5] = 11 and, [4.5] + 7 = 4 +7 = 11, , Ex. 15 : If {x} and [x] are the fractional, part function and greatest integer function, of x respectively. Solve for x, if {x + 1} + 2x, = 4[x + 1] − 6., , 4) Fractional part function:, Definition: For every real x, f (x) = {x} is defined, as {x} = x −[x], , Solution : {x + 1} + 2x = 4 [x + 1] − 6, , Illustrations:, , Since {x + n} = {x} and [x + n] = [x] + n, for, n ∈ I, also x = [x] + {x}, , f (4.8) = {4.8} = 4.8 − [4.8] = 4.8 − 4 = 0.8, f (−7.1) = {−7.1} = −7.1 − [−7.1], = −7.1 − (−8) = −7.1+8 = 0.9, f (8) = {8} = 8 − [8] = 8 − 8 = 0, , ∴, , {x} + 2({x} + [x]) = 4([x] + 1) − 6, , ∴, , {x} + 2{x} + 2[x] = 4[x] + 4 − 6, , ∴, , 3{x} = 4[x] − 2[x] − 2, , ∴, , 3{x} = 2[x] − 2, , ... (I), , Since 0 ≤ {x} < 1, , Fig. 6.46, , ∴, , 0 ≤ 3{x} < 3, , ∴, , 0 ≤ 2 [x] − 2 < 3, , ∴, , 0 + 2 ≤ 2 [x] < 3 + 2, , ∴, , 2 ≤ 2 [x] < 5, , ∴, , Domain = R and Range = [0,1), , ∴, 126, , 2 ≤ [x] < 5, 2, 2, 1 ≤ [x] < 2.5, , ( from I)

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But as [x] takes only integer values, , log10 10 < log10 23 < 2log1010, , [x] = 1, 2 since [x] = 1 ⇒ 1≤x<2 and [x] = 2 ⇒, 2≤x<3, , 1 < log1023 < 2 (⸪ log10, = 1), 10, Then [log10 23] = 1, hence Characteristic of, , Therefore x ∈ [1,3), , log10 23 is 1., , Note:, , The characteristic of the logarithm of a number N,, with 'm' digits in its integral part is 'm−1'., , 1), Property, f (x+y) = f (x) + f (y), f (x+y) = f (x) f (y), f (xy) = f (x) f (y), f (xy) = f (x) + f (y), 2), , Ex. 16 : Given that log10 2 = 0.3010, find the, number of digits in the number 2010., , f (x), kx, akx, xn, log x, , Solution : Let x = 2010, taking log10 on either, sides, we get, log10 x = log10 (2010) = 10log1020, = 10log10 (2×10) = 10{log102 + log1010}, , If n(A) = m and n(B) = n then, , = 10{log10 2 + 1} = 10{0.3010 + 1}, , (a) number of functions from A and B is, , = 10 (1.3010) = 13.010, That is characteristic of x is 13., , nm (b) for m≤n, number of one-one, functions is, , So number of digits in x is 13 + 1 = 14, , n!, ( n − m )!, , EXERCISE 6.2, , (c) for m>n, number of one-one functions is, 0, , 1), , (d) for m≥n, number of onto functions are, nm − nC1(n−1)m + nC2(n−2)m − nC3(n−3)m, + ... + (−1)n−1 nCn−1, , If f (x) = 3x + 5, g (x) = 6x − 1, then find, (a) (f+g) (x), (c) (fg) (3), , 2), , (e) for m<n, number of onto functions are 0., (f) number of constant fuctions is m., , (b) (f−g) (2), (d) (f /g) (x) and its domain., , Let f :{2,4,5}→{2,3,6} and g :{2,3,6}→{2,4}, be given by f ={(2,3), (4,6), (5,2)} and, g ={(2,4), (3,4), (6,2)} . Write down g ◦ f, , 3) Characteristic & Mantissa of Common, Logarithm log10 x :, , 3), , If f (x) = 2x2 + 3, g (x) = 5x − 2, then find, (a) f ◦ g, (b) g ◦ f, (c) f ◦ f, (d) g ◦ g, , 4), , Verify that f and g are inverse functions of, each other, where, , As x = [x] + {x}, log10 x = [log10 x] + {log10 x}, Where, integral part [log10 x] is called, Characteristic & fractional part {log10 x} is called, Mantissa., , (a) f (x) =, , Illustration : For log10 23,, , x−7, 4 , g (x) = 4x + 7, , (b) f (x) = x3 + 4, g (x) =, , log10 10 < log10 23 < log10 100, , 3, , x−4, , x+3, 2x+3, (c) f (x) = x−2 , g (x) = x−1, , log10 10 < log10 23 < log10 102, 127

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5), , Check if the following functions have an, inverse function. If yes, find the inverse, function., (a) f (x) = 5x2, (c) f (x) =, , 11) Solve the following for x, where |x| is modulus, function, [x] is greatest integer function, [x], is a fractional part function., , (b) f (x) = 8, , 6x−7, 3, , (d) f (x) =, , 4x + 5, , (e) f (x) = 9x + 8, x + 7 x < 0, (f) f (x) = , 8 − x x ≥ 0, 6), , (a) f (3), , 7), , (a) f (−4), (c) f (1), , x≤2, , then find, x>2, •, , x ≤ −3, −3 < x < 3 , then find, x≥3, , 9), , If f (x) = 4[x] − 3, where [x] is greatest integer, function of x, then find, , (g) {x}>4, , (h) {x} = 0, , (i) {x} = 0.5, , If f:A → B is a function and f (x) = y, where, x ∈ A and y ∈ B, then, , Range of f is f (A) = Set of Outputs = Set, of Images = Set of values of y for which y =, f (x) is defined = Projection of graph of, f (x) on Y-axis., Co-domain of f is B., , (b) f (0.5), ,  5, (c) f  −  (d) f (2π), where π = 3.14,  2, 10) If f (x) = 2{x} + 5x , where {x} is fractional, part function of x, then find, , (c) f (−1.2), , (f) [x + [x + [x]]] = 9, , Domain of f is A = Set of Inputs = Set of, Pre-images = Set of values of x for which, y = f (x) is defined = Projection of graph of, f (x) on X-axis., , (b) f (−3), (d) f (5), , If f (x) = 2|x| + 3x, then find, (a) f (2), (b) f (−5), , (a) f (−1), , (d) |x| ≤ 3, , Let's Remember, , (c) f (0), , 8), , (a) f (7.2), , (b) x2 + 7|x| + 12 = 0, , (j) 2{x} = x + [x], , (b) f (2), , 4 x − 2,, , If f (x) =  5,,  x2 ,, , , (b) |x−4| + |x−2| = 3, , (e) 2|x| = 5, , 3, ,  x 2 + 3,, If f (x) = , 5 x + 7,, , (a) |x+4|≥5 , , 1, (b) f  , 4, (d) f (−6), , 128, , •, , If f (x1) = f (x2) ⇒ x1 = x2 then f is one-one, and for every y ∈ B, if there exists x ∈ A, such that f (x) = y then f is onto., , •, , If f:A → B. g:B → C then a function, g ° f:A → C is a composite function., , •, , If f:A → B, then f−1:B → A is inverse, function of f., , •, , If f:R → R is a real valued function of real, variable, the following table is formed.

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Type of f, , Form of f, , Domain of f, , Range of f, , Constant function, , f (x) = k, , R, , k, , Identity function, , f (x) = x, , R, , R, , Square function, , f (x) = x2, , R, , [0, ∞ ) or R+, , Cube function, , f (x) = x3, , R, , R, , Linear function, , f (x) = ax + b, , R, , R, , Quadratic function, , f (x) = ax2 + bx + c, , R, ,  4ac − b 2 , ,∞, ,  4a, , , Cubic function, , f (x) = ax3 + bx2 + cx + d R, , Square root funtion, , f (x) =, , Cube root function, , f (x) =, , 3, , Rational function, , f (x) =, , p( x), q( x), , Exponential function, , R, , x, , [0, ∞), , [0, ∞) or R+, , x, , R, , R, , R − {x q(x) = 0}, , depends on p(x) and, q(x), , f (x) = ax, a > 1, , R, , (0, ∞), , Logarithmic function, , f (x) = loga x, a > 1, , (0, ∞) or R+, , R, , Absolute function, , f (x) = x , , R, , [0, ∞) or R+, , Signum function, , f (x) = sgn (x), , R, , {−1, 0, 1}, , Greatest Integer, function, , f (x) = [x], , R, , I (set of integers), , Fractional Part function f (x) = {x}, , R, , [0,1), , 3), , MISCELLANEOUS EXERCISE 6, (I) Select the correct answer from given, alternatives., 1), , 4), , If log (5x – 9) – log (x + 3) = log 2 then, x = ..............., , If log10(log10(log10x)) = 0 then x =, A) 1000, , B) 1010, , C) 10, , D) 0, , A) 4, −4, , B) 4 , , C) −4, , D) not defined , , The equation log x2 16 + log2x64 = 3 has,, A), B), C), D), , A) 3 B) 5 C) 2 D) 7, 2), , Find x, if 2log2 x = 4, , 5), , one irrational solution, no prime solution, two real solutions, one integral solution, , If f (x) =, A) x – 1, , 129, , 1, , then f (f{f (x)}] is, 1–x, B) 1 – x, , C) x, , D) –x

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6), , 7), , 8), , If f : R → R is defined by f (x) = x3 then f–1 (8), is euqal to :, A) {2} , B) {–2. 2}, C){–2} , D) (–2. 2), 2x + 1, Let the function f be defined by f (x) =, 1 – 3x, then f–1 (x) is:, A), , x–1, , 3x + 2, , B), , x+1, 3x – 2, , C), , 2x + 1, , 1 – 3x, , C), , 3x + 2, x–1, , Find whether following functions are onto, or not., i) f: Z→Z defined by f(x) = 6x–7 for all, x∈Z, ii) f: R→R defined by f(x) = x2+3 for all, x∈R, , 4), , Let f: R→R be a function defined by, f(x) = 5x3–8 for all x∈R, show that f is oneone and onto. Hence find f –1., , 5), , If f (x) = 2x2 + bx + c and f (0) = 3 and, f (2) = 1, then f (1) is equal to, A) –2, , 9), , 3), , B) 0, , C) 1, , D) 2, , 1, where [x] is greatest, [x]−x, integer function is, , 6), , A function f is defined as f(x) = 4x+5, for, –4 ≤ x < 0. Find the values of f(–1), f(–2),, f(0), if they exist., , 7), , A function f is defined as : f(x) = 5–x for, 0 ≤ x ≤ 4. Find the value of x such that, (i) f(x) = 3 (ii) f(x) = 5, , 8), , If f(x) = 3x4 – 5x2 + 7 find f(x–1)., , 9), , If f(x) = 3x + a and f(1) = 7 find a and f(4)., , The domain of , , A) R, , B) Z, , C) R−Z, , D) Q - {o}, , 10) The domain and range of f (x) = 2 − x − 5is, A) R+, (−∞, 1], , B) R, (−∞, 2], , C) R, (−∞, 2), , D) R , (−∞, 2], , 3x, +2,, 5, x∈R. Show that f is one-one and onto. Hence, find f –1., A function f: R→R defined by f(x) =, , +, , 10) If f(x) = ax2 + bx + 2 and f(1) = 3, f(4) = 42,, find a and b., , (II) Answer the following., 1), , Which of the following relations are, functions? If it is a function determine its, domain and range., i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5),, (12, 6), (14, 7)}, , 11) Find composite of f and g, i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}, g = {(3, 6), (4, 8), (5, 10), (6, 12)}, ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}, g = {(1, 1), (3, 27), (4, 64)}, , ii) {(0, 0), (1, 1), (1, –1), (4, 2), (4, –2),, (9, 3), (9, –3), (16, 4), (16, –4)}, , 12) Find fog and gof, i), , iii) {2, 1), (3, 1), (5, 2)}, , f(x) = x2 + 5, g(x) = x–8, , ii) f(x) = 3x – 2, g(x) = x2, , 2), , Find whether following functions are oneone., i) f:R→R defined by f(x) = x2+5, 5x + 7, ii) f: R−{3}→R defined by f(x) =, x–3, for x∈ R−{3}, , iii) f(x) = 256x4, g(x) =, 13) If f(x) =, , 2x – 1, 5, ,x≠, 2, 5x – 2, , Show that (fof) (x) = x., 130, , x

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14) If f(x) =, , x+3, 3 + 5x, , g(x) =, then show that, 4x – 5, 4x – 1, , 29) If log3 [log2(log3x)] = 1, show that x = 6561., 30) If f(x) = log(1–x), 0 ≤ x < 1 show that, , (fog) (x) = x., ,  1 , f,  = f(1–x) – f(–x),  1+ x , , x2 –4, 15) Let f: R – {2}→ R be defined by f(x) =, x–2, and g : R → R be defined by g (x) = x + 2. Ex, whether f = g or not., , 31) Without using log tables, prove that, 2, 1, < log103 <, 5, 2, , 16) Let f: R → R be given by f(x) = x + 5 for all, x ∈R. Draw its graph., , 32) Show that, , 17) Let f: R → R be given by f(x) = x3 + 1 for all, x ∈R. Draw its graph., , 8,  15 , 2,  32 , 7 log   +6 log   +5 log   +log  , 3,  16 , 5,  25 , = log 3, , 18) For any base show that, log (1+2+3) = log 1 + log2 + log3., , 33) Solve :, , log 2 x 4 + 4 log 4, , 19) Find x, if x = 33log32, 20) Show that,, , 34) Find value of, , 2, 2, log | x + 1 + x | + log | x + 1 − x | = 0, , 21) Show that, log, , a2, b2, c2, + log, + log, =0, bc, ca, ab, , 35) If, , 3 + log10 343, 1,  1 ,  49  1, 2 + log10   + log10  , 2,  25 ,  4  2, , logb, logc, loga, = y+z–2x = z+x–2y , show that, x+y–2z, , abc = 1., , 22) Simplify, log (logx4) – log (logx)., , 36) Show that, logy x3.logz y4.logx z5 = 60, , 23) Simplify, log10, , 2, =2, x, , log2b, log2c, log2a, 3 2, =, =, 6, 3k and a b c = 1, 4, find the value of k., , 35, 325, 28, 13, – log10, + log10, – log10, 324, 432, 45, 15, , 37) If, ,  a+b  1,  = (loga + logb), then show,  2  2, , 24) If log , , 38) If a2 = b3 = c4 d 5, show that logabcd =, , that a=b, , 47, ., 30, , 39) Solve the following for x, where x is, modulus function, [x] is greatest interger, function, {x} is a fractional part function., , 25) If b =ac. prove that, log a + log c = 2log b, 2, , 26) Solve for x, logx (8x – 3) – logx4 = 2, , a) 1 < x − 1< 4, , 27) If a2 + b2 = 7ab, show that,, , c) x2 − x −6= x + 2, , c) x2 − 9+ x2 − 4= 5, ,  a+b  1, 1, = 2 log a + 2 log b,  3 , , d) −2 < [x] ≤ 7, , log , , e) 2[2x − 5] − 1 = 7, , f) [x2] − 5 [x]+6 = 0, , 1,  x− y 1, 28) If log ,  = 2 log x + 2 log y, show that,  5 , x2 + y2 = 27xy., , g) [x − 2] + [x + 2] + {x} = 0,  x   x  5x, h)   +   =, 2 3 6, 131

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40) Find the domain of the following functions., x2 + 4x + 4, a) f ( x) = 2, x + x−6, 1, b) f ( x) = x − 3 +, log(5 − x), c), , f ( x) = 1 − 1 − 1 − x 2, , d), , f ( x) = x !, , e), , f ( x) =5− x Px −1, , f), , f ( x) = x − x + 5 − x, , g), , f ( x) = log( x 2 − 6 x + 6), , 42) Find (f ° g) (x) and (g ° f) (x), f (x) = ex, g (x) = log x, , b), , f ( x) =, , x, , g ( x) = x, x +1, 1− x, , 43) Find f (x) if, a) g(x) = x2 + x, = 4x2 − 10x + 4, (b) g(x) = 1 +, , 44) Find (f ° f) (x) if, x, (a) f ( x) =, 1 + x2, (b) f ( x) =, , f ( x) = 1 + 2 x + 4 x, , v, , v, , 132, , v, , − 2 and (g ° f) (x), , x and f [g(x)] = 3 + 2 x + x., , 2, , 41) Find the range of the following functions., x, a) f ( x) = x − 5, b) f ( x) =, 9 + x2, 1, c) f ( x) =, d) f ( x) = [ x ] − x, 1+ x, e), , a), , 2x +1, 3x − 2

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7, , LIMITS, We will study functions of x, a real variable, and, a, b, c etc will denote constants. x → a implies that, x takes values as near a as possible. So in this case, we have to consider x going nearer a from either, 1, 1, 1, 1, side. So, x = a – , a + , a – , a +, ......, 2, 2, 4, 4, , Let's Study, •, •, •, •, •, , Meaning – Definition of Limit, Calculation of various limits, Limits of Trigonometric Functions, Limits of Exponential and Logarithmic, Functions, Limit at Infinity and Infinite limit, , We will illustrate with an example., Consider the function f(x) = x + 3, , Introduction:, Calculus is an important branch of, mathematics. The concept of limit of a function, is a fundamental concept in calculus., , Take the value of x very close to 3 but not equal, to 3., The following table shows that as x gets nearer to, 3, the corresponding values of f(x) also get nearer, to 6., (I), , 7.1.1 Definition of Limit:, Let's :Learn, , x, f(x), , 7.1.1 LIMIT OF A FUNCTION :, Suppose x is a variable and a is a constant., If x takes values closer and closer to ‘a’ but, not equal to ‘a’, then we say that x tends to a., Symbolically it is denoted as x → a., , x approaches to 3 from left, 2.5, 2.6, …, 2.9, 2.99, 5.5, 5.6, …, 5.9, 5.99, , (II), x approaches to 3 from right, x, , 3.6, , 3.5, , …, , 3.1, , 3.01, , f(x), , 6.6, , 6.5, , …, , 6.1, , 6.01, , From the table we see that as x → 3 from either, side, f(x) → 6., , Fig. 7.1, , This idea can be expressed by saying that the, limiting value of f(x) is 6 when x approaches 3,, at x = 3, which is the limiting value of f(x) as, x→3, , You can observe that the values of x are very, near to ‘a’ but not equal to ‘a’., When x > a and x takes values near a, for example,, 1, 1, 1, x = a + , x = a + , x = a + ... etc; we say that, 2, 8, 4, x  a+ (x tends to a from larger values). When, x < a and x takes values near a, for example,, 1, 1, 1, x = a – , x = a – , x = a – ... etc. then we, 2, 4, 8, –, say that x a (x tends to a from smaller values)., , Observe that if P(x) is a polynomical in x, then, lim P(x) = P(a), for any constant ‘a’., x→a, We are going to study the limit of a rational, P( x), function f(x) =, as x  a., Q( x), Here P(x) and Q(x) are polynomials in x., 133

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We get three different cases., , SOLVED EXAMPLES, , (1) Q(a) ≠ 0,, (2) Q(a) = 0 and P(a) = 0, (3) Q(a) = 0 and P(a) ≠ 0, , Strategy : Steps for verifying the ∈ - δ definition., Consider ∈ > 0 given, substitute the values of f(x), and l in |f(x) – l|< ∈ and proceed to find the value, of δ. We may have to manipulate the inequalities. , , In case (1) the limit of f(x) as x → a is, P( x) P(a), lim, =, ., x→a Q( x), Q(a), , Ex. 1. Consider the example f(x) = 3x + 1, take, a = 0 and l = 1, We want to find some δ > 0 such that,, , In Case (2) (x – a) is a factor of P(x) as, well as Q(x). So we expresse P(x) and Q(x) as, P(x) = (x – a)P1(x) and Q(x) = (x – a)Q1(x), , 0 < |x – 0|< δ impies that, |(3x + 1)– 1| <∈, , ∈, i.e. if 3|x| <∈ i.e. if |x| <, 3, ∈, So, we can choose δ=, 3, ∈, (If fact any δ ≤, will do.), 3, ∴ 0 < |x – 0| < δ ⇒ |f(x) – l|< ∈, if |3x| <∈, , P( x) ( x − a ) r P1 ( x), =, Now =, (x – a ≠ 0), Q( x) ( x − a) s Q1 ( x), If r = s then lim, x→a, , P1 (a ), P( x), =, Q1 (a ), Q( x), , If r > s then lim, x→a, , P( x), =0, Q( x), , ∴ lim (3x + 1) = 1, x→0, , If r < s then we proceed to case (3)., In case (3), if Q(a) = 0 and P(a) ≠ 0, lim, x→a, , P( x), does not exist., Q( x), , 7.1.2 Definition of Limit :, We need to confirm that f(x) is very near to, l (or as near as expected). This is expressed by, |f(x) – l| < ∈ for any ∈ > 0. Here ∈ can be arbitrarily, small to ensure that f(x) is very near l. If this, condition is satisfied for all x near enough, then, we can say that f (x) → l as x → a, the fact that x is, near enough a is expressed by 0 < |x – a| < δ where, δ > 0. This δ can be very small and depends upon, f(x) and ∈., , Fig. 7.2, Ex. 2. f(x) = x2, Here take a = 3 and l = 9, We want to find some δ > 0 such that, 0 < |x – 3| < δ implies |x2 – 9|< ∈, , Hence the Definition, , i.e. 3 – δ < x < 3 + δ ⇒ |x2 – 9|< ∈, , If given ∈ > 0 there exists δ > 0 such that, |f(x) – l |< ∈ for all |x – a |< δ, then we say, that f(x) → l as x → a., , δ can be chosen as small as we like, and take δ < 1, 134

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From the above example we notice that the, limits of f(x) as x → a+ or a– can be different. This, induces us to define the following., , Then, 3 – δ < x < 3 + δ ⇒ 2 < x < 4 or, , 5<x+3<7, We want |(x + 3) (x – 3)| < ∈, But |(x + 3) (x – 3) | < 7 |x – 3|, , 7.1.3 One Sided Limit: xlim, f(x) and xlim, f(x);, →a+, →a−, if they exist are called one sided limits., , So 7 |(x – 3)| < ∈ ⇒ |x2 – 9| < ∈, ∈, , |(x – 3)| < δ ⇒ |x2 – 9| < ∈, 7, ∈, So we choose δ = min { ,1}, 7, then |(x – 3)| < δ ⇒ |f (x) – 9| < ∈, , 7.1.4 Left hand Limit: If given ∈ > 0, there, exists δ > 0 such that for |f(x) – l| < ∈ for all, x with a – δ < x< a then, , * Note that, we want to get rid of factor, |x + 3| Hence we have to get its lower bound., , 7.1.5 Right hand Limit : If given ∈ > 0 there, exists δ > 0 such that for |f(x) – l| < ∈ for all, x with a < x < a + δ then, lim f(x) = l, , If δ =, , lim f(x) =l, , x→a−, , Ex. 3. f(x) = [x], 2 < x< 4 where [x] is a greatest, integer function., We have seen the f(x) = [x], 2 ≤ x ≤ 4, , x→a+, , 7.1.6 Existence of a limit of a function at a, point x = a, , Note that, [x] = 2 for 2 ≤ x < 3, = 3 for 3 ≤ x < 4, Let us study the limits of f(x) as x, x  2.7, , , , If lim+ f(x) = lim− f(x) = l, then limit of the, function f(x) as x   , a exists and its value is l., And if lim+ f(x) ≠ lim− f(x) then lim, f(x) does, x→a, x→a, x→a, not exist., , lim f(x) = 3,, , x→3+, , lim− f(x) = 2., But for x < 3, f(x) = 2. So, x→, 3, , Example:, Find left hand limit and right hand limit for the, following function., , V, , 1, If we take l = 3, then for ∈ = and any δ >0, 2, 3 − δ < x < 3 ⇒ f(x) = 2 and |f(x) – l| = 1 ∈, ,  3 x + 1 if x < 1, , f ( x) = , 7 x 2 − 3 if x ≥ 1, , , If we take l = 2, then 3< x <3 + δ ⇒ f(x) = 3,, V, , |f(x) – 2| = 1, , x→a, , x→a, , 3 and, , ∈, , ∴ lim, f(x) does not exist., x→3, , Solution : Right hand limit, lim f(x), for x > 1, x→1, , lim f(x), Consider x→2, .7, , i.e. lim+ f(x) = lim (7x2 – 3) = 4, x→1, , x→1, , Consider a = 2.7 We see that for 2 < x < 3,, f(x) = 2., , Left hand limit, lim f(x), for x < 1:, x→1, , lim f(x) = lim (3x + 1) = 4, , If we choose δ = 0.3,, , x→1−, , x→1, , Since left and right-hand limits are equal, the, two-sided limit is defined, and lim f(x) = 4., , then 2 < 2.7 − δ < 2.7 < 2.7 + δ < 3, and f(x) = 2 is a constant., , x→1, , ∴ lim f(x) = 2, , lim, lim, Note : lim, x → a f(x) means x → a − f(x) = x → a + f(x), , x→2.7, , 135

