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MATHEMATICS, , Visit https://telegram.me/booksforcbse for more books., , MTG Learning Media (P) Ltd., New Delhi | Gurugram

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Price : ` 200, Edition, , : 2022, , Published by, , : MTG Learning Media (P) Ltd., New Delhi, , Corporate Office : Plot 99, Sector 44 Institutional Area, Gurugram, Haryana-122 003, Phone : 0124 - 6601200 Web: mtg.in Email: [email protected], Registered Office : 406, Taj Apt., Ring Road, Near Safdarjung Hospital, New Delhi-110 029, , Information contained in this book has been obtained by mtg, from sources believed to be reliable. Every effort has been made to, avoid errors or omissions in this book. In spite of this, some errors might have crept in. Any mistakes, error or discrepancy noted may, be brought to our notice which shall be taken care of in the next edition. It is notified that neither the publishers nor the author or seller, will be responsible for any damage or loss of action to anyone, of any kind, in any manner, therefrom., , © MTG Learning Media (P) Ltd. Copyright reserved. No part of this publication may be reproduced, stored in a retrieval system or, transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission, of the Publisher., All disputes subject to Delhi jurisdiction only., Visit www.mtg.in for buying books online.

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SYLLABUS, MATHEMATICS (Code No. 041), COURSE STRUCTURE, CLASS XII (2021 - 22), TERM - II, Time : 2 Hours, , Max Marks : 40, , S. No., , Units, , Marks, , III., , Calculus, , 18, , IV., , Vectors and Three Dimensional Geometry, , 14, , VI., , Probability, , 8, Total, , 40, , Internal Assessment, , 10, , Total, , 50, , Unit-III : Calculus, , dy/dx = f(y/x). Solutions of linear differential equation of the type:, , 1. Integrals, , dy/dx+py = q,where p and q are functions of x or constant., , Integration as inverse process of differentiation. Integration of a, , Unit-IV : Vectors and Three Dimensional Geometry, , variety of functions by substitution, by partial fractions and by, , 4. Vector Algebra, , parts, Evaluation of simple integrals of the following types and, , Fundamental Theorem of Calculus (without proof).Basic properties, , Vectors and scalars, magnitude and direction of a vector. Direction, cosines and direction ratios of a vector. Types of vectors (equal,, unit, zero, parallel and collinear vectors), position vector of a point,, negative of a vector, components of a vector, addition of vectors,, multiplication of a vector by a scalar, position vector of a point, dividing a line segment in a given ratio. Definition, Geometrical, Interpretation, properties and application of scalar (dot) product, of vectors, vector (cross) product of vectors., , of definite integrals and evaluation of definite integrals., , 5. Three Dimensional Geometry, , 2. Applications of Integrals, , Direction cosines and direction ratios of a line joining two points., Cartesian equation and vector equation of a line, coplanar and, skew lines, shortest distance between two lines. Cartesian and, vector equation of a plane. Distance of a point from a plane., , problems based on them., dx, , ∫ x 2 ± a2 , ∫, px + q, , dx, 2, , x ±a, , ∫ ax 2 + bx + c dx, ∫, , 2, , ,∫, , dx, 2, , a −x, , 2, , px + q, 2, , ax + bx + c, , ,∫, , 2, , dx, , ax + bx + c, , dx, , ,∫, , dx , ∫ a2 ± x 2 dx ,, , 2, , ax + bx + c, , ∫, , ,, , x 2 − a 2 dx, , Applications in finding the area under simple curves, especially, lines, parabolas; area of circles /ellipses (in standard form only) (the, region should be clearly identifiable)., 3. Differential Equations, , Unit-VI : Probability, , Definition, order and degree, general and particular solutions, , 6. Probability, , of a differential equation. Solution of differential equations by, method of separation of variables, solutions of homogeneous, differential equations of first order and first degree of the type:, , Conditional probability, multiplication theorem on probability,, independent events, total probability, Bayes’ theorem, Random, variable and its probability distribution., , Internal Assessment, , 10 Marks, , Periodic Test, , 5 Marks, , Mathematics Activities: Activity file record +Term end assessment of one activity & Viva, , 5 Marks

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CONTENTS, Chapter 1, , Integrals, , Chapter 2, , Applications of Integrals, , 22-40, , Chapter 3, , Differential Equations, , 41-58, , Chapter 4, , Vector Algebra, , 59-77, , Chapter 5, , Three Dimensional Geometry, , 78-97, , Chapter 6, , Probability, , 1-21, , 98-116, , Practice Papers 1-3, , 117-136, , *As per the Circular Issued by CBSE on July 05, 2021, Special Scheme of Assessment, , for Board Examination Class XII for the Session 2021-22 is as follows :, Term II Examination/Year-end Examination :, • At the end of the second term, the Board would organize Term II or Year-end Examination based on the rationalized, syllabus of Term II only (i.e. approximately 50% of the entire syllabus)., •, , This examination would be held around March-April 2022 at the examination centres fixed by the Board., , • The paper will be of 2 hours duration and have questions of different formats (case-based/ situation based, open, ended- short answer/ long answer type)., • In case the situation is not conducive for normal descriptive examination a 90 minute MCQ based exam will be, conducted at the end of the Term II also., •, , Marks of the Term II Examination would contribute to the final overall score., , To cope up with ongoing unpredictable pandemic situation, this book contains chapterwise objective as well as subjective, questions., In Objective Section, each question carry 1 mark and in Subjective Section, each VSA carry 1 mark, SA I carry 2 marks,, SA II carry 3 marks and LA carry 5 marks., *As per the CBSE Term-II 2021-2022 curriculum, for latest information visit www.cbse.gov.in

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CHAPTER, , 1, , Integrals, Recap Notes, , INDEFINITE INTEGRAL, h, , Integration is the inverse process of differentiation., i.e., d F(x) = f (x) ⇒ ∫ f (x) dx = F(x) + C ,, dx, where C is the constant of integration., Integrals are also known as antiderivatives., , Some Standard Integrals, h, , ∫ dx = x + C ,, , where ‘ C ’ is the constant of, , integration, n, , xn+1, + C , where n ≠ –1, n+1, , ∫ cot x dx = log|sin x| + C, ∫ sec x dx = log|sec x + tan x| + C, , h, h, , x π, = log tan  +  + C, 2 4, , , , ∫ cosec x dx, , h, , , , = log|cosec x – cot x| + C, x, = log tan + C, 2, , Properties of Indefinite Integral, (i), , ∫ f ′(x)dx = f (x) + C, , (ii), , ∫ f (x)dx = ∫ g(x)dx + C ,, , h, , ∫x, , h, , ∫e, , h, , ∫a, , h, , ∫ x dx = log e |x|+ C , where x ≠ 0, , h, , ∫ sinx dx = − cos x + C, , h, , ∫ cosx dx = sin x + C, , h, , ∫ sec, , h, , ∫ cosec, , h, , ∫ sec x tan x dx = sec x + C, , ∫ f (ax + b)dx, , ax + b = t, , h, , ∫ cosec x cot x dx = −cosec x + C, , ∫ f ( g(x))g′(x)dx, , g(x) = t, , h, , ∫, , = sin −1 x + C = – cos–1x + C,, 1−x, where |x| < 1, dx, –1, –1, ∫ 1 + x 2 = tan x + C = – cot x + C, 1, –1, –1, ∫ 2 dx = sec x + C = – cosec x + C,, x x −1, where |x| > 1, , ∫, , , h, h, , , h, , x, x, , dx =, , dx = e x + C, dx =, , 1, , 2, , ax, + C , where a > 0, log e a, , (iii) ∫ [ f (x) + g(x)] dx = ∫ f (x) dx + ∫ g(x) dx, (iv), , ∫ k ⋅ f (x) dx = k∫ f (x) dx, k being any real number., , METHODS OF INTEGRATION, Integration by Substitution, The given integral ∫ f (x) dx can be transformed, into another form by changing the independent, variable x to t by substituting x = g(t)., , h, , x dx = tan x + C, 2, , x dx = −cot x + C, , dx, , 2, , ∫ tan xdx = log|sec x| + C = –log |cos x| + C, , f and g are indefinite integrals, , with the same derivative., , Integrals, , f ′(x), dx, f (x), n, , ∫ ( f (x)), or, , ∫, , f(x) = t, , f ′(x)dx, , ∫ (px + q), , Substitution, , cx + d dx, , px + q, dx, cx + d, , f(x) = t, px + q = A(cx + d) + B., Find A and B by equating, coefficients of like powers, of x on both sides.

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CBSE Board Term-II Mathematics Class-12, , 2, 1, , ∫ (px + q), , cx + d, 1, , ∫ (px 2 + qx + r ), ∫, , ∫, , cx + d, , 1, 2, , ( px + q) cx + dx + e, , 1, 2, , 2, , ( px + q) cx + d, px + q, , ∫ ax 2 + bx + c dx, ∫, , px + q, ax 2 + bx + c, , ∫ (px + q), , Integrals of Some Particular Functions, , dx or cx + d = t2, , dx, , or, , px + q =, x=, , 1, and then c + dt2 = u2, t, , (px + q), =A, , dx, , 1, t, , d, (ax 2 + bx + c ) + B, dx, , or, , ax 2 + bx + c dx, , Integration using Trigonometric Identities, h, , When the integrand consists of trigonometric, functions, we use known identities to convert it into, a form which can be easily integrated. Some of the, identities useful for this purpose are given below :, 2 x, (i) 2 sin   = (1 − cos x), 2, x, (ii) 2 cos 2   = (1 + cos x), 2, , h, , (iii) 2 sin x cos y = sin (x + y) + sin (x – y), (iv) 2 cos x sin y = sin (x + y) – sin (x – y), (v) 2 cos x cos y = cos (x + y) + cos (x – y), (vi) 2 sin x sin y = cos (x – y) – cos (x + y), Some Special Substitutions, , Expression, , Substitution, x = a sinq or a cosq, , a2 − x 2, a2 + x 2 or (a2 + x 2 ), , x = a tanq or a cotq, x = a secq or a cosecq, , x 2 − a2, a−x, or, a+x, , a+x, a−x, , x, or, a−x, , a−x, x, , x, or, a+x, , a+x, x, , a−x, x−b, or, x−b, a−x, or (a − x)(x − b), , x = a cos2q, x = a sin2q or a cos2q, x = a tan2q or a cot2q, , 1, , a+x, , ∫ a2 − x 2 dx = 2a log a − x, , (ii), , ∫, , (iii), , ∫, , (iv), , ∫, , (v), , ∫ x 2 − a2 dx = 2a log x + a + C, , (vi), , ∫ x 2 + a2 dx = a tan, , dx, , dx, , 1, , (i), , 1, 2, , a −x, , x, dx = sin −1   + C, a, , 2, , 1, 2, , x − a2, 1, 2, , x + a2, , +C, , dx = log x + x 2 − a 2 + C, dx = log x + x 2 + a 2 + C, , 1, , 1, , 1, , 1, , x−a, , −1, , x, +C, a, , Integration by Partial Fractions, h, , If f ( x ) and g ( x ) are two polynomials such that, deg f(x) ≥ deg g(x), then we divide f(x) by g(x)., \, , h, , f (x), Remainder, = Quotient +, g(x), g(x), , If f(x) and g(x) are two polynomials such that the, degree of f(x) is less than the degree of g(x), then, f (x), dx by decomposing f (x), we can evaluate ∫, g(x), g(x), into partial fraction., Form of the Rational, Function, px + q, ,a ≠ b, (x − a)(x − b), , 2, , x = a cos q + b sin q, , A, B, +, x−a x−b, , px + q, , A, B, +, x − a (x − a)2, , px + q, , A, B, C, +, +, x − a (x − a)2 (x − a)3, , (x − a)2, , 3, , (x − a), , px 2 + qx + r, (x − a)(x − b)(x − c ), px 2 + qx + r, 2, , (x − a) (x − b), px 2 + qx + r, , 2, , Form of the Partial, Fraction, , 2, , (x − a)(x + bx + c ), where x2 + bx + c can not, be factorised further, , A, B, C, +, +, x−a x−b x−c, A, B, C, +, +, 2, x − a (x − a) (x − b), A, Bx + C, +, x − a x 2 + bx + c

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3, , Integrals, , Integration by Parts, h If u and v are two differentiable functions of x, then, , h, , b, , du, , ⋅ vdx  dx .,  dx ∫, , , , ∫ (uv) dx = u ⋅ ∫ vdx  − ∫ , st, , In order to choose 1 function, we take the letter, which comes first in the word ILATE., I – Inverse Trigonometric Function, L – Logarithmic Function, A – Algebraic Function, T – Trigonometric Function, E – Exponential Function, h, , x, , ( f (x) + f ′(x))dx = e x f (x) + C, , INTEGRALS OF SOME MORE TYPES, (i), (ii), (iii), , ∫, , a2, x 2, x, a2 − x 2 dx =, a − x 2 + sin −1   + C, a, 2, 2, , ∫, , x 2 − a 2 dx =, , x 2 2 a, 2, 2, x − a − log x + x − a + C, 2, 2, , ∫, , x 2 + a 2 dx =, , x 2, a2, x + a 2 + log x + x 2 + a 2 + C, 2, 2, , 2, , DEFINITE INTEGRAL, h, , F(x) be an integral of f(x), then, , Let F(x) be integral of f(x), then for any two values of, the independent variable x, say a and b, the difference, F(b) – F(a) is called the definite integral of f(x) from, , h, , ∫ f (x)dx ., a, , Here, x = a is the lower limit and x = b is the upper, limit of the integral., , SOME PROPERTIES OF DEFINITE INTEGRALS, b, , First Fundamental Theorem : Let f(x) be a continuous, function in the closed interval [a, b] and let A(x) be, the area function. Then A′(x) = f(x), for all x ∈ [a, b]., , b, , (i), , ∫a f (x)dx = ∫a f (t)dt, , (ii), , ∫a f (x)dx = −∫b f (x)dx, , b, , a, , In Particular, b, , a, , ∫a f (x)dx = 0, , c, , b, , (iii), , ∫a f (x)dx =∫a f (x)dx +∫c, , (iv), , ∫a f (x)dx = ∫a f (a + b − x)dx, , (v), , ∫0 f (x)dx = ∫0 f (a − x)dx, , (vi), , ∫−a f (x)dx =  2, , (vii), , FUNDAMENTAL THEOREM OF CALCULUS, h, , a, , When definite integral is to be found by substitution,, change the lower and upper limits of integration. If, substitution is t = f(x) and lower limit of integration, is a and upper limit is b, then new lower and upper, limits will be f(a) and f(b) respectively., , b, , a to b and is denoted by, , ∫ f (x)dx = F(b) − F(a), , EVALUATION OF DEFINITE INTEGRAL BY, SUBSTITUTION, , Integral of the type, , ∫e, , Second Fundamental Theorem : Let f ( x ) be a, continuous function in the closed interval [a, b] and, , (viii), , b, , b, , a, , a, , 0, , , a, , , , 2a, , ∫0, , 2a, , ∫, , 0, , a, , ∫0, , f (x)dx , where a < c < b, , , if f (−x) = − f (x), f (x)dx , if f (−x) = f (x), , a, , a, , 0, , 0, , f (x)dx = ∫ f (x)dx + ∫ f (2 a − x)dx, ,  a, 2 f (x)dx , if f (2 a − x) = f (x), f (x)dx =  ∫, 0, , if f (2 a − x) = − f (x), 0,

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Practice Time, OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1., , Evaluate :, , ∫ (3 sin x − 2 cos x + 4 sec, , (a), (b), (c), (d), , 2, , 2, , x − 5 cosec x ) dx, , –3 cosx – 2 sinx + 4 tanx + 5 cotx + C, 3 cosx + 2 sinx + 4 tanx + 5 cotx + C, –3 cosx + 2 sinx – 4 tanx – 5 cotx + C, – 3 cosx – 2 sinx – 4 tanx – 5 cotx + C, , x, −x 2, Evaluate : ∫ (2 + 2 ) dx, 1, (a), (22x − 2−2x ) + C, 2 log 2, 1, (22x − 2−2x ) + 2x + C, (b), 2 log 2, 1, (22x + 2−2x ) + 2x + C, (c), 2 log 2, , 7., , 2., , 4., , Evaluate :, , (a), , 1,  17 , log  , 5, 2, , ∫, , 2x, , 2, , +1, , dx, ,  17 , (c) log  , 5, 5., , 2, , x, ∫ xe dx is equal to, 2, , ex, +C, (a) −, 2, ex, (c), +C, 2, 6., , (b), , −2, +C, x, x, cos − sin, 2, 2, 2, (d), +C, x, x, cos − sin, 2, 2, , 4, , ∫, , (b), , (x 2 + x ), 2x + 1, , (a) 57 − 5 5, , 1, (22x + 2−2x ) + C, 2 log 2, sin2 x − cos2 x, dx ., 3. Find the value of ∫, sin2 x cos2 x, (a) tanx – cotx + C, (b) –tanx + cotx + C, (c) tanx + cotx + C, (d) –tanx – cotx +C, x, , Evaluate:, , 2, , (d), , 4, , 2, +C, x, x, cos + sin, 2, 2, −2, (c), +C, x, x, cos + sin, 2, 2, (a), , 1, 5, log  ,  17 , 2, , 5, (d) log  ,  17 , 2, , ex, (b), +C, 2, ex, (d) −, +C, 2, cos x, Evaluate : ∫, dx, 3, x, x, , cos + sin, , 2, 2, , 57 − 5, 5, 57 − 5 5, (d), 5, , (b), , (c), , 57 + 5 5, 5, , 8., , Evaluate :, , (a), , dx, , 1, , 2x, , 1, , x, , 22, 3, (log 2), , x, , 22 2x x, ∫ 2 2 2 dx, , (b), , +C, , (c), , 22 + C, (log 2)2, , 9., , Evaluate : ∫ 2(x + 3)dx, , (a), , 2x, +C, log 2, , (b), , (c), , 2( x + 3), +C, log 2, , (d), , 10. Evaluate :, −1, 32, 1, (b), 32, (a), , (c), , {, {, {, , (d), , ∫ sin, , 3, , 1, , (log 2), , 2x, , 22, (log 2)4, 23, +C, log 2, , 2( x − 3), +C, log 2, , x cos3 x dx, , }, }, }, , 1 −3, 1, cos 2x + cos 6x + C, 32 2, 6, , (d) None of these, , ∫, , 22 + C, , 1, , −3, 1, cos 2x + cos 6x + C, 2, 6, −3, 1, cos 6x + cos 2x + C, 2, 6, , 11. Evaluate :, , x, , 3, , (x − 3)(5 − x ) dx, , +C

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5, , Integrals, , (a), , 1, 1, ( x − 4 ) ( x − 3)(5 − x ) + cos−1 ( x − 4 ) + C, 2, 2, , (b), , 1, 1, ( x − 4 ) ( x − 3)(5 − x ) + sin−1 ( x − 4 ) + C, 2, 2, , 1, 1, ( x − 3)(5 − x ) + sin−1 ( x − 4 ) + C, 2, 2, (d) None of these, (c), , π/4, , 12. Evaluate :, , ∫, , (a) (1 – log2) , 1, (c), (1 − log 2) , 2, 13. Evaluate :, −3, (a) 3, +C, sin x, (c), , cot x, , −2, , (b), , 3, , +C, sin x, 14. Evaluate :, , sin3 x, , (a) p, , ∫x, , 2, , dx, , ∫ esin x + 1, , (b) 0, , 19. Evaluate :, (a), (b), (c), (d), , +C, , loge(10x, loge(10x, loge(10x, loge(10x, , (ax + b)−2dx, , ∫, , (c) 3 p, , 10x + x10, , ∫ sin x +, , 1, , (a), , 1, x π, log tan  +  + C, 2 6, 2, , (b), , 1, x, log tan + C, 2, 2, , , 1 , b2, − 2b log( ax + b)  + C, ax, +, b, −, , ax + b, a3 , , , (b), , , 1 , b2, − 2b log( ax + b)  + C, ax + b +, , ax + b, a3 , , , (c), , 1, x π, log tan  −  + C, 2 6, 2, , (c), , , 1 , b2, + 2b log( ax + b)  + C, ax + b +, , ax + b, a3 , , , (d), , π, 1, , log tan  x −  + C, , 2, 6, , (c) e + 1 −, , 16. Evaluate :, (a), , 3, x , sin −1  , a, 2, , (c), , 2, x , cos−1  , a, 3, , 17., , ∫, , x, 3, , 3/ 2, 3/ 2, , 2, , a −x, , 3, , 2 2 4, −, π, π, , (a), , +c, , 2, x , sin −1  , a, 3, , +c, , (d), , 3, x , cos−1  , a, 2, , 3/ 2, 3/ 2, , ∫, , 3 cos x, , dx, , dx, 1 − 2x − x 2, , 1, , 1 + x , +c, sin −1 ,  2 , 2, , −1  1 + x , +c, (c) sin ,  2 , , dx, , (b), , 3− 4x, dx, Evaluate : ∫ e, 0, , 22. Evaluate :, , 2 2 4, (b) e − 1 −, +, π, π, (d) e − 1 +, , dx, , 21. Evaluate :, , }, , 2 2 4, +, π, π, , π, 2, , sec2 x, ∫ 2 + tan x dx, (a) log|tanx| + C, (b) log|2 – tanx| + C , (c) log|2 + tanx| + C (d) none of these, , , 1 , b2, (d), + 2b log( ax + b)  + C, ax + b −, , ax + b, a3 , , 1, 15. Evaluate : ∫ e x + sin πx dx, 4, 0, 2 2 4, (a) e + 1 +, +, π, π, , (d), , 10x 9 + 10x log e 10, , (a), , {, , is, , – x10) + C, + x10) + C, + x 9) + C, – x 9) + C, , 20. Evaluate :, , (d) None of these, , 1/3, , 2π, 0, , (b) (1 + log2), 1, (d), (1 + log 2), 2, , ∫ 3 sin x dx, , 1 5 3, (e − e ), 4, (d) −1 ( e−5 − e3 ), 4, (b), , 18. The value of, , tan3 x dx, , 0, , −1 5 3, (e − e ), 4, 1 −5 3, (c), (e − e ), 4, (a), , (b), (d), , (a x + bx )2, , 1, , 2, , 1, , +c, , 23. Evaluate :, , +c, , a, b,  ,  , (a) b + a + 2x + C , a ≠ b, a, b, log, log, b, a, , x, , ∫, x, , a x bx, , dx, , log(1 + x ) + c, , 1 + x , +c, log ,  2 , 2