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7.1.7 ALGEBRA OF LIMITS:, It is easy to verify the following., Let f(x) and g(x) be two functions such that, , , , lim [f(x) ± g(x)] = lim f(x) ± lim g(x), x →a, , x →a, , x →a, , Ex. 2 : lim, [(y2 – 3)(y + 2)], y→2, , = l ± m, 2., , x →a, ,  6+ x − 7− x , , , Ex. 3 : lim, x →3 , x, , , , lim [k f(x)]= = k × lim f(x) = kl, where ‘k’, x →a, , x →a, , is a constant, 4., , = (22 – 3)(2 + 2) = (8– 3)(4) = 5 × 4 = 20, , x →a, , =l×m, , , 3., , = lim, [(y2 – 3) [y + 2], y→2, , lim [f(x) × g(x)] = lim f(x) × lim g(x), x →a, , lim, x →a, , =, , , , lim f(x) = l and lim g(x) = m, then, x→a, x→a, 1., , (3) × (3 + 1) × (6 + 1), 6, 3× 4 × 7, =, 6, = 14, , , , lim f ( x ) l, f ( x), =, = x →a, where m ≠ 0, g ( x), lim g ( x ) m, , =, , lim, x →3, , (, , Note:, lim k = k , where k is a constant, , 2), , lim x = a, , 3), , lim x n = a n, , 4), , lim r x = r a, , 5), , If p(x) is a polynomial,, , x→a, , x→a, , x→a, , While evaluating limits, we must always, check whether the denominator tends to zero, and, if it does, then whether the numerator also tends, to zero. In case both tend to zero we have to study, the function in detail., , 6+3 − 7−3, 3, , =, , 9− 4, 3− 2 1, =, = 2, 3, 3, ,  (1 − x ) , , , == lim, x →1 ( x − 1) × x, , ,  − ( x − 1) , , , == lim, x →1 ( x − 1) × x, , , , SOLVED EXAMPLES, , (3) × (3 + 1) × (6 + 1), 6, , ), ,  1− x ,  x , = lim , , x →1,  x −1 , , , , then lim p ( x) = p (a ), , =, , 7−x, , 1 ,  −1 , x , Ex. 4 : lim, x →1,  x −1 , , , , x→a, , , , lim ( x ), , (, , =, , x→a, , n,  n(n + 1)(2n + 1) , Ex. 1 : lim  ∑ r 2  = lim , , n, →, 3, 6, n →3, , ,  r =1 , , x →3, , x →3, , x →a, , 1), , ), , 6 + x − lim, , 1,  −1 , 1, = − lim  [As, = −x g1, x−1 ≠ 0], == lim, , , , x →1, x →1 x, 1,  , x, 1, 1,  −1 , = lim   = − lim   = −, x →1, x →1 x, 1,  , x, = –1, 136

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77.1.8 Theorem: Prove that, , Ex. 5 : Discuss the limit of the following function, as x tends to 3 if, , xn − an, lim, = na n −1 where n∈N, a > 0, x→a x − a, ,  x 2 + x + 1, 2 ≤ x ≤ 3, f(x) = , 3< x ≤ 4,  2 x + 1,, , Proof : We know, an − bn = (a − b)(an−1 + an−2b +, an−3b2 + ... +bn−1) for n ∈ N., , Solution: As f(x) is defined seperately for x ≤ 3, and x > 3, we have to find left hand limit (when x, ≤ 3) and right hand limit (when x > 3) to discuss, the existance of limit of f(x) as x→3., ,  xn − an , lim , =, x→a,  x−a , n −1, n−2, n −3 2, n −1, , , lim  ( x − a ) ( x + x a + x a + ...... + a ) , x→a, x−a, , , , For the interval 2 ≤ x ≤ 3; f(x) = x2 + x + 1, ∴ lim− f(x) = lim− (x2 + x + 1) = (3)2 + 3 + 1, x→3, , x→3, , =, , , = 9 + 3 + 1 = 13, -----(I), Similarly for the interval 3 < x ≤ 4;, f(x) = 2 x + 1, , =, =, =, , lim f(x) = lim+ (2x + 1) = (2 × 3) + 1, , x→3+, , x→3, , =6+1=7, , -----(II), , From (I) and (II), lim− f(x) ≠ lim+ f(x), x→3, , x→3, , Note : The above limit can also be found by using, the substitution x − a = h., , ∴ lim f(x) does not exist., x→3, , x − a = h ∴ x = a + h and x→a ⇒ h→0. Use, binomial theorem to expand (a + h)n, simplify, and apply the limit to get the result, , Ex. 6 : For a given ∈ > 0, find δ > 0 such that, whenever |x − a| < δ, we have |f(x) − l| < ∈ so that, lim (4x + 3) = 7, x→1, Solution : We want to find δ so that, ,  xn − an , n −1, lim , =na, x→a,  x−a , , lim (4x + 3) = 7, x→1, , Verify : If n < 0 say n = – m then, , Here a = 1, l = 7 and f(x) = 4x + 3, Consider ∈ > 0 and |f(x) − l| < ∈, , −m, −m, n, n, lim  x − a  lim  x − a , x→a ,  = x→a  x − a , , ,  x−a , , |(4x + 3) − 7| < ∈, , = – ma–m–1, , if |(4x − 4)| < ∈, i.e. if |4(x − 1)| < ∈, i.e. if |x − 1| <, We can have δ ≤, ⇒ |f(x) − 7| < ∈, , lim (xn−1 + xn−2a + xn−3a2 + ... +an−1), x→a, for [x − a ≠0], an−1 + an−2a + an−3a2 + ... +an−1, an−1 + an−1 + an−1 + ... +an−1 (n terms), nan−1, , Note : The above theorem can also be verified if, p, n is a fraction say n =, where q ≠ 0. Then, q, , ε, 4, , ε, so that |(x − 1)| < δ, 4, , p,  qp, q, lim  x − a  lim  x − a, x→a ,  = x→a ,  x−a ,  x−a, , n, , 137, , n, , , p, , p a q −1, = q, , 

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 2x − 2 , Ex. 4 : Evaluate lim  3, , x →1,  26 + x − 3 , Solution: Put 26 + x = t3, ∴ x = t3 – 26 , As x 1, t 3, , SOLVED EXAMPLES,  x 4 − 625 , lim, Ex. 1 : Evaluate x→5  x − 5 , , , ,  2x − 2 , lim  3, ∴ lim, , x→1, x →1,  26 + x − 3 , ,  x 4 − 625 ,  x 4 − 54 , lim, lim, Solution : x→5  x − 5  = x→5  x − 5 , , , , , , (, , ), , , , , , xn − an, 4−1, = na n −1 , lim, = 4(5) .....  x →a, x−a, , , ,  2 t 3 − 26 − 2 , , lim , = lim, 3 3, x →3, , t→3 , 3, t, −, , , , , = 4(125) = 500, ,  2 t 3 − 33, , = lim, x →3,  t − 3, , (, ,  x 7 − 128 , Ex. 2 : Evaluate lim, x→2 , , 5,  x − 32 ,  x 7 − 128  lim, Solution : lim, x→2 ,  = x→2, 5,  x − 32 , , , , , = lim, x→5 , , , , , , , = 2 × 3(3)3–1, = 54, , , , , , , , , EXERCISE 7.1, , ... [As x→2, x − 2 ≠ 0], , Q.I Evaluate the following limits :, , 6, , , , =, , 7(2), 5(2) 4, , , , xn − an, = na n −1 , ...  lim,  x→a x − a, , , , , , =, , 7(2) 2, 5, , , , =, , 28, 5, ,  x n − 4n , lim, Ex. 3 : If x →4 ,  = 48 and n∈N, find n.,  x−4 , ,  z+6, lim , , z →−3,  z , , 2., ,  y 5 + 243 , lim  3, , y →−3,  y + 27 , , 3., ,  1 1  ,  z + 5  , , lim  , z →−5, +, 5, z, , , , , , n, , 1., , ∴ n(4)n–1 = 3(4)3–1, ∴n=3, , 1., , Q.II Evaluate the following limits :, , x −4 , Solution : Given lim ,  = 48, x→4,  x−4 , ∴ n(4)n-1 = 48, n, , , ,  t 3 − 33 , = 2 lim , , t →3,  t −3 , ,  x 7 − 27 ,  5 5,  x −2 , x7 − 2 7, x−2, x 5 − 25, x−2, , ) , , .… by comparing, 138, ,  2x + 6 , = lim, , , x →3,  x 

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Q.IV In the following examples, given ∈ > 0,, find a d > 0 such that whenever, x-a < d, we, must have f(x)-l < ∈, , 2., ,  x −3 − 2−3 , lim , , x→2,  x−2 , , 3., ,  x3 − 125 , lim  5, , x →5 x − 3125, , , , 4., ,  x − 1, x −a , If lim , = lim , , ,, x →1,  x − 1  x→a  x − a , 4, , 3, , 3, , 2., , , lim , x →7 , , , 3., , 4., , 5., , (, , 3, , )(, , 3, , 7., , 8., ,  x 3 − 343 , lim , , x →7,  x− 7, , 9., ,  x + x 3 + x5 + ... + x 2 n −1 − n , lim , , x →1, x −1, , , , x→2, , lim( x 2 + x + 1) = 3, x →1, , METHOD OF FACTORIZATION:, P( x), . We consider lim, x → a f(x)., Q( x), , If lim Q(x) = m ≠ 0, x→a, ,  P( x) , then lim f(x)= lim, , x, →, a, x→a,  m , 2), , If lim, x → a Q(x) = 0, then (x − a) divides Q(x). In, , such a case if (x − a) does not divide P(x) then, lim f(x) does not exist., x→a, , (3) Further if lim, P(x) is also 0, then (x − a) is a, x→a, factor of both P(x) and Q(x)., ,  3 1+ x − 1+ x , lim , , x →0, x, , , , 3, 3, , , 2, 2, z, +, 2, −, a, +, 2, (, ), (, ), , , lim, z →a , , z−a, , , , 4., , 1), , ), ,  (1 − x )8 − 1 , lim , , 2, x →0,  (1 − x ) − 1 , ,  2y − 2 , lim , , y →1 3 7 + y − 2, , , , lim( x 2 − 1) = 3, , lim (3 x + 2) = −7, , x →−3, , lim, Let’s check lim, x → a Q(x) and x → a P(x)., ,  x k − 5k , If lim ,  = 500 , find all possible, x →5,  x −5 , values of k., , 6., , 3., , f(x) =, , x+37 , , , x−7, , , x−37, , 2., , x→2, , P(x) and Q(x) are polynomials in x such that, , Q.III Evaluate the following limits :, ,  x + x 2 + x3 + ......... + x n − n , lim , , x →1, x −1, , , , lim(2 x + 3) = 7, , 7.2, , find all possible values of a., , 1., , 1., , lim  P ( x) / ( x − a )  ., So lim, f(x), =, x→a, x→a,  Q( x) / ( x − a) , Factorization of polynomials is a useful tool, to determine the limits of rational algebraic, expressions., SOLVED EXAMPLES, Ex.1 :, ,  z ( 2 z − 3) − 9 , Evaluate lim  2, , z →3,  z − 4z + 3 , , Solution: If we substitute z = 3 in numerator, and denominator,, 139

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we get z(2z – 3) – 9 = 0 and z2 – 4z + 3 = 0, So (z  –  3) is a factor in the numerator and, denominator.,  z ( 2 z − 3) − 9 , lim  2, , z →3,  z − 4z + 3 , ,  2 z 2 − 3z − 9 , lim, = z →3  2, ,  z − 4z + 3 , , , ,  ( z − 3) ( 2 z + 3) , , , = = lim, z →3,  ( z − 3) ( z − 1) , , , ,  ( 2 z + 3) , , , = = lim, z →3,  ( z − 1) , , , , , ==, , , , = =, , ...[As z→3, z − 3 ≠ 0], , , ,  1, , 2, −, , , = lim, x →1 x − 1, ( x − 1) ( x + 1) , , , , ,  1+ x − 2 , , , = lim, x →1 ( x − 1) ( x + 1), , , , , , , , x −1, , , = lim, x →1 ( x − 1) ( x + 1), , , , , ,  1 , lim, = x →1  x + 1 , ) ,  (, , , 1, =, 2, , 2 ( 3) + 3, 3 −1, 9, 2, , ...[As x→1, x − 1 ≠ 0], ,  x3 + x 2 − 5 x + 3 , Ex.4: Evaluate lim , , x →1, x2 −1, , , , ,  3, ( x − 8 x 2 + 16 x)9 , Ex.2: Evaluate lim , 18, x→4 , 2, , x, −, x, −, 12, , , , (, , Solution : In this case (x – 1) is a factor of the, numerator and denominator., To find another factor we use synthetic division., Numerator: x3 + x2 – 5x + 3, , ), ,   x x − 4 2 9 , ) ,   (, Solution : lim, 18, 18 , x→4 ,  ( x − 4 ) ( x + 3) , , 1 1, , 1, 1, , 1, , 2, , –5, 2, –3, , 3, –3, 0, ,  ( x − 4 ) x9 , lim, , = x→4 , 18, 18,  ( x − 4 ) ( x + 3) , , ∴ x3 + x2 – 5x + 3 = (x– 1)(x2 + 2x –3), ,  x9 , lim, , 18, = x→4 ,  ( x + 3) , ,  x3 + x 2 − 5 x + 3 , lim , , x →1, x2 −1, , , , 18, , , , Denominator: x2 – 1 = (x + 1)(x –1), , ...[As x→4, x − 4 ≠ 0], , 49, = 18, 7, 2 ,  1, +, Ex.3: Evaluate lim, , x →1 x − 1 1 − x 2 , , ,  1, , 2, Solution : lim , +, , x →1 x − 1, (1 − x ) ( x + 1) , , 140, , (, , ) , , =, ,  ( x − 1) x 2 + 2 x − 3, lim , x →1,  ( x + 1) ( x − 1), , =, ,  x2 + 2x − 3 , lim , , x →1,  x + 1  ...[As x→1, x −1 ≠ 0], , =, , 1+ 2 − 3, = 0, 1+1, , 

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Q.II Evaluate the following limits :, ,  1 − 3 x2 + 1 , Ex.5: Evaluate lim, , x→0 , 2, , , x, , , Solution : Put, , 3, , 3 2,  (1 − t ) , lim  1 − x + 1  = lim  3, , x→0, t, →, 1, 2,  (t − 1) , , , x, , , , , −(t − 1), , = lim, , , 2, t →1 (t − 1)(t + t + 1), , , ,  −1 , = lim  2, , t →1 t + t + 1, , , , , =, , −1, 1, =−, 1+1+1, 3, , 1., , 2., 3., 4., , 5., , 9x ,  1, lim , − 3, x →3 x − 3, x − 27 , , , 3., ,  x3 − 4 x 2 + 4 x , lim , , x→2, x2 −1, , , , 4., ,  ( x + ∆x )2 − 2 ( x + ∆x ) + 1 − x 2 − 2 x + 1, lim , ∆x →0 , ∆x, , , 6., , EXERCISE 7.2, Q.I, , 2., , (, , ) , ,  x2 + x 2 − 4 , lim, 5. x → 2  x 2 − 3 x 2 + 4 , , , , ...[As t→1, t − 1 ≠ 0], , , ,  u 4 − 1, lim  3 , u →1 u − 1, , , , x 2 + 1 = t, x2 + 1 = t3, , ∴ x2 = t3 − 1, as x→0, t→1, , , , 1., , , , x3 − 7 x + 6, lim  3, , 2, x → 2 x − 7 x + 16 x − 12, , , , Q.III Evaluate the Following limits :, 1., ,  1 − 8 y3 , lim , 3, 1, y→  y − 4 y , 2, , 2., , 1,  x−2, , − 3, lim  2, 2, , x →1 x − x, x − 3x + 2 x , , , 3., ,  x 4 − 3x 2 + 2 , lim  3, x →1 x − 5 x 2 + 3 x + 1 , , , ,  −2 x − 4 , lim  3, 2,  x + 2 x , , 4., ,  x+2, x−4, +, lim  2, 2, x →1 x − 5 x + 4, 3 x − 3x + 2, , ,  x 2 + 2 x − 15 , lim  2, , x →3,  x − 5x + 6 , , 5., , 1, 1, , , lim  2, + 2, 2, x → a x − 3ax + 2a 2, 2 x − 3ax + a , , , Evaluate the following limits :,  z 2 − 5z + 6 , lim , , 2, z →2,  z −4 ,  x+3 , lim  2, x →−3 x + 4 x + 3 , , ,  5 y3 + 8 y 2 , lim  4, , y →0 3 y − 16 y 2, , , x →−2, , 141, , (, , ), , , , , , , 

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7.3 METHOD OF RATIONALIZATION:, ,  b+ z − b−z, b+ z + b−z , ×, = = lim, , , z →0, z, b+ z + b−z , , , If the function in the limit involves a square, root or a trigonometric function, it may be possible, to simplify the expression by multiplying and, dividing by its rationalizing factor., ,  (b + z ) − (b − z ), , 1, ×, = lim, , , z →0, z, b+ z + b−z , , , SOLVED EXAMPLES,  1+ x −1 , , , Ex. 1. Evaluate : lim, x →0 , x, , ,  1+ x −1 , , , Soluton : lim, x →0 , x, , , , 1,  2z, , lim  ×, , z →0, b+ z + b−z ,  z, , =, , 2, , , lim , , z →0,  b+ z + b−z , , =, ,  1+ x −1 1+ x +1 , ×, = lim, , , x →0 , x, 1 + x + 1 , , , =, , , 1+ x −1, = lim , x →0 , x 1+ x +1, , , =, , (, , , , = lim, x →0 , x, , , (, , ), , , , , , , , , 1+ x +1 , , x, , =, , ...[As z→0, z ≠ 0], , 2, b+0 + b−0, 2, 2 b, 1, b, , , , x 2 + x − 20, lim, Ex. 3. Evaluate x →4  2, , 2,  x − 7 − 25 − x , , ), , , , x 2 + x − 20, lim, Solution : x →4  2, , 2,  x − 7 − 25 − x , , 1, , , = lim, ...[As x→0, x ≠ 0], x →0  1 + x + 1 , , , , , x 2 + x − 20, x 2 − 7 + 25 − x 2, ×, , = lim, 2, 2, x→4 , x 2 − 7 + 25 − x 2,  x − 7 − 25 − x, , 1, 1, =, 1+ 0 +1 2, , 1, 1, , 2, b, +, z, −, b, −, z, (, ) ( ) 2 , Ex. 2. Evaluate lim , z →0 , , z, , , 1, 1, , , 2, 2, b, +, z, −, b, −, z, (, ), (, ), , , Solution : lim, z →0 , , z, , , , =, , =, ,  b+ z − b−z , lim , , z →0, z, , , 142, , =, ,  ( x − 4 ) ( x + 5) x 2 − 7 + 25 − x 2, lim , x→4, x 2 − 7 − 25 + x 2, , , , (, , ) , , =, ,  ( x − 4 ) ( x + 5) x 2 − 7 + 25 − x 2, lim , x→4, 2( x 2 − 16), , , , (, , ) , , =, ,  ( x − 4 ) ( x + 5) x 2 − 7 + 25 − x 2, , lim , x→4, 2( x − 4)( x + 4), , , , (, , ) , , , , , , , , , , , , , , , , , 

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=, ,  ( x + 5), lim , x→4, , , , (, , ), , x 2 − 7 + 25 − x 2 , , , 2( x + 4), , , , 5., , ...[As x→4, x − 4 ≠ 0], =, , (4 + 5), , (, , 42 − 7 + 25 − 42, 2( 4 + 4), , Q.III Evaluate the Following limits :, , ) = (9)(3 + 3) = 27, 2(8), , 8, , EXERCISE 7.3, Q.I, 1., , 2., , 3., , 4., , Evaluate the following limits :,  6 + x + x2 − 6 , lim , , x →0, x, , ,  2x + 3 − 4x − 3 , lim , , x →3, x2 − 9, , ,  1− y2 − 1+ y2 , lim , , y →0, y2, , , , 2., ,  2+ x − 6− x , lim , , x→2, x− 2, , , , 4., ,  a+ y − a, lim , , y →0,  y a + y , , 2., ,  1 + x2 − 1 + x , lim , , x →0,  1 + x 3 − 1 + x , , 3., ,  x 2 + x − 20 , lim , , x→4,  3x + 4 − 4 , , 4., , 3 − 5 + z , lim , , z →4 1 − 5 − z, , , , 5., , 3, 1, , lim , − , x →0 x 9 − x, x, , , 1) lim sinx = sina, x→a, , 2) lim cosx = cosa, x→a, , Using these results and trigonometric identities,, we solve some examples., , , , x2 − 4, lim , , x→2,  x + 2 − 3x − 2 ,  1+ 2 + x − 3 , , lim , x→2, x−2, , , ,  x2 + x x − 2 , lim , , x →1, x −1, , , , Let’s Learn :, To evaluate the limits involving trigonometric, functions, we state -, ,  a + 2 x − 3x , lim , , x→a,  3a + x − 2 x , , 3., , 1., , 7.4 LIMIT OF A TRIGONOMETRIC , FUNCTION :, , Q.II Evaluate the following limits :, 1., ,  x2 + 9 − 2x2 + 9 , , , = lim, 2, 2, x →0 , ,  3x + 4 − 2 x + 4 , , Evaluation of limits can be done by the, method of Factorization, Rationalization or, Simplification as the case may be. While solving, examples based on trigonometric functions we, can use trigonometric identities., Squeeze theorem (Also known as Sandwich, theorem), 143

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,   πx , πx,  sin ,  sin ,  π,  π, = – 2 lim   45   × × lim   180   ×, x →0,  π x  45 x →0 ,  180, πx, , , , , ,  180 , 45 , ,  sin x 2 (1 − cos x 2 ) 1 + cos x 2 , ×, = lim, , x →0, x6, 1 + cos x 2 , ,  sin x 2 (1 − cos 2 x 2 ), , 1, ×, = lim, , x →0, x6, 1 + cos x 2 , , , ,  sin pθ, π, π, × (1) ×, ......  lim , 45, 180,  θ →0  pθ, , = −2 × (1) ×, ,  sin x 2 .sin 2 x 2, , 1, lim, ×, = x →0 , 6, 2, 1 + cos x , x, , , π, π,  π , ×, = −8 , = −2 × 4 ×, , 180 180,  180 , , ,  3 2, 1, sin x, , , ×, = lim, 2, x →0 , 2 3, , 1, +, cos, x,  (x ), , , Activity-1 lim, x →0, , tan x − sin x, sin 3 x, , Solution : lim, x →0, , tan x − sin x, sin 3 x, , 3, ,  sin x 2 , 1, , , × lim , = lim, , , , 2, 2, 2, x →0,  x  x →0  1 + cos x , .....[As x→ 0, x2 → 0], = (1), , 3, , 2, , , ,  sin x × 1 − sin x , lim, , �, = x →0 , , , sin 3 x, , , , 1, 1, 1, =, =, 1 + cos 0 1 + 1 2,   sin θ , 2, ..... lim ,  = 1, here θ = x , θ →0,   θ , , ,  1, , −� , sin x ,  cos x , = lim, x →0, sin 3 x, ,  cos 5 xο − cos 3 xο , lim, , Ex. 8. Evaluate : x →0 , x2, , , , 1 ,  (1 − cos x), × 2 , = lim, x →0 , sin x ,  cos x, ,  cos 5 xο − cos 3 xο , lim, , Solution : x →0 , x2, , , , =, , (1 cos x) 1, cos x, 1, , lim, x, , 0, ,  −2 sin 4 xο sin xο , lim, , = x →0 , x2, , , , = lim, x 0, ,  sin 4 xο sin xο , ×, = – 2 lim, , , x →0, x ,  x, ,  1, , 1, ×, , = lim, x → 0 cos x, (1 + cos x) , , , ,  πx ,  πx ,  sin 4  180  sin  180  , , × , , , = – 2 lim, x →0, x, x, , , , , , , ,  1, , 1, ×, = , � ) ,  cos� (1 + cos, , , ,  πx ,  πx ,  sin  45  sin  180  , = – 2 lim, × , , x →0,  , , , x, x, , , , 1 1, = ×, 1 �, 1, =, �, , =, , 147, , �, , 1, , ×, , (1 cos x), co, , 1, (1 +, � ), , 1, 1 cos ), ,  ,  = 1,  

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 3 − tan x , lim , , π, Ex. 4. Evaluate x →  π − 3 x , 3, Solution : Put, , =, , π, π, − x = t, ∴ x = – t,, 3, 3, As x →, , , , 3 − tan t, 1,  3−, = lim, 1 + 3 tan t, 3 t →0 , t, , , , EXERCISE 7.5, , , , , , , , , , , , , , , , ,  3 + 3 tan t − 3 + tan t, 1, lim, = t →0 , 1 + 3 tan t, , 3, , t, , , , , , , , , , =, , , , 1, 4 tan t, lim , , 3 t →0  t (1 + 3 tan t ) , , =, , , , 4, 1,  tan t , lim , lim, , , , 3 t →0  t  t →0  (1 + 3 tan t ) , , ), , 4, 3, , π, ,t→0, 3, , , , , 1, π, , lim  3 − tan  − t  , 3 t →0 ,  3 , , , t, , , , ,  3 − tan(π / 3) − tan t, 1, , = lim, 1 + tan(π / 3) tan t, 3 t →0 , , t, , , (, , =, , , ,  3 − tan ,  3 − tan x , lim ,  = lim   π, , π, π − 3 x  x → π  3  − x  , x→ , 3, 3, ,  3, , , , =, , 4, 1, tan θ, , , .......  lim, = 1, (1) ×, 3,  θ →0 θ, , 1 + 3 tan 0, , I), , Evaluate the following, , 1), , , ,  cos ec x −1 , lim , 2 , π,  , x→   π, −, x, 2, , ,  2,  , , , , 2), , lim, , 3), ,  5 + cos x − 2 , lim , , 2, x →π,  (π − x), , , 4), ,  cos x − 3 sin x , lim , , π, π − 6x, x→ , , 6, , 5), ,  1 − x2 , lim , , x →1,  sin π x , , x→a, , sin x − sin a, 5, x−5a, , II) Evaluate the following, 1), , 2), , 3), , 150, ,  2 sin x −1 , lim , , π, x→  π − 6 x , 6,  2 − cos x − sin x , lim , , 2, π, x→ , , (4x −π ), 4 , ,  2 − 3 cos x − sin x , lim , , 2, π, x→ , x, −, π, 6, , (, ), 6 

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4), , 5), 7.6, ,  sin, lim , xa , , , xa, , SOLVED EXAMPLES, , , , ,  5x  1, Ex. 1. Evaluate : lim , , x 0,  sinx , , cos 3 x 3 cos x, (2 x )3, , lim, x, ,  x   sin  a  , , 2, ,  5x  1, Solution : lim , , x 0,  sinx , Divide Numerator and Denominator by x, , LIMITS OF EXPONENTIAL AND, LOGARITHMIC FUNCTIONS :, ,  5x  1 , , , = lim  x , x 0,  sinx ,  x , , Let's :Learn, We use the following results without proof., 1), ,  e 1 , lim ,   log e  1, x 0,  x , , 2), ,  a x 1 , lim ,   log a (a > 0, a ≠ 0), x 0,  x , ,  5x  1, lim , , x 0,  x , =,  sinx , lim , x 0,  x , , x, , =, , log 5, 1, , 1, , 3), , lim 1  x  x  e, , 4), ,  log 1  x , lim , x 0, x, , , 5), ,  e px  1 , lim ,   1 , (p constant), x 0, px, , , , 6), ,  a px  1 , lim ,   log a , (p constant), x 0,  px , , 7), ,  log 1  px  , lim ,   1, (p constant), x 0, px, , , , 8), , lim 1  px  px  e , (p constant), , ,  sin , .... lim ,  0,  , , , x 0, , ,  1, , , ,  ax 1 , , , 1, ,, lim,  log a , , , , x 0, ,  x , , , = (log5), ,  5 x - 3x , Ex. 2. Evaluate : lim , , x 0,  x ,  5 x  3x , Solution : Given lim , , x 0,  x , =, ,  5 x  1  3x  1 , lim , , x 0, x, , , , =, ,  5 x  1  (3x  1) , , lim , x 0, x, , , , 1, , x 0, , =, 151, , , , 5, lim, x 0, , , , x, ,  – lim 3  1, , 1, x, , x, , x 0, , x

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=, , log5 – log3, ,  a x −1 , ..... lim ,  = log a, x→a,  x , , 3 1, ×, , , , 2 2 3, , , 3x, 3, x, , , lim 1 +  ,  x →0 , 2  , , , =, −5 1, ×, 2, 2 3, , , −, 5, x, 5, x, −,  , lim 1 +, , →, x, 0,  , 2  , , , ,   5 , = log   ,   3 , , 1, ,  5x  x, 1+ , Ex. 3. Evaluate : lim, x →0 , 6, , , , , 3, , =, , e6, e, ,  5x , 1+ , Solution : lim, x →0 , 6, , , 1, x, , 8, ,  log 4 + log(0.25 + x) , Ex. 5. Evaluate : lim , , x →0, x, , , 5, , Solution :, , 5, , = (e)6, ,  log 4 + log(0.25 + x) , lim , , x →0, x, , = lim  log [ 4(0.25 + x) ] , x →0, x, , , , 1, ,  log (1 + 4 x ) , = lim, , , x →0, x, , , ,  3x + 2  3 x, Ex. 4. Evaluate : lim , x →0 2 − 5 x , , , 1, ,  log (1 + 4 x ) , = 4 × lim, , , x →0, 4x, , , , 1, ,  3x + 2  3 x,  2 + 3x  3 x, = lim , Solution: lim, , , x →0 2 − 5 x, x →0 2 − 5 x , , , , , , , , 4, , 6, = e=, e3, , , , 1, , 6, 6, x, 5, , ,  1+, , = lim,  5x , , x →0 , 6, , , , , , , , 1, , , kx = e, ....., , lim, 1, +, kx, (, ), −5,  x →0, , 6, ,   3x  ,  2 1 + 2  , , = lim  , x →0,  2 1 − 5 x  , ,  , 2  , , , , = 4(1) ..... lim  log(1 + px)  = 1, px,  ,  x →0 , , 1, 3x, , =4,  e 2 x + e −2 x − 2 , , , Ex. 6. Evaluate : lim, x →0, x sin x , , , 1, 1 3, , , x, 3, x, , ,  1 +  , , 2  , = lim  , 1 , x →0,   5x  x ,  1 − 2  , , , , Solution :,  e 2 x ( e 2 x + e −2 x − 2 ) ,  e 2 x + e −2 x − 2 , , lim ,  = lim , x →0, , x sin x  x →0 , e 2 x x sin x, , , ,  e 4 x + 1 − 2e 2 x , = lim  2 x, , x →0,  e x sin x , 152

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 ( 3x − 1) ( 7 x − 1) , , , 2, x, ,  ....[ As x → 0, x 2 ≠ 0], = lim, x →0 , x log(1 + x) , , , x2, , , ,  ( e 2 x ) 2 − 2e 2 x + 1 , = lim , , 2x, x →0,  e x sin x ,  (e 2 x − 1) 2 1 , = lim , × 2x , x →0,  x sin x e , ,  3x − 1 7 x − 1 , ×, , x , = lim  x, x →0,  log(1 + x) , , , x, ,  (e 2 x − 1) 2, , , , 2, 1, x, = lim , × 2x , x →0, e ,  x sin x, , , 2, x, , , , =, , ....[As x→0, x ≠ 0, x2≠0], , log (1 + x ) , , ax −1, = log a, lim, ....  lim, = 1, x →0, x,  x →0 x, , , 2, ,   e2 x − 1  , lim,  × 4, , x →0,  1 ,  2 x , , =, × lim  2 x , x, →, 0,  sin x , e , lim , , x →0,  x , =, , = log 3.log 7, Activity-3, , , , (1) 2 × 4 1, sin θ, e kx − 1, × ο .....  lim, = 1, lim, = 1, θ →0 θ, 1, e,  x →0 kx, , ,  8x − 4 x − 2 x + 1, Evaluate : lim, , , x →0, x2, , , , =4, , , ,  21x − 7 x − 3x + 1 , lim, Ex. 7. Evaluate : x →0  x log(1 + x) , , , x, , x, , x, , x, , lim, x, , x, , , , x, ,  ( 3x − 1) ( 7 x − 1) , , = lim , x →0,  x log(1 + x) , , 153, , x, , − 4x − 2x + 1, , x2, , , x, , 4x 2x 1, x2, , 0, , (2, x, , 1), , 2, , (2 x 1). (4 x 1), lim, x 0, x2, , , ,  7 x ( 3x − 1) − ( 3x − 1) , , = lim , x →0 , , x log(1 + x), , , , 4x, , ), , 0, , lim, ,  7 .3 − 7 − 3 + 1 , = lim, , , x →0,  x log(1 + x) , x, , (4× �, = lim , x →0, , , , ,  21 − 7 − 3 + 1 , Solution : lim , , x →0,  x log(1 + x) , x, , log 3.log 7, 1, , , ,  (2 x − 1) ,  (4 x − 1) , = lim , lim, ×, , , , x →0,  x  x →0  x , , , , =

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Activity-4:, , Q.II Evaluate the following limits :, ,  e x − sin x − 1 , lim, Evaluate : x →0 , , x, , , , 1), ,  3x + 3− x − 2 , lim , , x →0,  x.tanx , , 2), , 3 + x  x, lim , x →0 3 − x , , ,  5x + 3  x, lim , x →0 3 − 2 x , , , , , ,  (e x − 1) −� , = lim , , x →0, x, , , , , , sin x , �, = lim  −, x →0 x, � , , , 3), , , , � ,  sin x , = lim   − lim , x →0 x,   x →0  x , , 4), ,  log ( 3 − x ) − log (3 + x) , lim , , x →0, x, , , , , , =, , 5), ,  4 x + 1 x, lim , x →0 1 − 4 x , , , , 6), ,  5 + 7 x  3x, lim , x →0 5 − 3 x , , , , 1, , 2, , −1, , , , = 1−1, , , , =, , 1, , 1, , EXERCISE 7.6, Q.I, , Q.III Evaluate the following limits :, , Evaluate the following limits :, , 1), , , , ax − bx, lim , , x →0 sin ( 4 x ) − sin ( 2 x ), , , , 2), , 3, , , 2x −1, , lim  x, x →0  (3 − 1).sinx.log (1 + x ) , , , , 3), , 15 x − 5 x − 3x + 1 , lim , , x →0, x . sinx, , , , 4), ,  6 x + 5 x + 4 x − 3x +1 , lim , , x →0, sin x, , , , 4), ,  (25) x − 2 ( 5 ) x + 1 , lim , , x →0, x . sinx, , , , 5), ,  8sin x − 2 tan x , lim , , 2x, x →0,  e −1 , , 5), ,  (49) x − 2 ( 35 ) x + ( 25 ) x , lim , , x →0,  sin x.log(1 + 2 x) , , 1), ,  9 x − 5x , lim  x, , x →0,  4 −1 , , 2), ,  5 x + 3x − 2 x − 1 , lim , , x →0, x, , , , 3), ,  ax + bx + cx − 3 , lim , , x →0, sin x, , , , 154, , (, , )

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7.7 LIMIT AT INFINITY :, (FUNCTION TENDING TO INFINITY), , Definition : A function f is said to tend to limit ‘l’, as x tends to – ∞ if for given ∈ > 0, there exists, a positive number M such that |f(x) – l| < ∈, for, all x > M, , Let's :Learn, , f ( x) = l, ∴ xlim, →− ∞, , 7.7.1 Limit at infinity :, , Note : Whenever expression is of the, form ∞, ∞ , then divide, by suitable power of, x to get finite limits of numerator as well as, , 1, x, Observe that as x approaches to ∞ or –∞ the value, of f(x) is shown below,, , Let us consider the function f ( x ) =, , i) Observe the following table for f ( x ) =, x, f(x), , 1, , 10, , 100, , 1000, , denominator., , 1, x, , 10000, , 100000, , …, , 1 0.1 0.01 0.001 0.0001, , 0.00001, , …, , 7.7.2 Infinite Limits :, , 1, ., x, Observe the behavior of f(x) as x approaches zero, Let us consider the function f(x) =, , We see that as x assumes larger and larger, 1, values,, assumes the value nearer and nearer, x, to zero., , from right and from left., i) Observe the following table for f ( x ) =, x=, f(x), , 1, x, , ∴ lim = 0, x →∞, , f ( x) = lim+, ∴ xlim, → 0+, x →0, , ∴ lim f ( x) = l, x →∞, , x, f(x), , 0.1 0.01 0.001 0.0001 0.00001, 10 100 1000 10000 100000, , …, …, , We see that as x assumes values nearer 0, but, 1, greater than 0,, assumes the values larger and, x, larger., , Definition : A function f is said to tend to limit ‘l’, as x tends to ∞ if for given ∈ > 0, there exists a, positive number M such that |f(x) – l| < ∈, ∀ x in, the domain of f for which x > M, , ii) Observe the following table for f ( x ) =, , 1, 1, , 1, x, , 1, x, , 1, →∞, x, , ii) Observe the following table for f ( x ) =, , -1 -10 -100 -1000 -10000 -100000 …, -1 -0.1 -0.01 -0.001 -0.0001 -0.00001 …, , x=, f(x), , We see that as x assumes values which tend, 1, to −∞,, assumes the value nearer and nearer to, x, zero., , 1, x, , -1 -0.1 -0.01 -0.001 -0.0001 -0.00001 …, -1 -10 -100 -1000 -10000 -100000 …, , We see that as x assumes values nearer to, 1, 0, but less than 0,, assumes the values which, x, tends to – ∞, , 1, x, , =0, ∴ xlim, →−∞, , f ( x) = lim−, ∴ xlim, → 0−, x →0, , 155, , 1, → −∞, x

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SOLVED EXAMPLES, ,  ax +b , Ex. 1. Evaluate : lim, x →∞  cx + d , , , , , , , ,  ( x 2 + 3 x − x)( x 2 + 3 x + x) , , , = lim, x →∞, ( x 2 + 3x + x), , , ,  x 2 + 3x − x 2 ,  2, , = lim, x →∞,  x + 3x + x , , --- as x → ∞, 1 → 0, x, , , , 3x, lim , , 2, = x →∞,  x + 3x + x , , a, c, , , , 3x, , , , lim , 3, = x →∞  x 1 + + x , x, , , , , , 10 x 2 + 5 x + 3 , Solution : lim  2, , x →∞,  5 x − 3x + 8 , , , , 3, , , lim , 3 , = x →∞  1 + + 1 , x, , , , , , Divide by x2 to get finite limits of the, numerator as well as of the denominator,, , =, , 1, = 0), x, ,  x 2 + 3x − x , Solution : lim, x →∞ , , , 10x 2 +5x +3 , Ex. 2. Evaluate : lim,  2, , x →∞,  5x -3x +8 , , =, , 10, 5, , x →∞, , x →∞, , b, , lim  a + , x →∞, x, , =,  d, lim c + , x →∞, x, , , =, , =, , ( lim, , Ex. 3. Evaluate : lim  x 2 + 3 x − � x , ,  ax + b ,  x , , , = lim, x →∞ cx + d, , ,  x , , a+0, c+0, , 10 + 0 + 0, 5−0+0, , = 2, ,  ax + b , Solution : lim, x →∞  cx + d , , , , =, , =, ,  10 x 2 + 5 x + 3 , , , 2, lim  2 x, , x →∞,  5 x − 3x + 8 , , , x2, , 3, 1+ 0 +1, , =, , 5 3, , lim 10 + + 2 , x →∞, x x , ,  3 8, lim 5 − + 2 , x →∞, x x , , , 156, , =, , 3, 1+1, , =, , 3, 2, , 1, = 0), x →∞ x, , ( lim

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EXERCISE 7.7, I, , Let's Remember, , Evaluate the following :, , Note : For limits of trigonometric functions,, angle is supposed to be in radian., ,  ax3 + bx 2 + cx + d , lim, 1) x →∞  3, , 2,  ex + fx + gx + h , , 1), , lim sin x = 1, x→0, x, , 2), , lim sin x = sin a, x→a, , 3), , lim cos x = cos a, x→0, , 4), , lim tan x = 1, x→0, x, , 5), , lim sin kx = k, x→0, x, ,  7 x2 + 2x − 3 , 1) lim , , 4, x →∞,  x +x+2 , , 6), , lim tan kx = k, x→0, x, ,  x 2 + 4 x + 16 − x 2 + 16 , 2) lim, x →∞ , , , 7), , lim x.sin  1  = 0, x→0,  , x, , 3) lim  x 4 + 4 x 2 − x 2 , x →∞ , , , 8), , 2, lim 1–cosp x = p, x→0, 2, x2, , 9), , 2, 2, lim cosm x – cosn x = n − m, x→0, 2, x2, , , , x3 + 3x + 2, lim, , 2) x →∞ ,  ( x + 4 ) ( x − 6 ) ( x − 3) ,  7 x2 + 5x − 3 , 3) lim  2, , x →∞ 8 x − 2 x + 7, , , II, , Evaluate the following :, , III Evaluate the following :, , (, , )(, , ), ,  3 x 2 + 4 4 x 2 − 6 (5 x 2 + 2) , 1) lim , , x →∞, 4 x6 + 2 x4 − 1, , ,  ( 3 x − 4 )3 ( 4 x + 3 ) 4 , 2) lim , , 7, x →∞, , , ( 3x + 2 ), 3) lim  x, x →∞ , , (, , ax–1 = log a, " a > 0, 10) lim, x→0, x, 1, 11) lim, x→0 (1 + x ) x = e, , log (1 + x) = 1, 12) lim, x→0, x, , ), , x +1 − x , , ,  xn − an , lim, 13) x →a ,  = nan–1 for a > 0, −, x, a, , , ,  ( 2 x − 1)20 ( 3 x − 1)30 , , , 4) lim, 50, x →∞, , , ( 2 x + 1), ,  ex −1 , lim, 14) x→0 ,  =1,  x , ,  x2 + 5 − x2 − 3 , 5) lim , , x →∞,  x 2 + 3 − x 2 + 1 , , 1,   =0, 15) lim, x→∞  x , , 157

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 k , p  = 0 for k, p ∈ R and p > 0, x , , 16) lim, , x→∞, , 7), , 1, 17) As x → 0,  x  → ∞,  , ,  3 cos x + cos 3 x , lim , =, π, ( 2 x − π )3 , x→ , 2, A), , x, ,  a  = 0, if a < b, 18) lim, , x→∞ , b, , 8), , MISCELLANEOUS EXERCISE - 7, , 3, 2, , B), , 1, 2, , 1, 2, , C) –, , D), , 1, 4, ,  15 x − 3x − 5 x + 1 , lim , =, x→ 0, sin 2 x, , , A) log 15 , B) log 3 + log 5, C) log 3.log 5, D) 3 log 5, 1, , I), , 1), , Select the correct answer from the given, alternatives., ,  x 4 −16 , lim  2, =, x→2 x − 5 x + 6, , , A) 23, , 2), , 4), , 5), , 56, 3, , C) –32, , D) –16, , B), , 112, 3, , C), , 121, 3, , D), , , ,  3cos x −1 , lim, =, 11) x → π  π, , −x , 2, ,  2, A) 1, B) log 3, , 28, 3, , , , 1, 1, lim  2, + 2, =, x →3 x − 11x + 24, x − x−6 , , 2, 2, 7, 7, A) −, B), C), D) −, 25, 25, 25, 25,  x + 4 −3 , lim , =, x →5  3 x − 11 − 2 , , , −2, 2, 5, A), B), C), 9, 7, 9, , D), , B), , 1, 2, , , , (3sin x −1)3, lim, 13) x→ 0  (3x −1) ⋅ tan x ⋅ log(1 + x)  =, , , A) 3 log 3 , B) 2 log 3, , 2, 9, , C) (log 3)2 , C), , 1, 3, , D), ,  5 x −3 − 4 x −3 , lim, 14) x → 3  sin( x − 3)  =, , , , 1, 4, , A) log 5 – 4, ,  5 sin x − x cos x , lim , =, 2, x →0, 2, x, −, 3, x, tan, , , A) 0, , B) 1, , C) 2, , π, , C) 3 2, ,  x.log (1 + 3 x) , , =, 3x, 2, 12) lim, x→ 0,  (e −1) , 1, 1, 1, A) 9, B) 3, C), e, e, 9, ,  tan 2 x − 3 , lim , =, 3, π, x →  sec x − 8 , 3, A) 1, , 6), , B) 32, , C) e9, ,  log(5 + x) − log(5 − x) , , =, 10) lim, x→ 0, sin x, , , 3, 5, 1, A), B) –, C) –, 2, 2, 2, ,  x 7 + 128 , lim  3, =, x →−2,  x +8 , , A), , 3), , 9), ,  3 + 5x  x, lim ,  =, x→ 0 3 − 4 x, , , A) e3, B) e6, , C), , D) 3, 158, , log 5, , log 4, , D) (log 3)3, , 5, 4, log 5, D), 4, B) log, , D) e–3, , D), , 2, 5, , D) 3 log 3, , D), , 1, 3

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 (2 x + 3)7 ( x − 5)3 , lim, =, 15) x → ∞ , (2 x − 5)10, , , A), , 3, 8, , B), , 1, 8, , C), , 1, 6, , D), , 1, 4, ,  ab x − a x b , 13) lim  2, , x →1,  x −1 , , II) Evaluate the following., , (, , 2), 3), 4), 5), ,  (1 − x ) − 1 , lim , , 3, x →0,  (1 − x ) − 1 , x →0, ,  f (r + h) − f (r ) , If f(r) = πr2 then find lim , , h →0, h, , ,  x , lim , 2, x →0,  x + x , ,  x cos a − a cos x , 16) lim, , x→a , x−a, , 2, lim  (sin x − cos x) , π, 17) x→ , , 4  2 − sin x − cos x , , Find the limit of the function, if it exists,, at x = 1, 7 − 4 x for, 2,  x + 2 for, , 6), ,  ( 2 x + 1)2 ( 7 x − 3)3 , 15) lim , , 5, x →∞, , , (5x + 2), , lim [ x ] ([*] is a greatest integer function.), , f(x) = , ,  22 x − 2 − 2 x + 1 , 18) lim , , 2, x→1,  sin ( x − 1) , , x <1, x ≥1, ,  4 x −1 − 2 x + 1 , 19) lim , , 2, x→1,  ( x − 1) , , Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x., , , , 20) lim  x − 1 , x→1,  log x , , Determine the value of lim f ( x ), x →3, , 7), ,  secx 2 −1 , lim , , 4, x →0,  x, , , 8), ,  e x + e− x − 2 , lim , , x →0,  x.tanx , , 9), , ), , 2, , , 5x − 1, , 14) lim  x, x →0  ( 2 − 1)log (1 + x ) , , , , 5, , 1), ,  log x − log 2 , lim , , x→2, x−2, , , 12), ,  1− cos x , 21) lim, , x→0 , , x, , , 22) lim, x→1, , , , x (6 x − 3 x ), lim , , x →0 cos ( 6 x ) − cos( 4 x ), , , , (2n 1) x n n 2, , x 3x 2 5 x3, x 1, , 23) lim, x→0, ,  a3 x − a 2 x − a x + 1, 10) lim , , x →0, x.tanx, , , ,  1 ,  x2 ,  x4 ,  x2 ,  x 4   ,  12 1 − cos   − cos   + cos   cos    ,  2 ,  4 ,  2 ,  4   ,  x , ,  sinx − sina , 11) lim , x→a,  x − a , , 4 x +3, ,  8 x 2 + 5 x + 3  8 x −1, 24) lim, , 2, x→∞ ,  2x − 7x − 5 , , v, , v, 159, , v

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8, , CONTINUITY, 8.1.1 CONTINUITY OFA FUNCTION AT A, POINT, , Let's Study, •, , Continuity of a function at a point., , •, , Continuity of a function over an interval., , •, , Intermediate value theorem., , We are going to study continuity of functions of, real variable so the domain will be an interval in, R. Before we consider a formal definition of a, function to be continuous at a point, let’s consider, various functions that fail to meet our notion of, continuity. The functions are indicated by graphs, where y = f(x), , Let's Recall, •, , Different types of functions., , •, , Limits of Algebraic, Trigonometric,, Exponential and Logarithmic functions., , •, , Left hand and Right hand limits of functions., , 8.1 CONTINUOUSAND DISCONTINUOUS, FUNCTIONS, The dictionary meaning of the word continuity is, the unbroken and consistent existence over a, period of time. The intuitive idea of continuity is, manifested in the following examples., , Fig. 8.1, , Fig. 8.2, , (i) An unbroken road between two cities., (ii) Flow of river water., (iii) Railway tracks., (iv) The changing temperature of a city during a, day., , Fig. 8.3, , In winter the temperature of Pune rises from, 140C at night to 290C in the afternoon. This, change in the temp is continuous and all the, values between 14 and 29 are taken during 12, hours. An activity that takes place gradually,, without interruption or abrupt change is called a, continuous process. There are no jumps, breaks,, gaps or holes in the graph of the function., , The function in figure 8.1 has a hole at x = a., In fact f(x) is not defined at x = a., The function in figure 8.2 has a break at x = a., For the function in figure 8.3, f(a) is not in the, continuous line., , 160