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CBSE Board Term-II Mathematics Class-12, , 6, x, , x, , a, b,  ,  , (b) b + a + 2x + C , a ≠ b, a, a, log, log, b, b, x, , (a), (b), , x, , (c)  a  +  b  + 2x + C , a ≠ b,  ,  , b, a, , (c), , (d) None of these, π/2, , 24. Find the value of, (a) 0, , (c), , 1, , ∫, , |sin x |dx ., , −π/2, , (b) 1, , 25. Evaluate :, (a), , ∫, , (c) 2, , x tan −1 x, , 0 (1 + x, , 2 3/ 2, , ), , 4−π, , (d) 3, , 4+π, 2 2, , (d) None of these, , 4 2, , 26. Evaluate :, , x 2 − 3x + 2, , (c), , 3, , (d) log  x +  − x 2 − 3x + 2 + C, , 2, , (c), , ex, , (x − 2)3, ex, , 2, , (x − 2), , 3, (a) x + x + C, 3, , (c), , (b), , +C, , 28. Evaluate :, , x3, + x2 + C, 3, , 29. Evaluate :, , (d), , x −4, , +C, , (d), , −e x, , +C, (x − 2)3, −e, , x, , +C, , (x − 2)2, , x3 − x2 + x − 1, dx, ∫, x −1, (b) x3 + x + C, (d), , , ∫ 5x, , 3, , (c) log |sec x| –, , (b), , 3, , (c) log  x −  − x 2 − 3x + 2 + C, , 2, , (a), , ∫ tan x tan 2x tan 3x dx, , 1, log |sec 3x| – log |sec x| + c, 3, 1, (b) log |sec 3x| –, log |sec 2x| + c, 2, , (a), , 3, , (b) log  x −  + x 2 − 3x + 2 + C, , 2, , x, ∫ (x − 2)3 ⋅ e dx, , 5x 4, 1, 7x 2, + 4+, + 2 x − 5 log|x |+ C, 4, 2, 2x, , x3, +C, 3, , + 2x −5 − 7x +, , 1, 21, , log, , 1, 2 21, 1, 21, 1, 2 21, , 21 − x − 4, , x, , +, , 5,  dx, x, , +C, , 21 + x + 4, , log, , log, , dx, , ∫ 5 − 8x − x 2, , 21 + x + 4, , +C, , 21 − x − 4, 21 − x − 4, 21 + x + 4, , log, , +C, , 21 − x − 4, , +C, , 21 + x + 4, , 32. Evaluate : ∫ [sin(log x ) + cos(log x )]dx, (a) xsin(logx) + C, (c) xcos(logx) + C, 33. Evaluate :, , ∫ sec, , (b) sin(logx) + C, (d) cos(logx) + C, , 2, , (7 − 4x )dx, , (a), , 1, tan(7 − 4 x ) + C, 4, , (b), , 1, tan(7 + 4 x ) + C, 4, , (c), , −1, tan(7 + 4 x ) + C, 4, , (d), , −1, tan(7 − 4 x ) + C, 4, , 34. Evaluate :, 1, , 1, 1, log |sec 3x| + log |sec 2x|+ c, 2, 2, , 1, 1, log |sec 3x| – log |sec 2x| – log |sec x| + c, 3, 2, , 31. Evaluate :, , 3, , (a) log  x +  + x 2 − 3x + 2 + C, , 2, , 27. Evaluate :, , 5x 4, 1, 7x 2, +, +, + 2 x + 5 log|x |+C, 4, 2, 2x 4, , 30. Evaluate :, , (d), , dx, , ∫, , 5x 4, 1, 7x 2, −, −, + 2 x + 5 log|x |+C, 4, 2, 2x 4, , (a), , dx, , (b), , 2 2, 4−π, , (d), , 5x 4, 1, 7x 2, −, −, + 2 x − 5 log|x |+C, 4, 2, 2x 4, , (a), , x3, ∫ x + 2 dx, , x3, − x 2 − 4 x − 8 log|x + 2|+C, 3

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7, , Integrals, , (b), (c), (d), , x3, − x 2 + 4 x − 8 log|x + 2|+C, 3, , 2, 3,  5 ,  5 , (a), +, +C, 2, 3, loge   loge  , 5, 5, , x3, + x 2 + 4 x − 8 log|x + 2|+C, 3, , 2, 3,  5 ,  5 , +, +C, (b), 5, 3, loge   loge  , 2, 5, , x3, + x 2 + 4 x + 8 log|x + 2|+C, 3, , 35. Evaluate :, , sin x, , ∫ 1 + sin x dx, , (a) secx – tanx + C, (c) secx + tanx + C, 36. Evaluate :, , π, , (b) secx + tanx + x +C, (d) secx – tanx + x + C, , 10, , ∫x, , sin7x dx, , −π, , (a) 1, (c) –1, 37. Evaluate :, (a) a x log a +, , ∫ (e, , (b) 2, (d) 0, x log a, , ), , (b) a x log a + (a + 1)x a +1 + a a x + c, (c), , ax, x a +1, +, + aax + c, log a a + 1, , (d) None of these, 38. Evaluate :, , ∫, , x, , 2 +3, 5x, , x, , x, , x, , x, , x, , x, , 2, 3,  5 ,  5 , (c), −, +C, 2, 3, loge   loge  , 5, 5, (d) none of these, , + ea log x + ea log a dx, , x a +1 a a, +, +c, a +1 x, , x, , 39. Evaluate :, (a) 0, (c) 1, , 2, , ∫0 (x − [x ]) dx, , (b) –1, (d) 2, , 1, , 40. Evaluate :, , ∫, , (x 4 − x ) 4, , 5, , x5, , 5, , 4 , 1 4, (a), 1−,  +C, , 15 , x3 , , 1 4, −4 , (b), 1−,  +C, , 15 , x3 , , 2 , 1 4, (c), 1−,  +C, , 15 , x3 , , 1 4, −2 , (d), 1−,  +C, , 15 , x3 , , (3x + 4)4, +c, (a), 12, , 3(3x + 4)4, +c, (b), 4, , 5, , dx, , dx, , 5, , Case Based MCQs, Case I : Read the following passage and answer, the questions from 41 to 45., Integration is the process of finding the antiderivative of a function. In this process, we are, provided with the derivative of a function and, asked to find out the function (i.e., Primitive), Integration is the inverse process of, differentiation., Let f (x) be a function of x. If there is a function, d, g(x), such that, (g(x)) = f (x), then g(x) is, dx, called an integral of f (x) w.r.t x and is denoted, by, , ∫ f (x )dx =, , g(x) + c, where c is constant of, , integration., 41., , 3, , ∫ (3x + 4), , dx is equal to, , (c), 42., , 3(3x + 4)2, +c, 2, (x + 1)2, , ∫ x(x 2 + 1) dx, , (d), is equal to, , (a) log|x| + c, (c) –log|x2 + 1| + c, 2, , (b) log|x| + 2 tan–1x + c, (d) log|x(x2 + 1)| + c, , x dx is equal to, , 43., , ∫ sin, , (a), , x sin 2x, +, +c, 2, 4, , (c) x +, , 3(3x + 4)2, +c, 4, , sin 2x, +c, 2, , (b), , x sin 2x, −, +c, 2, 4, , (d) x −, , sin 2x, +c, 2

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CBSE Board Term-II Mathematics Class-12, , 8, 44., , ∫ tan, , 2, , x dx is equal to, , (a) tan x + x + c, (c) – tan x + x + c, 45., , dx, , ∫ sin 2 x cos2 x, , (b) – tan x – x + c, (d) tan x – x + c, , is equal to, , (a) 2 tan 2x + c, (c) –2 cot 2x + c, , (b) –2 tan 2x + c, (d) 2 cot 2x + c, , Case II : Read the following passage and answer, the questions from 46 to 50., When the intergrand can be expressed as a, product of two functions, one of which can be, differentiated and the other can be integrated,, then we apply integration by parts., If f(x) = first function (that can be differentiated), and g(x) = second function (that can be, integrated), then the preference of this order, can be decided by the word “ILATE”, where, I, stands for Inverse Trigonometric Function, L stands for Logarithmic Function, A stands for Algebraic Function, T stands for Trigonometric Function, E stands for Exponential Function, then, , d, ∫ f (x )g(x )dx = f (x )∫ g(x )dx −∫  dx f (x )∫ g(x )dx  dx, 46., , ∫ x sin3x dx =, , (a), , x cos 3x sin 3x, −, +c, 3, a, , x cos 3x sin 3x, (b) −, +, +c, 3, 9, (c) x cos 3x + sin 3x + c, 3, 9, x cos 3x sin 3x, −, +c, (d) −, 3, 9, 47., (a), (b), (c), (d), 48., , ∫ log(x + 1) dx =, , log (x + 1) – x + c, x log(x + 1) – x + c, x log(x + 1) – log (x + 1) + x + c, x log(x + 1) + log (x + 1) – x + c, , ∫ tan, , (a) x tan, , −1, , −1, , x dx =, 1, x + log|1 − x 2 |+ c, 2, , 1, 2, (b) − log|1 + x |+ c , 2, , 1, (c) −x tan −1 x − log|1 + x 2 |+ c, 2, 1, −1, 2, (d) x tan x − log|1 + x |+ c, 2, 49., , ∫x, , 2 3x, , e, , dx =, , e3x, (a), (9x2 + 6x + 2) + c, 9, (b), , e3x, (9x2 – 6x + 2) + c, 9, , (c), , e3x, (9x2 + 6x + 2) + c, 27, , (d), , e3x, (9x2 – 6x + 2) + c, 27, , 50., , ∫ ( f (x )g′′(x ) − f ′′(x )g(x )) dx =, , (a) f(x)g′(x) – f ′(x)g(x) + c, (b) f(x)g′(x) + f ′(x)g(x) + c, (c) f ′(x)g(x) – f(x)g′(x) + c, f (x ), (d), +c, g ′(x ), , Case III : Read the following passage and, answer the questions from 51 to 55., Let f be a continuous function defined on the, closed interval [a, b] and F be an antiderivative, b, , of f, then ∫ f (x )dx =| f (x )|ba = F (b) − F (a), a, , This result is very useful as it gives us a metod, of calculating the definite integral easily. Here,, we have no need to write integration constant, c because if, we will write F(x) + c, instead of, f(x), we get, b, , b, , ∫a f (x )dx =| f (x ) + c|a = F (b) + c − F (a) − c = F (b) − F (a), 51. Evaluate :, , π/2, , ∫, , cos 2x dx, , π/4, , (a), , 1, 4, , 52. Evaluate :, (a), , 1, 2, , 1, 2, , (b), 2, , (c) −, , 1, 4, , (d), , −, , 1, 2, , dx, , ∫ x2, , 1, , (b) 1, , (c) 2, , (d) –1

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9, , Integrals, , 53., , dx, ∫−1 2x + 3 is equal to, 0, , 3, , ∫ (x − 1)(x − 2)(x − 3)dx is equal to, , 1, , 3, 2, , (b) log 3 – log 1, , log 3, 2, , (d) log 3 + log 1, , (a) log, (c), , 54., , (a) 3, 55., , (b) 2, 5 x, , ∫4 e, , (c) 1, , (d) 0, , dx equals, , (a) e5 – e4 (b) e4 – e5 (c) e9, , (d) e20, , Assertion & Reasoning Based MCQs, Directions (Q.-56 to 60) : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose, the correct answer out of the following choices :, (a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion., (b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion., (c) Assertion is correct statement but Reason is wrong statement., (d) Assertion is wrong statement but Reason is correct statement., 56. Assertion :, , ∫ sin 3x cos 5x dx =, , rational by the substitution t =, , − cos 8x cos 2x, +, +C, 16, 4, , Reason : 2cos A sin B = sin( A + B) − sin( A − B), , 59. Assertion :, , 2π, , ∫ sin, , 3, , x, 1 + x3, , 3, , x dx = 0, , 0, , 57. Let F(x) be an indefinite integral of sin2 x., , Reason : sin3 x is an odd function., , Assertion : The function F(x) satisfies F(x + p), = F(x) for all real x., , 60. Assertion : The value of, , π/2, 0, , Reason : sin2(x + p) = sin2x for all real x., 58. Assertion : I =, , dx, , 1, , ∫0 3, , 1 + x3, , =, , 2−1/ 3, , ∫0, , dt, , 1 − t3, , Reason : The integrand of the integral I becomes, , ∫, , Reason : If n is even, then, , sin6 x dx =, , π/2, , ∫, , 5π, ., 16, , sinn x dx equals, , 0, , n −1 n − 3 n − 5, 1 π, ⋅, ...... ⋅, ⋅, n n−2 n−4, 2 2., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1., , 1 , , Write the antiderivative of  3 x +, ., , x, , 2., , Evaluate :, , 3., 4., 5., , ∫ cos, , −1, , (sin x )dx, , Write the value of, Write the value of, Evaluate :, , ∫, , sec2 x, , ∫ cosec2 x dx., ∫, , 2 − 3 sin x, cos2 x, , (log x )2, dx, x, , dx., , dx, , 6., , Find :, , ∫ 9 + 4x 2, , 7., , Find :, , ∫x, , 8., , If, , a, , 1, , 4, , log x dx, π, , ∫ 4 + x 2 dx = 8 , find the value of a., , 0, , 9., , Write the value of, , 1, , ex, , ∫ 1 + e2x dx., , 0, , 10. Find the value of, , 4, , ∫ x − 5 dx. , , 1

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11, , Integrals, , Integrating by parts, we get, , OBJECTIVE TYPE QUESTIONS, 1., , 4, (x 2 + x), 2, ∫ 2x + 1 dx = (x + x) ⋅ 2x + 1  2 − ∫ (2x + 1) ⋅ 2x + 1 dx, 2, 2, , I = ∫ (3 sin x − 2 cos x + 4 sec2 x − 5 cosec2 x) dx, 2, , 4, , ⇒ I = 3∫ sin x dx − 2 ∫ cos x dx + 4∫ sec x dx − 5∫ cosec x dx, 2, , ⇒, , I = –3 cosx – 2sin x + 4 tan x + 5 cot x + C, , 2., , (b) : We have, ∫ (2 x + 2 −x )2 dx = ∫ (2 2x + 2 −2x + 2) dx, , 3., , (c) : We have,, , ∫, , sin 2 x cos2 x, , (a) : Let I = ∫, , x, , dx, , ⇒, , I=, , 5., , 1, 2, , 17, , ∫, , 5, , ⇒ x dx =, , 2xdx = dt, , 1, dt, 2, , 1,  17 , dt 1, 17 1, = [log t]5 = [log 17 − log 5] = log  , 2, 2, 5, t 2, , 2, , x, x, cos + sin, , 2, 2, , x, x, Put t = cos + sin, 2, 2, ⇒, , ∫, , =, 7., , cos 2 (x/2) − sin 2 (x/2), , {cos(x/2) + sin(x/2)} 3, , x, x, , ⇒ 2 dt =  cos − sin  dx, 2, 2, , , cos(x/2) − sin(x/2), x, x, , cos + sin, , 2, 2, , 2, , dx = 2 ∫, , 1, , t2, , dt, , −2, −2, +C =, +C, cos(x/2) + sin(x/2), t, 4, , (x2 + x), (d) : We have, ∫, dx, 2 2x + 1, , (log 2), , dt =, , 1, , 3, , (log 2), , 1, , t +C =, , 3, , (log 2), , 22, , 2x, , +C, , (c) : ∫ 2( x + 3) dx = ∫ 2 x ⋅ 2 3 dx = 8 ∫ 2 x dx, , I=, , }, , ⇒ I = ∫ −{x 2 − 8x + 16 − 16 + 15} dx, , (c) : We have,, dx = ∫, 3, , 3, , 11. (b) : Let I = ∫ (x − 3)(5 − x) dx = ∫ −x 2 + 8x − 15 dx, , 1 t, et, ex, e dt = + C =, +C, ∫, 2, 2, 2, cos x, , 1, , {, , 2, , ∫, , x, , 1, (2 sin x cos x)3 dx, 8∫, 1, 1 3 sin 2x − sin 6x, ⇒ I = ∫ sin 3 2x dx ⇒ I = ∫, dx, 8, 8, 4, 1, 3, 1, ⇒ I=, − cos 2x + cos 6x + C, 32 2, 6, , ⇒, , dt, Put x = t ⇒ 2xdx = dt ⇒ x dx =, 2, , 6., , 2x, , = t ⇒ 2 2 2 2 2 x (log 2)3 dx = dt, , 10. (c) : Let I = ∫ sin 3 x cos 3 x dx, , (b) : Let I = ∫ xe x dx, , I=, , x, , x, 2(x + 3), = 8⋅ 2 +C =, +C, log 2, log 2, , 2, , ∴, , 2x, , dx, , Also, x = 2 ⇒ t = 5 and x = 4 ⇒ t = 17, ∴, , 2x, , (a) : Let I = ∫ 2 2 2 2 2 x dx, , ⇒ I=∫, 9., , 2, 2x +1, , Put x2 + 1 = t, , 4, 1, = (60 − 6 5 ) − ⋅ (2x + 1)5/2  2, 5, , Let 2 2, , = tanx + cotx + C, 4., , 2, , 8., , = ∫ (sec2 x − cosec 2 x )dx, 4, , = (60 − 6 5 ) − ∫ (2x + 1)3/2 dx, ,  57 − 5 5 , 57,  243, , = (60 − 6 5 ) − , − 5 5  =  − 5  = , ,  5, ,  5,  , 5, , 2 2x, 2 −2x, +, + 2⋅x + C, (log 2) × 2, (log 2) (− 2 ), 1, =, (2 2x − 2 −2x ) + 2x + C, 2 log 2, =, , sin 2 x − cos2 x, , 4, , 4, , (a) : Let, , ⇒ I = ∫ −{(x − 4)2 − 12 } dx = ∫ 12 − (x − 4)2 dx, dx, , 1, 1, x − 4, ⇒ I = (x − 4) (x − 3)(5 − x) + sin −1 , +C,  1 , 2, 2, 12. (c) : Let I =, , π/4, , ∫, , tan 3 x dx =, , π/4, , 0, , =, , π/4, , ∫, , sec2 x tan x dx −, , 0, , ∫, , (sec2 x − 1)tan x dx, , 0, π/4, , ∫, , tan x dx, , 0, , Put tan x = t in first integral ⇒ sec2 x dx = dt, When x = 0, ⇒ t=0, x = p/4 ⇒ t = 1, , \, , 1, , π /4, , 0, , 0, , I = ∫ t dt −, , ∫, , 1, , π /4,  2, tan x dx =  t  − [ log|sec x|]0,  2 0