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8.1.2 DEFINITION OF CONTINUITY, , Hence f (x) is continuous at x = 0., , A function f(x) is said to be continuous at a point, x = a, if the following three conditions are, satisfied:, , Illustration 2 : Consider f (x) = x2 and let us, discuss the continuity of f at x = 2., f (x) = x2, , i. f is defined at every point on an open interval, containing a., , ∴ f (2) = 22 = 4, , ii. lim f (x) exists, , lim f (x) = lim (x2) = 22 = 4, x→2, x→2, , iii. lim f (x) = f (a)., , ∴ lim, f (x) = f (2) = 4, x→2, , x→a, , x→a, , Among the three graphs given above, decide, which conditions of continuity are not satisfied., , ∴ The function f (x) is continuous at x = 2., , The condition (iii) can be reformulated and the, continuity of f(x) at x = a, can be restated as, follows :, , Observe that f(x) = x3, x4, ... etc. are continuous, at every point. It follows that all polynomials are, continuous functions of x., , A function f(x) is said to be continuous, at a point x = a if it is defined in some, neighborhood of ‘a’ and if, , There are some functions, which are defined, in two different ways on either side of a point., In such cases we have to consider the limits of, function from left as well as right of that point., , lim [f (a + h) −f (a)] = 0., h→0, , Illustration 1. Let f (x) = |x| be defined on R., , 8.1.3 CONTINUITY FROM THE RIGHT, AND FROM THE LEFT, , f (x) = −x , for x < 0, = x , for x ≥ 0, , A function f (x) is said to be continuous from the, right at x = a if lim f (x) = f (a)., xa, , A function f (x) is said to be continuous from the, left at x = a if lim f (x) = f (a)., xa, , If a function is continuous on the right and also, on the left of a then it is continuous at a because, lim f(x) = f(a) = lim f (x) ., , xa, , Fig. 8.4, , Illustration 3: Consider the function f (x) =  x , in the interval [2, 4)., , Consider, lim f (x) = lim (−x) = 0, x0, , x→0, , lim f (x) = lim (x) = 0, , x0, , Note :  x  is the greatest integer function or floor, function., , x→0, , lim f (x) = lim f (x) = f (0) = 0, , x0, , xa, , x0, , 161

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Solution :, =, ,  x  ,, , for x ∈ [ 2, 4), , that is f (x) =, , 2 ,, , for x ∈ [ 2, 3), , =, , 3 ,, , for x ∈ [3, 4), , f (x), , lim f (x) = lim (5x − 4.5) = 15 − 4.5 = 10.5, x→3, , x3, , lim f (x) = 10.5, lim f (x) = x, 3, , x3, , ∴ lim, f (x) = f (3), x→3, , The graph of which is as shown in figure 8.5, , ∴ f(x) is continuous at x = 3., , Test of continuity at x = 3., For x = 3, f(3) = 3, , Fig. 8.5, lim f (x) = lim  x  = lim (2) = 2, and, x3, x3, , x3, , Fig. 8.6, , lim f (x) = lim  x  = lim (3) = 3, x3, x3, , x3, , 8.1.4 Examples of Continuous Functions., , lim f (x) ≠ lim f (x), , x3, , (1) Constant function, that is f(x) = k, is, continuous at every point on R., , x3, , ∴ f (x) is discontinuous at x = 3., , (2) Power functions, that is f(x) = xn, with, positive integral exponents are continuous at, every point on R., , Illustration 4 :, Consider f (x) = x2 +, , 3, for 0 ≤ x ≤ 3, 2, , (3) Polynomial functions,, P(x) = a0xn + a1xn-1+ a2xn-2 + ...... + an-1x + an, are continuous at every point on R, , = 5x − 4.5 for 3 < x ≤ 5;, For x = 3, f (3) = 32 +, , (4) The trigonometric functions sin x and cos x, are continuous at every point on R., , 3, = 10.5, 2, , (5) The exponential function ax (a > 0) and, logarithmic function logbx (for x, > 0, and b,, b ≠ 1) are continuous on R., , lim f (x) = lim (x2 + 3 ) = 32 + 3 = 10.5, x3, x→3, 2, 2, , 162

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(6) Rational functions are of the form, , Illustration 5: Consider, , P( x), ,, Q( x), , f (x) = x2 − x − 5, for −4 ≤ x < − 2., , Q(x) ≠ 0. They are continuous at every point, a if Q(a) ≠ 0., 8.1.5 PROPERTIES, FUNCTIONS:, , OF, , = x3 − 4x − 3, for −2 ≤ x ≤ 1., For x = -2, f (−2) = (−2)3 − 4 (−2) −3 = −3, , CONTINUOUS, , lim f (x) = lim (x2 − x − 5) = 4 + 2 − 5 = 1 and, x2, , x2, , If the functions ƒ and g are continuous at x = a,, then,, , lim f (x) = lim (x3 − 4x − 3) = −8 + 8 − 3 = −3, x2, , x2, , 1., , their sum, that is ( ƒ + g ) is continuous at, x = a., , ∴ lim f (x) ≠ lim f (x), , 2., , their difference, that is ( ƒ – g) or (g – f) is, continuous at x = a., , Hence lim f(x) does not exist., , 3., , the constant multiple of f(x), that is k.ƒ, for, any k ∈ R, is continuous at x = a., , 4., , their product, that is ( ƒ.g ) is continuous at, x = a., their quotient, that is f , if g(a) ≠ 0, is, g, continuous at x = a., , 5., 6., , x2, , x2, , ∴ the function f(x) has a jump discontinuity., 8.1.8 REMOVABLE DISCONTINUITY, Some functions have a discontinuity at some, point, but it is possible to define or redefine the, function at that point to make it continuous. These, types of functions are said to have a removable, discontinuity. Let us look at the function f(x), represented by the graph in Figure 8.1 or Figure, 8.3. The function has a limit. However, there is a, hole or gap at x = a. f(x) is not defined at x = a., That can be corrected by defining f(x) at x = a., , their composite function, f[g(x)] or g[f(x)],, that is fog(x) or gof(x), is continuous at x = a., , 8.1.6 TYPES OF DISCONTINUITIES, We have seen that discontinuities have several, different types. Let us classify the types of, discontinuities., , A function f(x) has a discontinuity at, x = a, and lim f (x) exists, but either f(a) is not, x→a, defined or lim f (x) ≠ f (a) . In such case we, x→a, define or redefine f(a) as lim f (x). Then with, x→a, new definition, the function f(x) becomes, continuous at x = a. Such a discontinuity is, called a Removable discontinuity., , 8.1.7 JUMP DISCONTINUITY, As in figure 8.2, for a function, both left-hand, limit and right-hand limits may exist but they, are different. So the graph “jumps” at x = a. The, function is said to have a jump discontinuity., A function f(x) has a Jump Discontinuity at, , If the original function is not defined at a and the, new definition of f makes it continuous at a, then, the new definition is called the extension of the, original function., , x = a if the left hand and right-hand limits, both exist but are different, that is, lim f (x) ≠ lim f (x) , , xa, , x2, , xa, , 163

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Illustration 6:, , 1, is the, x, function to be considered. It is easy to see that, f(x) → ∞ as x → 0+ and f(x) → −∞ as x → 0–., f(0) is not defined. Of course, this function is, discontinuous at x = 0., Observe the graph of xy = 1. y = f(x) =, , x 2  3 x  10, Consider f (x) =, , for x ≠ 2., x3  8, Here f(2) is not defined.,  2, , lim f(x) = lim  x  3 x  10 , 3, x→2, x→2,  x 8 ,   x  2   x  5, , = lim, x→2,   x  2 x2  2x  4, , , , , , , A function f(x) is said to have an infinite, discontinuity at x = a,, , , , , , , if lim f (x) = ±∞ or lim f (x) = ±∞ , xa, , xa, , Fig. 8.7 says, f(x) has an infinite discontinuity., , 7, 2+5,  x5 , = lim, =, =, , x→2  2, 4 + 4 + 4 12,  x  2x  4 , , 8.1.10 CONTINUITY OVER AN INTERVAL, So far we have explored the concept of continuity, of a function at a point. Now we will extend the, , 7,  x 2  3 x  10 , lim, ∴ x→2 ,  = 12, 3,  x 8 , , idea of continuity on an interval., , Here f(2) is not defined but lim f(x) exists., , Let (a, b) be an open interval. If for every, , Hence f(x) has a removable discontinuity., The extension of the original function is , , x ∈ (a, b), f is continuous at x then we say that, f is continuous on (a, b) ., , x 2  3 x  10, f(x) =, for x ≠ 2, x3  8, , Consider f defined on [a, b). If f is continuous, on (a, b) and f is continuous to the right of a,, lim f (x) = f(a) then f is continuous on [a, b), , x→2, , =, , 7, for x = 2, 12, , xa, , Consider f defined on (a, b]. If f is continuous, on (a, b) and f is continuous to the left of b, lim f (x) = f(b), then f is continuous on (a, b], , This is coninuous at x = 2., 8.1.9 INFINITE DISCONTINUITY, , xa, , Consider a function f continuous on the open, lim f (x) exists,, interval (a, b). If lim f (x) and x, b, xa, , , , then we can extend the function to [a, b] so that it, is continuous on [a, b]., SOLVED EXAMPLES, Ex. 1. : Discuss the continuity of the function, f (x) = |x − 3| at x = 3., Solution : By definition of a modulus function,, the given function can be rewritten as, , Fig. 8.7, 164

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= – (x – 3) if x < 3, , f(x), , = x – 3 , , and lim f (x) = 7, , x 2, , if x ≥ 3, , So lim f (x) = lim f (x) = 7 ⇒ lim, f (x) = 7, x→2, , Now, for x = 3, f(3) = 3 – 3 = 0. , lim f (x) = – (3 – 3) = 0 and, , Also, lim, f (x) = f (2) = 7 , x→2, , x3, , ∴ f(x) is continuous at x = 2., , lim f (x) =3 – 3 = 0, , x3, , Let us check the continuity at x = 4., , so, lim f (x) = lim f (x) = 0 ⇒ lim f (x) = 0, x3, , x 2, , x 2, , f(4) = (42 − 8) = 8, , x→3, , x3, , lim f (x) = lim (7) = 7 and, , x 4, , x→4, , lim f (x) = lim (x2 − 8) = 42 – 8 = 8, , x 4, , x→4, , lim f (x) ≠ lim f (x), , x 4, , x 4, , so lim f (x) does not exist., x→4, , and lim f (x) = f (3) = 0, , Since one of the three conditions does not hold at, x = 4, the function. Hence f (x) is discontinuous at, x = 4. Therefore the function f (x) is continuous, on it’s domain � , except at x = 4. There exists a, jump discontinuity at x = 4., , Therefore the function f(x) is continuous at x = 3., , \ f is discontinuous at x = 4., , Ex. 2 : Determine whether the function f is, continuous on the set of real numbers, , Ex. 3 : Test whether the function f (x) is continuous, at x = − 4, where, , Fig. 8.8, x→3, , where f(x) = 3x + 1, for x < 2, , f (x) =, , = 7, for 2 ≤ x < 4, = x2 −8, , = 8, for x = −4., , for x ≥ 4 ., , Solution : f(−4) = 8 (defined), , If it is discontinuous, state the type of discontinuity., ,  2, , lim f (x) = lim  x  16 x  48 , x4, x4, x4, , , , Solution : The function is defined in three parts, by, polynomial functions, and all polynomial functions, are continuous on their respective domains. Any, discontinuity, if at all it exists, would be at the, points where the definition changes. That is at, x = 2 and x = 4., ,  x  4   x  12  , lim  , = x4, , x4, , , = lim (x+12) ....... [∵ x + 4 ≠ 0], x4, , = − 4 + 12 = 8, ∴ lim f (x) = f (−4) = 8, , Let us check at x = 2., f (2) = 7 ( Given ), , x4, , ∴ by definition, the function f(x) is continuous at, , lim f (x) = lim (3x +1) = 3(2) + 1 = 7., , x 2, , x 2 + 16 x + 48, , for x ≠ −4, x+4, , x = −4., , x 2, , 165

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Ex. 4 : Discuss the continuity of f (x) =, on the interval [ −3, 3 ]., , Test of continuity of f at x = 0., For x = 0, f(0) = 0 , , 9 − a2 ,, , lim f (x) = lim  x  = lim (−1) = −1 and, x0, x0, , Solution : The domain of f is [−3, 3]., , x0, , [Note, f(x) is defined if 9 − x2 ≥ 0], , lim f (x) = lim  x  = lim (0) = 0, x0, x0, , x0, , Let x = a be any point in the interval ( -3, 3 ) that, is a ∈ ( −3, 3 )., lim f (x) = lim, x→a, , x→a, , =, , , , 9  x2, , lim f (x) ≠ lim f (x), , x0, , , , Therefore f (x) is discontinuous at x = 0., Test of continuity of f at x = 1., , 9 − a 2 = f (a), , For x = 1, f(1) = 1, , ∴ for a = 3, f(3) = 0 and for a = −3 , f(−3) = 0, , lim f (x) = lim  x  = lim (0) = 0 and, x1, x1, , Now, lim f(x) = f(3) = 0, , x1, , and lim f (x) =f(–3) = 0, , x1, , x3, , lim f (x) = lim  x  = lim (1) = 1, x1, x1, , x3, , lim f (x) ≠ lim f (x), , Thus f (x) is continuous at every point on (−3, 3), and also continuous to the right at x = −3 and to, the left at x = 3., , x1, , Hence the function f (x) =  x  is not continuous, at x = 0, 1 in the interval [−1, 2 )., , Ex. 5 : Show that the function f (x) =  x  is not, continuous at x = 0, 1 in the interval [−1, 2), , , , , x1, , Therefore f (x) is discontinuous at x = 1., , Hence, f (x) is continuous on [−3, 3 ]., , Solution :, , x0, , Ex. 6 : Discuss the continuity of the following, function at x = 0, where, , f (x) =  x  for x ∈ [ −1, 3), that is f (x) = −1 for x ∈ [ −1, 0), f (x) = 0 for x ∈ [ 0, 1), f (x) = 1 for x ∈ [ 1, 2), , 1, f (x) = x2 sin   , for x ≠ 0, x, = 0, for x = 0., Solution : The function f (x) is defined for all, x ∈ R., Let’s check the continuity of f (x) at x = 0., Given, for x = 0, f(0) = 0. , 1, we know that, −1≤ sin   ≤ 1 for any x ≠ 0, x, Multiplying throughout by x2 we get, 1, − x2 ≤ x2 sin   ≤ x2, x, , Fig. 8.9, , Taking limit as x → 0 throughout we get,, 166

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lim (−x2) ≤ lim  x 2 sin  1   ≤ lim (x2), x→0, x→0,  x   x→0, , , ,  sin   = 1 = lim  tan , , since lim,  0  ,  0 , , ,  , , , ,  1 , 0  lim  x 2 sin     0, x 0,  x , , , 2, 3, Ex. 8 : If f is continuous at x = 1, where, ∴ k =, , ∴ by squeeze theorem we get,, , f (x) =, , lim  x 2 sin  1   = 0, x→0,  x , , , sin( x ), x 1, , + a,, , = 2π ,, , lim f (x) = lim  x 2 sin  1   = 0, x→0, x→0,  x , , , =, , 1  cos( x ), ,  (1  x ) 2, , for x < 1, for x = 1, , + b, for x > 1,, , then find the values of a and b., , lim f (x) = f (0) = 0, , Solution : Given that f (x) is continuous at x = 1, , x→0, , ∴ f (x) is continuous at x = 0, , ∴ lim f (x) = lim f (x) = f (1), , Ex. 7 : Find k if f(x) is continuous at x = 0, where, , Now, lim f (x) =, , f (x) =, , xe x + tan x, ,, sin 3 x, , = k, , , , , x1, , ............ (1), , f (1), , lim  sin( x)  a  = 2p,  x 1, , , for x ≠ 0, , x1, , for x = 0, , Put x – 1 = t, x = 1 + t as x→1, t → 0, lim  sin  (1 t )  a  = 2p, , t →0 , t, , , , ∴ f (0) = lim f (x), x→0, , x, k = lim  xe  tan x , , x→0 ,  sin 3 x ,  x tan x , e  x , = lim  sin 3 x , x→0, , , x, , , , lim, t →0, , lim, t →0, ,  sin(   t )  a , ,  = 2p, t, , , ,   sin  t, ,  t, , , , ,  a  = 2p, ,  sin  t , − lim,  × π + lim, t →0 , t →0 (a) = 2p,  t , ,  tan x , , =, , x1, , x1, , Solution : Given that f(x) is continuous at x = 0,, , lim(e x )  lim , , ,  x ,  sin 3 x   3, lim , , x 0,  3x , , x 0, , , , , , x 0, , − (1) π + a = 2π ⇒ a = 3p,  lim  sin  ,   0   , , , , , 1+1 1, 2, =, × = as x → 0, 3x → 0, 1, 3, 3, , From (1), lim f (x) = f (1), x1, , 167, , , = 1, 

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Solution :, , lim  1  cos( x)  b  = 2p, , 2, x1 ,   (1  x), , , (1) f (x) =, , Put 1− x = θ ∴ x =1− θ , as x → 1, θ → 0, , Here f (x) is a rational function, which is continuous, for all real values of x, except for x = 6. Therefore , f(6) is not defined., , ,  1  cos( (1   )),  b  = 2p, ∴ lim, 2,  0 , , , , ,  1  cos(   ),  b  = 2p, ∴ lim, 2,  0 , , , , , 2, lim  x  3 x  18 , Now, lim, f, (x), =, x→6, x→6 , , x6, , , , ,  1  cos ,  b  = 2p, ∴ lim, 2,  0 ,  , , , 2  ,  2 sin  2, , ∴ lim,  0,  2, , , ,   ,  sin , 2 lim   2, p  0  , , 2, , p, ∴ 2 + b = 2p, ∴ a = 3p , b =, ,  ( x  6)( x  3) , , , = lim, x→6, x6, , , ,  ,  ,   b, = 2p, , , , ,  (x−6)≠ 0], = lim, x→6 (x +3) [, ∴ lim, f (x) = 9, x→6, Here f(6) is not defined but lim, f (x) exists., x→6, , 2, , ,   2,   ,    + lim, (b) = 2p,  0,  2, , , 2 (1) × p2 + b = 2p, p, 4, ∴b=, , Hence f (x) has a removable discontinuity., (2) g(x) = 3x + 1, for x < 3, = 2 – 3x, for x ≥ 3, ,  sin  , [ lim,  = 1],  0 ,   , , This function is defined by different polynomials, on two intervals. So they are continuous on the, open intervals (– ∞, 3) and (3, ∞)., , 3p, 2, , We examine continuity at x = 3., , 3p, 2, , For x = 3, g(3) = 2 – 3(3) = −7, , Ex. 9 : Identify discontinuities for the following, functions as either a jump or a removable, discontinuity on R., (1) f (x) =, , lim g (x) = lim (3x + 1) = 10 and, , x3, , x3, , lim g (x) = lim (2 − 3x) = −7, , x3, , x 2 − 3 x − 18, , , x−6, , (2) g (x) = 3x + 1,, = 2 − 3x,, , x 2 − 3 x − 18, x−6, , x3, , lim g (x) ≠ lim g (x) ∴ lim g (x) does not exist., x→3, , x3, , x3, , Hence g is not continuous at x = 3., , for x < 3, for x ≥ 3, , The function g (x) has a jump discontinuity at, x = 3. , , (3) h (x) = 13 − x2, for x < 5, = 13 – 5x, for x > 5, , (3) h (x) = 13 − x2 , for x < 5 ,, = 13 – 5x , for x > 5 , , 168

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but h(5) is not defined., , Solution : f (p / 2) = log 5 − e, , h(x) is continuous at any x < 5 and x > 5, ,  cos x  2  x  , , , , Now, lim f (x) = lim  5  e, ,  , , x, x, cot x, 2, 2 , , , , p, p, p, Let, − x = t, x =, − t as x →, ,t→0, 2, 2, 2, , = lim (13 − x2) = 13 − 25 = −12, , lim h(x), , x5, , x5, , and lim h(x) = lim (13 – 5x) = 13 − 25 = −12, x5, , x5, , So, lim h(x) = lim h(x) ∴ lim, h(x) = −12, x→5, x5, ,   ,  5cos 2 t   et, , lim f (x) = lim, t→0, ,  cot    t , x, 2, , , , 2 , , , x5, , But for x = 5, f(x) is not defined., So the function h(x) has a removable discontinuity., Note :, , , , , , , , ,  5sin t  et , lim, = t→0 , ,  tan t , , sin x = 1., We have proved that lim, x→0, x, Some standard limits are stated without proof., ,  5sin t  1  et  1 , lim, = t→0 , , tan t, , , , lim ex – 1 = 1, x→0, x, ,  (5sin t  1)  (et  1) , lim, = t→0 , , tan t, , , , lim a – 1 = log a, x→0, x, x, , 1, , lim (1 + t ) t = e, x→0, ,  5sin t  1 et  1 , , , , t, t , = lim, , t→0, tan t, , , , , t, , , , 1, , lim (1 − t ) t = e–1 = 1, x→0, e, log(1, +, x), lim, = 1,, x→0, x, lim log(1 – x) = –1, x→0, x, , [As t→0, t ≠ 0],  sin t 5sin t  1 et  1 , , , , , t, sin, t, t , = lim, t→0 , tan t, , , , , t, , , [sin t ≠ 0], , These can be proved using L' Hospital's rule, or, expressions in power series which will be studied, at advanced stage., Ex. 10 : Show that the function,  ,  x, , , 5cos x  e 2, f (x) =, cot x, , , for x ≠, , p, 2, , = log 5 − e , for x =, , p, 2, , (1).(log 5) − 1, =, 1, , ,   sin  ,  tan  , ,   1  lim, , , ,  lim,  0,  0,   ,   , , , x, ,   e 1 ,  a 1 ,  log a ,  1, lim , , , ,  lim, x 0,  x , ,  x 0  x , , lim f (x) = log5 - 1, x, , p, ., 2, Redefine the function so that it becomes, p, continuous at x =, ., 2, , , 2, , f (p / 2) is defined and lim f (x) exists , , has a removable discontinuity at x =, , x, , But lim f (x) ≠ f, x, , 169, , , 2, , p, 2, , , 2

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Ex. 12 :If f(x) is defined on R, discuss the, p, continuity of f at x =, , where, 2, , ∴the function f(x) has removable discontinuity., This discontinuity can be removed by redefining, f (p / 2) = log 5 – 1., , f ( x) , , So the function can be redefined as follows,  cos x  2  x ,  5  e , f (x) = , cot x, , , , , , p,  , for x ≠ 2, , , p, = log 5 − 1 , for x =, 2, , p, 5cos x  5 cos x  2, , for x ≠, 2,  2    2x , (3 cot x).log , , 2, , , 2 log 5, , =, , 3, , , ,  5cos x  5 cos x  2 , lim f (x) = lim , , , , 2    2x  , , x, x, , 2, 2, 3(cot x ) log , , , 2, , , , is continuous at x = 0 then find f (0), Solution : Given that f(x) is continuous at x = 0, 1, ,  3x  2  x, ∴ f (0) = lim, , x→0 ,  2  5x , , , ,  5cos x  5 cos x  2 , , = lim , x,    2x  , 2  3(cot x ) log  1 , , , 2  , , , 1, ,   3x   x,  2 1  2  , , lim, = x→0  ,  2 1  5 x  , ,  , 2  ,  , , Let, , , , , , , , , , x, , 2, , , ,  5sin t 5sin t  5 sin t  2, , = lim, t →0 , sin t,  3(tan t ) log 1  t  .5, , , 3, , e2, , p, p, − t as x →, ,t→0, 2, 2, ,  5sin t  5 sin t  2 , , , = lim, t →0,  3(tan t ).log 1  t  , , , , 5, , 2, , 2, 5x 5x, , lim  1   , 2  ,  x 0 , , , , −5, , 2, , =t ⇒ x=, , , ,  ,  ,  cos  2  t   cos  2  t , , , , 5, 5, 2, , , = lim, t →0 , , , 3 cot   t  .log 1  t , , , 2 , , , , , , 2, 2, , 3x  3 x, , lim  1   , 2  ,  x 0 , , , , e2, ,   2x, , ∴ lim f (x), , 3, , =, , p, ,, 2, ,    2 log 5, f   =, ,, 3, 2, , 1, , =, , p, ., 2, , Solution : Given that, for x =, , x, Ex. 11 : If f (x) =  3 x  2  , for x ≠ 0,,  2  5x , , 1, , 3x  x, ,  1  , 2 , lim, = x→0  , 1,  1  5x  x, , , 2 , , , , for x =, ,  lim 1  kx kx1  e , ,  x 0 , , , f(0) = e4, ,  , , , , .........[Multiply and divide by 5sin t ], 170