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CBSE Board Term-II Mathematics Class-12, , 12, 1, π, 1, =  − 0  − log sec + log|sec 0|= (1 − log 2), 2, , 4, 2, cot x, cos x, 13. (a) : Let I = ∫, dx = ∫, dx, 3, 1/3, sin x, sin x ⋅ sin x, cos x, =∫, dx = ∫ sin −4/3 x ⋅ cos xdx, sin 4/3 x, Put sinx = t ⇒ cosx dx = dt, ⇒ I = ∫t, , −4 /3, , t −1/3, −3, dt =, +C = 3, +C, −1 / 3, sin x, , {, , e x + sin, , 0, 1, , }, , 4, π , 4, 4, − cos x  = e − 1 −, +, π , 4 0, 2π π, , = e−1−, , 2 2 4, +, π, π, , 1, , 3, , a −x, , 3, , dx, , 3/2, , ⇒, , I=, , 2π, , ∫, , 0, , ∫, , 0, , dx, , esin x + 1, , esin( 2 π − x) + 1, , , , ∫ 1 ⋅ dx = 2π, , +1, , dx ...(ii), , I=p, , \, , 10x 9 + 10 x log e 10, 10 x + x 10, , 10x 9 + 10 x log e 10, 10 x + x 10, , dx, , dx = ∫, , dt, t, , 1, sin x + 3 cos x, , dx, , 1, dx, 2∫ 1, 3, sin x +, cos x, 2, 2, π, 1, 1, 1, , ⇒ I= ∫, dx = ∫ cosec  x +  dx, , π, , 2, 2, 3, sin x +, , , 3, , , , =, , 1, x π, log tan  +  + C, 2 6, 2, , dt, = log|t|+ C = log|2 + tan x|+ C, t, , =, , ∫, , a, , ∫, , 0, , dx, 1 − (x 2 + 2x), , dx, 2, , 2 − (1 + x), , =∫, , =∫, , dx, 2 − (x 2 + 2x + 1), , dx, 2, , ( 2 ) − (1 + x)2, , , , Put 1 + x = z ⇒ dx = dz, dz, , ∴ I=∫, , ( 2 )2 − z2, , 23. (a) : We have,, , ...(i), , ∵, , , sec2 x, dx, 2 + tan x, , 22. (c) : Let I = ∫, ,  3 − 4x , 17. (d) : We have, e 3− 4x dx =  e, , ∫,  −4  0, 0, 1, −1  −5 3 , = −  e 3− 8 − e 3− 0  =, e −e , 4, 4 , 18. (a) : Let I =, , I=∫, , ∴ I=∫, , 2, , 2, , dx, , e, , Put 2 + tanx = t ⇒ sec2xdx = dt, , +c, , 2π, , 2π, , 21. (c) : Let I = ∫, , 2  −1  t  , ,  3/2  , sin  3/2   + c = 2 sin −1  x, +c, , , , 3, a,  a3/2  , , 3, , 2, x, = sin −1  , a, 3, , 0, , esin x, , sin x, , 0, , ⇒ I=, , 3 1/2, x dx = dt, Put x3/2 = t ⇒, 2, 2, dt, 2, dt, \ I= ∫, = ∫, 3, /, 2, 3, 2, 3 (a )2 − t 2, 3, a −t, =, , ∫, , , , x, ∵ ∫ cos ecxdx = log tan 2 + C , , , , x, , 16. (b) : Let I = ∫, , 2I =, , 20. (a) : Let I = ∫, , πx, dx, 4, , = [ e x ]0 +, , e − sin x + 1, , 2π, , = loget + C = loge(10x + x10) + C., , 2, 1 , , = 3  ax + b − b − 2b log(ax + b)  + C, , , a, ax + b, 1, , 0, , ⇒ I=, , Adding (i) and (ii), we get, , ∴, , 2, 1 , , = 3  t − b − 2b log t  + C, , a , t, , ∫, , ∫, , dx, , Put 10x + x10 = t, ⇒ (10x loge10 + 10x9)dx = dt, , dx, (ax + b)2, 1, Put ax + b = t ⇒ dx = dt, a, 2, 1 (t − b), 1 , b 2 2b , ∴ I= 3∫, dt = 3 ∫  1 +, −  dt, 2, t , a, t, a , t2, , 15. (b) : We have,, , 2π, , I=, , 19. (b) : Let I = ∫, , x2, , 14. (a) : Let I = ∫, , ⇒, , a, , , f (x)dx = ∫ f (a − x)dx , , 0, , x, , x, , = sin −1, , ∫, , z, −1  1 + x , + c = sin , +c, 2 , 2, , (ax + b x )2, axb x, , a, b, , x, , dx = ∫, b, a, , a2x + b 2x + 2ax b x, axb x, , x, ,  a   b , , = ∫    +   + 2  dx =, +, + 2x + C , a ≠ b,  b   a , , b, a, log, log, a, b, , dx

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CBSE Board Term-II Mathematics Class-12, , 14, 2 x + 3x, , 38. (a) : Let I = ∫, 2x, , 5, , x, ,  cos 3x ,  cos 3x , = x −,  − 1⋅  −,  dx + c, , 3  ∫ , 3 , , dx, x, , 3x, , x, , 2, 3, dx = ∫   dx + ∫   dx, 5, 5, 5x, , ⇒, , I =∫, , ⇒, , 2, 3,  ,  , 5, 5, I=, +, +C, 2, 3, log e   log e  , 5, 5, , 5x, , x cos 3x 1, + ∫ cos 3x dx + c, 3, 3, x cos 3x 1 sin 3x, x cos 3x sin 3x, \I =−, + ⋅, +c = −, +, +c, 3, 3, 3, 3, 9, =−, , dx + ∫, x, , x, , 47. (d) : Let I = ∫ log (x + 1) dx = ∫ log(x + 1)⋅ 1dx, , 2, , 2, , 2, , 0, , 0, , 0, , 39. (c) : Let I = ∫ (x − [x])dx = ∫ x dx − ∫ [x]dx, , = x log (x + 1) − x + log (x + 1) + c, , 2 2, , 1, 2, 1, 2, x, 4, =   − ∫ [x] dx − ∫ [x]dx = − ∫ 0 dx − ∫ 1 dx,  2 0, 2, 0, , 1, , 0, , 48. (d) : Let I = ∫ tan −1 x dx = ∫ tan −1 x ⋅ 1 dx, , d, = tan −1 x ∫ 1 dx − ∫  (tan −1 x) ∫ 1 dx  dx, ,  dx, 1, = x tan −1 x − ∫, (x) dx, 1 + x2, 1, 2x, dx, = x tan −1 x − ∫, 2 1 + x2, , 2, , 1, , 40. (a) : Let I = ∫, , x5, , 1, , ⇒, , I=∫, , Put 1 −, \, , 1, , 1 4, , x 1 − 3 , , x , x, , x3, , 5, , =t ⇒, , 1, , dx = ∫, , x, , 4, , dx = dt, 4, , 1, , I, , 3x, , 5, , 5, , 3, , ∫ (3x + 4), , 42. (b) : Let I =, 1, , 2, , , , dx =, , 4, , 4, , (3x + 4), (3x + 4), +c =, +c, 4⋅3, 12, , (x + 1)2, , ∫ x(x 2 + 1) dx, , ∫  x + x 2 + 1  dx, , =, , ∫, , x 2 + 1 + 2x, x(x 2 + 1), , dx, , = log |x| + 2 tan–1x + c, , 1 − cos 2x, dx, 2, 1, sin 2x , x sin 2x, = x −, +c, +c = −, 2, 2 , 2, 4, , 43. (b) :, , 44. (d) :, , ∫ sin, , 2, , ∫ tan, , 2, , x dx = ∫, , x dx =, , ∫ (sec, , 2, , x − 1) dx, , = tan x – x + c, 4, dx, dx, 45. (c) : Let I = ∫, = ∫, 2, 2, 2, 4 sin x cos 2 x, sin x cos x, , , 2, = 4∫ cosec 2x dx = –2 cot 2x + c, , 46. (b) : Let I =, , 1, log|1 + x 2 |+ c, 2, , 49. (d) : Let I = ∫ x 2 e 3x dx, , 3, x, , I = x tan −1 x −, , \, , dx, , 1, 1 4, 4, 1, I = ∫ t 4 dt = × t 4 + C =  1 −  4 + C., , 3, 3 5, 15, x3 , , 41. (a) :, , =, , dx, 1 4, ,  1 − 3 , x, , II, , I, , 1, , = 2 – 0 – [x] 1 = 2 –[2 – 1] = 2 – 1 = 1., (x 4 − x) 4, , II, , I, , 1, = log(x + 1) ⋅ x − ∫, ⋅ x dx, x+1, x+1, 1, = x log (x + 1) − ∫, dx + ∫, dx, x+1, x+1, , ∫ xI sin II3x dx, ,  d, , = x ∫ sin 3x dx − ∫  (x).∫ sin 3x dx  dx,  dx, , , II, , 3x, , e , e, = x2 , − ∫ 2x, dx, , 3,  3 , =, ,  e 3x ,  e 3x , x 2 e 3x, − (2x) , + (2) , , +c, 3,  9 ,  27 , I=, , \, , e 3x, (9x2 – 6x + 2) + c, 27, , 50. (a) : Let I = ∫ ( f (x)g″(x) − f ″(x)g(x)) dx, = ∫ f (x)g″ (x) dx − ∫ g(x) f ″ (x) dx, I, , II, , = f(x)g′(x) –, , II, , I, , ∫ f ′(x)g′(x) dx − g(x) f ′(x) + ∫ g′(x) f ′(x) dx, , = f(x)g′(x) – g(x)f ′(x) + c, 51. (d) : We have, I =, , π /2, , ∫, , cos 2x dx, , π /4, π /2, , , , 1,  sin 2x , =−, =,  2  π/4, 2, , 52. (a) : Let I =, , 2, , dx, , ∫ x2, 1, , 53. (c) : We have I =, , 2, , 1,   −1  , ⇒ I =   =, , ,  x 1 2, 0, , ∫, , −1, 0, , dx, 2x + 3, ,  log 3 log 1  log 3,  log(2x + 3) , =, =, −, =, ,  −1  2, , 2, 2 , 2

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CBSE Board Term-II Mathematics Class-12, , 16, 1, , ex, , Let I = ∫, , 9., , 0 1+ e, , 15. Let I =, , dx, , 2x, , Also let,, , Put ex = t ⇒ ex dx = dt, Also, x = 0 ⇒ t = e0 = 1, and x = 1 ⇒ t = e1 = e, e, , dt, , I=∫, , ∴, , ( 2), 1 1+t, , e, , = [tan −1 t]1 = tan −1 e − tan −1 1, ,  e − 1, = tan −1 ,  1 + e , 4, , ∫ x − 5 dx, , 10. Let I =, , ,  x2, = −∫ (x − 5) dx =  −, + 5x , ,  2, 1, , =−, , 4, , ⇒ x + 1 = A(x + 3) + B(x + 2), Putting x = –3 in (i), we get, –B = –3 + 1 = –2 ⇒ B = 2, Putting x = –2 in (i), we get, A = –2 + 1 = –1, −1, 1, \ I= ∫, dx + 2 ∫, dx, (x + 2), (x + 3), , 1, , Integrating by parts, we get, , , 1, d, = sin −1(2x)x − ∫ , (2x) ⋅ x  dx,  1 − 4x 2 dx, , , 11. Let I = ∫ ( 1 − sin 2x ) dx, , = x sin −1(2x) − ∫, , 2, , = ∫ cos x + sin x − 2 sin x cos x dx, , = x sin −1(2x) + ∫, , = ± ∫ (cos x − sin x) dx, Since,, , I = (sin x − cos x) dx = –(cosx + sinx) + C, , 12. Let I = ∫, =∫, =∫, , cos 2x + 2 sin 2 x, , cos 2 x, cos 2 x − sin 2 x + 2 sin 2 x, cos 2 x, , 2, , 2, , cos x + sin x, 2, , cos x, , dx = ∫, , dx, , cos 2 x, , =, , =, , dx, , =, , dx, , dt, 4 t, , Integrating by parts, we get, d, , I = tan −1 x ∫ xdx − ∫  (tan −1 x)∫ xdx  dx,  dx, , , dx, , = ∫ sec2 x dx = tan x + C, 13. Let I = ∫, , 1 − 4x 2, , 17. Let I = ∫ x ⋅ tan −1 x dx, , dx, 1, , 2x, , (Putting 1 – 4x2 = t ⇒ –8xdx = dt), 2, = x sin −1(2x) + (t)1/2 + C, 4, 1, = x sin −1(2x) +, 1 − 4x 2 + C, 2, , π, π, < x < , so we get, 4, 2, , ∫, , 1, 2∫, , dx, , = (tan −1 x), , 1, x2, x2, −∫, dx, 2, 2, (1 + x ) 2, , 5, − 2x − x 2, 2, , =, , 1, dx, 1, dx, =, ∫, 2∫ 7, 2, 2, 2, − 1 − 2x − x,  7, 2, 2, ,  − (x + 1), 2, , x 2 tan −1 x 1 , 1 , − ∫ 1 −,  dx, 2, 2 , 1 + x2 , , =, , x 2 tan −1 x 1, 1, 1, − ∫ dx + ∫, dx, 2, 2, 2 1 + x2, , =, , x 2 tan −1 x x 1, − + tan −1 x + C, 2, 2 2, , 5 − 4x − 2 x, , 2, ,  2, , 1, 1,  x + 1, sin −1, sin −1  (x + 1) + C, +C =, , ,  7, , 2, 7, 2, , , 2 , , 14. We have,, =∫, , dx, , dx, , dx, , ∫ x 2 + 4x + 8 =∫ x 2 + 4x + 4 + 4, , (x + 2)2 + (2)2, , =, , ...(i), , 16. Let I = ∫ sin −1(2x) dx = ∫ 1 ⋅ sin −1(2x) dx, , 16 . 1, 1, 1 15, + 5 4 + − 5 = −8 + 20 − 5 + = 7 + =, 2, 2, 2, 2 2, , 2, , (x + 1), A, B, =, +, (x + 2)(x + 3) (x + 2) (x + 3), , = –log(x + 2) + 2log (x + 3) + C, , 1, , 4, , (x + 1), , ∫ (x + 2)(x + 3) dx, , 1, x+2, tan −1 , +C,  2 , 2, , 1, x, = (1 + x 2 )tan −1 x − + C, 2, 2, 18. Let I =, , 2, , 1, , 1 , , ∫  x − 2x 2  e, , 2x, , dx, , 1, , Putting 2x = y ⇒ 2dx = dy, As x → 1 ⇒ y → 2 and x → 2 ⇒ y → 4

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17, , Integrals, 4, 1 2 2  y,  −  e dy =, 2 ∫  y y 2 , , I=, , \, , 2, , 4, , 1, , 1, , ∫  y − y 2  e, , , 2, , 4, , 2, , y, , = ∫ sin 3x sin x sin 2x dx, , dy, , 1, (cos 2x − cos 4x) sin 2x dx, 2∫, 1, 1, = ∫ sin 2x cos 2x dx − ∫ cos 4x sin 2x dx, 2, 2, =, , 2, , , e e, 1, 1, 1, , =  ey ⋅  = e4 − e2 =  − 1, , , y 2 4, 2, 2 2, , 19. Let I =, , 1, , ∫ tan, , 1 − 2x , ,  dx, 1 + x − x2 , , −1 , , 0, , 1, 1, sin 4x dx − ∫ (sin 6x − sin 2x) dx, ∫, 4, 4, 1, 1, 1, = ∫ sin 4x dx − ∫ sin 6x dx + ∫ sin 2x dx, 4, 4, 4, =, , 1, ,  (1 − x) − x , = ∫ tan −1 ,  dx,  1 + x(1 − x) , 0, , 1, , −1, , I = ∫ [tan (1 − x) − tan, , −1, , x] dx ...(i), , 0, 1, , I = ∫ [tan −1 x − tan −1(1 − x)] dx ...(ii), 0, , , Using property,, , , a, , ∫, , 0, , a, , , f (x)dx = ∫ f (a − x)dx , , 0, , =, , 1  − cos 4x (− cos 6x) (− cos 2x) , −, +,  + C, 4 , 4, 6, 2, , =, , 1  cos 6x cos 4x cos 2x , +C, −, −, 4  6, 4, 2 , , 23. Let I = ∫ ( tan x + cot x ) dx,  sin x, cos x , sin x + cos x, = ∫, +, dx, dx = ∫, sin x , sin x cos x,  cos x, , Adding (i) and (ii), we get, 1, , 2 I = ∫ [tan −1(1 − x) − tan −1 x + tan −1 x − tan −1(1 − x)] dx = 0, , ⇒, , 0, , I=0, 0, , (1 + tan x), 20. Let I = ∫, dx =, (, − π 1 − tan x), 4, , =, , sin x , , 1+, , cos x  dx, ∫, sin x , −π  1 −, , cos, x, 4, 0, , 0, , cos x + sin x, dx, − π cos x − sin x, , ∫, , 4, , Put cosx – sinx = t ⇒ –(sinx + cosx) dx = dt, −π, When x = 0, t = 1, when x =, ,t = 2, 4, ∴ I=, , 1, , ∫, , 2, , −, , dt, =, t, , 2, , ∫, , 1, , 2, dt, = [log t], 1, t, , 1, log 2, 2, sin(x − a), sin(x + a − 2 a), 21. Let I = ∫, dx = ∫, dx, sin(x + a), sin(x + a), , = 2∫, , sin x + cos x, sin x + cos x, dx = 2 ∫, dx, 2 sin x cos x, sin 2x + 1 − 1, , = 2∫, , sin x + cos x, sin x + cos x, dx = 2 ∫, dx, 1 − (1 − sin 2x), 1 − (sin x − cos x)2, , Put sin x – cos x = t ⇒ (cos x + sin x) dx = dt, dt, = 2 sin −1 t + C, \ I = 2∫, 2, 1−t, = 2 sin −1(sin x − cos x) + C, 1, 1, (2x + 5) −, x+2, 2 dx, 24. Let I = ∫, dx = 2, ∫ 2, x 2 + 5x + 6, x + 5x + 6, =, , Put x2 + 5x + 6 = t ⇒ (2x + 5) dx = dt, ⇒, , = log 2 − log 1 =, ,  sin(x + a)cos 2 a − cos(x + a)sin 2 a , = ∫,  dx, sin(x + a), , ⇒, , I = cos 2 a∫ dx − sin 2 a∫, , cos(x + a), dx, sin(x + a), , Put sin (x + a) = t ⇒ cos(x + a)dx = dt, dt, ⇒ I = cos 2 a∫ dx − sin 2 a∫, t, = x cos 2a – sin 2a log|sin(x + a)| + C, 22., , ∫ sin x sin 2x sin 3x dx, , 1, 1, dx, (x 2 + 5x + 6)−1/2 (2x + 5)dx − ∫, 2, 2∫, 2, x + 5x + 6, , I=, , 1 −1/2, 1, t, dt − ∫, ∫, 2, 2, , dx, 2, , 5, , 1, −, x+, , 2, 2, 2, , 2, , +C, , 2, , 5, 5, 1 t 1/2 1, , , 1, − log  x +  +  x +  −   + C, , , , , 2, 1, 2, 2, 2, 2, 2, 1, 5, = x 2 + 5x + 6 − log x + + x 2 + 5x + 6 + C, 2, 2, =, , 25. Let I = ∫, =, , 5, ( 2 x + 4) − 7, dx = ∫ 2, dx, x 2 + 4x + 10, x 2 + 4x + 10, 5x + 3, , 5, 2x + 4, dx, dx − 7 ∫, 2, 2 ∫ x 2 + 4x + 10, x + 4x + 10, , = I1 + I2 (say) , , ...(1)

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CBSE Board Term-II Mathematics Class-12, , 18, 5, 2x + 4, dx, 2 ∫ x 2 + 4x + 10, Put x2 + 4x + 10 = t ⇒ (2x + 4)dx = dt, where I 1 =, , ∴, , 5, 5 t 1/2, I 1 = ∫ t −1/2 dt = ⋅, =5 t, 2, 2 (1 / 2), , , and I 2 = −7 ∫, = −7 ∫, , = 5 x 2 + 4x + 10 + C 1 , dx, , ...(2), , x 2 + 4x + 10, dx, ...(3), , = −7 log|x + 2 + x 2 + 4x + 10 |+ C 2 , , From (1), (2) and (3), we get, , I = 5 x 2 + 4x + 10 − 7 log|x + 2 + x 2 + 4x + 10 | + C ,, where C = C1 + C2, , =∫, =∫, , sin x − cos x, sin x − cos x, dx =, ∫ 1 + sin 2x − 1 dx, sin 2x, , sin x − cos x, , sin 2 x + cos 2 x + 2 sin x cos x − 1, sin x − cos x, (sin x + cos x)2 − 1, , dx, , ⋅ dx, , Put sin x + cos x = t ⇒ (cos x – sin x) dx = dt, −dt, ∴ I=∫, = − log|t + t 2 − 1|+ C, 2, t −1, (where t = sin x + cos x), = − log|sin x + cos x + sin 2x |+ C, 27. Let I = ∫, 2, , Let x = t, \, , (x 2 + 3)(x 2 − 5), , (x 2 + 1)(x 2 + 4), (x 2 + 3)(x 2 − 5), =, , Let, , (x 2 + 1)(x 2 + 4), , =, , dx, , (t + 1)(t + 4), (t + 3)(t − 5), , 2, , (x + 3)(x − 5), , x, , dx = ∫ dx +, , x, , 2, , (x + 1)(x − 1), , (x 2 + 1)(x − 1), , =, , Ax + B, x2 + 1, , dx, , +, , C, x−1, , ...(1), , , ⇒ x = (Ax + B)(x –1) + C(x2 + 1), ...(2), 2, Comparing coefficients of x , x and constant terms, we, get, A + C = 0; B – A = 1; –B + C = 0, Solving these, we get, 1, 1, 1, A = − ,C = , B =, 2, 2, 2, \ From (1), we get, 1, − (x − 1) 1 1, x, 2, =, + ⋅, 2 x−1, (x 2 + 1)(x − 1), x2 + 1, 1, 1, x, 1, 1 1, =− ⋅ 2, + ⋅, + ⋅, 2 x + 1 2 x2 + 1 2 x − 1, ∴ I=−, , 1 dx, 1, 2x, 1, dx, dx + ∫ 2, +, 4 ∫ x2 + 1, 2 x +1 2∫x−1, , 1, 1, 1, ⇒ I = − log|x 2 + 1|+ tan −1 x + log|x − 1|+ C 1, 4, 2, 2, , =, , [− cos(3x + 1)], [ − cos(3x + 1)] dx, − ∫ 2e 2x ⋅, 3, 3, , −e 2 x cos(3x + 1) 2 2x, + ∫ e cos(3x + 1)dx, 3, 3, , −e 2 x cos(3x + 1) 2  2x, +  e ∫ cos(3x + 1)dx, 3, 3,  d,  , − ∫  (e 2x ) ⋅ ∫ cos(3x + 1)dx  dx ,  dx,  , , =, , 7t +19 = A(t – 5) + B(t + 3), 27, Putting t = 5, we get B =, 4, 1, Putting t = – 3, we get A =, 4, t 2 + 5t + 4, 1, 27, \, = 1+, +, (t + 3)(t − 5), 4(t + 3) 4(t − 5), 2, , 28. Let I = ∫, , = e2 x, , 7t + 19, A, B, =, +, (t + 3)(t − 5) t + 3 t − 5, , (x 2 + 1)(x 2 + 4), , x− 5, 1,  x  27, tan −1 , +, log, +C,  3  8 5, 4 3, x+ 5, ,  2x, , = e 2 x ∫ sin(3x + 1) dx − ∫  d(e ) ⋅ sin(3x + 1)dx  dx,  dx ∫, , , t 2 + 5t + 4, 7t + 19, = 1+, (t + 3)(t − 5), (t + 3)(t − 5), , I=∫, , =x+, , 1, , 29. Let I = ∫ e 2 x sin(3x + 1) dx, , ⇒, , ⇒, , x− 5, 1,  x  27, tan −1 , log, +C,  + ×, , 4 2 5, 4 3, 3, x+ 5, , Let, , (x + 2)2 + ( 6 )2, , 26. Let I = ∫, , =x+, , =, , −e 2x cos(3x + 1) 2 2x, + e sin(3x + 1), 3, 9, , −, , , , 1, 1, dx, 4 ∫ (x 2 + 3), , +, , 27, 1, dx, ∫, 2, 4 (x − 5), , =, ∴, , 4 2x, e sin(3x + 1)dx, 9∫, , −e 2x cos(3x + 1) 2 2x, 4, + e sin(3x + 1) − I + C 1, 3, 9, 9, I+, , 4, −e 2x cos(3x + 1) 2 2x, I=, + e sin(3x + 1) + C 1, 9, 3, 9

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CBSE Board Term-II Mathematics Class-12, , 20, ∞, , ⇒, , I = π∫, , 0, , ⇒, , I=, , ⇒, , I=, , Adding (1) and (2), we get, , 1, , (t + sin α)2 + cos 2 α, , dt, 2I =, , ∞, , π ,  t + sin α  , tan −1 ,  cos α   0, cos α , , 4, , 34. Let I = ∫ (|x − 1|+|x − 2|+|x − 4|) dx, Also, let f(x) = |x – 1| + |x – 2| + |x – 4|, We have three critical points x = 1, 2, 4., , −x + 5, if 1 ≤ x < 2, ∴ f (x) = , x + 1, if 2 ≤ x < 4, 1, , 1, , 2, , ∴ I = ∫ f (x)dx = ∫ f (x)dx +∫ f (x)dx, 2, , 2, , 4, , 4, , , ,  x2,  x2, = ∫ (−x + 5) dx + ∫ (x + 1) dx =  −, + 5x  +  + x , 1  2, 2,  2, 1, 2,  4,   1,   16,  4, , =  − + 10  −  − + 5  +  + 4  −  + 2 ,  2,   2,   2,  2, , = 8−, , 35. Let I =, , ⇒, , I=, , π, 3, , ⇒, , I=, , π, 6, , ∫, π, 6, , ∫ (1 +, , ∫, , π, 6, , =, , π, 3, , ⇒, , I=, , ∫, , π, 6, , I=, , π π, , π π, , + − x + sin, + −x, 3 6, , 3 6, , b, , ∫, a, , dx, , b, , , f (x)dx = ∫ f (a + b − x)dx , , a, , π, , π, , − x + sin, cos, −x, 2, , 2, , , π, 6, , 6x + 7, 6x + 7, dx = ∫, dx, 2, (x − 5)(x − 4), x − 9x + 20, , 3(2x − 9), , I=∫, , 2, , x − 9x + 20, , dx + ∫, , 34, 2, , x − 9x + 20, , I=∫, , 3, dx, dt + 34∫, 2, t, 81, 9, , + 20 −, x−, , 4, 2, , t 1/2, + 34∫, 1/2, , dx, 2, , 9, 1, , −, x−, , 2, 4, dx, 2, , 9, , 1, −, x−, , 2, 2, , 2, , sin x, dx ...(2), sin x + cos x, , 2, , +C, , 9, , = 6 x 2 − 9x + 20 + 34 log  x −  + x 2 − 9x + 20 + C, , 2, , 37. Let I = ∫, Let, , dx, , dx, , Put x2 – 9x + 20 = t in first integral, , 2, , π, 6, , ∫, , π, 12, , 9, 9, , , 1, = 6 t + 34 log  x −  +  x −  −  , , , 2, 2, 2, , π π, , cos, + −x, 3 6, , , π, , −x, cos, 2, , , π, 3, , ∴, , cos x, dx ...(1), cos x + sin x, , π, 3, , ∫, , I=, , =3, , , ∵, , , ⇒, , ⇒, , tan x ), , sin x ,  1 +, , cos x , , cos, , 2I, , π, π π π, =  −  = ⇒ 2I =, 3 6 6, 6, , = 3∫ t −1/2 dt + 34∫, , dx, , dx, , π, 3, , ⇒, , ∴, , 9, 9 23, + 12 − 4 = 16 − =, 2, 2, 2, π, 3, , 6, , π /3, = [x], π /6, , , d, Let 6x + 7 = A  (x 2 − 9x + 20) + B,  dx, , \ 6x + 7 = A[2x – 9] + B, Equating the coefficients of like terms from both sides,, we get, 2A = 6 and – 9A + B = 7, ⇒ A = 3 and, – 9(3) + B = 7 ⇒ B = 7 + 27 = 34, , (x − 1) − (x − 2) − (x − 4), if 1 ≤ x < 2, f (x) = , (x − 1) + (x − 2) − (x − 4), if 2 ≤ x < 4, , 4, , cos x + sin x, dx = ∫ dx, cos x + sin x, π, , ∫, , 36. Let I = ∫, , 1, , 2, , π, 3, , π, 6, , π , π π, , −1, −1,  − α ,  tan ∞ − tan (tan α) ⇒ I =, cos α, cos α  2, , 4, , π, 3, , x2 + 1, , (x − 1)2 (x + 3), , x2 + 1, 2, , (x − 1) (x + 3), , =, , dx, , A, B, C, +, +, 2, x − 1 (x − 1), x+3, , x2 +1 = A(x – 1)(x + 3) + B(x + 3) + C(x –1)2 ...(1), 1, Put x = 1 in (1), we get B =, 2, Put x = –3 in (1), we get C = 5, 8, 3, Put x = 0 in (1), we get A =, 8, ⇒

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21, , Integrals, x2 + 1, , ∴, , (x − 1)2 (x + 3), , 3 1, 1, 1, 5 1, ⋅, + ⋅, + ⋅, 8 x − 1 2 (x − 1)2 8 x + 3, , =, , ∴, , x +1, , I=∫, , 2, , (x − 1) (x + 3), , dx =, , 3, 1, dx, dx, + ∫, ∫, 8 (x − 1) 2 (x − 1)2, , = 2 ⋅ sin −1 1 = 2 ⋅, , 5 dx, 8∫x+3, , We know that sin −1 x + cos −1 x =, , 2, , 2, , x dx, , 2, 2, 0 a cos x + b sin x, , π, , π, , ⇒ I=∫, , 2, , 4, −1, 1, , = x −  x ⋅ cos −1 x − ∫ x ⋅, ⋅, dx + C, π, 1 − x 2 x , , I=, , 0, , 1, , f ( π − x) =, , ⇒, , f ( π − x) =, , ∫, , 0, , = 2, = 2, ,  sin x, +, , cos x, π /4, , 2 sin θ, ⋅ 2 sin θ cos θ dθ + C, π ∫ cos θ, 2, x − ∫ (1 − cos 2θ) dθ + C, π, sin 2θ , 2, +C, x − θ −, π, 2 , 2, x − [ θ − sin θ cos θ] + C, π, 2, x − [ sin −1 x − x 1 − x ] + C, π, x−, , ∫, , 0, π /4, , ∫, , ∴, , sin x + cos x, , dx, , 0, , π, 2, 2, , 2, , 2, , 2, , 2, , 1, , a cos (π − x) + b 2 sin 2 (π − x), 1, , = f (x), , a cos x + b 2 sin 2 x, , π /2, , ∫, , ∫, , 2, , , , a cos x + b sin x , 2, , 0, , π /2, 0, , 2, , dx, , 2, , 2, , sec2 xdx, , a + b 2 tan 2 x, , Put tan x = t ⇒ sec2xdx = dt., Also when x = 0 ⇒ t = tan 0 = 0., , π /4, , ∫, , I=, , ⇒ I=π, , sin x + cos x, cos x , dx,  dx = ∫, sin x cos x, sin x , 0, , (sin x + cos x), dx = 2, 2 sin x cos x, , a2 cos 2 x + b 2 sin 2 x, , a, 2a, , ,  using ∫ f (x)dx = 2 ∫ f (x) dx , if f (2 a − x) = f (x), , , 0, 0, , ( tan x + cot x ) dx, , π/, /4, , ...(2), , 0, , ⇒, , 0, , 0, , π, , sin 2 θ, 4, 2, ∴ I = x − x cos −1 x − ∫, ⋅ 2 sin θ cos θ dθ + C, π, π, 1 − sin 2 θ, , π /4, , ∫, , π, dx, ∫, 2, 2, 2 a cos x + b 2 sin 2 x, , Let f (x) =, , ∫, , a, , f (x)dx = ∫ f (a − x)dx , , 0, , (π − x)dx, , a cos 2 x + b 2 sin 2 x , Adding (1) and (2), we get, , π /4, , a, , , Using, , , π, 2, , π, − 2 cos −1 x, 4, I=∫ 2, dx = ∫ 1 ⋅ dx − ∫ 1 ⋅ cos −1 x dx, π /2, π, , 39. L.H.S. =, , , , 2, 2, 2, 2, 0 a cos ( π − x) + b sin ( π − x), , π, x = − cos −1 x, 2, , 4, x cos −1, π, 4, = x − x cos −1, π, 4, = x − x cos −1, π, 4, = x − x cos −1, π, 4, = x − x cos −1, π, , ...(1), , (π − x)dx, , ⇒ I=∫, , Put x = sin2q ⇒ dx = 2 sin q cos q dq, , =, , π, = R.H.S., 2, , π, , 40. Let I = ∫, , −1, x − cos −1 x, 38. Let I = sin, ∫ sin −1 x + cos−1 x dx, x ∈[0,1], , =x−, , 1 − t2, , −1, , = 2  sin −1 t  = 2 [sin −1 0 − sin −1 (−1)], −1, , 3, 1, 1, 5, = log|x − 1|− ⋅, + log|x + 3|+C 1, 8, 2 (x − 1) 8, , ∴, , dt, , ∫, , 0, , +, , ⇒ sin −1, , ∫, , 0, , ( tan x + cot x ) dx = 2, , 0, , Integrating both sides, we get, 2, , π/4, , sin x + cos x, 1 − (sin x − cos x)2, , 1 − (sin x − cos x)2, Let sin x – cos x = t, then (cos x + sin x) dx = dt, Also, x = 0 ⇒ t = – 1 and x = p/4 ⇒ t = 0., , And when x =, , dx, , \, , I=, , ∞, , dt, , π, π, ⇒ t = tan = ∞, 2, 2, , ∫ a 2 + b 2t 2, 0, , ⇒I=, , π, , b2, ∞, , ∞, , dt, , ∫  a 2, 0, , b, , ⇒, , I=, , π b, bt , tan −1   , 2  a, , a 0, b, , =, , I=, , π, π2, [, tan −1 ∞ − tan −1 0 ] =, 2 ab, ab, , 0, , , , + t2

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CHAPTER, , 2, , Applications of Integrals, Recap Notes, , INTRODUCTION, , h, , In geometry, we have learnt formulae to calculate, areas of various geometrical figures. Such formulae, of elementary geometry allow us to calculate areas of, many simple figures. However, they are inadequate for, calculating the areas enclosed by curves. For that we, shall need some concepts of integral calculus., , y, , y = f(x), x=a, , Area of shaded portion, as shown in figure, is given, by, y, y = f(x), , O x=a, , A=, , x=b, , x=c x=b, , A=, , O, , y = mx is, , A=, , b, , ∫ f (y)dy, a, , 8a2, , 3m3, , sq. units., , h, , The area of a region bounded by y2 = 4ax and its, 2, latus rectum is 8a sq. units., 3, , h, , The area of a region bounded by one arc of sinax or, 2, cosax and x-axis is, sq. units., a, x2 y 2, Area of region bounded by an ellipse 2 + 2 = 1, a, b, is pab sq. units., , h, , x, , c, , h, , y, y=b, x = f(y), y=a, , a, , x2 = 4by is 16ab sq. units., 3, The area of a region bounded by y 2 = 4ax and, , b, , Area of shaded portion, as shown in figure, is given, by, , b, , The area of a region bounded by y 2 = 4ax and, , a, , h, , c, , ∫ f (x) dx + ∫ f (x) dx, , h, , x, , ∫ f (x)dx, , x, , O, , Area Under Simple Curves, h, , Area of shaded portion, as shown in figure, is, given by, , h, , The area of a region bounded by y = ax2 + bx + c and, 3, , 2, 2, x-axis is (b − 4ac ) sq. units., 2, 6a

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Practice Time, OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1. The area bounded by the curve y = x2 + 4x + 5,, the axes of coordinates and minimum ordinate is, (a), , 3, , 2, sq. units, 3, , (b) 4, , 2, sq. units, 3, , 2 sq. units, (d) none of these, 3, x 2 y2, 2. Area of the ellipse, +, = 1 is, a 2 b2, (a) 4pab sq. units, (b) 2pab sq. units, (c), , 5, , (c), , pab sq. units, , 3., is, , πab, sq. units, 2, The area bounded by the curve 2x2 + y2 = 2, , (a) p sq. units, π, (c), sq. units, 2, , (d), , (b), , 2 π sq. units, , (d) 2p sq. units, , 4. Area enclosed by the circle x2 + y2 = a2 is, equal to, (a) 2pa2 sq. units, (b) pa2 sq. units, (c) 2pa sq. units, (d) pa sq. units, 5., , x 2 y2, = 1 is, +, 4, 9, (b) 3p sq. units, (d) none of these, , Area bounded by the ellipse, , (a) 6p sq. units , (c) 12p sq. units, , 6. The area enclosed between the curve, x2 + y2 = 16 and the coordinate axes in the first, quadrant is, (a) 4p sq. units, (b) 3p sq. units, (c) 2p sq. units, (d) p sq. units, 7., , x 2 y2, +, = 1 is, 25 9, (b) 15p sq. units, (d) 4p sq. units, , The area enclosed by the curve, , (a) 10p sq. units, (c) 5p sq. units, , 8. The area bounded by the curve y = f(x), the, x-axis and x = 1 and x = b is (b – 1) sin (3b + 4)., Then, f(x) is, , (a), (b), (c), (d), , (x – 1) cos (3x + 4), sin (3x + 4), sin (3x + 4) + 3(x – 1)⋅cos (3x + 4), none of these, , 9. The area of the region bounded by the parabola, y = x2 + 1 and the straight line x + y = 3 is given, by, 25, 45, (a), sq. units, (b), sq. units, 4, 7, 5, 9, (c), sq. units, (d), sq. units, 18, 2, 10. The area enclosed between the curve y2 = 4x, and the line y = x is, 4, 8, (a), sq. units, (b), sq. units, 3, 3, (c), , 2, sq. units, 3, , (d), , 1, sq. units, 2, , 11. The area bounded by the lines y = |x – 2|, x = 1,, x = 3 and the x- axis is, (a) 1 sq. unit, (b) 2 sq. units, (c) 3 sq. units, (d) 4 sq. units, 12. Area of the region bounded by the curve, y = x2 and the line y = 4 is, 32, (a) 11 sq. units, (b), sq. units, 3, 3, 43, 47, (c), sq. units, (d), sq. units, 3, 3, 13. Area lying between the parabola y2 = 4x and its, latus rectum is, 2, 1, (a), sq. units, (b), sq. units, 3, 3, (c), , 5, sq. units, 3, , (d), , 8, sq. units, 3, , 14. The area bounded by the curve y2 = x, line, y = 4 and y-axis is

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CBSE Board Term-II Mathematics Class-12, , 24, 64, sq. units, 3, , (a), , 16, sq. units, 3, , (b), , (c), , 7 2 sq. units, , (d) none of these, , 15. The area bounded by the curve x = 3y2 – 9, and the line x = 0, y = 0 and y = 1 is, (a) 8 sq. units, (b) 8/3 sq. units, (c) 3/8 sq. unit, (d) 3 sq. units, 16. Find the area above x-axis, bounded by the, curves y = 2kx, x = 0 and x = 2., k, , k, , (a), , 4 −1, k log e 2, , (b), , 2 −1, 2 log e 2, , (c), , 3−k, k log e 2, , (d), , −1 + 3k, 2 log e 2, , 17. Find the area enclosed by the parabola, y2 = x and the line y + x = 2 and the x-axis., 7, 5, (a), sq. units, (b), sq. units, 6, 6, 6, 4, (c), sq. units, (d), sq. units, 7, 7, , 18. The area bounded by the curve x2 + y2 = 1, in first quadrant is, π, π, (a), sq. units, (b), sq. units, 4, 2, π, π, (c), sq. units, (d), sq. units, 3, 6, 19. Area bounded by the curve y = cos x between, 3π, x = 0 and x =, is, 2, (a) 1 sq. unit, (b) 2 sq. units, (c) 3 sq. units, (d) 4 sq. units, 20. Area of the region bounded by the curve, π, y = tanx, line x =, and the x-axis is, 4, 1, log 2 sq. units, (a) log 2 sq. units, (b), 2, 1, (c), (d) 5 log 2 sq. units, log 2 sq. units, 3, 21. The area bounded by the curve y = sec2 x, y = 0, π, and |x| =, is, 3, (a) 3 sq. units, (b) 2 sq. units, (c) 2 3 sq. units, , (d) none of these, , 22. The area bounded by the curve x2 = 4y + 4, and line 3x + 4y = 0 is, , 25, sq. units, 4, 125, (c), sq. units, 16, (a), , 125, sq. units, 8, 125, (d), sq. units, 24, , (b), , 23. The area bounded by the x-axis, the curve, y = f(x) and the lines x = 1, x = b is equal to, b2 + 1 − 2 for all b > 1, then f(x) is, (a), , x −1, , (b), , (c), , x2 +1, , (d) x / x 2 + 1, , x +1, , 24. The area (in sq. units) enclosed between the, graph of y = x3 and the lines x = 0, y = 1, y = 8, is, 45, (a), (b) 14, 4, (c) 7, (d) none of these, 25. The area of the region bounded by the curve, y = 16 − x 2 and x-axis is, (a) 8p sq. units, (b) 20p sq. units, (c) 16p sq. units, (d) 256p sq. units, 26., y=, (a), (c), , Area of the region bounded by the curve, cos x between x = 0 and x = p is, 2 sq. units, (b) 4 sq. units, 3 sq. units, (d) 1 sq. unit, , 27. The area of the region bounded by parabola, y2 = x and the straight line 2y = x is, 4, (a), sq. units, (b) 1 sq. unit, 3, 1, 2, (c), sq. unit, (d), sq. unit, 3, 3, 28. The area of the region bounded by the curve, π, y = sin x between the ordinates x = 0, x = and, 2, the x-axis is, (a) 2 sq. units, (b) 4 sq. units, (c) 3 sq. units, (d) 1 sq. unit, 29. The area of the region bounded by the, x 2 y2, +, = 1 is, 25 16, (a) 20p sq. units, (c) 16p2 sq. units, ellipse, , (b) 20p2 sq. units, (d) 25p sq. units, , 30. The area of the region bounded by the circle, x2 + y2 = 1 is, (a) 2p sq. units, (b) p sq. units, (c) 3p sq. units, (d) 4p sq. units

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25, , Applications of Integrals, , Case Based MCQs, Case I : Read the following passage and answer, the questions from 31 to 35., In a classroom, teacher explains the properties of, a particular curve by saying that this particular, curve has beautiful up and downs. It starts at 1, and heads down until p radian, and then heads, up again and closely related to sine function and, π, both follow each other, exactly, radians apart, 2, as shown in figure., , y, B(1, 3), , C(3, 2), x′, , A(–1, 0), O, , y, , –2p −3π –p, , 2, , −π 0, 2, , π, 2, , p, , 3π 2p, 2, , x, , y′, , 31. Name the curve, about which teacher, explained in the classroom., (a) cosine, (b) sine, (c) tangent, (d) cotangent, 32. Area of curve explained in the passage from, π, 0 to, is, 2, 1, 1, (a), sq. unit, (b), sq. unit, 3, 2, (c) 1 sq. unit, (d) 2 sq. units, 33. Area of curve discussed in classroom from, π, 3π, is, to, 2, 2, (a) –2 sq. units, (b) 2 sq. units, (c) 3 sq. units, (d) –3 sq. units, 34., 3π, 2, (a), (c), , Area of curve discussed in classroom from, to 2p is, 1 sq. unit, 3 sq. units, , D, , x, , y′, y = sin x, , x′, , E, , (b) 2 sq. units, (d) 4 sq. units, , 35. Area of explained curve from 0 to 2p is, (a) 1 sq. unit, (b) 2 sq. units, (c) 3 sq. units, (d) 4 sq. units, Case II : Read the following passage and answer, the questions from 36 to 40., Location of three houses of a society is represented, by the points A(–1, 0), B(1, 3) and C(3, 2) as shown, in figure., , 36. Equation of line AB is, 3, 3, (a) y = (x + 1), (b) y = (x − 1), 2, 2, 1, 1, (c) y = (x + 1), (d) y = (x − 1), 2, 2, 37. Equation of line BC is, (a) y =, , 1, 7, x−, 2, 2, , (b) y =, , 3, 7, x−, 2, 2, , (c) y =, , −1, 7, x+, 2, 2, , (d) y =, , 3, 7, x+, 2, 2, , 38. Area of region ABCD is, (a) 2 sq. units, (b) 4 sq. units, (c) 6 sq. units, (d) 8 sq. units, 39. Area of DADC is, (a) 4 sq. units, (c) 16 sq. units, , (b) 8 sq. units, (d) 32 sq. units, , 40. Area of DABC is, (a) 3 sq. units, (c) 5 sq. units, , (b) 4 sq. units, (d) 6 sq. units, , Case III : Read the following passage and, answer the questions from 41 to 45., Ajay cut two circular pieces of cardboard and, placed one upon other as shown in figure. One of, the circle represents the equation (x – 1)2 + y2 = 1,, while other circle represents the equation x2 + y2 = 1.