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Activity 1 :, ,  (5sin t )2  1  2  5sin t , , lim, = t →0 ,  3(tan t ) log 1  t  (5sin t ) , , , , Discuss the continuity of f(x), , =, , , , (5sin t  1) 2, 2, , ,  sin t, 2, , , sin, t, = lim, t →0 , sin t ,  3(tan t ).log(1  t ).(5 ) , , for x = 5, ........... (I), ,  log x  log 5 , f ( x)  lim , ∴ lim, , x 5, x 5, x 5, , , So we can multiply and divide the numerator by, sin2t, , put x – 5 = t ∴ x = 5 + t. As x → 5, t → 0, , log(, , lim f ( x)= lim, , ,   sin t, 2,  1   sin t  ,  5,  ,   sin t   t  , ,  , , = lim, t →0,   tan t  log 1  t  sin t , .5,  3, , ., t, t, , , , , , , 2, , og 5, t, , 5, , t, , log, , = lim, t, , 5, t, , [Dividing Numerator and Denominator by t2 as, t ≠ 0], , t, log 1, lim, 5, = t, t, , 2, ,   5sin t  1  , sin t 2, lim ,    lim , ,  t 0  sin t   t 0  t ,  1 , , = , × lim,  sin t , t →0,  tan t ,  log(1  t ) , 5, , 3 lim , .lim, , t 0  t  t 0 , t, , (log 5)2 × (1), 1, ×, 3(1)(1), 50, , 1, 5, , Solution. : Given that f(5) = , , t is small but t ≠ 0. Hence sin t ≠ 0, , =, , log x − log 5, for x ≠ 5, x −5, , where f(x) =, , , , (5sin t  1) 2, , , = lim, sin t , t →0 , t, , t, ), 3, (tan, ).log, 1, .(, 5,  , , , , = lim, t, , ,   sin  ,  tan  ,  1,   1, lim, , , , lim,  0,  0, , , , , , , , , , ,  x , lim  log(1  x)   1, lim  a  1   log a , x 0  x , ,  x 0 , x, , , , , , , =  1, , (log 5) 2, ∴ lim f (x) =, x, 3, 2, ,  lim f ( x) , ,  , ∴ lim f (x) ≠ f  , x, 2, 2, , lim f ( x)  f (5), , x 0, , 1, �, , 1, �, , log 1, , t, 5, , 1, , t, ,   log 1  px   , lim, ,   1, px,  x 0 ,  , ........... (II), , ∴ from (I) and (II), x 5, ,  , ∴ f(x) is discontinuous at x =  , 2, , ∴ The function f(x) is continuous at x = 5., 171

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8.1.11 THE INTERMEDIATE VALUE, THEOREM FOR CONTINUOUS, FUNCTIONS, Theorem : If ƒ is a continuous function on a closed, interval [a, b], and if y0 is any value between f(a), and f(b) then y0 = f(c) for some c in [a, b]., , So by intermediate value theorem there has to be, a point c between 1 and 2 with f (c) = 0., Hence there is a root for the equation, x3 − x −1 = 0 between 1 and 2., , EXERCISE 8.1, 1), , Examine the continuity of, , (i) f (x) = x3 + 2x2 − x − 2 at x = − 2., p, 4, p, p, = cos x, for x >, , at x =, 4, 4, 2, x −9, (iii) f (x) =, , for x ≠ 3, x −3, (ii) f (x) = sin x, for x ≤, , Fig. 8.10, , = 8 for x = 3, , Geometrically, the Intermediate Value Theorem, says that any horizontal line y = y0 crossing the, Y-axis between the numbers ƒ(a) and ƒ(b) will, cross the curve y = f (x) at least once over the, interval [a, b]. The proof of the Intermediate Value, Theorem depends on the completeness property, of the real number system and can be found in, more advanced texts. The continuity of ƒ on the, interval is essential. If ƒ is discontinuous at even, one point of the interval, the conclusion of the, theorem may fail., , 2), , Examine whether the function is continuous, at the points indicated against them., , (i) f (x) = x3 − 2x + 1 ,, , if x ≤ 2, , = 3x − 2 , if x > 2 , at x = 2., x 2  18 x  19, (ii) f (x) =, , for x ≠ 1, x 1, = 20 for x = 1 , at x = 1, , Illustration 1 : Show that there is a root for the, equation x3 − x −1 = 0 between 1 and 2, , (iii) f (x) =, , Solution : Let f (x) = x3 − x −1 . f(x) is a polynomial, function so, it is continuous everywhere. We say, that root of f(x) exists if, , =, 3), , f (x) = 0 for some value of x., , x, + 2 ,, tan 3 x, , for x < 0, , 7, , for x ≥ 0 , at x = 0., 3, , Find all the points of discontinuities of , f(x) =  x  on the interval (−3, 2)., , For x = 1, f (1) = 13 − 1− 1 = −1 < 0 , so, f (1) < 0, , 4), , For x = 2, f(2) = 23 − 2 − 1 = 5 > 0 so,, f(2) > 0, 172, , Discuss the continuity of the function, f(x) = |2x + 3| , at x = −3/2

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5), , Test the continuity of the following functions, at the points or interval indicated against, them., , = 3 + 8x , for x > −3., (iv) f(x), , 1, 3, , x − 1 − ( x − 1), , for x ≠ 2, x−2, , (i) f (x)=, =, , (iii) f (x) = x2 − 3x − 2 , for x < −3, , 1, , , 5, , , 7), , for x = 2, , at x = 2 , x3  8, x  2  3x  2, , (ii) f (x)=, , , = - 24, , (iii) f (x) = 4x + 1,, =, , , , for x ≤ 8/3., , 59 − 9 x, , for x > 8/3, at x = 8/3., 3, , , , 1, , 9  3(243  5 x) 5, , = 2 , , 8), , , for x ≠ 0 , , , , for 0 < x < 3; x≠2, , = 12 , for x = 2, 2 − 2 x − x2, =, , x−4, , for 3 ≤ x < 4, , at x = 2 , 6), , Identify discontinuities for the following, functions as either a jump or a removable, discontinuity., (i), , Discuss the continuity of the following, functions at the points indicated against, them., , (i) f (x) =, , for x = 0 , at x = 0, , x 2  8 x  20, (v) f (x) =, 2 x 2  9 x  10, , , , x2 1, (iii) f (x) = 3, for x ≠ −1., x 1, , 1, , (iv) f (x) =, , 1 − cos 2 x, , for x ≠ 0. , sin x, , 3 sin 2 x  2 cos x(1  cos 2 x), (ii) f (x) =, , for x ≠ 0., 2 1  cos 2 x, , for x = 2 at x = 2, , (27  2 x) 3  3, , = 3 – cos x for x > π, , Show that following functions have , continuous extension to the point where f(x), is not defined. Also find the extension, , (i) f (x) =, , for x ≠ 2, , , , = 4 + sin x , for x < π, , (ii) f (x) = x2 + 3x − 2 , for x ≤ 4, , 3, , , 4, , p, p, , at x =, ., 3, 3, for x ≠ 0, , , , for x = 0 , at x = 0., , = 1 , , ,, , 4 x  2 x 1  1, (iii) f (x) =, ,, 1  cos 2 x, , for x ≠ 0, , (log 2) 2, =, ,, 2, , for x = 0 , at x = 0., , Which of the following functions has, a removable discontinuity? If it has a, removable discontinuity, redefine the, function so that it becomes continuous., (i) f (x) =, , = 5x + 3 , for x > 4., , for x =, , e1/ x  1, (ii) f (x) = 1/ x, ,, e 1, , 9), , x 2  10 x  21, f (x) =, ., x7, , =, , 3  tan x, p, ,x≠, 3,   3x, , e5 sin x − e 2 x, , for x ≠ 0, 5 tan x − 3 x, , = 3/4 , for x = 0 , at x = 0., 173

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(ii) f (x) = log(1+3x) (1+5x) for x > 0, =, , 32 x − 1, ,, 8x − 1, , , , , , for x < 0 , at x = 0., , 1, , x, (iii) f (x) =  3  8 x  , for x ≠ 0. ,  3  2x , , (iv) f (x) = 3x + 2 ,, , for −2< x ≤ 6. , , x3 − 8, ,, x2 − 4, = 3 ,, , (v) f (x) =, , e3( x − 2 ) − 1, =, ,, 2( x − 2) 2, , (iv) For what values of a and b is the function, f (x) = ax + 2b + 18 , for x ≤ 0 , = x2 + 3a − b , for 0 < x ≤ 2, = 8x – 2 ,, for x > 2, , continuous for every x ?, , for −4 ≤ x ≤ −2, , = 2x – 3 ,, , sin 2 x, − a ,, for x > 0, 5x, = 4 for x = 0, = x2 + b − 3 , for x < 0, is continuous at x = 0, find a and b., , (iii) If f (x) =, , for x > 2, for x = 2 , , (v) For what values of a and b is the function, , for x < 2, , x2 − 4, f (x) =, , for x < 2, x−2, , 2, , , 10) (i) If f (x) =, , 2  sin x  3, p, , for x ≠, ,, 2, 2, cos x, , is continuous at x =, (ii) If f (x) =, , = ax2 − bx + 3 , for 2 ≤ x < 3 , = 2x – a + b , for x ≥ 3 , continuous for every x on R?, 12) Discuss the continuity of f on its domain,, where, ,  , p, then find f   ., 2, 2, , cos 2 x  sin 2 x  1, 3x 2  1  1, , f (x) = |x + 1| ,, for x ≠ 0,, , = |x − 5| ,, , is continuous at x = 0 then find f(0)., , 13), , 4 x   4  x  2, for x ≠ p ,, ( x   )2, is continuous at x = p, then find f(p) ., , (iii) If f (x) =, , =, 14), , 5 x  5 x  2, , for x ≠ 0, x2, = k for x = 0, is continuous at x = 0, find k., , (ii) If f (x) =, , p, 4, , (sin x  cos x)3  2 2, p, , for x ≠, 4, sin 2 x  1, 3, ,, 2, , for x =, , p, ., 4, , Determine the values of p and q such that, the following function is continuous on the, entire real number line., f (x) = x + 1 ,, = x2 + px + q ,, , 174, , for 2 < x ≤ 7 ., , Discuss the continuity of f(x) at x =, where,, , f (x) =, , 24 x  8 x  3x  1, 11) (i) If f (x) = x, , for x ≠ 0, 12  4 x  3x  1, = k , for x = 0, is continuous at x = 0, find k., , for −3 ≤ x ≤ 2, , for 1 < x < 3, for |x − 2| ≥ 1.

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15) Show that there is a root for the equation, Let's Remember, , 2x3 − x − 16 = 0 between 2 and 3., 16) Show that there is a root for the equation, , Continuity at a point, , x − 3x = 0 between 1and 2., 3, , A function f (x) is continuous at a point a if and, only if the following three conditions are satisfied:, , 17) Activity : Let f(x) = ax + b (where a and b, are unknown), , , (1) f (a) is defined, (2) lim, f (x) exists, and, x→a, , = x2 + 5 for x ∈ R, , (3) lim, f (x) = f (a), x→a, , Find the values of a and b, so that f(x) is, , Continuity from right : A function is continuous, from right at a if lim f (x) = f (a), , continuous at x = 1. (Fig. 8.11), , x a, , Continuity from left : A function is continuous, from left at b if lim f (x) = f (b), x b, , Continuity over an interval :, Open Interval : A function is continuous over an, open interval if it is continuous at every point in, the interval., Closed Interval : A function f (x) is continuous, over a closed interval [a,b] if it is continuous at, every point in (a,b),and it is continuous from , right at a and from left at b., , Fig. 8.11, , Discontinuity at a point :, A function is discontinuous at a point or has a pont, of discontinuity if it is not continuous at that point, , 18) Activity : Suppose f(x) = px + 3 for a ≤ x ≤ b, = 5x2 − q for b < x ≤ c, Find the condition on p, q, so that f(x) is, continuous on [a,c], by filling in the boxes., , Infinite discontinuity :, An infinite discontinuity occurs at a point a if, , f(b) =, , lim f (x) = ±∞ or lim f (x) = ±∞, , xb, , ∴ pb + 3 =, ∴p=, , b, , x a, , x a , , lim f(x) =, , Jump discontinuity :, , –q, , A jump discontinuity occurs at a point ‘a’ if , , is the required condition., , lim f (x) and lim f (x) both exist, but, , x a , , x a, , lim f (x) ≠ lim f (x), , x a , , 175, , x a

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Removable discontinuity :, A removable discontinuity occurs at a point a if, f (x) is discontinuous at a, but lim f (x) exists and, , 2, (4) f (x) = x 2  7 x  10 , for x ∈ [−6, −3], x  2x  8, (A) f is discontinuous at x = 2., (B) f is discontinuous at x = -4., (C) f is discontinuous at x = 0., (D) f is discontinuous at x = 2 and x = -4., , x→a, , f (a) may or may not be defined., Intermediate Value Theorem :, Let f be continuous over a closed bounded interval, [a,b]. If z is any real number between, , (5) If f (x) = ax2 + bx + 1, for |x −1| ≥ 3 and, , = 4x + 5, for -2 < x < 4, is continuous everywhere then,, , f (a) and f (b), then there is a number c in [a,b], satisfying f (c) = z., , (A) a =, , MISCELLANEOUS EXERCISE-8, , (6), , (A) , , 1, 2, , (3) If f (x) =, , 10, e2, ,  , p, , then f   =, 4, 4, , 10, e4, , x, , x, , 8, 3, , (B), , 8, 15, , (C) -, , 8, 15, , (D), , 20, 3, , 1, 2, , (C) -, , 1, 4, , (D), , (A) 6 (B) 4 (C) (log2)(log4) (D) 3log4 , 12 x  4 x  3x  1, (8) If f (x) =, , for x ≠ 0 is, 1  cos 2 x, continuous at x = 0 then the value of f(0) is, log12, (A) , , (B) log2.log3 , 2, log 2.log 3, (C), , (D) None of these. , 2, , 1, 4, ,  sin 2 x  tan 5 x, , (B), , x, , = k , for x = 0,, is continuous at x = 0, then value of ‘k’ is, , , for x ≠ 0, (e 2 x  1) 2, is continuous at x = 0, then f(0) is, (A) , , 1, , b = -3 , 2, , 32 x  8 x  4 x  1, (7) f (x) =, , for x ≠ 0, 4 x  2 x 1  1, , 1  2 sin x, p, , for x ≠, 4,   4x, , (B) -, , (D) a =, , 16  1 9  1, f (x) =, , for x ≠ 0,  27  1 32  1, , (A) , , 2, , is continuous at x =, , 1, , b = 3, 2, , = k , for x = 0, is continuous at x = 0, then ‘k’ = , , (D) lim f (x) = 2log3, , (2) If f (x) =, , 1, , b = - 3, 2, , x, , p, (1) f (x) =, for x ≠, 2, p, = log 2 ,, for x =, 2, p, (A) f is continuous at x =, 2, p, (B) f has a jump discontinuity at x =, 2, (C) f has a removable discontinuity, x, , (B) a = -, , (C) a = -, , (I) Select the correct answer from the given , alternatives., 2cot x  1, ,,   2x, , 1, , b = 3, 2, , (C), , 5, 4, , (D), , 5, 2, 176

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4, , 4  5x  x, (9) If f (x) = ,  , for x ≠ 0 and f(0) = k, is,  4  7x , , (5) f (x) =, , x 1, ,, 2x  x 1, , for x ≠ −1, , 2, , = 0 for x = -1 at x = -1., , continuous at x = 0, then k is, 3, , (A) e7, , (B) e3, , (C) e12, , (D) e 4, , (6) f (x) =  x  1, , Where [*] is greatest integer function. , , (10) If f(x) =  x  for x∈(−1,2) then f is, discontinuous at, (A) x = -1, 0, 1, 2, , , (B) x = -1, 0, 1 , , (C) x = 0, 1 , , (D) x = 2 , , (7) f (x) = 2x2 + x + 1, for |x − 3| ≥ 2 , = x2+3 , for 1 < x < 5 , , (II) Discuss the continuity of the following, functions at the point(s) or on the interval , indicated against them., , (2) f (x) = x2 + 5x + 1 , for 0 ≤ x ≤ 3, = x3 + x + 5 , for 3 < x ≤ 6, , x 2 − 3 x − 10, =, , for 6 < x ≤ 9, x −5, , (3) f (x) =, , (2) f (x) = 2x2 −2x + 5 , for 0 ≤ x ≤ 2, 1 − 3x − x 2, =, , for 2 < x < 4 , 1− x, , =, , = 7 for x = 5, cos 4 x − cos 9 x, , for x ≠ 0, 1 − cos x, 68, , , , at x = 0 on   x , 15, 2, 2, , sin 2  x, (4) f (x) =, , , 3(1  x) 2, , x2 + x + 1, , for x ∈ [ 0, 3 ), x +1, , 3x  4, ,, x2  5, , for x ∈ [ 3, 6]., , (IV) Discuss the continuity of the following, functions at the point or on the interval, indicated against them. If the function, is discontinuous, identify the type of, discontinuity and state whether the, discontinuity is removable. If it has a, removable discontinuity, redefine the, function so that it becomes continuous., , x 2 − 25, = x − 5 , for 4 ≤ x ≤ 7 and x ≠ 5, , f (0) =, , (III) Identify discontinuities if any for the, following functions as either a jump or a, removable discontinuity on their respective, domains., (1) f (x) = x2 + x - 3 , for x ∈ [ -5, -2 ), = x2 - 5 , for x ∈ ( -2, 5 ], , x 2 − 3 x − 10, (1) f (x) =, , for 3 ≤ x ≤ 6, x ≠ 5, x −5, = 10 , for x = 5 , , (3) f (x) =, , for x ∈ [−2, 2), , ( x  3)( x 2  6 x  8), (1) f (x) =, x 2  x  12, (2) f (x) = x2 + 2x + 5,, , for x ≠ 1, , , , x ,  2 sin 2  ,  2 , =, for x = 1, at x = 1., x , 3  4 cos 2 , ,  2 , 177, , = x3 − 2x2 − 5,, , for x ≤ 3 , for x > 3

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(V) Find k if following functions are continuous, at the points indicated against them., , , , 3, 2 x4, , (1) f (x) =  5 x  8 , , for x ≠ 2 ,  8  3x , = k , for x = 2 at x = 2., 45 x  9 x  5 x  1, (2) f (x) =, ,, (k x  1)(3x  1), =, , 2, ,, 3, , (VII) Find f(a), if f is continuous at x = a where,, 1  cos( x), (1) f (x) =, , for x ≠ 1 and ,  (1  x) 2, , at a = 1., , for x ≠ 0, , (2) f (x) =, , for x = 0, at x = 0, , (VI) Find a and b if following functions are, continuous at the points or on the interval , indicated against them., 4 tan x  5 sin x, (1) f (x) =, ,for x < 0, ax 1, 9, =, ,, for x = 0, log 2, =, , (2) f (x) = ax2 + bx + 1 , for |2x − 3| ≥ 2, 5, 1, = 3x + 2 , for, <x< ., 2, 2, , 11x  7 x.cos x, ,, bx 1, , 1  cos[7( x   )], , for x ≠ p at, 5( x   ) 2, a = p., , (VIII) Solve using intermediate value theorem., (1) Show that 5x − 6x = 0 has a root in [1, 2], (2) Show that x3 − 5x2 + 3x + 6 = 0 has at, least two real roots between x = 1 and, x = 5., , for x > 0., , v, , v, , 178, , v

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9, , DIFFERENTIATION, When we speak of velocity, it is the speed with, the direction of movements. In problems with, no change in direction, words speed and velocity, may be interchanged., , Let's :Learn, •, , The meaning of rate of change., , •, , Definition of derivative and the formula, associated with it., , •, , Derivatives of some standard functions., , •, , Relationship between, Differentiability., , Continuity, , The rate of change in a function at a point with, respect to the variable is called the derivative of, the function at that point. The process of finding a, derivative is called differentiation, , and, , 9.1.2 DEFINITION OF DERIVATIVE AND, DIFFERENTIABILITY, Let f(x) be a function defined on an open interval, containing the point ‘a’. If, , Let's Recall, •, , Real valued functions on R., , •, , Limits of functions., , •, , Continuity of a function at a point and over, an interval., ,  f (a   x)  f (a )  exists, then f is said to, lim , ,  x 0, x, , , be differentiable at x = a and this limit is said to, be the derivative of f at a and is denoted by f ′(a)., , 9.1.1 INTRODUCTION :, , We can calculate derivative of ‘f  ’ at any point x in, the domain of f., , Suppose we are travelling in a car from, Mumbai to Pune. We are displacing ourselves, from the origin (Mumbai) from time to time. We, know that the average speed of the car, =, , Let y = f ( x) be a function. Let there be a small, increment in the value of ‘x’, say δ x , then, correspondingly there will be a small increment, in the value of y say δ y ., , Total distance travelled, Time taken to travel that distance, ,  y   y  f ( x   x), , But at different times the speed of the car can be, different. It is the ratio of a very small distance, travelled, with the small time interval required to, travel that distance. The limit of this ratio as the, time interval tends to zero is the speed of the car, at that time. This process of obtaining the speed, is given by the differentiation of the distance, function with respect to time. This is an example, of derivative or differentiation. This measures, how quickly the car moves with time. Speed is, the rate of change of distance with time., ,   y  f ( x   x)  y, ,  y  f ( x   x)  f ( x), , .... [  y  f ( x)], , As δ x is a small increment and δx ≠ 0, so dividing, ,  y f ( x   x)  f ( x), , x, x, Now, taking the limit as δ x → 0 we get, throughout by δ x , we get, ,  y ,  f ( x   x)  f ( x) ,  lim , lim , , ,  x 0  x, x, ,   x 0 , , 179

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If the above limit exists, then that limiting value, is called the derivative of the function and it is, dy, = f '( x), symbolically represented as,, dx, so, ,  f ( x  h)  f ( x)  dy, f '( x)  lim ,  = dx = f '( x), h 0, h, , , The derivative of y = f(x) with respect to x at x = a, by method of first principle is given by, , dy, = f '( x), dx, ,  f (a  h)  f (a )   dy , f (a )  lim ,  =  dx  x  a, h 0 , h,   , , We can consider the graph of f(x) i.e., {(x, y) / y = f(x)} and write the differentiation in, terms of y and x, , 9.1.4 DERIVATIVES OF SOME STANDARD, FUNCTIONS, (1) Find the derivative of x n w. r. t. x. for n∈N, , NOTE : (1) If y = f(x) is a differentiable function,   y  dy, of x then lim , and, ,  x 0   x ,  dx, , Solution :, ,  f ( x   x)  f ( x) , lim ,   f '( x),  x 0, x, , , , Let f ( x) = x n, , f ( x + h) = ( x + h) n, , (2) Let δ x = h , Suppose that, , By method of first principle,, ,  f ( a  h)  f ( a ) , lim ,  exists. It implies that, h 0 , h, , ,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, , ,  f ( a  h)  f ( a ) ,  f ( a  h)  f ( a ) , = lim , lim , , , h 0 , h, h, ,  h 0 , ,  ( x  h) n  x n , f ( x)  lim , , h 0, h, , , ,  f (a  h)  f (a )  is called the Left Hand, lim ,  Derivative (LHD)at x = a, h 0 , h, , ,  x n  nC1 x n 1h  nC2 x n  2 h 2  .......  h n  x n , = lim , , h 0, h, , , ,  f ( a + h) − f ( a ) , lim+ ,  is called the Right Hand, h →0 , h,  Derivative (RHD)at x = a, ,  nC1 x n 1h  nC2 x n  2 h 2  nC3 x n 3 h3  .....  h n , , h, , , , = hlim, 0 , , Generally LHD at x = a is represented as f ′(a-), or L f ′(a), and RHD at x = a is represented as, f ′(a+) or R f ′(a), ,  h( nC1 x n 1  nC2 x n  2 h  nC3 x n 3 h 2  .....  h n 1 ) , , h, , , , = hlim, 0 , ,  nx, = hlim, 0, , 9.1.3 DERIVATIVE BY METHOD OF FIRST, PRINCIPLE., The process of finding the derivative of a function, from the definition of derivative is known as, derivatives from first principle. Just for the sake, of convenience δx can be replaced by h., , n 1, ,  nC2 x n  2 h  nC3 x n 3 h 2  .....  h n 1, , , , ... (as h →0 , h ≠ 0), ∴ if f(x) = xn, f (x) = nxn−1, , , If f(x) is a given function on an open interval, then, the derivative of f(x) with respect to x by method, of first principle is given by, 180