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CBSE Board Term-II Mathematics Class-12, , 26, , x2 + y2 = 1, , 41. Both the circular pieces of cardboard meet, each other at, 1, 1, 1, (a) x = 1 (b) x =, (c) x =, (d) x =, 4, 2, 3, 42. Graph of given two curves can be drawn as, , (a) x′, , y, , x=, , 1, 2, , (x – 1)2 + y2 = 1, , x, , O, , 43. Value of, , 1/2, , ∫, , 1 − (x − 1)2 dx is, , 0, , (a), , π, 3, −, 6, 8, , (b), , π, 3, +, 6, 8, , (c), , π, 3, +, 2, 4, , (d), , π, 3, −, 2, 4, , 44. Value of, , 1, , ∫, , 1 − x 2 dx is, , 1/2, , (b) x′, , (x – 1)2 + y2 = 1, , y′, x=−, , y, , 1, 2, , x2 + y2 = 1, , x, , O, , 2, , +, 2, , y, , +y, , 1), , 2, , x, , =1, , –, (x, , 2, , =, 1, , (c) x′, , O, , (d) None of these, , π, 3, +, 2, 4, , (b), , π, 3, +, 6, 8, , (c), , π, 3, −, 6, 8, , (d), , π, 3, −, 2, 4, , 45. Area of hidden portion of lower circle is, , , , (a)  2π + 3  sq. units,  3, 2 , , y′, 1, y x=, 2, , (a), , (1, 0), , x, , y′, , , , (b)  π − 3  sq. units, 3, 8 , π, 3  sq. units, (c)  +, , 3, 8 , , , (d)  2π − 3  sq. units,  3, 2 , , Assertion & Reasoning Based MCQs, Directions (Q. 46-50) : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose, the correct answer out of the following choices :, (a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion., (b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion., (c) Assertion is correct statement but Reason is wrong statement., (d) Assertion is wrong statement but Reason is correct statement., 46. Assertion : The area of the region bounded, by the curve y 2 = 4x and the line x = 3 is, 8 3 sq. units., Reason : The area of the region bounded by, 9, the curve x2 = 4y and the line x = 4y – 2 is, 8, sq. units., 47. Assertion : The area of the smaller region, x 2 y2, bounded by the ellipse, +, = 1 and the line, 9, 4, 3, x y, + = 1 is ( π − 2) sq. units., 2, 3 2, , Reason : Formula to calculate the area of the, 2, 2, smaller region bounded by the ellipse x + y = 1, a 2 b2, ab, x y, and the line + = 1 is, ( π − 2) sq. units., 4, a b, , 48. Assertion : The area bounded by the parabola, y 2 = 4ax and the line x = a and x = 4a is, , 56a 2, sq. units., 3, Reason : The area bounded by the curves, y = 3x and y = x2 is 9.5 sq. units.

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27, , Applications of Integrals, , 49. Assertion : The area bounded by the curves y2 =, 8a, 4a2(x – 1) and lines x = 1 and y = 4a is, sq. units., 3, Reason : The area enclosed between the parabola, 4, y = x2 – x + 2 and the line y = x + 2 is, sq. units., 3, , 50. Assertion : The area bounded by the curve, y = 2cosx and the x-axis from x = 0 to x = 2p is, 8 sq. units., Reason : The area bounded by the curve, y = sinx between x = p and x = 2p is 4 sq. units., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), Find the area between the curve, y = 4 + 3x – x2 and x-axis., x 2 y2, 2. Find the area of the ellipse 2 + 2 = 1., 4, 9, 3. Find the area of the region bounded by, y = |x|, x ≤ 5 in the first quadrant., 1., , 4. Find the area of the smaller region bounded, by x2 + y2 = 9 and the line x = 1., 5. Find the area of the region bounded by the, curve y = x + 1 and the lines x = 2 and x = 3., 6. Find the area of the region bounded by the, curve x = 2y + 3 and the lines y = 1 and y = –1., , 7. Find the area lying in the first quadrant, and bounded by the circle x2 + y2 = 4 and the, lines x = 0 and x = 2., 8. Using integration, find the area of the, region enclosed by the curves y = logx, x-axis, and ordinates x = 1, x = 2., 9. Find the area bounded by the curves y = sin x,, the line x = 0 and the line x = 2p., 10. Find the area bounded by the curve y2 = 9x, and the lines x = 1, x = 4 and y = 0 in the first, quadrant., , Short Answer Type Questions (SA-I), 11. Find the area bounded by the lines y = ||x | – 1|, and the x-axis., , 16. Find the area bounded by the curve, y = x |x | , x-axis and the lines x = –3 and x = 3., , 12. Find the area of the region bounded by the, curve y2 = x and the lines x = 1, x = 4 and the, x-axis., , 17. Find the area of region bounded by the, curve y2 = 4x and the lines x = 2, x = 4 and the, x-axis., , 13. Find the area of the region bounded by, y2 = 9x, x = 2, x = 4 and the x-axis in the first, quadrant., , curve y = 4 − x 2 and x-axis., , 14. If y = 2 sin x + sin 2x for 0 ≤ x ≤ 2p,, then find the area enclosed by the curve and, x-axis., 15. Find the area of triangle whose two vertices, formed from the x-axis and line y = 3 – |x|., , 18. Find the area of the region bounded by the, 19. Using integration, find the area of region, bounded between the line x = 2 and the parabola, y2 = 8x., 20. Draw the region lying in first quadrant and, bounded by y = 9x2, x = 0, y = 1 and y = 4. Also,, find the area of region using integration., , Short Answer Type Questions (SA-II), 21. Find the area of region bounded by y = x, , 22. Draw the graph of curve y = |x + 1|., , and y = x., , Hence, evaluate, , 2, , ∫ |x + 1|dx., , −4

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CBSE Board Term-II Mathematics Class-12, , 28, 23. Find the area of the region bounded by the, 2, , curve y = 1 − x , line y = x and the positive x-axis., 24. Find the area of the region in the first, quadrant enclosed by the x-axis, the line y = x, and the circle x2 + y2 = 32., 25. Find the area bounded by the ellipse, x 2 y2, +, = 1 and the ordinates x = ae and x = 0,, a 2 b2, where b2 = a2(1 – e2) and e < 1., , 26. Find the area of the region bounded by the, parabola y2 = 2x + 1 and the line x – y – 1 = 0., 27. Find the area of smaller region bounded by, x 2 y2, x y, the ellipse, +, = 1 and the line, + = 1., 16, 9, 4 3, 28. Find the area bounded by the curve, y = 2x – x2 and the straight line y = –x., 29. AOB is a positive quadrant of the ellipse, x 2 y2, +, = 1, where OA = a, OB = b. Find the, a 2 b2, , area between the arc AB and chord AB of the, ellipse., 30. Find the area of the triangle formed by the, , (, , ), , tangent and normal at the point 1, 3 on the, circle x2 + y2 = 4 and the x-axis., 31. Draw the region bounded by y = 2x – x2 and, x-axis and find its area using integration., 32. Determine the area under the curve, y = a2 − x 2 included between the lines x = 0, and x = a., 33. Find the area of the region bounded by, y = |x – 1| and y = 1., 34. If the area bounded the curve y2 = 16x and, , 2, , then find the value of m., 3, 35. Find the area enclosed between the parabola, 4y = 3x2 and the straight line 3x – 2y + 12 = 0., line y = mx is, , Long Answer Type Questions (LA), 36. Find the area bounded by lines y = 4x + 5,, y = 5 – x and 4y = x + 5., 37. Using integration, find the area of the, region bounded by the curves :, y = |x + 1| + 1, x = – 3, x = 3 and y = 0., , (b) : We have, y = x2 + 4x + 5 = (x + 2)2 + 1, , \, , Required area =, , 39. Find the area of the region bounded by the, parabola y = x2 and y = |x|., , 40. Find the area bounded by the circle x2 + y2 = 16, and the line 3y = x in the first quadrant, using, integration., , 2., , OBJECTIVE TYPE QUESTIONS, 1., , 38. Using integration, find the area bounded by, the curve x2 = 4y and the line x = 4y – 2., , (c) : Total area, A = 4 × Area in first quadrant, (0, b), x′, , 0, , 2, ∫ (x + 4x + 5) dx, , O, , −2, , y′, , 0, ,  x3, , =  + 2x 2 + 5x , 3,  −2, , , , =2+, , 8 14, 2, =, = 4 sq. units, 3 3, 3, , 0, , 0, , O, y′, , x, , b 2, a − x 2 dx, a, a, , 4b  x 2, a2, x, a − x 2 + sin −1  = πab sq. units, = , , a 2, 2, a 0, , (0, 5), x′, , a, , = 4 × ∫ y dx = 4∫, , y, , ,  8, = −  − + 8 − 10 , 3, , , , a, , 3., , (b) : We have, 2x2 + y2 = 2, , ⇒, , x2 y 2, +, = 1, which is an ellipse, 1, 2

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29, , Applications of Integrals, , Area bounded by the ellipse, , Q, , x2, , a2, Required area = π 2 sq. units., , \, , +, , y2, , 1 1 , 8 9, , =  2 − −  −  −4 − 2 +  = sq. units, , 2 3 , 3 2, , = 1 is pab, , b2, , 4. (b) : We have, x2 + y2 = a2, which is a circle with, centre (0, 0) and radius a., \ Required area = 4 × Area in the first quadrant, a, , = 4∫ a2 − x 2 dx , 0, , a, , x 2, a2, x, a − x 2 + sin −1 , = 4, 2, 2, a 0, , x′, , O, , =, , y′, , x2, , +, , y2, , = 1 is pab sq. units., a2 b 2, \ Required area = p × 2 × 3 = 6p sq. units., 6. (a) : Given curve is a, y, circle with centre (0, 0), (0, 4), and radius 4., \ Required area, x′, , 4, , = ∫ 16 − x 2 dx, 0, , 4, , 0, , 0, 4, 2, , (4, 0), , O, , 4, , x, 16, x, =  16 − x 2 + sin −1  = 4π sq. units, 2, 4 0, 2, , 8., , (c) : Given,, , Required area, , \, 2, , 2, , 3, , 3, , ,  x2, , x2 , = ∫ (2 − x) dx + ∫ (x − 2) dx =  2x −  +  − 2x , 2, , , , 2 1, 2, 1, 2, , x2 y 2, 7. (b) : We have, +, = 1, which is an ellipse, 25 9, Here, a = 5 and b = 3, x2 y 2, Since, area of region bounded by the ellipse 2 + 2 = 1, a, b, is pab., Required area = p (5) (3) = 15p sq. units, , ...(i), ...(ii), , O, x, , y′, , \, , 32, 8, − 8 = sq. units, 3, 3, , 11. (a) : We have, y = – x + 2 " x < 2, y = x – 2 " x ≥ 2, and x = 1, x = 3, , (a) : Here a2 = 4 and b2 = 9., , Since, area of ellipse, , 4, , = ∫ ( 4x − x)dx = ∫ (2x 1/2 − x) dx , ,  x 3/2 x, 4, 42, = 2, −  = (4 3/2 ) −, 2,  3 / 2 2 0 3, ,  a2  π, = 4   = πa2 sq. units,  2  2, 5., , 10. (a) : We have, y2 = 4x...(i), and y = x...(ii), \ Required area, , x, , 2, , y=, , Here, a = 1 and b =, , 1   9 4 , , , , = (4 − 2) −  2 −   +   −  − (6 − 4), , 2   2 2 , , , 1 1, = + = 1 sq. unit, 2 2, 12. (b) : We have, y = x2...(i), and y = 4, ...(ii), , b, , ∫ f (x) dx = (b − 1)sin(3b + 4), 1, , x′, , x, , Area function = ∫ f (x) dx = (x − 1)sin(3x + 4), , y′, , 1, , \, , On differentiating, we get, f(x) = sin (3x + 4) + 3(x – 1)⋅ cos (3x + 4), 9. (d) : We have, y = x2 + 1, and x + y = 3, Solving (i) and (ii), we get, x2 + x – 2 = 0 ⇒ x = –2, 1, \ Required area, =, , 1, , ∫ {3 − x − (x, , −2, , 2, , Required area, 2, , ...(i), ...(ii), , 2,  , 32, x3 , = 2 ∫ (4 − x 2 )dx =  2  4x −   =, sq. units, , 3 0 3, , 0, , 13. (d) : We know that the area of region bounded by, 8, the parabola y2 = 4ax and its latus rectum is a2 sq. units., 3, , + 1)} dx, 1, , , x2 x3 , =  2x −, − , , 2, 3  −2, , O, , x′, y′, , (1, 0)

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CBSE Board Term-II Mathematics Class-12, , 30, Here, a = 1, therefore required area =, 14. (b) : We have, y2 = x,, which is a parabola with, vertex (0, 0) and line y = 4, \ Required area, 4, , =, , ∫y, 0, , 2, , 18. (a) : We have, x2 + y2 = 1, which is a circle with, centre (0, 0) and radius = 1., Required area, , 8, sq. units, 3, , y, , 1, , y2 =x, , (0, 4), , y=4, x, , O, , dy, , 1, , x, 1, x, =  1 − x 2 + sin −1 , 2, 1 0, 2, , y, , 3y2 = x + 9, y=1, , π /2, , 0, , 1, , 16. (a) : Required area =, , 3π /2, , ∫, , cos x dx, , π /2, , = [ sin x ], , = ∫ (3y 2 − 9)dy = [y 3 − 9y ]10 = |1 – 9| = 8 sq. units, 0, , ∫, , cos x dx +, , 0, , y′, , Required area, , π /2, , =, , x, , O, , –9, ,  1 π π, 1 =  ×  = sq. units,  2 2 4, , 19. (c) : We have, y = cosx, whose graph is shown, 3π, below, between x = 0 and x =, , 2, \ Required area, , 15. (a) : We have, x = 3y2 – 9 ⇒ 3y2 = x + 9, , \, , 0, , 1, =  sin −1, 2, , 4, ,  y3 , 64, =  =, sq. units,  3 0 3, , x′, , = ∫ 1 − x 2 dx, , 3π /2, , + [ sin x ], , π /2, , = 1 + |(– 1 – 1)| = 1 + 2 = 3 sq. units, π, 20. (b) : We have, y = tanx and x =, 4, , 2, , ∫ y dx, 0, , O, , Required area, , \, O, , =, , π /4, , ∫, , 0, , 2, , 2, , kx , , = ∫ 2 kx dx =  2, ,  k log e 2  0, 0, , = log 2 =, , 1, , 2, , 0, , 1, , 1, , 2, , , 22  , 1, 2, = (1 − 0) +  2 × 2 −  −  2 − , , , , 3, 2, 2, 2, 3 4 + 12 − 9 7, = +2− =, = sq. units, 3, 2, 6, 6, , 1, log 2 sq. units, 2, , ,4, , y2=x, , y, , π, π, ,−, 3, 3, , ,4, , 4, , ), , (0, 2) (1, 1, , (2, 0), , x′, , x, , (4, –2), ,  x 3/2  , x2 , =,  +  2x − , 2 1,  3/2  0 , , 1, + log 1, 2, , y, , O, , = ∫ x dx +∫ (2 − x) dx, , = − log, , 21. (c) : We have, y = sec2x and y = 0 and x =, , 2 2k, 1, 4k − 1, =, −, =, k log e 2 k log e 2 k log e 2, 17. (b) : The given line and, parabola meet at the points, (1, 1) and (4, –2)., \ Required area, , π /4, , tan x dx = [ − log cos x ]0, , x, , O, y′, , Required area, , \, =, , π /3, , ∫, , − π /3, , π /3, , sec2 x dx = [ tan x ]− π/3 = 2 3 sq. units, , 22. (d) : We have, x2 = 4y + 4, and 3x + 4y = 0, Solving (i) and (ii), we get x = –4, 1, , ...(i), ...(ii)

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31, , Applications of Integrals, 26. (a) : We have, y = cosx, y, , y=cosx, , 1, , x, , –1, , \, , Required area, , =, ,  3x x 2, , ∫−4  − 4 − 4 + 1  dx, , Required area, , \, , 1, , 3, 1, 125, 45 5, = − (1 − 16) − (1 + 64) + 5 =, =, sq. units, −, 8, 12, 24, 8 12, 23. (d) : We have,, , , , /2, , O, , b, , b2 + 1 − 2, , ∫ f (x) dx =, , = 2, , π /2, , ∫, , 0, , cos x dx = 2 [ sin x ]0π/2 = 2 sq. units, , 27. (a) : We have 2y = x ... (i), a straight line, and, y2 = x ...(ii), a parabola with vertex (0, 0)., Solving (i) and (ii), we get x = 0 and x = 4., , 1, , On differentiating w.r.t. b, we get, f (b) =, , 2b, 2, , 2 b +1, , x, , ⇒ f (x) =, , 2, , x +1, , 24. (a) : Given curve is y = x3 or x = y1/3, , 4, , 4, ,  2 3/2 x 2 , x, 2, 16, , −  = ×8−, = ∫  x −  dx =  x, 2, 4 , 3, 4,  3, 0, 0, , y=8, , y=1, x′, , Required area, , \, , y, , y = x3, , x, , O, , =, , 16 − 12 4, = sq. units, 3, 3, , 28. (d) : We have, y = sinx, 0 ≤ x ≤, , π, 2, , y, y′, , 1, , Required area, , \, , 8, , 8,  y 4/3  3 4/3 4/3, = ∫ y 1/3 dy = ,  = 8 − 1 ,  4/3  1 4, 1, , =, , π/2, , O, , x, , –1, , Required area, , \, , 3, 3, 45, × (16 − 1) = × 15 =, sq. units, 4, 4, 4, , 25. (a) : We have, y = 16 − x 2 ⇒ y2 = 16 – x2, ⇒ x2 + y2 = 42, which is a circle with centre (0, 0) and, radius 4 units., , =, , π /2, , ∫, , 0, , π /2, , sin x dx = [− cos x]0, , = −[0 − 1] = 1 sq. unit, , 29. (a) : Area of the region bounded by the ellipse, x2, , y2, , = 1 is pab sq. units, a2 b 2, \ Required area = p × 5 × 4 = 20p sq. units, x′, , +, , 30. (b) : We have, x 2 + y 2 = 1, a circle with centre, (0, 0) and radius 1., , O, , y, , y′, , Required area, , \, =, , 4, , ∫, , −4, , x2+y2=1, 4, , x 2, x, 42, 4 − x2 +, sin −1 , 4 2 − x 2 dx = , 2, 2, 4  −4, , = 8 sin −1(1) − 8 sin −1(−1) =, , 8π 8π, +, = 8π sq. units, 2, 2, , x′, , (–1, 0), , O, , y′, , (1, 0), , x