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(2) Find derivative of sin x w. r. t. x., ,  tan( x  h)  tan x , f ( x)  lim , , h 0 , h, , , Solution :, ,  sin( x  h) sin x ,  cos( x  h)  cos x , , = lim , h 0 , h, , , , , , , Let f ( x) = sin x, , f ( x + h=, ) sin( x + h), By method of first principle,, ,  sin( x  h).cos x  cos( x  h) sin x , , , cos( x  h).cos x, , = lim , , h 0 , h, , , , , , ,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, ,  sin( x  h)  sin x , f ( x)  lim , , h 0 , h, , ,  2x  h , h,  2cos  2  .sin  2  , , ,  , lim , h 0 , h, , , , , , , =, , =, , =, , , h, sin   , ,  2x  h ,  2   1 , 2 lim cos , lim ,  , , h 0,  2  h 0  h   2 ,  2 , , , , sin h, , , , h0  h.cos( x  h).cos x , , =, , h 0 , , , lim, , lim, , , , , , , , 1,  sin h , , ,  . hlim, , 0, h ,  cos( x  h).cos x , , , 1 ,  sin , = (1). , ,  ........  lim, 2, , 0,  ,  cos x , , ,  ,   1,  , , ∴ if f(x) = tan x, f (x) = sec2x, , , h, sin   , , h, ,  2   1 , 2 lim cos  x   lim ,  , h 0, 2  h 0  h   2 , ,  2 , , , ,  sin p, 1, 2 cos x.(1)   .......  lim ,  0, 2,  p, , , , , =, , , (4) Find the derivative of sec x w. r. t. x., Solution:, Let f ( x) = sec x, ,  ,   1,  , , f ( x + h=, ) sec( x + h), From the definition,, , ∴ if f(x) = sin x, f (x) = cosx, ,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, , , (3) Find the derivative of tan x w. r. t. x., Solution:, ,  sec( x  h)  sec x , f ( x)  lim , , h 0 , h, , , Let f ( x) = tan x, , f ( x + h=, ) tan( x + h), , 1, 1 , ,  cos( x  h)  cos x , , lim , h 0 , h, , , , , , , From the definition,,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, , 181

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=, ,  cos x  cos( x  h) ,  cos( x  h).cos x , , lim , h 0 , h, , , , , , , ,  h,  log 1  x    1 , ,  , = lim ,  , h 0 , h,  x, , , x, , , , =, , ,  2x  h ,  h  ,  2 sin  2  .sin  2  , , ,  , lim , h 0 , h.cos( x  h).cos x, , , , , , , 1, = 1.  , x, , ,  log(1 + x)  , ,  = 1,  xlim, x,  ,  →0 , , ∴ if f(x) = log x, f (x) =, , 1, x, , , , h,  2x  h  ,  sin  2  ,  sin  2    1 , ,   lim ,  , , , = 2 hlim,  , , 0, h, , 0, h, ,  cos( x  h).cos x , ,  2 , (6) Find then derivative of, , ,  2 , , , , , , Solution:, ,  sin p  , 2 sin x, 1, .(1).   .......  lim , =,   1,  0, cos 2 x, 2,  p  , Let f ( x) = a x, , ∴ if f(x) = sec x, f (x) = secx , tanx, , w. r. t. x. (a > 0), , f ( x + h) =, a x+h, , (5) Find the derivative of log x w. r. t. x.(x > 0), , From the definition,, , Solution:, ,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, , , Let f ( x) = log x, , f ( x + h=, ) log( x + h), ,  a( xh)  a x, f ( x)  lim , h 0, h, , , From the definition,,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, , , , , , ,  a x (a h  1) , = hlim, , , 0, h, , , ,  log( x  h)  log( x) , f ( x)  lim , , h 0 , h, , ,  ah 1 , = a x lim , , h 0,  h , , , ,  h,  xh,  log 1  x  ,  log  x  , , , ,    lim , , = hlim, 0 , h, , 0, h, h, , , , , , , , , , , , , , ,  a x 1 ,  log a , f ( x)  a x log a.........  lim , , x 0,  x , , , , ∴ if f(x) = ax, f (x) = ax.loga, , , ,  h,  xh,  log 1  x  ,  log  x  , , , ,    lim , lim , =, h 0 , h, , 0, h, h, , , , , , , , , , , , 182

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Try the following, , , , h, = lim , , h 0 h ( x  h  x ), , , , (1) If f ( x) = 1n , for x ≠ 0, n∈N, then prove that, x, , f ( x)  , , 1, , , = hlim, ...[As h0, h ≠ 0], , 0  x  h  x , , , n, , x n1, , (2) If f(x) = cos x, then prove that, f ′( x) = − sin x, , =, , (3) If f(x) = cot x, then prove that, , f(x+h) = cos (2(x+h)+3) =cos ((2x+3)+2h), , (4) If f(x) = cosec x, then prove that, f ′( x) = − cos ecx.cot x, , From the definition,, , (5) If f ( x) = e x , then prove that f ′( x) = e x, ,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, , , SOLVED EXAMPLES, ,  cos[(2 x  3)  2h]  cos(2 x  3) , f ( x)  lim , , h 0 , h, , , Ex. 1. Find the derivatives of the following from, the definition,, =, , x (ii) cos (2x+3) (iii) 4 x (iv) log(3x−2), , Solution :, (i) Let f ( x) = x, f ( x + h) =, , =, , x+h, , , ,  2(2 x  3)  2h ,  sin(h) ,  2 sin , 2, , , , lim , h 0 , h, , , , , ,  2 sin(2 x  3  h) sin(h) , , h, , , h 0 , , , lim, ,  sin h , == lim[−2sin(2 x + 3 + h)] lim , h →0, h →0  h , , , From the definition,,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, ,  xh  x , f ( x)  lim , , h 0, h, , , , 2 x, , (ii) Let f(x) = cos (2x + 3), , f ′( x) = − cos ec 2 x, , (i), , 1, , =, , ,  sin , [2 sin(2 x  3)](1).......  lim ,   0  , , f (=x)  2 sin(2 x  3), (iii) Let f ( x) = 4 x, ,  xh  x, xh  x , , , , = hlim, , 0, h, x  h  x , , , f ( x + h) =, 4 x+h, , , , xhx, = lim , , h 0 h ( x  h  x ), , , , From the definition, , 183, ,  ,   1,  

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 f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, ,  4 xh  4 x , f ( x)  lim , , h 0, h, , , , =, , f ( x) , ,  4 x (4h  1) , lim, = h 0 , , h, , , ,  3  ,  log(1  px)  , (1)  ,   1, ,  lim, x, , 0, px,  3x  2  ,  , , 3, 3x  2, , Ex. 2. Find the derivative of f(x) = sin x, at, x =π, , h, ,  4 1 , x, = 4 hlim, , , 0,  h , , Solution:, , f=, ( x) sin, =, x sin, =, π 0, , , ,  a x 1 , x, , f ( x)  4 log 4......  lim ,  log a , , x 0,  x , , , , f(π) = sin π = 0, , f (  h)  sin(  h)   sin h, From the definition,, , f ( x) log(3x − 2), (iv) Let =, ,  f ( a  h)  f ( a ) , f (a )  lim , , h 0 , h, , , f ( x +=, h) log[3( x + h) −=, 2] log[(3 x − 2) + 3h], From the definition,, ,  f (  h)  f ( ) , f ( )  lim , , h 0 , h, , ,  f ( x  h)  f ( x ) , f ( x)  lim , , h 0 , h, ,  log[(3 x  2)  3h]  log(3 x  2) , f ( x)  lim , , h 0 , h, , ,  (3 x  2)  3h  ,  log  3 x  2  , , , , = hlim, 0 , h, , , , , , ,  sin h ,   sin h  0 , = hlim, , ,   hlim, 0 , , 0, h, h , , , , , ,  sin h , =  lim , h 0  h , , , ,  sin  , , , ,   1, , = –11  lim,  0    , , Ex. 3. Find the derivative of x 2 + x + 2 , at, x=–3, , , 3h  , ,  log 1 + 3 x − 2  , , , == lim , h →0 , h, , , , , , , Solution :, Let f ( x) = x 2 + x + 2, , , 3h  , ,  log 1 + 3 x − 2    3 , ,  ×, == lim , , , h →0 , 3h,   3x − 2 , , , 3x − 2, , , , For x = – 3, f(– 3) = (–3)2 – 3 = 9 – 3 + 2 = 8, f (3  h)  (3  h) 2  (3  h)  2,  h 2  6 h  9  3  h  2  h 2  5h  8, , From the definition,, 184

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 f ( a  h)  f ( a ) , f (a )  lim , , h 0 , h, , ,   f ( a  h)  f ( a )  , lim  h ,  hf '(a),    lim, h 0, h,   h 0,  , ,  f (3  h)  f (3) , f (3)  lim , , h 0 , h, , , lim [f(a + h) − f(a)] = 0[f’(a)] = 0, h→0, ∴ lim, f(a + h) = f(a) , h→0, ,  h  5h  8  8 , = hlim, , 0 , h, , , 2, , This proves that f(x) is continuous at x = a., Hence every differentiable function is continuous., ,  h  5h , = lim , , h 0,  h , 2, , Note that a continuous function need not be, differentiable., This can be proved by an example., ,  h(h  5) , = lim , , h 0 , h , , Ex.: Let f ( x) = x be defined on R., , (h  5)......[h  0, h  0], = hlim, 0, , f(x) = - x, , for x < 0, , = x, , for x ≥ 0, , , , f (3)  5, , Consider, lim f ( x)  lim ( x)  0, x 0, , 9.1.5 RELATIONSHIP BETWEEN, DIFFERENTIABILITY AND, CONTINUITY, , x 0, , lim f ( x)  lim ( x)  0, , x0, , x 0, , For, x = 0, f(0) = 0, , Theorem : Every differentiable function is, continuous., Proof : Let f(x) be differentiable at x = a., Then, f (a )  lim  f (a  h)  f (a )  ........... (1), , h 0 , h, , , , lim f ( x)  lim ( x)  f (0)  0, , x0, , x 0, , Hence f(x) is continuous at x = 0., Now, we have to prove that f(x) is not differentiable, at x = 0 i.e. f ′(0) doesn’t exist., i.e. we have to prove that,, , we have to prove that f(x) is continuous at x = a., i.e. we have to prove that lim f ( x)  f (a ), Let x = a + h, x→a, h→0, ,  f ( 0  h )  f ( 0) ,  f ( 0  h )  f ( 0) ,  lim , lim , , , h 0 , h, h,  h 0 , , , We need to show that, , We have, L. H. D. at x = 0, is f ′(0−), , x a, , lim f(a + h) = f(a), h→0, , =, , The equation (1) can also be written as,  f ( a  h)  f ( a ) , lim , f '(a ),   lim, h 0, h, ,  h 0, , =, ,  f ( 0  h )  f ( 0) , , h, , , h0 , , , lim, ,  h , lim    lim (1), h 0  h  h 0, , As h→0, h ≠ 0, , f ′(0−) = −1, , Multiplying both the sides of above equation by, h we get, , Now, R. H. D. at x = 0, is f ′(0+), , 185, , ……….. (I)

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=, ,  f ( 0  h )  f ( 0) , lim , , h 0 , h, , , =, , h, lim    lim (1), h 0  h  h 0, f ′(0+) = 1……….. (II), Therefore from (I) and (II), we get, ,  f ( 0  h )  f ( 0) ,  f ( 0  h )  f ( 0) ,   hlim, , , , 0, h, h, , , , , Solution : Given that f ( x)  ( x  2) x  2, That is f ( x)  ( x  2) 2 for x < 2, = ( x − 2) 2 for x ≥ 2, , 2, 5, ,  f (2  h)  f (2) , Lf (2)  lim , , h 0 , h, , , 2, 3, ,  (2  h  2) 2  (2  2) 2 , = lim , , h 0, h, , , , Solution :, 2, , x) (3 x − 2) 5, Given that, f (=, ,  h 2 , = hlim, ,   lim  h , 0 ,  h  h 0, ,  f ( a  h)  f ( a ) , f (a )  lim , , h 0 , h, , , Lf ′(2) = 0, , 2, Note that f   = 0, 3, ,  f (2  h)  f (2) , Rf ′(2) = lim , , h 0, h, , , , 2 ,  2 , f   h   f   , , 2, 2,  , 3, For, x  , f     lim   3 , , h, , 0, 3, h, 3, , , , , , =, , 2, , 5, 2, , , , ,  3  h   2,    3  , lim  , h 0, h, , , , , 2, , 5, 2, , 3, h, , 2, , , lim , h 0 , h, , , , , , , , , f ( x) =( x − 2) x − 2 at x = 2, , SOLVED EXAMPLES, , =, ,  1, lim  3, 3 h 0  5, h, 2, 5, , Ex. 2. Examine the differentiability of, , Though f(x) is continuous at x = 0, it is not, differentiable at x = 0., , is differentiable at x =, , =, , , , , , , , ∴ f(x) is not differentiable at x = 2, 3, , h0 , , , Ex. 1. Test whether the function f ( x)  (3 x  2), ,  52, , 2,   3 5 lim  h, , h 0  h, , , , , , This limit does not exist., , f (0 )  f (0 ) that is, , lim, , =, , 2, , 5, (3h), lim , h 0  h, , , ,  (2  h  2) 2  (2  2) 2 , lim, = h 0  , , h, , ,  h2 , lim, h, = h0    hlim, ,  h  0, , , , , , , , , , Rf ′(2) = 0, So, Lf ′(2) = Rf ′(2) = 0, Hence the function f(x) is differentiable at x = 2., , , , , , , 186

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Ex. 3. Show the function f (x) is continuous at, x = 3, but not differentiable at x = 3. if, f(x) = 2x + 1, for x ≤ 3, = 16 − x 2, , ∴ f(x) is not differentiable at x = 3., Hence f(x) is continuous at x = 3, but not, differentiable at x = 3., Ex. 4. Show that the function f(x) is differentiable, , for x > 3., , Solution : f(x) = 2x + 1, , for x ≤ 3, , = 16 − x 2, , for x > 3., , ) x2 + 2 ., at x = − 3 where, f ( x=, Solution :, For x = −3, f (3)  lim  f (3  h)  f (3) , h 0 , h, , , For x = 3 , f(3) = 2(3) + 1 = 7, ,  (3  h) 2  2  11 , = lim , , h 0, h, , , , lim f ( x)  lim(2 x  1)  2(3)  1  7, , x3, , x3, , lim f ( x)  lim(16  x 2 )  16  (3) 2  7, , x3, ,  9  6h  h 2  2  11 , = lim , , h 0, h, , , , x3, , lim f ( x)  lim f ( x)  f (3)  7, , x3, , x3, ,  h 2  6h , , , , = lim, h 0,  h , , f(x) is continuous at x = 3.,  f (3  h)  f (3) , Lf (3)  lim , , h 0 , h, ,  2(3  h)  1  7 ,  6  2h  1  7 ,  lim , = hlim, , , , 0 , h, h,  h 0 , , ,  h  6, = lim, h 0, f (3)  6, , f (3) exists so, f(x) is differentiable at x = − 3., , 3  h)  1  7 ,  6  2h  1  7 ,   hlim, , , , 0, h, h, , , , , EXERCISE 9.1, (1) Find the derivatives of the following w. r. t. x, by using method of first principle., ,  2h , = hlim,   lim (2), 0 ,  h  h 0, Lf ′(3) = 2, , ……………… (1), ,  f (3  h)  f (3) , Rf (3)  lim , , h 0 , h, , =, , (a) x 2 + 3 x − 1, , (b) sin (3x), , (c) e 2 x +1, , (e) log(2 x + 5), , (d) 3x, , (f) tan (2x +3) (g) sec ( 5x − 2), (h) x x, ,  16  (3  h) 2  7 ,  16  9  6h  h 2  7 ,  lim , lim , , , h 0, h, h, ,  h 0 , , , (2) Find the derivatives of the following w. r. t. x., at the points indicated against them by using, method of first principle, ,   h (6  h ) ,  (6  h ) , = lim ,   hlim, h 0 , h,  0, Rf ′(3) = − 6, , (h→0, h ≠ 0), , (a), , ……………… (2), , from (1) and (2), Lf ′(3) ≠ Rf ′(3), 187, , 2 x + 5 at x = 2 (b) tan x at x = π / 4, , (c) 23 x+1 at x = 2, , (d) log(2 x + 1) at x = 2, , (e) e3x − 4 at x = 2, , (f) cosx at x = 5π, 4

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(3) Show that the function f is not differentiable, at x = −3,, where f(x) = x2 + 2 for x < − 3, , = 2 − 3x, , Let there be a small increment in the value of, x ,say δ x ,then u changes to ( u + δ u ) and v, changes to ( v + δ v ) respectively, correspondingly, y changes to ( y + δ y ), , for x ≥ −3, ,  y   y, , u   u , , , ,  v   v, , (4) Show that f ( x) = x 2 is continuous and, differentiable at x = 0., , , ,  y , , u   u , ,  v   v - y, , (5) Discuss the continuity and differentiability of, ,  y , , u   u , ,   v   v  - u  v , , (i) f ( x) = x x at x = 0, , , , (ii) f ( x) =(2 x + 3) 2 x + 3 at x = − 3/2, , , ,   y  u v, , (6) Discuss the continuity and differentiability of, f(x) at x = 2, , As δ x is small increment in x and δ x ≠ 0 ,, dividing throughout by x we get,, , f ( x)   x  if x ∈ [0, 4) . [where [*] is a, greatest integer ( floor ) function], ,  y u v u v, , , , x, x, x x, Taking the limit as δ x → 0 , we get,, , (7) Test the continuity and differentiability of, f(x) = 3x + 2 if x > 2, = 12 − x 2 if x ≤ 2 at x = 2., ,  y ,  u v , ,   lim, , , x, , 0, , x x , ,  x 0 , x, , lim, , (8) If f(x) = sin x – cos x if x ≤ π / 2, = 2 x − π + 1 if x > π / 2 . Test the continuity, and differentiability of f at x = π / 2, ,  y ,  u ,  v , lim ,  lim    lim  , ,  x 0   x   x 0   x   x 0   x , , ...... (I), , Since u and v are differentiable function of x, , (9) Examine the function, ,   v  dv,   u  du,  lim   , and lim   , ….. (II),  x0   x  dx,  x0   x  dx, , 1, f ( x) = x 2 cos   , for x ≠ 0, x, , = 0 ,, , y  u  v, , ∴, , for x = 0, , for continuity and differentiability at x = 0., ,   y  du dv [From (I) and (II)], , ,  dx dx, ,  x 0 , x, , lim, , i.e. dy  du  dv, , dx, , 9.2 RULES OF DIFFERENTIATION, , dx, , dx, , 9.2.1. Theorem 1. Derivative of Sum of functions., , 9.2.2 Theorem 2. Derivative of Difference of, functions., , If u and v are differentiable functions of x such, , If u and v are differentiable functions of x such, , dy du dv, +, that y= u + v , then =, dx dx dx, , that y= u − v , then, , Proof: Given that, y= u + v where u and v are, , dy du dv, , , dx dx dx, , [Left for students to prove], , differentiable functions of x, 188

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= –x5 cosecx.cotx + 5x4 cosecx +, 1, x × (sec2x) +, tan x, 2 x, 3), , Ex. 2) If f(x) = p tan x + q sin x + r, f(0) = −4 3 ,,  ,  , f   = −7 3 , f’   = 3 then find p, q, 3, 3, and r., ,  ex  5 , Given that y =  x, , e 5, , Solution :, Given that f(x) = p tan x + q sin x + r, , f’(x) = p sec2 x + q cos x, , Differentiate w.r.t.x., dy, d  ex  5 , =, , , dx, dx  e x  5 , =, , (e x  5) , , f(0) = −4 3, , d x, d, (e  5)  (e x  5)  (e x  5), dx, dx, (e x  5) 2, , put x = 0 in (1), f(0) = p tan 0 + q sin 0 + r = r ∴r = −4 3,  , f   = −7 3 ,, 3, , (e x  5).(e x )  (e x  5).(e x ), =, (e x  5) 2, ,  ,  , ∴ from (1) p tan   + q sin   + r = −7 3, 3, 3, , e 2 x  5(e x )  e 2 x  5(e x ), =, (e x  5) 2, =, , 4), , p 3+q, , 10 e x, (e x + 5) 2, ,  ,  , ∴ from (2), psec2   + qcos   = 3, 3, 3, q, 4p +, = 3 ∴ 8p + q = 6, ...(4), 2, , Differentiate w.r.t.x.,, dy, d  x sin x , =, , , dx, dx  x  sin x , =, , 3, − 4 3 = −7 3 ∴ 2p + q = −6 ...(3), 2, ,  , f’   = 3, 3, , x sin x, y = x + sin x, , ( x  sin x) , , (4) − (3) gives 6p = 12 ∴ p = 2, put p = 2 in (3),, we get q = −10 ∴ p = 2, q = −10 and r = −4 3, , d, d, ( x sin x)  ( x sin x )  ( x  sin x), dx, dx, ( x  sin x) 2, , 9.2.5 Derivatives of Algebraic Functions, , d,  d, , ( x  sin x)   x (sin x)  sin x ( x)   ( x sin x).(1  cos x), dx,  dx, , =, ( x  sin x) 2, , Sr.No., 01, 02, , f (x), c, xn, , =, , ( x  sin x).( x cos x  sin x)  ( x sin x).(1  cos x), ( x  sin x) 2, , 03, , 1, xn, , =, , x 2 cos x  x sin x  x sin x cos x  sin 2 x  x sin x  x sin x cos x, ( x  sin x) 2, , 04, , x, , x cos x + sin x, ( x + sin x) 2, 2, , =, , ...(1), ...(2), , 2, , 191, , f '(x), 0, nxn-1, −, , n, x n +1, , 1, 2 x

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9.2.6 Derivatives of Trigonometric functions, Sr.No., 01, 02, 03, 04, 05, 06, , y = f (x), sin x, cos x, tan x, cot x, sec x, cosec x, , 6), , dy/dx = f '(x), cos x, − sin x, sec2 x, − cosec2 x, sec x tan x, − cosec x cot x, , (III) Differentiate the following w.r.t.x., , 9.2.7 Derivatives of Logarithmic and, Exponential functions, Sr.No., 01, 02, 03, , y = f (x), log x, ex, ax, , 3, , y = logex log x3, , dy/dx = f '(x), 1/x, ex, ax log a, , 1), , y = x2 x +x4logx, , 2), , y = exsecx − x 3 logx, , 3), , y = x4 + x x cos x − x2ex, , 4), , y = (x3 − 2) tan x − x cos x + 7x.x7, , 5), , y = sinx logx + ex cosx − ex x, , 6), , y = ex tanx + cos x log x −, , 5, , (IV) Differentiate the following w.r.t.x., 1), , x2  3, y= 2, x 5, , 2), , y=, , 3), , xe x, y=, x + ex, , EXERCISE 9.2, (I), , Differentiate the following w.r.t.x, 4, 3, , 1), , y = x + e − sin x, , 2), , y=, , 3), , y = log x − cosec x + 5x −, , 4), 5), 6), , x, , x + tan x − x3, , 7, 3, , 4, 5, , y = x + 5x −, , 4), , 3, 3, , x2, , 5, , 2, x x − logx + 77, 3, 4, y = 3 cotx − 5ex + 3logx − 3, x4, , 5), 6), , y=, , (V) (1), , 2), , y = x3 logx, , 3), , y = (x2 + 2)2 sin x, , 4), , y = ex logx, , 5), , 5e x − 4, 3e x − 2, , If f(x) is a quadratic polynomial such, that f(0) = 3, f '(2) = 2 and f '(3) = 12, then find f(x), , (II) Differentiate the following w.r.t.x., y = x5 tan x, , x log x, x + log x, , y=, , 2, , x5, , x 5, x 5, , x 2 sin x, y=, x + cos x, , y = 7x + x7 −, , 1), , x 5x, , (2), ,  , If f(x) = a sin x−b cos x, f '   =, 4,  , and f '   = 2, then find f(x)., 6, , (VI) Fill in the blanks. (Activity Problems), (1), , 3, 2, , y = ex.tanx, diff. w.r.t.x., , y = x exlogx, 192, , 2