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CBSE Board Term-II Mathematics Class-12, , 32, Required area, , \, , 40. (b) : Area of DABC = Area of region ABCD – Area, of DACD = 8 – 4 = 4 sq. units, 41. (b) : We have, (x – 1)2 + y2 = 1, , 1, , 1, , 1, x, 2, −1 , 2, = 4∫ 1 − x dx = 4  1 − x + sin x , 2, 0, 2, 0, 1 π, = 4 × × = π sq. units, 2 2, 31. (a) : Here, teacher explained about cosine curve., π /2, , 32. (c) : Required area =, π /2, = [ sin x ]0 = sin, , ∫, , 3π /2, , From (i) and (ii), we get, , 1 − (x − 1)2 = 1 − x 2, , 3π /2, cos x dx = [ sin x ]π/2, , (x – 1)2 = x2 ⇒ 2x = 1 ⇒ x =, , 42. (c) :, , =, , =, , ∫, , cos x dx +, , ∫, , π /2, , 0, , 2π, , cos x dx +, , ∫, , cos x dx, , 3π /2, , = 1 + 2 + 1 = 4 sq. units, 36. (a) : Equation of line AB is, 3−0, 3, y–0=, (x + 1) ⇒ y = (x + 1), 1+1, 2, 37. (c) : Equation of line BC is, 2−3, y–3=, (x − 1), 3−1, 1, 1, −1, 7, ⇒ y=− x+ +3 ⇒ y=, x+, 2, 2, 2, 2, 38. (d) : Area of region ABCD, = Area of DABE + Area of region BCDE, =, , 1, , ∫, , −1, , =, , 3, , 3, 7,  −1, (x + 1) dx + ∫  x +  dx,  2, 2, 2, 1, , 1, , 1, , 2π, , 3π /2, , 3π /2, , 2, , x′, , O, , x, , (1, 0), , cos x dx = [ sin x ]3π/2, , = sin 2p – sin 3π = 0 – (–1) = 1 sq. unit, 2, 35. (d) : Required area, π /2, , y, , ∫, , 34. (a) : Required area =, , x, , +y, , +, , 2π, , 2, , 1, 2, , 2, , 2, , 3π, π, − sin, = |–1 – 1| = |–2|, 2, 2, = 2 sq. units, [Since, area can’t be negative], = sin, , x=, , y, , =1, , 1, 2, 1), , π /2, , ⇒, , –, (x, , ∫, , 1 − x 2 ...(ii), , Also, x2 + y2 = 1 ⇒ y =, , cos x dx, , 0, , π, − sin 0 = 1 – 0 = 1 sq. unit, 2, , 33. (b) : Required area =, , y = 1 − (x − 1)2 ...(i), , ⇒, , 3, ,  −x 2 7 , , 3  x2, + x,  + x + ,  −1  4, 2 2, 2 1, , 3 1, 1,   −9 21 1 7 , + −, + 1 − + 1 +  +, , , ,  4, 2 2, 2, 2 4 2 , = 3 + 5 = 8 sq. units, =, , 2−0, (x + 1), 39. (a) : Equation of line AC is y – 0 =, 3+1, 1, ⇒ y = (x + 1), 2, 3, 3,  x2 1 , 1, \ Area of DADC = ∫ (x + 1) dx =  + x ,  4 2  −1, 2, −1, 9 3 1 1, = + − + = 4 sq. units, 4 2 4 2, , y′, , 43. (a) :, , 1/2, , ∫, , 1 − (x − 1)2 dx, , 0, , 1/2, , 1, x −1,  x − 1 , =, 1 − (x − 1)2 + sin −1 ,  1   0, 2,  2, , 1 1 −1  1   1 , 11 ,  − 1 1 − + sin  −  −  −  (0), 22 , 4 2, 2, 2, 1, − sin −1( −1), , 2,  −1 3 1 π, 1 π − 3 π π, = ⋅, − ⋅ +0+ ⋅  =, − +, 2 2, 8, 12 4, 4 2 2 6, =, , =, , π, 3, −, 6, 8, , 44. (c) :, , 1, , ∫, , 1/2, , 1, , 1, x, , 1 − x 2 dx =  1 − x 2 + sin −1 x , 2, 2, 1/2, , 1, 1, 1 1,  1, = 0 + sin −1 (1) −, 1 − − sin −1  ,  2, 2, 4, 4 2, =, , π, 3 π π, 3, −, −, = −, 4, 8 12 6, 8, , 45. (d) : Required area, , 1/2, , ∫, , , = 2, , , , 1 − (x − 1)2 dx +, , 0, , 1, , ∫, , 1/2, , , 1 − x 2 dx , , , π, 3 π, 3, = 2 −, + −, , 8, 6, 8 , 6, π, 3   2π, 3, −, = 2 −,  sq. units,  = , 4 , 3, 2 , 3

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33, , Applications of Integrals, 46. (b) : Assertion : We have, y2 = 4x and x = 3., \ Required area, y, 3, , 3, , y2 = 4x, , = 2 ∫ y dx = 2 ∫ 2 x dx, 0, , 3/2  3, , Reason : The intersection points of given curves are, (0, 0) and (3, 9)., , 0, , x′, , x, = 4, ,  3 / 2 0, 8, = (3 3 ) = 8 3 sq. units, 3, , x, , O, , x=3, , y′, 2, , x, Reason : We have, x = 4y ⇒ y =, 4, x+2, ., and x = 4y –2 ⇒ y =, 4, The point of intersection of given curves are A(2, 1), 1, , and B  −1,  ., , 4, 2, , Required area =, , \, , Required area =, , 2, , x+2,  dx −, 4 , , ∫ , , −1, 3 2, , 2, , 2, , ∫, , −1, , x2, dx, 4, , =, , , 1  x2, 1 x ,  + 2x  −  ,  −1 4  3  −1, 4 2, , =, , 1, 3 1, 15 3 9, ×9=, − = sq. units,  6 +  −, 4, 2, 12, 8 4 8, , 2, , ) dx, , 3, ,  3x 2 x 3 , 27, =, −  =, = 4.5 sq. units,  2, 3 0 6, 49. (d) : Assertion : On solving y 2 = 4a 2(x – 1) and, y = 4a, we get x = 5, y, , O, , y=4a, (1, 0)A, , Required area =, , \, , x, , x=1, , y′, , x=5, , 5, , ∫ ( 4a − 2 a, , x − 1 ) dx, , 1, 3/2  5, , , (x − 1), =  4ax − 2 a ⋅, 3 /2, , , 47. (a) : Clearly, reason is correct statement., Now, we have, equation of ellipse, , 2, , 0, , x′, , \, , 3, , ∫ (3x − x, , 16a, sq. units,  =, 3, 1, , Reason : Given, parabola y = x2 – x + 2 and the line, y = x + 2 intersects each other at points (0, 2) and (2, 4)., y, , y=x+2, , 2, , y, x, x y, +, = 1 and line + = 1, 3 2, 9, 4, \ Here, a = 3, b = 2, ab, \ Required area = (π − 2), 4, 3×2, 3, =, (π − 2) = (π − 2) sq. units, 4, 2, 48. (c) : Assertion :, , (2, 4), (0, 2), x′, , x, , O, , Required area =, , \, , y=x2–x+2, , 2, , y′, , ∫ (x + 2) − (x, 0, , 2, , − x + 2) dx, , 2, , 2,  −x 3, , 8, 4, = ∫ (−x 2 + 2x) dx = , + x 2  = − + 4 = sq. units,  3, 0, 3, 3, 0, , 50. (c) : Assertion : We have, y = 2cosx, Let us draw the graph of 2cosx between 0 to 2p., y, 4a, , Required area = 2 ∫, a, , =, , 4a, ,  x 3/2 , 4ax dx = 4 a , ,  3 / 2 a, 2, , 8 ( 3/2, 56a, a 8a, − a 3/2 ) =, sq. units, 3, 3, , x, , O, y, , /2, , 3 /2, , x

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CBSE Board Term-II Mathematics Class-12, , 34, , =, , 4. We have, x2 + y2 = 9, and x = 1., \ Required area, , Required area, , \, , π /2, , ∫, , 2 cos x dx +, , 3π /2, , ∫, , 2 cos x dx +, , π /2, , 0, π /2, 0, , = 2 [ sin x ], , 2π, , ∫, , 2 cos x dx, , 3π /2, 3π /2, π /2, , + [ 2 sin x ], , 2π, 3π /2, , + [ 2 sin x ], , 3π, 3π , π, π, , , , , − sin  + 2 sin 2 π − sin , = 2 sin − 0  + 2 sin, 2, 2, 2, 2, , , , , = 2 + 2 × 2 + 2 = 2 + 4 + 2 = 8 sq. units, Reason : We have y = sinx, Let us draw a graph of sinx., (0, 1), , 3, , 2, = 2  ∫ 9 − x dx ,  1, , , x′, , 3, , x, 9, x, = 2  9 − x 2 + sin −1 , 2, 2, 3 1, , We have, y = x + 1, which is a straight line, , 5., , =, , x′, , Required area, 2π, , ∫ sin x dx, , y′, , ,  x2, 9,  4, , = ∫ (x + 1 ) dx =  + x  =  + 3  −  + 2 , , , , , , , 2, 2, 2, 2, 2, , SUBJECTIVE TYPE QUESTIONS, 1. We have, y = 4 + 3x – x2, a parabola with vertex at,  3 25  .,  , , 2 4, Putting y = 0, we get x2 – 3x – 4 = 0, ⇒ (x – 4)(x + 1) = 0 ⇒ x = –1 or x = 4, , 4, , 4, , ∫ (4 + 3x − x, , 2, , 15, 7, − 4 = sq. units, 2, 2, 6. We have x = 2y + 3, a straight line, =, , ) dx, , –1.5, , −1, , , 3x 2 x 3 , 125, sq. units, =  4x +, −  =, , 2, 3  −1, 6, x2 y 2, 2. Since, area of the ellipse 2 + 2 = 1 is pab., a, b, 3. We have, y = –x, if x < 0, y = x, if x ≥ 0, , Required area, , \, =, , 1, , ∫ (2y + 3) dy, , −1, , 1, , =  y 2 + 3y  −1, , = (1 + 3) – (1 – 3) = 4 + 2 = 6 sq. units, , Required area = p × 4 × 9 = 36p sq. units., , \, , 3, , 3, , = |– 1 – 1| = 2 sq. units, , Required area =, , Required area, , \, , π, , = |–cos2p + cosp|, , \, , 1, –1 O, , 2π, , = [ − cos x ]π, , y′, , 3 1, 9, 1, 9, = 2  sin −1 −, 8 − sin −1 , 2, 3 2, 2, 3, 1,  −1, , = − 8 + 9  sin 1 − sin −1 , , 3, , (0, –1), , \, , O, , ... (i), ...(ii), , 2, , 7., 2, , Required area = ∫ y dx, , = ∫ 4 − x 2 dx, 0, , 0, , , 2, , x, 4, x, =, 4 − x 2 + sin −1 , 2, 2, 2 0, Required area, , \, 5, , 5, ,  x2 , 25, = ∫ x dx =   =, sq. units,  2 0 2, 0, , = [0 + 2 sin–1 (1)] – [0 – 0], π, = 2 = π sq. units, 2, 8., , 2, , Required area = ∫ log xdx = [ x log x − 1]12, 1

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35, , Applications of Integrals, = 2log 2 – 1 = log 4 – log e, 4, = log, sq. units, e, , 2 [ 3/2, 2, 14, 4, − 1] = [8 − 1] = sq.units., 3, 3, 3, 13. The given parabola is y2 = 9x. It is symmetrical about, positive x-axis., y, , (), , =, , y, 1, , 9., x, , 4, , Required area = ∫ y dx, , ∴, , 2, , 4, , x, , O, , 2, , = ∫ 3 x dx , , x, , 2, , –1, , Required area, 2π, , 0, , π, , π, , = − cos π + cos 0+|− cos 2 π + cos π|, = 1 + 1 + |–1 –1| = 2 + |– 2| = 2 + 2 = 4 sq. units, , 4, , 4, ,  x 3/2 , = ∫ 3 x dx = 3 , ,  3 / 2 1, 1, , y, , x′, , y2=9x, x, , O, , 14. We have, y = 2sin x + sin 2x, 0 ≤ x ≤ 2p, Putting y = 0, we get 2sin x + sin 2x = 0, y, ⇒ 2sin x + 2sin x cos x = 0, ⇒ 2sin x (1 + cos x) = 0, 1, 3π, ⇒ sin x = 0 or cos x = –1, π 2 2π, x′, ⇒ x = 0, p, 2p, O, π/2, \ Required area, = 2 ∫ (2 sin x + sin 2x)dx, , y′, , π, , 11. We have, y = |x – 1|, if x ≥ 0, , cos 2x , , = 2  −2 cos x −, , 2  0, cos 2 π  , cos 0  , , = −2   2 cos π +,  −  2 cos 0 +, , , 2, 2  , , , (x − 1), if x ≥ 1, and, =, −(x − 1), if x < 1, , 1, 1, 1 , 1 , , , = −2   −2 +  −  2 +   = −2  −2 + − 2 − , , , , , , 2, 2, 2, 2 , , , y = |– x – 1| = | – (x + 1)|, = |x + 1| if x < 0, , = –2[–4] = 8 sq. units, , y′, , = 2(43/2 – 1) = 2(8 – 1), = 14 sq. units, , (x + 1), if x ≥ −1, =, −(x + 1), if x < −1, , x=4, , x′, y′, , 1, , Required area = 2 ∫ (1 − x) dx, , \, , 0, , 0, , 1, , −3, , 12. Since the given parabola y2 = x is symmetrical about, positive x-axis, y, , y =x, 2, , x, , x, y, 4, , Required area = ∫ y dx, , ∴, , 1, , 4, , 4, , 4, ,  x 3/2 , = ∫ x dx = ∫ x 1/2 dx = , ,  3 / 2 1, 1, 1, , 15. We have, y = 3 – |x|, ⇒ y = 3 + x, " x < 0, and y = 3 – x, " x ≥ 0, \ Required area = area of shaded region, 0, , x2 , = 2 ∫ (3 + x) dx = 2  3x + , , 2  −3, , , x2 , 1, = 2  x −  = 2 × = 1 sq. unit, , 2 0, 2, , x, , –1, , π, 0, , x=1, , y, , = 2 [ 4 3/2 − 2 3/2 ] = 2[8 − 2 2 ] = (16 − 4 2 ) sq. units., , 2π, , = ∫ (sin x) dx+| ∫ sin x dx|= [ − cos x ]0 + [ − cos x ]π, , 10. We have, y2 = 9x and, lines x = 1, x = 4, \ Required area, , x, , 4, , 4,  x 3/2 , = 3∫ x 1/2 dx = 3 , ,  3 / 2 2, 2, , y, , π, , y2 = 9x, , 9, , = −2  −9 + , , 2 , , −9, = 9 sq. units , 2, 16. The equation of the curve is, = −2 ×, ,  x 2 , x ≥ 0, y =x x =, 2, − x , x < 0, , ...(i), ...(ii)

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CBSE Board Term-II Mathematics Class-12, , 36, \ Required area = 2(Area of region shaded in first, quadrant), 3, , 20. The rough sketch of the curve y = 9x2, x = 0, y = 1, and y = 4 is as shown in the figure., y, , 3, ,  x3 , = 2 ∫ x dx = 2 ×   = 2 × 9 = 18 sq. units,  3 0, 2, , (0, 4), , 0, , 17. Since the given curve represented by the equation, y2 = 4x is a parabola, y, y2 = 4x, \ Required area, , (0, 1), O, , x′, , 4, , = ∫ ydx, 2, , y′, , , , 4, , = ∫ 2 x dx =, 2, , 4, , 4 1, 2 x2, , ∫, , x, , x, , dx, , 2, , =, , −2, , 2  3/2, 2, 14, 4, − 1 = (8 − 1) =, sq.units, 9, 9, 9, , 21. We have, curves y = x and y = x., y, , , 1), , A (1, , C(0, 2), , ⇒ x 2 + y 2 = 4, which is, circle with centre (0, 0) and, radius 2 units, \ Required area, 2, , 1, , 4, y, 1 2, , dy =  (y )3/2 , 3, 3 3, 1, , y, , 18. We have, y = 4 − x 2, , ∫, , 4, 1, , y, , 4, , The required area of shaded region = ∫ x dy, =∫, ,  3, 3, 3, x2 , 4, 4, = 2   = (4) 2 − (2) 2  = (8 − 2 2 ) sq. units, 3, 3,  3,  , , ,  2 2, , =, , x, , x′, , B(–2, 0) O, , A(2, 0), , x, , O, , x′, , x, , y′, , 2, , x,  x , 4 − x 2 dx = , 4 − x 2 + 2 sin −1   ,  2   −2, 2, ,  2,  2 , = , 4 − 4 + 2 sin −1   ,  2 , 2, , , y′, , The points of intersection of y = x and y = x are, O(0, 0) and A(1, 1)., 1, ,  −2,  −2  , −, 4 − (−2)2 + 2 sin −1   ,  2  , 2, π, π, , = 1 × 0 + 2 × + 1 × 0 + 2 ×  = 2 π sq. units, 2, 2, , , The required area of shaded region = ∫ ( y 2 − y1 ) dx, 0, , where y 2 = x and y1 = x, 1, , ∴, , ∫(, , Required area =, , ), , x − x dx, , 0, , 1, ,  2x 3/2 x 2 , 2 1 1, =, −,  = − = sq.unit, 3, 2, ,  0 3 2 6, , 19. The rough sketch of the parabola y2 = 8x and line, x = 2 is as shown in the figure., , 22. We have, y = | x + 1|, −(x + 1), x < − 1, y =,  (x + 1), x ≥ − 1, The graph of the curve y = | x + 1| is shown in figure., ∴, , y, , y=, 2, , 2, , 0, , 0, , x = –4, , The area of shaded region = 2 ∫ y dx = 2 ∫ 2 2x dx, 2, , = 4 2∫, , 0, , 2, , x′, , , 2, xdx = 4 2  x 3/2 , 0, 3, , 2, 32, = 4 2 ×  2 3/2 − 0  =, sq.units, 3, 3, , –(, , x+, , (x +, y=, , 1), (–1, 0) O, , (–4, 0), , 1), x= 2, , (2, 0), , y′, , ∴, , 2, , −1, , −4, , −4, , ∫ |x + 1|dx =, , ∫, , −(x + 1)dx +, , 2, , ∫ (x + 1) dx, , −1, , x

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37, , Applications of Integrals, −1, , 2, , 4, , 4 2, , 0, , 4, ,  x2, ,  x2, , =−, + x + , + x,  2,  −4  2,  −1, , ∴ Required area =, ,  1,   16,   4,  1, , = −  − 1 − , − 4  +  + 2  −  − 1, , , , , , , , , 2, 2,  2,   2, , 4, x, x , 32, x, =  +, 32 − x 2 + sin −1, 2, 4 2  4,  2  0  2, ,  1 8 8 1 9 9, = − − −  +  +  = + = 9,  2 2 2 2 2 2, 23. We have, curve y = 1 − x 2 ⇒ x2 + y2 = 1 and line, y=x, The rough sketch of the curve and line y = x is shown, in the figure., y, , y=, , x, , A, O, , x′, , 1,1, 2 2, x2 + y2 = 1, , 1 ,0 (1, 0), 2, , x, , ∫ xdx + ∫, , 32 − x 2 dx, , 2 4, , 2, ,   32, 4 2  4, 32, 4 , −, = 8 +   0 + sin −1, 32 − 16 + sin −1, , , , 2, 2, 2, 4 2, 4 2 , , , 1 , , = 8 + 16 sin −1 1 − 8 − 16 sin −1, 2 , , 16π, π, =8+, − 8 − 16   = 4π sq.units, 4, 2, , 2, 2, 25. The rough sketch of ellipse x + y = 1 and the line, 2, 2, a, b, x = 0 and x = ae is shown in the figure., y, , y′, , The intersection points of line y = x and x2 + y2 = 1 are, 1 ,  1, ,, O(0, 0) and A , .,  2, 2, ∴, , Required area =, , 1, 2, , ∫, , x dx +, , 1, , ∫, , 1, , y, , ae, , The required area of shaded region = 2 ∫ y dx ,, , 1,  x2  2  x, 1, , =   +  1 − x 2 + sin −1 x , 2,  1, 2,  2  0, , b 2 2, 2b, a − x ∴ Area =, a, a, ae, 2b  x 2, a2, 2, −1 x , a −x +, =, sin, , , 2, a , a  2, 0, , where y =, , 2, , 1  1, 1 1, 1 ,   1, = +   0 + sin −1 1  − , 1 − + sin −1, , , , 4 , 2, 2 2, 2  , 2 2, , ae, , ∫, , 0, , a2 − x 2 dx, , 0, , 2b , ae , ae a 2 − a 2 e 2 + a 2 sin −1  − 0, 2 a , a, b, =  ae a2 (1 − e 2 ) + a2 sin −1 e , , a , =, , 1 1 π 1 1 π, = + × − − × , 4 2 2 4 2 4, =, , x, , 1 − x 2 dx, , 1, 2, , 0, , x, , 1 π 1 π π, + − − = sq.unit, 4 4 4 8 8, , 24. We have, x2 + y2 = 32, ...(i), and y = x...(ii), Solving (i) and (ii), the intersection points are, O(0, 0) and A(4, 4) in first quadrant., The rough sketch of the circle x 2 + y 2 = 32 and line, y = x is shown in the figure., , = ab  e 1 − e 2 + sin −1 e  sq.units, , , 26. The rough sketch of the parabola y2 = 2x + 1 and line, x – y – 1 = 0 is as shown in the figure., y, (4, 3), , y, , x′, , (4, 4), A, O, x′, , y′, , B, (4, 0) (4 2, 0), , O, , x, (0, –1), , x, , y′, , The intersection points of y2 = 2x + 1 and x – y = 1 are, (0, –1) and (4, 3).

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CBSE Board Term-II Mathematics Class-12, , 38, The required area of shaded region, =, , 3, , ∫ (x1 − x2 )dy, , −1, , where x1 = y +1 and x2 =, Area =, , ∴, , 3, , 0, 3, , 2, , y −1, 2, ,  y2 − 1 ,   dy, 2  , , , , ∫ (y + 1) − , , −1 , , 3, , 3, , x2 y 2, +, = 1 and the line, 16, 9, x y, + = 1 is shown in the figure., 4 3, , 27. The rough sketch of ellipse, , Y, , (0, 3), (4, 0), O, , X, , a, , b, b 2, , = ∫, a − x 2 − (a − x)  dx, a, , a, =, , 0, , ( 16 − x, , 2, , a, , 0, , ), , 4, , 28. The curve y = 2x – x 2 represents a parabola, opening downwards and cutting x-axis at (0, 0) and, (2, 0). Clearly, y = –x represents a line passing through, the origin and making 135° with x-axis. A rough sketch, of the two curves is shown in the figure. The region, whose area is to be found is shaded in figure. The two, curves intersect each other at (0, 0) and (3, –3)., ,  b, b  a2, a2 , =  sin −1 1 −  a2 − ,  a, 2, a 2, ab π, 1, ab,  , =, − ba   = ( π − 2 ) sq.units, 2 4, 2 2, 3y = 4, , y, , =, , 1, , ∫x, 0, , 4, , 4−x, dx, 3, 1, , 3 dx + ∫, , (1, 3), , x, (3, –3), , y = –x, , (4, 0), , x, , 31. We have, y = 2x – x2 ⇒ – y = x2 – 2x, ⇒ –y + 1 = x2 – 2x + 1 ⇒ –(y – 1) = (x – 1)2, Clearly it represents a parabola opening downwards, whose vertex is (1, 1) and cuts x-axis at (0, 0) and (2, 0)., The rough sketch of the curve is given below :, y, , x, , x, y, , (2, 0), , O, , 3 3 3, 1 1 , 1, , +, = 2 3 sq. units, 4(4 − 1) − (16 − 1) =, = 3× +, , 2, 2, , , 2, 2, 3, , y = 2x – x2, O, , 4, ,  x2 , 1 , x2 , = 3  +,  4x −  ,  2 0, 2 1, 3, , y, , y, , a, , b x 2, a, x, b, x2 , a − x2 +, = , sin −1  −  ax − , 2, 2 0, a 2, a 0 a , , 1, , − 4 + x dx, , 3 x, x, x2 , 16, sin −1 − 4x +, =  16 − x 2 +, , 4  2, 2, 4, 2 , 0, 3, = 0 + 8 sin −1 1 − 16 + 8 − 0 , 4, 3, π,  3, = 8 × − 8 = [ 4π − 8] = 3(π − 2) sq.units, 4, 2,  4, , x, , a, , 2, , and equation of normal at (1, 3 ) is y = x 3., \ Required area, , 3, 3, Required area = ∫  16 − x 2 − (4 − x) dx, 4, 4, , , 0, , 4, , a, , b, b, a2 − x 2 dx − ∫ (a − x) dx, ∫, a, a, 0, , 4, , 3, 4∫, , 3, , 3, x3 , = ∫ (3x − x 2 ) dx =  x 2 − , 3 ,  2, 0, 0, 27 27 9, =, −, = sq.units, 2, 3 2, 29. Required area, , 30. The tangent on x 2 + y 2 = 4 at (1, 3 ) is x +, , Y, , =, , }, , 0, ,  9 27 9   1 1 3  16, = −, + − + − =, sq.units, 2, 6, 2 2 6 2 3, , X, , {, , = ∫ 2x − x 2 − (−x) dx, ,  y 2 y 3 3y ,  y2, y3 1 , −, +, =, +y−, + y = , , 6, 2 , 2, 6, 2 , ,  2, , −1, −1, , ∴, , Required area, , \, , 3, , \, ∴, , 2, , Required area = ∫ ydx, 2, , 0, , 2, , = ∫ (2x − x )dx, 0

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39, , Applications of Integrals, 2, , 3x 2 3x + 12, =, 4, 2, , , x3 , 8 4, =  x 2 −  = 4 − = sq.units, 3 , 3 3, , 0, , y, , 32. We have given the equation of curve, y = a 2 − x 2 ⇒ y2 = a2 – x2, ⇒ y2 + x2 = a2, a circle with centre (0, 0) and radius a, Thus the required area = area of the shaded region, a, , = ∫ a2 − x 2 dx , , x, , x, , y, , 0, , y, , a, , x 2, a2, x, a − x 2 + sin −1 , =, 2, 2, a 0, , x, , (–a, 0), x=0, ,  πa2, , a2, a2, = 0 + sin −1(1) − 0 − sin −1(0) =, , , 2, 2, 4, , (a, 0), x=a, , x, , y, , 33. We have, y = x – 1, if x – 1 ≥ 0, , , y = – x + 1, if x – 1 < 0, y, y=x–1, , y=1–x, (0,1), , O, , y=1, x, , (2,1), (1, 0), , ⇒, ⇒, \, , x2 – 2x – 8 = 0 ⇒ (x + 2) (x – 4) = 0, x = –2, 4, 4,  3x + 12 3 2 , Required area = ∫ , − x  dx,  2, 4 , 3 4, , −2, , 3, x, =  x 2 + 6x − , 4, 4  −2, 64   3, 8,  3 × 16, =, +6×4−  −  ×4−6×2+ ,  4, 4  4, 4, = 27 sq. units., 36. We have, y = 4x + 5, y = 5 – x and 4y = x + 5, The rough sketch of the lines is shown in the figure., , Required area, , \, 2, , 1, , y, 2, , , = ∫ 1 dx −  ∫ (1 − x) dx + ∫ (x − 1) dx ,  0, , 0, 1, 1, , 2, , , x2   x2, 1 1, 2 , = [ x ]0 −  x −  −  − x  = 2 − − = 1 sq. unit, 1, , 2 0  2, 2 2, , =, , 16 /m2, , ∫, , ( 16x − mx)dx =, , 0, , y, , 2, 3, 16 /m2, , ⇒, , ,  x2 , 2 3/2, − m  , 4× x, 3,  2   0, , , ⇒, , 1  512, 8 64 m 256 2,  2, − 128 =, × 3−, = ⇒ 3, 4,  3, , 3, 3 m, 2 m, 3, m, , ⇒, , m=4, , =, , 2, 3, , Here, equation of line AB is y = 4x + 5,, equation of line AC is y = 5 – x and, x +5, equation of line BC is y =, 4, The intersection point of line AB and AC is at A(0, 5)., Similarly, the intersection point of line AB and line BC is at, B(–1, 1) and the intersection point of line AC and BC is, at C(3, 2)., \, , Required area =, , 0, , 3, , ∫ (4x + 5)dx + ∫ (5 − x)dx −, , −1, , ⇒, , Solving (i) and (ii), we get, , 2, , 3x, ...(i), 4, 3x + 12, y=, ...(ii), 2, , 35. Given equations are y =, and 3x – 2y + 12 = 0, , x, , x, , 34. We have, y 2 = 16x, a, parabola with vertex (0, 0) and, line y = mx., \ Required area, , =  2x 2 + 5x , , 0, −1, , 0, , 2 3, , 3, , ∫, , −1, , 3, , , , 1  x2, x, +  5x −, + 5x ,  − , 2 , 4  2, ,  −1, 0, , 9  1  9, ,  1, , = 0 − (2 − 5) +  15 −  −   + 15  −  − 5  , ,  2, , 2  4  2, , x+5, dx, 4

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CBSE Board Term-II Mathematics Class-12, , 40, 21 1  39 9 , −  + , 2, 4 2, 2, 21, 21, 15, =3+, −6 =, −3=, sq.units, 2, 2, 2, , 1, 1, 1, = [4 − 1] + [2 + 1] − [8 + 1], 2, 12, 8, , 37. Given, y = |x + 1| +1, , 39. The given curves are y = x2...(i), , =3+, , =, , x + 2 if x ≥ −1, y=,  −x if x < −1, , \, , We now draw the lines : y = 0, x = 3, x = –3 and, y = x + 2 if x ≥ - 1, y = - x if x < – 1, Lines (i) and (ii) intersect each other at (– 1, 1), , 3 3 3 3 1, 1 3 3 9, + − =, + 1 −  =   = sq. units, 8 2 4 2  4, 2 2 4 8, ,  x , if x ≥ 0, y = |x| = , , −x , if x < 0, , ...(ii), , ...(i), ...(ii), , y, , 1,, , (–, , 1), x, , x, , Their points of intersection are A(1, 1), O(0, 0) and, B(–1, 1)., \ Required area, , y, , Required area =, , \, , 2  −1, , −1, , 3, , −3, , −1, , ∫ (−x)dx + ∫ (x + 2)dx, , 1, 3 1,  2,  1 1 1, = 2 ∫ (x − x 2 )dx = 2  x − x  = 2  −  = sq.unit., 2 3 3, 2, 3 0, 0, , 3, , ,  2, , = −  x  +  x + 2x ,  −1,  2  −3  2, , 40. We have curves, y =, , 1, 1, = − (1 − 9) + (9 − 1) + 2(3 + 1), 2, 2, = 4 + 4 + 8 = 16 sq. units., , 2, , 2, , and x + y = 16, , 38. The given curve is x2 = 4y...(i), and line is x = 4y – 2, ...(ii), y, , x2 = 4y, , y, x=4, , Required area = Area of region OBAO, = area DOBC + area of region BCAB, 2 3, , ∫, , x, 3, , 4, , ∫, , dx +, , 16 − x 2 dx, , 2 3, 2 3, , 4, , 16, x,  x , +  16 − x 2 + sin −1   ,  4  2 3, 2, 2, , Solving (i) and (ii), we get (x + 2) = x2, ⇒ x2 – x – 2 = 0 ⇒ (x – 2) (x + 1) = 0 ⇒ x = 2, –1, Thus the points of intersection of the given curve and, 1, line are A  −1,  and B (2, 1), , 4, , =, , 2, , x+2,  dx −, 4 , , ∫ , , −1, , =, , x′, , Required area, , \, , 2 2, , 2, , ∫, , −1, , 2, , x, dx =, 4, , 2, , x, , 1, , ∫  4 + 2 −, , −1, , 3 2, , 1 x, 1 2, 1 x,   + [x]−1 −  , 4  2 , 2, 4  3 , −1, −1, , 2, , x, 4, , y′, , ,  dx, , , ...(ii), , (−2 3, − 2)., ,  x2 , =, ,  2 3  0, , y, , x ...(i), , \, , 0, , x, , x, , 3, , Curves (i) and (ii) intersect each other at (2 3, 2) and, , =, , –2, , 1, , 8π, π, = 2 3 +8  −2 3 −, 2, 3, =, , 12π − 8π 4π, =, sq.units, 3, 3, , 

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CHAPTER, , 3, , Differential Equations, Recap Notes, , DIFFERENTIAL EQUATION, An equation involving an independent variable, a, dependent variable and the derivatives of the dependent, variable is called differential equation., h A differential equation involving derivatives, of the dependent variable with respect to only, one independent variable is called an ordinary, differential equation., h A differential equation involving derivatives with, respect to more than one independent variables is, called a partial differential equation., , Order and Degree of a Differential Equation, The order of highest derivative appearing in a, differential equation is called order of the differential, equation., h The power of the highest order derivative appearing, in a differential equation, after it is made free, from radicals and fractions, is called degree of the, differential equation., Note : Order and degree (if defined) of a differential, equation are always positive integers., h, , h, , h, , METHODS OF SOLVING DIFFERENTIAL, EQUATIONS, h, , h, , dx, , are functions of x only (or constants) is called a linear, differential equation of the first order., , SOLUTION OF A DIFFERENTIAL EQUATION, h, , Solution of a differential equation is a function, , g (x , y ), , f(x, y) and g(x, y) are homogeneous functions of, the same degree in x and y, then put y = vx and, , dy, dv, so that the dependent variable y is, =v+x, dx, dx, , dy f (x , y ), =, dx g (x , y ), , LINEAR DIFFERENTIAL EQUATIONS, dy, An equation of the form, + Py = Q where P and Q, dx, , Equation Reducible to Homogeneous Form : If, the equation is of the form dy = f (x , y ) , where, , A differential equation of the form, , where, f (x, y) and g (x, y) are homogeneous functions of, x and y of the same degree., , Equation in Variable Separable Form : If the, differential equation is of the form f (x) dx = g(y) dy,, then the variables are separable and such equations, can be solved by integrating on both sides. The, solution is given by, , ∫ f (x) dx = ∫ g(y) dy + C, where C is an arbitrary constant., , HOMOGENEOUS DIFFERENTIAL EQUATIONS, h, , of the form y = f (x) + C which satisfies the given, differential equation., General Solution : The solution of a differential, equation which contains a number of arbitrary, constants equal to the order of the differential, equation., Particular Solution : A solution obtained by giving, particular values to arbitrary constants in the, general solution., , h, , changed to another variable v, then apply variable, separable method., Solution of Linear Differential Equation : A, differential equation of the form, , dy, + Py = Q ,, dx, , where P and Q are functions of x (or constants) can, be solved as :, , 1. Find Integrating Factor (I.F.) = e ∫, 2. , The solution of the differential equation is, P dx, , y(I.F.) = ∫ Q (I.F.) dx + C , where C is constant, , of integration.

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CBSE Board Term-II Mathematics Class-12, , 42, , Practice Time, , OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1. The order of the differential equation whose, solution is y = acosx + bsinx + ce–x is, (a) 3 , (b) 2, (c) 1 , (d) none of these, 2. The order of the differential equation whose, general solution is given by, , y = (C1 + C2 )cos ( x + C 3 ) − C4, , e x + C5, , where C1, C2, C3, C4, C5 are arbitrary constants, is, (a) 5, (b) 4, (c) 3, (d) 2, 3. The differential equation of all circles of, radius a is of order, (a) 2 , (b) 3, (c) 4 , (d) none of these, 4. The differential equation of all parabolas, whose axis of symmetry is along x-axis is of, order, (a) 3 , (b) 1, (c) 2 , (d) none of these, 5. The degree of differential equation of all, curves having normal of constant length c, is, (a) 1 , (b) 3, (c) 4 , (d) none of these, 6. The order and degree respectively of the, differential equation, 3, ,   dy  2  2 d2 y, is, 1 +,  =,   dx  , dx 2, , (a) 2, 4 , (c), 7., y=, (a), (b), (c), (d), 8., y=, , 3, not defined, , 3, , (b) 3, 2, (d) 2, 2, , The differential equation whose solution is, Ae3x + Be–3x is given by, y2 – 3y1 + 3y = 0, xy2 + 3y1 – xy + x2 + 3 = 0, y2 – 9y = 0, (y1)3 – 3y(xy2 – 3y) = 0, The differential equation satisfied by, , A, + B is (A, B are parameters), x, , (a) x2y1 = y , (c) xy2 + 2y1 = 0, , (b) xy1 + 2y2 = 0, (d) none of these, , 9. The differential equation whose solution, represents the family xy = Aeax + Be–ax is, 2, , (a), ,  d2 y , dy, x, +2, = xy, 2, dx,  dx , , (b), ,  d2 y , dy, x, +2, = a2xy, 2, dx,  dx , , (c), , x, , 2, , d2 y, 2, , +2, , dy, = a2xy, dx, , dx, (d) none of these, , 10. The differential equation which represent, the family of curves y = aebx, where a and b are, arbitrary constants, is, (a) y′ = y2 , (b) y′′ = y y′, (c) y y′′ = y′ , (d) y y′′ = (y′)2, 11. The differential equation having solution as, y = 17ex + ae–x is, , (a) y′′ – x = 0 , (c) y′ – y = 0 , , (b) y′′ – y = 0, (d) y′ – x = 0., , 12. The order of the differential equation of all, tangent lines to the parabola y = x2, is, (a) 1, (b) 2, (c) 3, (d) 4, 13. The order of the differential equation whose, general solution is given by y = (A + B) cos (x + C), + Dex is, (a) 4, (b) 3, (c) 2, (d) 1, 14. The differential equation of all circles in the, first quadrant which touch the coordinate axes, is of order, (a) 1 , (b) 2, (c) 3 , (d) none of these, 15. The degree of the differential equation, 2/ 3, ,  d2 y , dy, = 0 is,  2  +4−3, dx,  dx , (a) 2 , (b) 1, (c) 3 , (d) none of these

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43, , Differential Equations, , 16. The solution of the differential equation, 2, , 2, , dy x + y + 1, satisfying y(1) = 1, is, =, dx, 2xy, , (a), (b), (c), (d), , x3 x5, +, + ..., dx − dy, 3! 5!, , is, =, 2, 4, dx, + dy, x, x, 1+, +, + ..., 2! 4 !, , x+, , (c), , (b) 2ye2x = C·e2x – 1, (d) none of these, , dy, 6x 2, ; y (1) = π., =, dx 2 y + cos y, , 2, , dy, dy, ,, 19. For the differential equation x + 2 y = xy, dx, dx, , order is 1 and degree is 1, solution is ln(yx2) = C – y, order is 1 and degree is 2, solution is ln(xy2) = C + y, , 1 − y2, , dy, , 20. The differential equation dx =, y, determines a family of circle with, (a) variable radii and fixed centre (0, 1), (b) variable radii and fixed centre (0, –1), (c) fixed radius 1 and variable centre on x-axis, (d) fixed radius 1 and variable centre on y-axis, 21. If y′ = y + 1, y (0) = 1, then y (ln 2) =, (a) 1, (b) 2, (c) 3, (d) 4, 22. The general solution of the differential, equation x(1 + y2) dx + y (1 + x2) dy = 0 is, (a) (1 + x2) (1 + y2) = 0, (b) (1 + x2) (1 + y2) = C, (c) (1 + x2) = C (1 + y2), (d) (1 + y4) = C (1 + x2), dy, 23. If dx = y sin 2x, y(0) = 1, then solution is, , (c), , y = cos2x , , (b) x3 + y3 = c, (d) x2 – y2 = c, , dy y, =, is, dx x, , k, x , y = k x , y=, , 26. If, , (b) y = k log x, (d) log y = k x, , x+y, dy, =, , y(1) = 1, then y =, dx, x, , (a) x + ln x , (c) xex – 1 , , (b) x2 + x ln x, (d) x + x ln x, , 27. The order and degree respectively of the, , Which of the following option is correct?, (a) Solution is y2 – sin y = – 2x3 + C, (b) Solution is y2 + sin y = 2x3 + C, (c) C = π + 2 2, (d) C = p2 + 2, , 2, y = esin x , , equation, (a), , 18. Given the differential equation, , (a), , dy x 2, is, =, dx y2, , 25. The general solution of the differential, , 17. Solution of the differential equation, , (a), (b), (c), (d), , equation, , (a) x3 – y3 = c , (c) x2 + y2 = c , , a hyperbola, a circle, y2 = x(1 + x) – 10, (x – 2)2 + (y – 3)2 = 5xy, , (a) 2ye2x = C·e2x + 1, (c) ye2x = C·e2x + 2, , 24. The general solution of the differential, , (b) y = sin2x, 2, , (d) y = ecos, , x, , 1, , 1, , 2, differential equation d y +  dy  4 + x 5 = 0, are, dx 2  dx , , (a) 2 and not defined (b) 2 and 2, (c) 2 and 3 , (d) 3 and 3, 28. Integrating factor of, , xdy, − y = x 4 − 3x is, dx, , (a) x , (c), , 1, , x, , (b) log x, (d) – x, , 29., y(1), (a), (c), , The number of solutions of, , when, =, dx x − 1, = 2 is, none , (b) one, two , (d) infinite, , dy, , y +1, , 30. Which of the following is a second order, differential equation?, (a) (y′)2 + x = y2, (b) y′y′′ + y = sin x, 2, (c) y′′′ + (y′′) + y = 0 (d) y′ = y2, 31. Integrating factor of the differential, dy, equation, + y tan x − sec x = 0 is, dx, (a) cos x , (b) sec x, cos x, (c) e, , (d) esec x, 32. The solution of the differential equation, dy 1 + y2, is, =, dx 1 + x 2, , (a) y = tan–1x , (c) x = tan–1y , , (b) y – x = k(1 + xy), (d) tan (xy) = k