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f ( x), f '( x) p,  lim, , then lim, xa g ( x), x  a g '( x ), q, , dy d x, = (e tan x), dx dx, d, d, tan x  tan x �, dx, dx, , �, , If lim, g'(x) = 0, then provided, x→a, lim f '(x) = 0, we can study lim f '(x), x→a, x → a g'(x), using the same rule., ,  tan x �, , � �, , ex, (2), , sin x, x2 + 2, diff. w.r.t.x., ,  sin x , Ex. 1 : lim,  2 , x 0,  x , Here f(x) = sinx, lim, f(x) = 0 and, x→0, , y=, , d, d, (sin x)  sin x �, dy, dx,  dx, 2, 2, ( x  2), dx, , g(x) = x2, lim, g(x) = 0, x→0, , �, , , , , 3), , ��, , f '(x) = cos x, lim, f '(x) = cos 0 = 1 ≠ 0, x→0, ,  sin x �, ( x  2) 2, , g'(x) = 2x, lim, g'(x) = 2(0) = 0, x→0, , 2, , Since lim, g'(x) = 0. L’ Hospital Rule cannot, x→0, be applied., , � �, ( x 2  2) 2, , y = (3x2 + 5) cos x, ,  x 2  7 x  10 , lim, Ex. 2 : x 2  x 2  2 x  8 , , , , Diff. w.r.t.x, , Here f(x) = x2 − 7x + 10, lim, f(x) = 0 and g(x), x→2, = x2 + 2x − 8 lim, g(x) = 0, x→2, , dy d,  (3 x 2  5) cos x , dx dx, ,   3x 2  5, , d, , dx, ,   3x  5 , 2, , ∴, 4), , dx,   3x 2  5 , dy, , d, dx, ,   cos x , ,   cos x ,  , , f '(x) = 2x −7, lim, f '(x) = 2(2) − 7 = −3 ≠ 0, x→2, , , , g'(x) = 2x +2, lim, g'(x) = 2(2) + 2 = 6 ≠ 0,, x→2, So L ‘Hospital’ s rule is applicable., , , ,  cos x, ,  x 2  7 x  10 , ∴ lim  2, , x2,  x  2x  8 , , Differentiate tan x and sec x w.r.t.x. using, the formulae for differentiation of u and 1, v, v, respectively., , 1,  2 x  7  3,  lim , , , , x2, 2,  2x  2  6, , Brief idea of L’ Hospital Rule, , Let's Remember, , Consider the functions f(x) and g(x),, If lim, f(x) = 0 and lim, g(x) = 0 and if, x→a, x→a, , •, , f(x) is differentiable at x = a if, Lf  ′(a) = Rf  ′(a), , lim f '(x) = p and lim g'(x) = q where q ≠ 0, x→a, x→a, 193

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•, , Derivative by First Principle :, , 2), , f(x + h) – f(x), f  ′(x) = lim, h, h→0, •, , Derivatives of standard functions :, y = f(x), c(constant), xn, 1, xn, x, sin x, cos x, tan x, cot x, sec x, cosecx, loge x, loga x, ex, ax, u±v, uv, u, v, , dy, = f ′(x), dx, 0, nxn–1, – nn+1, x, 1, 2 x, cos x, –sin x, sec2 x, –cosec2 x, sec x tan x, –cosecx cot x, 1, x, 1, x log a, ex, ax log a, u′ ± v′, uv′ + u′v, vu′ – uv′, v2, , 3), , 1), , 4), , (C), , 1, 2 x, , (B), (D), , ax  c, (B) (cx  d ) 2, , (C), , ac  bd, (cx  d ) 2, , (D), , 5), , (B)  (4 x  5)2, , 5, , (D) , , 7 cos x, (4 sin x + 3) 2, 7 cos x, (4 sin x  3) 2, , 13, (4 x  5) 2, , (B), , 23 cos x, (4 sin x + 3) 2, , (D) , , 15 cos x, (4 sin x  3) 2, , Suppose f(x) is the derivative of g(x) and g(x), is the derivative of h(x)., If h(x) = asin x + bcos x + c then f(x) + h(x) =, (A) 0, , 6), , 15, , 15, , 5 sin x  2, dy, , then, =, 4 sin x  3, dx, , If y =, , (C) , , ad  bc, (cx  d ) 2, , 3x + 5, dy, , then, =, 4x + 5, dx, , If y =, , (A), , x4, dy, If y =, , then, x2, dx, 1, x+4, , ab  cd, (cx  d ) 2, , (C)  (4 x  5)2, , Select the appropriate option from the, given alternative., , (A), , (A), , (A)  (3x  5)2, , MISCELLANEOUS EXERCISE-9, I), , ax + b, dy, If y = cx + d , then, =, dx, , (B) c (C) −c (D) −2(asin + b cosx), , If f(x) = 2x + 6 for 0 ≤ x ≤ 2, , = ax2 + bx for 2 < x ≤ 4, is differentiable at x = 2 then the values of, a and b are., , x, ( x + 2) 2, , 3, (A) a   , b  3, 2, , x, ( x + 2) 2, , a, (C)=, 194, , 1, =, ,b 8, 2, , a, (B)=, , 3, =, ,b 8, 2, , 3, (D) a   , b  8, 2

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7), , If f(x) = x2 + sin x + 1 for x ≤ 0, , = x2 − 2x + 1, , (4) Determine all real values of p and q that, ensure the function, , for x ≤ 0 then, , f ( x=, ) px + q for, x ≤ 1, , (A) f is continuous at x = 0, but not, differentiable at x = 0, , πx , = tan ,  , for 1 < x < 2 is differentiable at,  4 , x =1., , (B) f is neither continuous nor differentiable, at x = 0, , (5) Discuss whether the function, , (C) f is not continuous at x = 0, but, differentiable at x = 0, , f ( x) = x + 1 + x − 1 is differentiable ∀x ∈ �R, (6) Test whether the function, , (D) f is both continuous and differentiable, at x = 0, 8), , f ( x=, ) 2 x − 3 , for x ≥ 2, , = x-1, for x < 2, , x, x, x, x, , ,  ....   x  1 ,, If, f(x) , 50 49 48, 2, then f '(1) =, 50, , (A) 48, , 49, , (B) 49, , 48, , (C) 50, , 2, , is differentiable at x = 2., , (D) 51, , (7) Test whether the function, , f ( x=, ) x 2 + 1 , for x ≥ 2, , II), , = 2x+1, for x < 2, , (1) Determine whether the following function is, differentiable at x = 3 where,, , f ( x=, ) x2 + 2, = 6x − 7, , ,, , ,, , is differentiable at x = 2., , for x ≥ 3, , (8) Test whether the function, , for x < 3., , f ( x=, ) 5 x − 3x 2 for x ≥ 1, , (2) Find the values of p and q that make, function f(x) differentiable everywhere on R, , f ( x)= 3 − x, , ,, , = px 2 + qx ,, , = 3-x , for x < 1, is differentiable at x = 1., , for x < 1, , (9) If f(2) = 4, f ′(2) = 1 then find, , for x ≥ 1 ., , (3) Determine the values of p and q that make, the function f(x) differentiable on R where, , f ( x) = px3 ,, = x2 + q ,, , lim, ,  xf (2) − 2 f ( x) , , x−2, , x →2 , , , for x < 2, , ex, dy, 10) If y =, find, when x = 1., x, dx, , for x ≥ 2 ., , v, , v, , 195, , v

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ANSWERS, , 1. COMPLEX NUMBER, , EXERCISE 1.1, , Q.6, , 4 + 6i, , Q.7, , i) −i, , ii) 1, , Q.1, , i), , 0, , ii), , 11i, , iii) i, , iv) 1, , Q.2, , i), , (3−i), , ii), , (3+i), , v) −i, , vi) −1, , iii) − 5 + 7 i, , iv), , 5i, , vii) 1, , viii) 0, , v) − 5i, , vi) ( 5 + i), , vii) ( 2 − 3 i), , viii) cosq − i sinq, , i) 2i, , ii) 0, , Q.3, , i), , a = 3, b =, , 1, 2, , iii) a = −3 , b = 7, , Q.4, , v) a =, , 3, 2, ,b=, 13, 13, , vi) a =, , 3, -1, ,b=, 2, 2, , i) a = −4, b = −3, iii) a =, , viii) a =, ix) a =, , Q.10 − 1, Q.11 1, , a=5,b=0, , Q.13 Yes, its value = −2 ∈ R, , iv) a = ±2 , b = ±6, , Q.14 2i, Q.16 0, Q. 24 i) x = 1, y = 2, , ii) x = −2, y = 2, , iii) x = 1, y = 2, , iv) x + y = 3, , v) x + y = 9, , ii) a = 0, b = 1, , 1, -7, 3, -1, ,b=, iv) a =, ,b=, 2, 2, 10, 10, , v) a = −1, b = 0, vi) a =, , ii), , Q.9, , vi) a =, , EXERCISE 1.2, , -8, ,b=0, 29, , Q.1, , -1, -1, , b=, 4, 4, 11, 2 3, ,b=, 19, 19, Q.2, , 23, 15, ,b=, 13, 13, , i), , ii) ± (4+3i), , iii) ± (2− 3 i), , iv) ± ( 5 + 2 i), , v) ± ( 3 − i), , ii) ± ( 5 + 2 i), , i), ii), , 196, , ± (1−3i), , 1  7i 1  7i, ,, 8, 8, 3  5i 3  5i, ,, 4, 4

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Q.3, , Q.4, , iv) 2 + 3i , 2 - 3i, , iv), , 1 , 7, 7,  i sin,  cos, 4, 4, 2, , 7p,  1 e4i, ;, , 2, , , i, , − 2i, 2, , v), , 1 , 3, 3,  i sin,  cos, 4, 4, 2, ,  1 e 3p4 i, ;, 2, , , vi), , 3, 3, , 2  cos,  i sin, 4, 4, , , ;, , , , i), , 3, 3i, +, 2, 2, , ii), , (1 − i), , iii), , -7 3 7i, 2, 2, , iv), , 1 + 3i, 2, , v), , 1  i 3, 2, , vi), ,  3 i, 2, , 7  11i 7  11i, ,, iii), 6, 6, i) −5i, 2i, , ii), , iii) − 2i, , iv) − 2i, , i) −1 + 2i, 3 − i, , ii) 3 2 , 2i, , iii) 2 + 3i, 3 − 4i, , iv) 1 − i,, , 4 -2i, 5 5, , Q. 5 i) 7, , ii) 2, , iii) 7, , iv) 200 - 264i, , Q.5, , v) 6, EXERCISE 1.3, Q.1, , i), ii), , 1  5 , 74 ,  tan  , 7, , i ,, 2, , Q. 6, , 1 3π, ,, 2 4, , Q. 7, , 2 (cos θ + i sin θ) where, tan θ =, ,  2, tan 1 , ,  3, , θ = nπ, n∈Z, , Q.4, , 2, 2, ,  i sin, i) 2  cos, 3, 3, , , 3 1, 3 1, EXERCISE 1.4, , 3π, 1  15 , iii) 17,  tan   iv) 3 2 ,, 4, 8, π, 11π, v) 4 2 , 4, vi), 6, π, vii) 3,0, viii) 2 ,, 4, π, ix) 2,, x) 5 2 , tan -1 7, 3, Q.2, , 3p i, , 2e 4, , Q.1, , Q.3, , i), , 1, , ii), , 1, , iii) 1, , iv) 1, , i), , ii), , −1, , iii) −1, , 0, , iv) 0, , v) 1, Q.6, , 2p i, ,  ; 2e 3, , , i), , x2 + y2 = 100, , ii), , (x−3)2 + y2 = 4, , iii) (x−5)2 + (y+6)2 = 25, iv) x + 2 = 0, , 3, 3  3p i, ,  i sin, ii) 1  cos,  ; e2, 2, 2 , , , v), , y = −x, , vi) 2y − 3 = 0, , iii) 1 (cos π + i sin π) ; e pi, , Q.7, , i) Cos2θ + i Sin2θ, iii) 1, , 197, , ii) Cos11θ + i Sin11θ

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Q.8, , i) −4+4i, , ii), , iii) −8 + 8 3 i, , iv) 512 3 + 512i, , −8i, 6), , 17(cos θ + i sin θ ( where tanθ =, , MISECLLANEOUS EXERCISE - 1, , 15, 8, , ii) 37 ,   tan 1  1  ,,  , ,  6 , , (I), Q. No., , 1, , 2, , 3, , 4, , 5, , Ans, , B, , D, , A, , C, , B, , Q. No., , 7, , 8, , 9, , 10, , Ans, , B, , A, , D, , D, , 37 (cos   i sin  ) where tan  , , i) (3 + 8i), iii) (14 − 5i), v) −30 + 10i, , iv) 1, 5, , −35 45, − i, 26 26, , −4 + 0i, 15, iv), − 10i, 2, 1 7, + i, vi), 2 2, 1, 15, +, i, 4, 4, , ix) −i, , x), , 40 56, + i, 25 25, , i), , ii) x = 17, y = 19, , x = 2, y = 1, , =, x, iii), , 3, 28, =, ,y, 61, 61, , v), , ii), , viii), , vii), , vii) 1,, , 12), , ii), , 38 (cos   i sin  ) where tan  , ,  2, ;, 6, , 38 eiq, , i), , 5), , i) ± (3+5i), , ii) ± (4−i), , 16), , i) −i, , iii) ±, , iv) ± (3+3i), , 17), , 1 9, + i, 4 4, , 20), , 2, , vi) ±, ,  , , , , 1 cos  i sin , 4 , 4, 4, , p, , , , 2  cos  i sin  ; 2e 3 i, 3, 3, , , 4), , v) ± (2−i), , 0, , , , , , , 2  cos  i sin , 2, 2, 2, , , i), , i) −i, , 3 + i), , ii), , , ,   , 1 cos 5  i sin  5  , 4,  4 , , , 9) =, x 1=, , y 2, , 3), , 1, , 0, , 2,, , , ,, 4, , vi) 3, 3 , 3  cos 3  i sin 3 , 2, 2, 2 , , , iv) x = 4, y = −2, ii), , 1, 6, , iii) 1,  , 1 cos   i sin  , 3 , 3, 3, , (II) 1), , 2), ,  15 , , i) 17, tan−1   ,, 8, , iii) 3(cosθ + i sin θ) where tan θ = − 3 ;, 3eiq, , 2 (2+i), , 198, , ii) 0, , iii) −1

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2. SEQUENCES AND SERIES, EXERCISE 2.1, 1) i), ii) and iii) are G.Ps iv) and v) not G.Ps., 1, 5 , 5, ii) t6 = 7 iii) a = −7 iv) r = 3, , i) 2(3)n−1, 2) i) t7 =, , ii) (−5)n−1, , 1, 81, , 3) t10 = 510, , n 1, , iii), , 4, 4) x  , , 6) 3, 6, 12 or 12, 6, 3, 7), , 1 1, 1 1, , , 3, 27 or 27, 3, ,, 27 3, 3 27, or, , 11) 800, , 3, 3, 12) 80   ft , 80   ft, 4, 4, , 1,−2,4,−8,16, n, , 6, , ii) 4, , iii) 200(1.1)n, , ii), , 7 , 1 , , 9n   1  n  , , 81 ,  10  , , , , ii) 3, , , , , , 3) i) 5, , , , n, , , , , , 15 10, 3  1, 2 , , 12) 20.1 Lac, , 15) 10 years, , iii) −, 2) i), , 7, 9, , 3) 4, n, , 2  1 , ii) 1   , 9   10 , , , , , , ii), , 1) i) 1, , n, p  q , 1     if q < p, ii), p  q   p  , ,  5 ,  5, 5  1 , , n, , 11) i) 6138 `, , 2, , 7, 1 , 1  n  iv), iii), , 9  10 , , 4) i) 635, , 4 , 1 , , 9n   1  n  , , 81 ,  10  , , EXERCISE 2.3, , n, , p 2  q ,    1 if q > p, and, q  p  p , , , 266, 2) i), 243, , 6) i), , iii) 4(5)n−1, , EXERCISE 2.2, , n, 1) i) 3  2  1, , 8, 10 10n  1  9n , , 81 , , 8) tn = 4(3)n−1, , 13) i) 6, −3 ii) 3(2)19; −3 iii) 3(2)n−1; 3(−1)n−1, , 15) i) 10, , ii), , 5  1 , 7) i), 1   , 9   10 , , 8) 1, 2, 4, 8, 16, , 14) i) 200(1.1)3 ii) 200(1.1)10, , 3, 10 10n  1  9n , , 81 , , 4, 5, =, ,r, 25, 2, , =, 5) t1, , 9, , 5) i), ,  1, , , 6) i) 4, 7) i) 2, , 3, ii), 5, , ii) 2046, 199, , ii) 6, 9, 4, , iv) does not exist, ii), , 22, 9, , iii), , 11, 6, , 5), , 4) ii) -, , 1, 4, , ii), , iii), 4 2, 2 -1, , 8, 3, , 106, 45, , v) 90, iv), , 2296, 45, , 15 15 15, , , ..., 4 16 64, , iv), , 2, 3, , 8) 25 m

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EXERCISE 2.4, , 8), , 9) 48, , 1) (i) and (iii) are H.P. ; (ii) is not H.P., 2) i), iii), , 1, 1, ;, 3n - 1 23, , MISCELLANEOUS EXERCISE - 2, , 1, 1, ;, 2n + 2 18, , ii), , (I), , 1 1, ;, 5n 40, , 3) 5, , 24, 5, , 4), , 5) 60, , 7) −3, 9, , 1, D, , 3 3, ,, 11 10, , 6), , 8) 4, 16, , ii), , 2 (1 − n x n ), , +, , 1− x, , 9) 4, 9, , 1− 3 −2, , n, , +, , 1− x, 3n −1, 2, , iii), , 3 − 3n, 4, , 75, ii), 16, , 16, 2) i), 9, , n −1, , (1 − x ), iv) 3, , ), , 2, , [, , [, , 1), 3), , 6, n  n  3, 4, 2, , 4), , 5), , 3, , 6), , 2n  n  1 (2n  1), 3, , 16, , 2), , 5, A, , 211, 81, , 6, C, , 7, C, , 8, A, , 9, D, , 10, C, , 3, 4, , 3), , 2, 10 10n  1  9n , , 81 , , 9), , 2, 1 , 1  n , 3  10 , , 10), , n, 10n 2  27 n  1, , 6, , n ( n + 1) 3n 2 17 n 26, , 12, n n+ n+2, 12), 18, , 3, iii), 16, , 13), , n  n  1  2n  1, 24, ,  n  n  1  2n  1 n(n  1) , 14) 6 , , , 6, 2 , , 15) 2n(n+1)(n+3)(n+4), 16), , n (2n2 + 2n + 1), 2) __, 2, +1, 12, , 8), , 11), , −1 2n 1, , EXERCISE 2.6, n  4n 2  9n  1, , 4, C, , 49, 5, 4) a = 5 ; r =, 5) 5,10,20 or 20,10,5, 7, 1, 1, 6), , 1, 3, 9, 27 or 27, 9, 3, 1,, 3, 3, , 2, , 3 (1 −, , 3, A, , 1) 3072, , 2 x (1 − x n −1 ), , (1 − x ), , 2, C, , II), , EXERCISE 2.5, , 1) i), , 8.n (n + 1)(2n + 1) 3.n (n + 1), +, 6, 2, , n 2 (n + 1) 2 +, , 4, , 2, , 5, 36, , 17) 2364, , +2, , 20) 5, , + 48 + 41, , 24) 2, 7) 2485, , 7, 18) 9, , 7, 21), 15, , 1, 22), 3, ,   2 8 , 25) 2187 1    ,   3  , , 27) 10, 20 (28) A.P., 200, , 19) 1275, , 32), , 4, 45, ,   2 n , 23) 2 1    ,   3  , 26) 1, 33), , 35, 16

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3. PERMUTIONS AND COMBINATIONS, 4), , EXERCISE 3.1, 1), , 50 ways, , 2), , 12, , 3), , i) 25, , ii) 20, , 4), , i) 100, , ii) 48, , 5), , 5), , i) 28, , ii) 1, , iii) 3003, , iv) 6435, , i) 1848, , ii) 43/14, , iii) 5, , iv) 6, , v) 8, 6), , i) 11, , ii) 11, , 125, , iii) 7, , iv) 8, , 6), , 124, , v) 5, , 7), , 31, , 8), , 90, , 9), , 225, , 10), , 24, , 11), , 276, , 12), , 207, , 13), , 12, , 14), , 216, , 10), , 2), , 3), , i) 40320, , ii) 3628800, , iii) 3628080, , iv) 24, , i) 665280, , ii) 2, , iii) 720, , iv) 12, , v) 84, , vi) 29, , vii) 57.93, , viii) 20160, , i) , , n+3, n-2, iv) (3n +2)n!, , n2 +1, v), (n + 1)!, , (n 2 + 1), vi), (n + 1)!, , vii) 0, , viii), , ii), , 1, (n + 2)!, , EXERCISE 3.3, EXERCISE 3.2, , 1), , i) (2n + 1)(2n + 2), (n  1), iii), n (n  2)!, , 10 !, 4!, , 9!, iii), 5!, , ii) 35 × 5!, iv) 54 × 4!, 201, , 1), , n=9, , 2), , m = 6, n = 2, , 3), , r=6, , 5), , a) 2401, , b) 840, , 6), , a) 30240, , b) 151200, , c) 43200, , d) 5040, , 7), , 12 ! ×13!, 7, , 8), , a) 1440, , b) 720, , c) 7!, , d) 240, , e) 120, , f) 120

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9), , 144, , 10), , a) 1296, , 11), , 100, , 12), , 720, , 13), , 5), , 14 !, 2, 2.8!, , 6), , a) 5! × 2! = 240, , 46800, 20800, , 7), , 7! × 8P6, , 8) 144, , 14), , 243, , 9), , 10) 12.13!, , 15), , i) 2880, , 9!, , 4!, , 16), , i) 120, , 17), , i) 720, , 4), , b) 360, a) 120,, , b) 600, , ii) 5040, ii) 48, ii) 144, , iii) 72, iii) 288, , b) 2400, , EXERCISE 3.6, , iv) 144, , EXERCISE 3.4, , 1), , a) 1365, , 2), , a) n = 2, , b) 3160, , c) 16C5, , d) 19C15, , b) n = 7, , c) n = 9, 1), , i) 120, , ii) 60480, , 3), , r=4, , iii) 30240, , iv) 5040, , 4), , a) n = 10, r = 3, , b) n = 10, r = 4, , 5), , r=8, , 6) 126, , 7), , 39200, , 8) 120, , 9), , 12 , , 10) 190, , 11), , n, , 12), , 190, , 13), , a) 45, , b) 40, , 14), , a) 220, , b) 216, , 15), , 151200, , 16), , i) n = 20, , ii) n = 4, 3, iv) n = r, , v) 302400, 2), , 1260, , 3), , a) 70, , 4), , 13!, 5! 4 ! 4 !, , 5), , 12 !, 2 !3!2 !, , 6), , 11!, 4!2!2!, , a) 405720, , 7), , 210, , 8) 60, , 9), , 10 !, 2 !3!2 !, , 10) 1260, 1230, , iii) n = 1, 2, , 11), , 180, , 12) 144, , v) n = 6, , 13), , 36, 84, , 15), , a) 1800, , b) 37, , b), , 8!, 2!2!, , 14) 180, 60, b) 72, EXERCISE 3.5, , 1), 3), , 7! = 5040, a) 2.23!, , 2) 20!, 2.18!, b) 21.22!, 202, , C2 − n;, , a) 35, , b) 90, , c) 54, , 17), , x = r!, , 18) r = 7, , 19), , 14161, , 20), , a) 2508, , 21), , 16 , , 22) 2275, , 23), , 36873 ; 6885, , 24) 425, , 25), , 51051, , 26), , a) 84, , b) 1646, , b) 126, , c) 5973, , d) 20

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MISCELLANEOUS EXERCISE - 3, (I), 1, C, , 2, A, , 3, B, , 4, D, , 5, C, , 6, B, , 7, D, , 8, B, , 9, C, , 10, D, , (II), 1), , 45 , , 2) 120, , 3), , 720 ; AINMRE, , 4) 990, , 5), , 360 , , 6) 5541965, , 30 !, 7 ! 10 ! 13!, , 7), , 15 , , 8), , 9), , 127, , 10) 9C3 + 9C4 + 9C5, , 11), , 4095, , 12) 48, , 13), , 1680, , 14) 63, , 15), , 20 !, 8! 7 ! 5!, , 16) 896, , 17), , 60, , 18), , i) 66, , ii) 11, , iii) 220, , 4. METHOD OF INDUCTION AND BINOMIAL THEOREM, EXERCISE 4.1, Hints :, , 6) i) 970.299, , ii) 0.6561, , 7) i) 16, , ii) 16, , 8) 1.1262, , 2) p(n) = 3 + 7 + 11 + ... + (4n − 1) = n (2n + 1), , 9) 1.051, , 5) p(n) = 13 + 33 + 53 + ... + (2n − 1)3 =, n2 (2n2 − 1), , 10) 0.5314, , 7) p(n) = 1.3 + 3.5 + 5.7 + ... + (2n −1) (2n + 1), , EXERCISE 4.3, , n, (4n2 + 6n − 1), 3, 1, 1, 1, 1, , ,  ... , , 9) p(n) =, 3.5 5.7 7.9, (2n  1)(2n  3), n, =, 3(2n + 3), =, , v), , 10500, x3, 13, C .34 .49, , 2) i) 16x8 + 96x6 + 216x4 + 216x2 + 81, 60 12 1,  , x2 x4 x6, , iv), , 55a16, 9, , a5, , v), , -105, 8192, , 3) i), , 1792, 9, , 3) i) 32 3, , ii) 1364, , iii) 405, , 5) i) 108243216, , ii) 1.61051, , v) 10500000, 203, , ii) 84480x2, , 9, , iii) 48620, , ii) 145 5 − 229 2, , ii) 64x6 − 192x4 + 240x2 − 160 +, , iii), , 2) i) 122472 2, , EXERCISE 4.2, 1) i) 49 + 20 6, , 1) i) 4032x10, , ii) 700000, 5, iv), 16, , ii) −96096, iv) 84, , iv) 55

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4) i) 924, , ii) 35x5, 35x2, , iii) 1120x4, , 1,  , , 2 x 6 x2, iv) 5 2 1 , ,  ... , 5, 25, , , , iv) −252, , v) −462x9 and 462x2, , 1,  , , x 2x2, v) 5 3 1  ,  ... ,  5 25, , , 5) k = 5, 6) 91854, , 4) i) 9.9499, , ii) 5.0133, , 7) m = 8, , iii) 2.0025, , iv) 0.9057, , v) 1.0625, , EXERCISE 4.4, , MISCELLANEOUS EXERCISE - 4, , 1) i) 1  4 x  10 x  20 x  ..., 2, , 3, , (I), , x x 2 5 x3, - ..., ii) 1 - - 3 9 81, , 1, B, , iii) 1 + 3 x 2 + 6 x 4 + 10 x 6 + ..., x 3 x 2 11x 3, iv) 1  , ,  ..., 5 25 125, ,  3b 6b 2 10b3, , 2) i) a 1   2  3  ..., a, a, a, , ,  4b 10b 2 20b3, , ii) a 4 1   2  3  ..., a, a, a, , , 1, , , b, 3b 2, 7b3, , ,  ..., iii) a 4 1 , 2, 3,  4a 32a 128a, , , iv) a, , v) a, , 1, , 3, , , b, 1 4a, , , 5, A, , 6, D, , 7, D, , 8, D, , 5), , 16 x 4 16 x 2, 9, 81, , 6 2 , 81, 9, x 16 x 4, , 6), , 27 4 6, x y, 2, , 7), , 1760, x3, , 8) i) −20, , ii), , -63 x5, 8 y5, , iii) 280x8y6 and 560x6y8, , , 5b 2, 15b3, +, + ..., 2, 3, 32a 128a, , , 189 6, 21, x and - x 3, 16, 8, 9) i) 378, , ii) 153, , 10) i) 2268, , ii) 7920, , 12) ± 2, , 13) 2, , 9, , 7, x x 2 5 x3, 18) 1 + + +, + ..., 3 6 54, , 15) 2, , iv), , , , b 2b 14b, 1  3a  9a 2  81a 3  ..., , , 2, , 4, A, , 9, B, , 10, D, , 4) 243x10 + 810x8y + 1080x6y2 + 720x4y3 +, 240x2y4 + 32y5, , 3, , 1, 4, , 3, C, , (II), , v) 1  x 2  x 4  x 6  ..., , −, , 2, C, , 3, , 14), , 3) i) 1  8 x  40 x 2  ..., 3 x 27 x 2,  ..., ii) 1  , 2, 8,  x x2 , iii) 2 1   ... ,  2 4 , 1, 3, , 2, 3, 19) 1 + x + 5 x + 15 x + ..., 4 32 128, , 204

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2 x 6 x2, , 20) 5 1 , ,  ... , 5, 25, , , , , 1, 2, , 22) 0.2451, , 23) −80, , 24) a = 3, b = −2, c = 57, , 25) n = 9, , 26) n = 6 ; k = −2, , 21) 9.9833, , 5. SETS AND RELATIONS, 13) 9, , EXERCISE 5.1, , 14) i) (−8, 6], , 1) i) A = {M, O, V, E, N, T}, , ii) (−∞, −4) ∪ (5,∞), , 1 1 ,  20 , , iii) (−∞, 4) ∪ , ∞ iv)  , , 3 2,  3, , , ii) B {−1, 0, 1, 2, 3, 4}, iii) C = {3, 5, 7, ...}, , 15) i) (−7, 6], , ii) [2, 9], , iii) (−7, 3] ∪ [4, 9], , iv) [2, 3], , ii) {x | −3 ≤ x ≤ 3, x ∈ Z}, n, iii) {x | x = 2, , n ∈ N and n ≤ 7}, n +1, , v) [4, 6], , vi) { }, , vii) (3, 6], , viii) (−∞, 2) ∪ (9, ∞), , iv) {x | x = (−1)n−1 × (n−1), n ∈ N}, , ix) [2, 4), , iv) (−7, 2), , 2) i) {x | x ∈ W, x ∉ N}, , 3) A∪B∪C = {, , -1 3, -5, , -1, , , 3}, 3, 2 2, , EXERCISE 5.2, , 4) A∩B∩C = { }, 6) i) 45, , ii) 10, , 7) i) 132, , ii) 63, , 8) i) 1750, , ii) 250, , iii) 10, , iv) 25, , 1) x = 2 , y = −2, 2) x =, , iii) 1100, , 3) A×B = {(a,x), (b,x), (c,x), (a,y), (b,y), (c,y)}, , 9) 42, 10) i) 114, , 15, 1, ,y=, 2, 6, , ii) 38, , B×A = {(x,a), (x,b), (x,c), (y,a), (y,b), (y,c), , iii) 188, , A×A = {(a,a), (a,b), (a,c), (b,a), (b,b), (b,c),, (c,a) (c,b), (c,c)}, , 11) P(A) = {φ, {1}, {2}, {3}, {1,2}, {2,3},{1,3},, {1,2,3}}, , B×B = {(x,x), (x,y), (y,x), (y,y)}, , 12) i) {x | x ∈ R, −3<x<0}, , 4) P×Q = {(1,1), (1,4), (2,1), (2,4),(3,1), (3,4)}, , ii) {x | x ∈ R, 6≤x≤12}, , Q×P = {(1,1), (1,2), (1,3), (4,1),(4,2), (4,3)}, , iii) {x | x ∈ R, x<6}, , 6) {(0,10), (6,8), (8,6), (10,0)}, , iv) {x | x ∈ R, x<5}, iv) {x | x ∈ R, 2<x≤5}, iv) {x | x ∈ R, −3≤x<4}, 205

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8) i) R1 = {(2,4), (3,9), (5,25), (7,49), (11,121),, (13,169)}, , MISCELLANEOUS EXERCISE - 5, , Domain R1 = {2,3,5,7,11,13}, , I), , Range R1 = {4,9,25,49,121,169}, ii) R2 = {(1,1), (2,, , 1, 1, 1, 1, ), (3, ), (4, ), (5, ), 2, 3, 4, 5, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , C, , D, , D, , C, , A, , D, , C, , D, , D, , D, , II), , Domain R2 = {1,2,3,4,5}, , 1) i) A = {x / x = 10n, n ∈ N, n ≤ 5}, , 1 1 1 1, Range R2 = {1, , , , }, 2 3 4 5, , ii) B = {x / x is the vowel of English alphabet}, iii) C ={x / x is a day of a week}, , iii) R3 = {(1,3), (2,6), (3,9)}, , 2) i) {1,2,4,6,7,9,11}, ii) { }, iii) {1,10}, iv) {2,4,6,7,11}, v) {1,2,3,4,5,6,7,8,9,10,11,12}, , Domain R3 = {1,2,3}, Range R3 = {3,6,9}, , vi) {4,7}, , iv) R4 = {(1,4), (1,6), (2,4), (2,6)}, , 3) 230, , Domain R4 = {1,2}, , 4) 12, , (1,1), (1, 2), (1, 3), (2,1), (2, 2), , 5) i) A  A  , , (2, 3), (3,1), (3, 2), (3, 3), , , Range R4 = {4,6}, v) R5 = {(0,3), (1,2), (2,1), (3,0)}, , A×B = {(1,2),(1,4),(2,2),(2,4),(3,2),(3,4)}, , Domain R5 = {0,1,2,3}, , B×A = {(2,1),(2,2),(2,3),(4,1),(4,2),(4,4)}, , Range R5 = {3,2,1,0}, , B×B = {(2,2),(2,4),(4,2),(4,4)}, (A×B)∩(B×A) = {(2,2)}, , vi) R6 = {(1,4), (2,4), (3,4), (4,4), (5,4)}, , ii), , (1, 1, 1), (1, 1,1), (1,1, 1, , , , A  A  A  (1,1,1), (1, 1, 1), (1, 1,1),, , (1,1, 1), (1,1,1), , , , , Domain R6 = {1,2,3,4,5}, Range R6 = {4}, vii) R7 = {(1,5), (2,4), (3,3), (4,2), (5,1)}, , 6) i) Yes; D = {1}, R = {4,5,6}, , Domain R7 = {1,2,3,4,5}, , ii) Yes; D = {1,2,3}, R = {4,5,6}, , Range R7 = {5,4,3,2,1}, , iii) Yes; D = {1,2,3}, R = {4,5,6}, iv) No., , viii) R8 = {(1,3), (2,4), (3,5), (4,6)}, , 7., , Domain R8 = {1,2,3,4}, , i) D = {1,2,3,4}, R = {4}, ii) D = {-2,-1,0,1,2}, R = {0,1,2,3}, , Range R8 = {3,4,5,6}, , 8., , i) { }, , ii) A×A, , 9) reflexive, not symmetric and not transitive., 10) Yes, 206

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6. FUNCTION, e) injective and surjective, 3, 16), 16, , EXERCISE 6.1, 1) a) Yes, c) No, , b) No, , 2) a) No, , b) Yes, , c) No, , d) Yes, , 3) a) Yes, , b) No, , c) Yes, , d) Yes, , 17) a) 5 = log2 32, , 3, = log9 27, 2, , c) 1 = log23 23, , d), , 1, e) −4 = log3  ,  81 , g) ln 7.3890 = 2, , f) −2 = log10 0.01, 1, h) ln 1.6487 =, 2, , i) ln 6 = −x, , e) No, 4) a) 1, , b) 0 = log54 1, , b)19 c) -, , e) x2 + 3x + 1, 6, 5) a), 5, , 1 2, c) , 2 3, , b) ± 3, , 6) a) 0, ± 3, , b), , 1, d) x2 − x − l, 4, f) h + 1, , 1, = 5-2 c) 0.001 = 10−3, 25, , 18) a) 26 = 64 b), 1, d) 8 =  , 2, , d) 1,−2,3, , 3, , e) e0 = 1 f) e1 = e g), , 19) a) (5,∞), , 17 ± 33, 2, , b) (−∞,2) ∪ (3,∞), , 20) a) log p + log q − log r − log s, 1, 1, log x +, log y, 2, 3, , 7) 1) a = −2, b = 2, , b), ,  11 , 8) a) R;   ,   b) R − {2}; R − {1},  7, , , c) 3 ln a + 2 ln (a − 2)−, , c) (−5, ∞); R+, , d) R; R, , f) [3,7) ; [0,∞], , g) [−4,4] ; [0,4], , e) [2,5] ; [0,, , 3, ], 2, , 1, , 1, ,  x5 y 7 , 21) a) log , ,  z , , p, 9) a) A = s2 b) A =, 16, 10) a) A = pr2, , 1, ln (b2 + 5), 2, , , , d) 2  ln( x  2)  4ln(2 x  1)  ln( x  4)  ln(2 x  4) , 2, 3, , , 2, , pd 2, b) A = 4, , 1, = e−0.693, 2, , b) log, , (, , 3, , x−1 x, , ), ,  x2  4 , c) ln , 3,  ( x  5) , , c2, c) A = 4p, , 11) x(30 − 2x)2 ; (0,15), , 22), , 12) Not a function; f (0) has 2 values., 13) a) Injective but not surjective, , 5a + b, 2, , 25) a) 3, , b) neither injective nor surjective, c) neither injective nor surjective, d) injective but not surjective, 207, , 24) a =, b) 11,, , c) 8, , 15, ,b=9, 4, d) 1

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B, , EXERCISE 6.2, , B, , B, , C, , C, , A, , A, , B, , C, , B, , (II), 1), , a) 9x +4, d), , b) 0, , 1), , {(2,4), (4,2), (5,4)}, , 3), , a) 50x2 − 40x +11, c) 8x4 + 24x2 + 21, d) 25x − 12, , ii) Not a function, iii) Function ; {2,3,5}; {1,2}, 2), , i) not one one, , ii), , one one, , 3), , ii), , not onto, , 4), , i) not onto, 1, x  8 3, , −1, f (x) = , ,  5 , , b) f−1 doesn not exist, 3x + 7, c) f−1 (x) =, 6, , 5), , f−1 (x) =, , 6), , 1,−3, does not exist, , d) f−1 does not exist, , 7), , i) 2, , x -8, 9, f) f−1 does not exist, , 8), , 3x4 − 12x3 + 13x2 − 2x + 5, , 9), , a = 4, f(4) = 16, , b) 10x2 + 13, , a) f−1 does not exist, , e) f−1 =, , 3, , a) 22, , b) 7, , c) 3, , 7), , a) −18, , b) −14, , c) 5, , d) 25, , 8), , a) 10, , b) −5, , 9), , a) 25, , b) −3, , c) −15, , d) 21, , b) 1.75, , c) −4.4., , d) −30, , 10) a) −5, , 12) i) f°g = x2 − 16x +69, gof = x2 − 3, ii) f°g = 3x2 − 2, gog = 9x2 − 12x +4, iii) f°g = 256x2 , gof = 16x2, 15) f ≠ g, 19) 8 , , c) { }, , d) [−3,3], , f) 3 + r ; 0≤r<1, , g) { }, , h) N, Z, , i) n + 0.5 , n ∈Z, , 26), , MISCELLANEOUS EXERCISE - 6, , 5, , 6, , 7, , 8, , 9, , 3 1, ,, 2 2, , 37) −8, , 39) a) (−3,0)∪(2,5), , b) {−2,2,4}, , e) [, , 10, 208, , 33) 2, , 34) 3 , c) [−3,−2]∪[2,3], , (I), 4, , 22) log4, , 23) log105, , j) x = 0, , 3, , ii) 0, , 11) i) g°f = {(1,6),(2,8),(3,10),(4,12)}, ii) g°f = {(1,1),(2,64),(3,64),(4,27)}, , b) 1.5, 4.5, , 11) a) (−∞,−9], [1,∞), , 2, , 5( x - 2), 3, , 10) a = 3, b = −2, , 6), , 1, , i) Function ; {2,4,6,8,10,12,14};, , {1,2,3,4,5,6,7}, , 1 , 3x  5, ; R −  , 6, 6x 1, , 2), , 5), , c)238, , 13, ,7), 5, , d) (−2,8), f) [, , 9, ,5), 2

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g) x = 0, , e) (1,∞), , h) x = 6k , k∈Z, , 42) a) fog(x) = x = gof(x), , 40) a) R−{2,−3}, , b) fog(x) = x = gof (x), , b) [3,4) ∪ (4,5), c) [−1,1], , d) W, , e) {1,2,3}, , f) [0,1], , 43) a) f(x) = 2x − 3 or − 2x + 2, b) f(x) = x2 + 2, , g) (−∞,3− 3 ) ∪ (3+ 3 ,∞), , x, , 44) a), , 41) a) [0,∞), c) (0,1], , b) x, , 1 + 2x2, , 1 1, b) [ - -, ], 6 6, d) (−1,0], 7. LIMITS, EXERCISE 7.1, , I), , 1) –, , 1, 3, , 2) 15, , II) 1), , 2 3, 3, , III) 1), , n(n +1), 2, , 5) –, , 2) –, , 3) –, , 1, 25, , III) 1), , 3, 3, 3), 125, 16, 2, , 2), , 4) ±, , 3) 4, , 33 7, , 1, 6, , 6) 24, , 8) 294 7, , II) 1), , 7), , 1, 3) –, 2, , 4) 4, , 2) 0, , 3) 0, , 2) –2, , 3), , 4) 0, , 5) –, , 2) 2, , II) 1), , n2, m2, , 2) –, , III) 1), , 1, a 2 - b2, 2) –, 2, 4 2, c, , I), , 3, a2, , 1, 2, 1, 4), 3, , 1), , 1, 2 6, , 2) –, , 1, 18, , II) 1) –, 4), , 3) –1, , 4) –, , 3) 2, 1, 4, , 3), , 4), 209, , 4, , 1), , EXERCISE 7.3, I), , 3) 24, , m, n, , 5) 8, , 4) 2x – 2 5) –3, , 1, 2, , 2) 1, , 2, 1, 5) –, 2a, 3, 1, 3, , 5), , 4), , 1, 2, , 1, 2, , 3) 2 2 4) –3, , EXERCISE 7.5, , 6) Does not exist, III) 1) 3, , 7, 2, , 4), , 8 3, , 1), , I), , 3 a+2, 2, , 1, 4) –, 2, , 3 3, , 1, , EXERCISE 7.4, , 9) n2, , 1, 2) –, 2, , 2) –8 3), , 2, 3, , EXERCISE 7.2, 1, I) 1) –, 4, 4, II) 1), 3, , 2, , 2) 5a 5 .cos a, , 1, 2 3, , cos a, 2 a, , 5), , 2, π, , 2), , 1, 16 2, , 5) –, , 1, 2, , 3), , 1, 8, , 3), , 1, 36, , 1, 18

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EXERCISE 7.6, , I), , I), , 9, log  , 5, 1), log 4, 4) log (, , 2) log, , 40, ), 9, , II) 1) (log 3), , 15, 2, , 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15, C B A D C C C C A D B D C B B, , 3) log(abc), , II), , 5) log 2., 2) e, , 2, , 2, 3, , 6) e, , 5) e8, III) 1), , MISCELLANEOUS EXERCISE - 7, , 1, a, log, 2, b, , 4) (log 5)2, , 3) e, , 1), , 2, 4) –, 3, , 14/3, , 9), , 3, , (log 2), log 3, , 5), , 1   7 , log  , 2   5  , , 3) log 3.log 5, 2, , 1), , a, e, , 2) 1, , II) 1) 7, , 2) 2, , III) 1) 15, , 2), , 256, 81, , 3), , 8) 1, , (log 5) 2, 14), log 2, , (2) 2 (7)3 1372, =, 15), 16) a sin a + cos a, (5)5, 3125, 17) 2 2 18) (log 2)2 19) (log 2)2 20), , 7, 8, , 21) Does not exist, , 3) 2, 3), , 1, 2, , 1, –1, log2 10) 2(log a)2 11) cos a (12), 2, 10, , ab, b, 13), log, 2, a, , EXERCISE 7.8, I), , 2) Does not exist 3) 2πr, , 4) Does not exist 5) 3 6) 21 7), , 2, 3, , 2), , 5, 3, ,  3  30, 1, 4)   5) 4, 2, 2, , 23), , 1, , 256, , 22), , 1, 2, , n(n  1)(4n  1), 6, , 24) 2, , 8. CONTINUITY, EXERCISE 8.1, 1), , 2), , (i) Continuous at x = - 2 (ii) Continuous, p, at x =, (iii) Discontinuous at x = 3, 4, (i) Discontinuous., , (ii) Continuous., , (iii) Continuous, 3), , Discontinuous at x = – 2, x = -1, x = 0,, x = 1., , 4), , Continuous., , 5), , (i) Discontinuous., (ii) Continuous., (iii) Continuous, (iv) Continuous, (v) Discontinuous at x = 2, , 6), , (i) Removable , (iii) Jump , , 7), , (i) Extension = 0 i.e. f(0) = 0, , (ii) Jump, (iv) Removable, , 7, 2, 2, (iii) Extension = -2/3, i.e. f(−1) = −, 3, (i) Discontinuous (ii) Discontinuous, (iii) Continuous, (ii) Extension = 7/2, i.e. f(0) =, , 8), 210

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9), , 10), , (i) Removable, f(0) = 3/2, (ii) Removable, f(0) = 5/3, (iii) Removable, f(0) = e−2, (iv) Irremovable, (v) Irremovable, (i) −, , 1, , (ii) −, , 4 3, , 4, 3, , (3) Discontinuous at x = 0, (4) Continuous at x = 1, (5) Discontinuous, (6) Discontinuous at x = −1, 0, 1., (7) Continuous on R, except at x = 5., (iii) 4(log2)2, , (III) (1) Removable., (2) Jump, , 3, 11) (i), (ii) (log5)2 (iii) a = -18/5, b = 7., 2, (iv) a = 2, b = -4., (v) a = 1/2 and b = 1/2, , (3) Continuous., (IV) (1) Removable, , f(x) =, , 12), , Continuous, , 13), , Continuous, [Clue : (sin x + cos x)3 = [(sin x + cos x)2]3/2, , ( x  3)( x 2  6 x  8), x 2  x  12, , = −5 for x = −3, , = 2, for x = 4, (2) Irremovable, , = (1 + sin 2x)3/2, Let (1 + sin 2x) = t], (V), , 14) p = −3 and q = 4, , (1) e6, , (2) 125, , (VI) (1) a = 2 b = 4, 4, 27, (2) a = - , b =, 5, 5, , MISCELLANEOUS EXERCISE - 8, (I), 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , A, , D, , D, , B, , A, , B, , A, , B, , C, , C, , (II), , , , (VII) (1) f (1) =, , p, 2, , 49, (2) f (p) = 10, , (VIII) (1) f(1) < 0 and f(2)>0, , (1) Continuous on its domain except at, x=5, , (2) f(z) = 0; f(3)<0 and f(4)>0, , (2) Continuous on its domain except at, x=5, 9. DIFFERENTIATION, EXERCISE 9.1, 1), , 5), , (a) 2x + 3 b) 3 cos(3x) c) 2e2x+1 d) 3x log 3, , ii) Continuous and differentiable, , 2, f) 2 sec2 (2x + 3), 2x + 5, g) 5 sec (5x – 2) tan (5x – 2) h) 3 x, 2, a) 1 b) 2 c) 384 log 2 d) 2, 5, 3, 1, e) 3e2 f), 2, e), , 2), , i) Continuous and differentiable, , 211, , 6), , Neither continuous nor differentiable at x =2, , 7), , Continuous but not differentiable, , 8), , Continuous but not differentiable, , 9), , Continuous and differentiable

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EXERCISE 9.2, , (IV) (1) -, , 4 13, x  e x  cos x, 3, , (I) (1), , 1, , (2), ,  sec x  3 x, 2, , 2 x, , (4), , 7 x log 7  7 x 6  x , , (6), , 3 3, 3 cos ec x  5e   7, x, x4, , (4), , x + (log x) 2, ( x + log x) 2, , (V) (1), (2), , (2), , x 2 (1 + 3 log x), , (3), , ( x 2 + 2)[( x 2 + 2) cos x + 4 x sin x], , (4), , 1, , e x   log x , x, , , x, , , , x 5, , , , 2, , x 2 (1 + sin x + x cos x) + x sin 2 x, ( x + cos x) 2, 2e x, (3e x - 2) 2, f ( x)  5 x 2  18 x  3, f ( x)  ( 3  1) sin x  ( 3  1) cos x, , MISCELLANEOUS EXERCISE - 9, I., , 3, xe [1 + x log x + log x], 2, 2, 3 x (1 + 3 log x), , (1) C , , (2) D, , (3) C, , (4) B, , (5) B , , (6) D, , (7) A, , (8) C, , x, , (6), , II. (1), , (III), (1), , ex ( x2 + ex ), ( x + e x )2, , (6), , x 4 ( x sec 2 x + 5 tan x), , (5), , (3), , x, , 2, , (II) (1), , , , (5), , 1, x, , (5), , 5, , (2), 2, , 1, 9, + cos ecx.cot x + 5 x log 5 + 5, x, 2x 2, 7 43 4, 2, x + 1+ 7, 3, x5 x5, , (3), , 16 x, ( x 2 - 5) 2, , 3, 2, , 5, x + x 3 (1 + 4 log x), 2, 2, , Differentiable , , (2), , p = -3 ,, , (3), , p = 1/3 ,, , (4), , p =π /2 ,, , q = 5, q = -4/3, , q= (2 - π ) / 2, , (2) e x sec x(tan x  1)  x 3 (1  5 log x), 3, 3, 3, (3) 4 x3  x 2 sin x , x cos x  xe x ( x  2), 2, , (5) Not Differentiable, , (4), , ( x3  2) sec 2 x  3 x 2 tan x  x sin x  cos.x  x 6 .7 x (7  x log 7), , (5), , sin x,  1 2x ,  cos x log x  e x ( sin x  cos x)  e x , , x,  2 x , , , 2 x, , v, , Not Differentiable, , (7), , Not Differentiable. , , (8), , Differentiable , , (9), , 2. Hint : Add and subtract 2f(2) in, numerator., (10) e, 2, , (6) e x (sec 2 x  tan x)  cos x  sin x log x  5x  2 x log 5  1 , x, , (6), , , , v, 212, , v