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CBSE Board Term-II Mathematics Class-12, , 44, 33. The integrating factor of the differential, equation, (a), , x, , ex, , dy, 1+ y, +y=, is, dx, x, ex, (b), (c) xex, x, , 34. The solution of x, ex k, + , x x, , (a), , y=, , (c), , y = xex + k , , 35., y=, (a), (c), , (d) ex, , dy, + y = e x is, dx, , (b) y = xex + cx, ey k, x, =, +, (d), y y, , Differential equation having solution, Ax + B3 is of order, 3 , (b) 2, 1 , (d) not defined, , 36. Integrating factor of the differential equation, dy, cosx, + y sin x = 1 is, dx, (a) cos x (b) tan x (c) sec x (d) sin x, dy, − y = 1, y (0) = 1 is given by, dx, (a) xy = – ex , (b) xy = – e–x, (c) xy = –1 , (d) y = 2ex – 1, 37. Solution of, , 38. Integrating factor of the differential, dy, equation (1 − x 2 ) − xy = 1 is, dx, x, (a) – x (b), 1 + x2, (c), , 1 − x 2 , , (d), , 1, log (1 − x 2 ), 2, , Case Based MCQs, Case I : Read the following passage and, answer the questions from 39 to 43., A thermometer reading 80°F is, taken outside. Five minutes later, the thermometer reads 60°F. After, another 5 minutes the thermometer, reads 50 °F. At any time t the, thermometer reading be T °F and the, outside temperature be S °F., , It is known that, if the interest is compounded, continuously, the principal changes at the rate, equal to the product of the rate of bank interest, per annum and the principal. Let P denotes the, principal at any time t and rate of interest be r %, per annum., , 39. If l is positive constant of, proportionality, then, (a) l (T – S) , (c) lTS , , dT, is, dt, , (b) l (T + S), (d) –l (T – S), , 40. The value of T(5) is, (a) 30°F (b) 40°F (c) 50°F, , (d) 60°F, , 41. The value of T(10) is, (a) 50°F (b) 60°F (c) 80°F, , (d) 90°F, , 42. Find the general solution of differential, equation formed in given situation., (a) log T = St + c, (b) log (T – S) = –lt + c, (c) log S = tT + c, (d) log (T + S) = lt + c, 43. Find the value of constant of integration c, in the solution of differential equation formed, in given situation., (a) log (60 – S) , (b) log (80 + S), (c) log (80 – S) , (d) log (60 + S), Case II : Read the following passage and answer, the questions from 44 to 48., , 44. Find the value of dP ., (a), , Pr, 1000, , (b), , Pr, 100, , dt, , (c), , Pr, 10, , (d) Pr, , 45. If P0 be the initial principal, then find the, solution of differential equation formed in given, situation., (a), , P , rt, log   =,  P0  100, ,  P  rt, (b) log   =,  P0  10, , (c), , P , log   = rt,  P0 , , P , (d) log   = 100rt,  P0 , , 46. If the interest is compounded continuously, at 5% per annum, then in how many years will, ` 100 double itself ? (loge 2 = 0.6931), (a) 12.728 years, (b) 14.789 years, (c) 13.862 years, (d) 15.872 years

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45, , Differential Equations, , 47. At what interest rate will ` 100 double itself, in 10 years? (loge2 = 0.6931)., (a) 9.66% (b) 8.239% (c) 7.341% (d) 6.931%, 48. How much will ` 1000 be worth at 5%, interest after 10 years? (e0.5 = 1.648)., (a) ` 1648 (b) ` 1500 (c) ` 1664 (d) ` 1572, Case III : Read the following passage and answer, the questions from 49 to 53., In a college hostel accommodating 1000, students, one of the hostellers came in carrying, Corona virus, and the hostel was isolated. The, rate at which the virus spreads is assumed to, be proportional to the product of the number of, infected students and remaining students. There, are 50 infected students after 4 days., , dn is proportional to, dt, (a) n(1000 – n) , (b) n(1000 + n), (c) n(100 – n) , (d) n(100 + n), 50., , 51. The value of n(4) is, (a) 1 , (b) 50, (c) 100 , (d) 1000, 52. The most general solution of differential, equation formed in given situation is, 1,  1000 − n , (a), log ,  = λt + c, , 1000, n, (b), ,  n , log , = λt + c,  100 − n , , 1, n, , , log ,  = λt + c, , 1000, 1000 − n , (d) None of these, (c), , 53. The value of n at any time is given by, 1000, (a) n(t ) =, 1 + 999e −0.9906t, , 49. If n(t) denote the number of students infected, by Corona virus at any time t, then maximum, value of n(t) is, (a) 50, (b) 100, (c) 500, (d) 1000, , (b), , n(t ) =, , (c), , n(t ) =, , (d), , n(t ) =, , 1000, , 1 − 999e −0.9906t, 100, , 1 − 999e −0.996t, 100, 999 + e1000t, , Assertion & Reasoning Based MCQs, Directions (Q.-54 to 60) : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose, the correct answer out of the following choices :, (a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion., (b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion., (c) Assertion is correct statement but Reason is wrong statement., (d) Assertion is wrong statement but Reason is correct statement., 54. Assertion : The differential equation of all, circles in a plane must be of order 3., Reason : If three points are non-collinear, then, only one circle passes through these points., 55. Assertion : Order of the differential, equation whose solution is, y = c1ex + c2 + c3ex + c4 is 4., Reason : Order of the differential equation is, equal to the number of independent arbitrary, , constants mentioned in the solution of the, differential equation., 56. Assertion : y = a sin x + b cos x is a general, solution of y′′ + y = 0., Reason : y = a sin x + b cos x is a trigonometric, function., 57. Assertion : The elimination of four arbitrary, constants in y = (c1 + c2 + c3ec4)x results into a, differential equation of the first order x, , dy, = y., dx

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CBSE Board Term-II Mathematics Class-12, , 46, Reason : Elimination of n arbitrary constants, requires in general, a differential equation of, the nth order., , 58. Assertion : ‘x’ is not an integrating factor, dy, for the differential equation x, + 2y = ex., dx,  dy,  d, Reason : x  x + 2 y  = (x 2 y ) .,  dx,  dx, dy, 59. Assertion : x sin x, + (x + x cos x + sin x) y = sin x,, dx, , 1, π, 2, y   = 1 −, ⇒ lim y(x ) = ., 2, 3, x →0, π, Reason : The differential equation is linear, with integrating factor x(1 – cos x), dy, + xy = x3y3, x > 0, y ≥ 0 and, dx, 1, ., y(0) = 1, then y(1) =, 2, Reason : The differential equation is linear, with integrating factor ex., 60. Assertion : If, , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1., , Find the order and the degree of the, , 2, differential equation x, , 2., , 4, ,   dy  2 , = 1 +    ., dx 2   dx  , d2 y, , Write the sum of the order and degree of, 2, , 3, ,  2 , the differential equation  d y  +  dy  + x 4 = 0 ., 2,  dx , , 3., ,  dx , , Write the degree of differential equation, 2, , 4,  d2 y ,  dy , x , + x   = 0., , dx,  dx 2 , 3, , 4., , Find the differential equation whose solution, , A, is v = + B, where A and B are arbitrary constants., r, , , , 5. Find the differential equation whose solution, is y = mx, where m is an arbitrary constant., , 6., , Find the integrating factor of the differential, , equation ( y + 3x 2 ), 7., , dx, =x., dy, , Find the integrating factor of the differential, ,  e −2 x, y, equation , −,  x, x, , 8., ,  dx, =1., ,  dy, , Write the integrating factor of the differential, , equation (1 + x 2 ) + (2 yx − cot x ), 9., , dx, = 0., dy, , Write the solution of the differential equation, , dy, = 2− y ., dx, , 10. Find the solution of the differential equation, dy, = x 3 e −2 y ., dx, , Short Answer Type Questions (SA-I), 11. Find the differential equation whose solution, is y = e2x(a + bx), where ‘a’ and ‘b’ are arbitrary, constants., , 14. Find the general solution of the differential, , 12. Find the differential equation whose, solution is y = aebx + 5, where a and b are arbitrary, constants., , 15. Solve the differential equation, , + y2 + x2y2, given that y = 1 when x = 0., , 13. Solve the differential equation, , 16. Find the particular solution of the, , dy, + y = cos x − sin x ., dx, , dy = 0, given that y = 0 when x = 1., , equation xey/x dy = (yey/x + x2)dx, x ≠ 0, dy, = 1 + x2, dx, , differential equation (1 – y2)(1 + logx)dx + 2xy

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47, , Differential Equations, , 17. Find the particular solution of the, differential equation, that y =, , dy, x (2 log x + 1), =, , given, dx sin y + y cos y, , π, , when x = 1., 2, , 18. Find the particular solution of the, differential equation x (1 + y2) dx – y (1 + x2)dy = 0,, given that y = 1 when x = 0., , 19. Find the particular solution of the, differential equation xy, when x = 1., , dy, = ( x + 2) ( y + 2); y = −1, dx, , 20. Find the particular solution of the, dy, differential equation (x + 1), = 2e − y − 1; y = 0, dx, , when x = 0., , Short Answer Type Questions (SA-II), 21. Find the differential equation whose solution, is (x + a)2 + (y – a)2 = a2, where a is constant., 22. Find the differential equation whose, solution is x2 = 4ay, where a is constant., , 29. Solve the following differential equation, y2dx + (x2 – xy + y2)dy = 0, 30. Solve the differential equation, ( x 2 − 1), , dy, 2, , x ≠1, + 2xy =, 2, dx, x −1, , 23. Find the general solution of the differential, equation xex dy = (x3 + 2y ex)dx., , 31. Find the particular solution of the, , 24. Find the particular solution of the, , differential equation, , differential equation x dy = y − x tan  y  , given that, π, y=, at x = 1., 4, , 25. S o l v e, , the, , dx, , x , , differential, , equation, , y = 0 when x = 1, , dy, = 1 + x + y + xy, given that, dx, , 32. Solve the differential equation, , dy, π, + y cot x = 2 cos x , given that y = 0 when x = . , dx, 2, , xdy − ydx = x 2 + y2 dx , given that y = 0 when x = 1., , 33. If y(x) is a solution of the differential, , 26. S o l v e, , equation  2 + sin x  dy = − cos x and y(0) = 1,, , (1 + x 2 ), , the, , differential, , equation, , dy, + 2xy − 4 x 2 = 0, given y(0) = 0., dx, , 27. Find the particular solution of the, differential equation, y = 1 when x = 0., , dy, xy, =, , given that, dx x 2 + y2, , 28. Find the particular solution of the, differential equation ex tan y dx + (2 – ex) sec2y, π, dy = 0, given that y = when x = 0., 4, ,  1 + y  dx, , π, , then find the value of y   . , 2, 34. Find the particular solution of the following, differential equation x, that when x = 2, y = p., , dy,  y, − y + x sin   = 0, given, x , dx, , 35. Show that the differential equation, dy, y2, =, is homogeneous and also solve it., dx xy − x 2, , Long Answer Type Questions (LA), 36. Find the particular solution of the, dy, = ( x + 2 y ) , given that, differential equation ( x − y), dx, , y = 0 when x = 1., , 37. Find the particular solution of the differential, equation (tan –1x–y) dx = (1+x 2)dy, given that, y = 1 when x = 0., 38. Solve the following differential equation :,  1 + x 2 + y2 + x 2 y2  dx + xy dy = 0, , , , 39. Find the particular solution of the, differential equation ( x – y), , dy, = x + 2 y, given, dx, , that when x = 1, y = 0., , 40. Prove that x 2 – y 2 = C(x 2 + y 2 ) 2 is the, general solution of the differential equation, (x 3 – 3xy 2 )dx = (y 3 – 3x 2 y) dy, where C is a, parameter.

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CBSE Board Term-II Mathematics Class-12, , 48, , OBJECTIVE TYPE QUESTIONS, –x, , 1. (a) : y = a cos x + b sin x + ce is a three parameter, family of curve, it is a third order differential equation., 2., , (c) : The given equation can be rewritten as, y = Acos(x + C3) – Bex, , So, there are three independent variables, (A, B, C3)., Hence, the differential equation is of order 3., 3. (a) : The equation of the family of circles of radius a is, (x – h)2 + (y – k)2 = a2 which is a two parameter family, of curve. So, its differential equation is of order two., 4. (c) : The general form of the equation of parabola, whose symmetry is along x-axis is given by x = ay2 + b,, which is the solution of the differential equation of all, such type of parabolas which contains two arbitrary, constants. So, the order of the differential equation is 2., 2,  dy , (d) : We have, y 1 +   = c, dx, , (d) : The given differential equation can be written, 2, , 7. (c) : We have, y = Ae3x + Be–3x, Differentiating w.r.t. x, we get, y1 = 3Ae3x – 3Be–3x, Again differentiating w.r.t. x, we get, y2 = 9Ae3x + 9Be–3x = 9(Ae3x + Be–3x) = 9y, 8., , y ′′ 1, −, ( y ′ )2 = 0, y y2, , ⇒, , y y′′ = (y′)2, , 11. (b) : We have y = 17ex + ae–x ⇒ y′ = 17ex – ae–x, ⇒ y ′′ = 17 e x + ae − x ⇒ y ′′ = y ⇒ y ′′ − y = 0, , 12. (a) : The parametric form of the given equation is, x = t, y = t2., The equation of any tangent at t is 2xt = y + t2, On differentiating, we get 2t = y1, On putting this value in the above equation, we get, , 13. (b) : Given y = (A + B) cos (x + C) + Dex, This can be rewritten as,, y = k cos (x + C) + Dex, Now order of a differential is same as the number of, arbitrary unknowns present in the solution., Hence, order of the given differential equation is 3., 14. (a) : The equation of the family of circles which touch, both the axes is (x – a)2 + (y – a)2 = a2, where a is a parameter., This is one parameter family of curve. So, the order of, differential equation is one., , ⇒, ⇒, , 2, , y1 = – 2 ⇒ x y1 = – A, x, , dy, d 2 y dy, +x, +, = a2(Aeax + Be–ax), dx, dx 2 dx, , 1, y′ = b, y, , Again differentiating w.r.t. x, we get, , 2 /3, , ⇒, ...(i), , 2, , 3,  d2y , dy,  dy, , =3 −4 ⇒ , = 3, − 4, , 2, dx,  dx, ,  dx , , Degree of the differential equation is 2., , 16. (a) : Given,, , x, , Again differentiating w.r.t. x, we get, x2y2 + y1 2x = 0 ⇒ xy2 + 2y1 = 0, 9. (c) : Given xy = Aeax + Be–ax, Differentiating (i) w.r.t. x, we get, dy, y+ x, = a (Aeax – Be–ax), dx, Differentiating again w.r.t. x, we get, , (Using (i)), , 10. (d) : We have, y = aebx ⇒ ln y = ln a + bx, , \, , (c) : Given relation is y = A + B, A, , dy, 2, dx = a (xy), ,  d2y , 15. (a) : , 2,  dx , , y2 – 9y = 0, , Differentiating w.r.t. x, we get, , +2, , The order of this equation is 1., ,   dy  2 ,  d2y , 1 +    = , , dx , ,  dx 2 , Clearly, order and degree of given differential equation, are 2, 2 respectively., , ⇒, , dx, , 2, , 2, , Clearly, it is a differential equation of degree 2., , 3, , d2y, , y , xy 1 = y +  1  ⇒ 4xy 1 = 4y + y 12, 2, , 2, 2, ,  dy  ,  dy , ⇒ y 2 1 +    = c 2 ⇒ y 2   + y 2 = c 2, dx, dx, , , , 6., as, , x, , Differentiating w.r.t. x, we get, , where A = C 1 + C 2 and B = C 4 eC 5, , 5., , ⇒, , dy x 2 + y 2 + 1, =, dx, 2xy, , 2xydy = (x2 + y2 + 1)dx ⇒ 2xydy – y2dx = (x2 + 1)dx, xd(y2) – y2dx = (x2 + 1)dx, xd(y 2 ) − y 2 dx, x2, ,  y2   1 , 1 , , =  1 +  dx ⇒ d   = d  x − , x, x,  x2 , , Integrating both sides, we get, y2, 1, = x − +C, x, x, ⇒, , ( C2 ), , y2 = x2 – 1 + Cx ⇒ y 2 = x +, , Clearly, it represents a hyperbola., , 2, , −1−, , C2, 4

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CBSE Board Term-II Mathematics Class-12, , 50, dy 1 + y 2, dy, dx, ⇒, =, 32. (b) : dx =, 2, 2, 1+x, 1+ y, 1 + x2, On integrating both sides, we get, , 38. (c) : (1 – x2), \, , tan–1y = tan–1x + c ⇒ tan–1y – tan–1x = c, ⇒, ⇒, , 1, , y−x, y−x , tan , =c⇒, = tanc = k(say), 1 + xy,  1 + xy , y – x = k(1 + xy), –1 , , 33. (b) : dy + y = 1 + y, dx, x, , ⇒, , , , \, , I.F. = e, , 1, , ∫  1− x  dx, , =e, , x − log x, , I.F. =, , Now, solution is y⋅x =, ⇒, , yx = ex + k, , ex, , ∫x, , ⇒ y=, , =e e, , x log(x)–1, , =e e, , ⋅ x dx + k, , ex k, +, x x, , 35. (b) : y = Ax + B3, Differentiating w.r.t. x, we get, , dy, =A, dx, d2y, , =0, dx 2, which is a differential equation of order 2., dy, + ysinx = 1, dx, dy sin x, 1, dy, ⇒, + tanx ⋅ y = secx, +, ⋅y=, dx cos x, cos x, dx, I.F. = e∫tanxdx = elog(secx) = secx, , 36. (c) : cosx, , \, , 1, 2 2, ), , = 1 − x2, , 39. (d) : Given, at any time t the thermometer reading, be T°F and the outside temperature be S°F. Then, by, Newton’s law of cooling, we have, , is 50°F., , 42. (b): We have, dT = −λ(T − S), , dt, dT, 1, ⇒, = −λ dt ⇒ ∫, dT = −λ ∫ dt, T −S, T −S, ⇒ log(T – S) = – lt + c, , 43. (c) : Since, at t = 0, T = 80°F, \, , log(80 – S) = 0 + c ⇒ c = log(80 – S), , 44. (b) : Here, P denotes the principal at any time t, and the rate of interest be r% per annum compounded, continuously, then according to the law given in the, problem, we get dP = Pr, dt 100, dP Pr, =, 45. (a) : We have,, dt 100, 1, dP, r, r, =, dt ⇒ ∫ dP =, dt, P, 100 ∫, P 100, rt, ⇒ log P =, + C ...(1), 100, At t = 0, P = P0., ⇒, , Again differentiating w.r.t. x, we get, , ⇒, , = e log(1− x, , 41. (a) : Clearly from given information, value of T(10), , x –logx, , dy, + y = ex, dx, , = elogx = x, , log(1− x 2 ), , 1 −2x, dx, ∫, 1− x 2, , = e2, , 60°F, \ Value of T(5) = 60°F, , dy 1, ex, + y=, dx x, x, It is a linear differential equation with, 1, ∫ dx, e x, , x, dx, 1− x 2, , 40. (d) : Since, after 5 minutes, thermometer reads, , ex, = ex x–1 = x, 34. (a) : x, , = e2, , −∫, , dT, dT, ∝ (T − S ) ⇒, = −λ (T − S ), dt, dt, , dy, 1 y, dy, y 1, +y= +, ⇒, +y− =, dx, x x, dx, x x, dy , 1, 1, + 1 −  y =, dx , x, x, , ⇒, , I.F. = e, , dy, x, 1, dy, −, ⋅y=, – xy = 1 ⇒, dx 1 − x 2, dx, 1 − x2, , dy, dy, dy, –y=1 ⇒, =1+y ⇒, = dx, 1+ y, dx, dx, On integrating, we get, 37. (d) :, , log(1 + y) = x + c, Now, y(0) = 1 ⇒ log2 = c, \, , log(1 + y) = x + log2, , ⇒, , 1+ y, = ex ⇒ 1+ y = 2ex ⇒ y = 2ex – 1, 2, , \, , C = log P0, , So, log P =, ⇒, , rt, + log P0, 100, ,  P  rt ...(2), log   =,  P0  100, , 46. (c) : We have, r = 5, P0 = ` 100 and P = 2P0 = ` 200, Substituting these values in (2), we get, 5, log 2 =, t, 100, ⇒ t = 20 log e 2 = 20 × 0.6931 years = 13.862 years, 47. (d) : We have,, , P0 = ` 100, P = 2P0 = ` 200 and t = 10 years, Substituting these values in (2), we get, 10r, ⇒ r = 10 log 2 = 10 × 0.6931 = 6.931 %, log 2 =, 100

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51, , Differential Equations, 48. (a) : We have, P0 = ` 1000, r = 5 and t = 10, Substituting these values in (2), we get, P,  P  5 × 10,  P , =, ⇒ log , log , = 0.5 ⇒, = e 0.5, 100, 1000 , 1000,  1000 , , ⇒ P = 1000 × 1.648 = ` 1648, 49. (d) : Since, maximum number of students in hostel, is 1000., \ Maximum value of n(t) is 1000., 50. (a) : Clearly, according to given information,, , dn, = ln(1000 – n), where l is constant of proportionality., dt, 51. (b) : Since, 50 students are infected after 4 days., \, , n(4) = 50., , 52. (c) : We have, dn = λn(100 − n), dt, dn, ⇒ ∫, = λ∫ dt, n(1000 − n), ⇒, , 1 , 1, 1, +  dn = λ∫ dt, , 1000 ∫  1000 − n n , , ⇒, , 1  log(1000 − n), , + log n = λt + c, , , , −1, 1000, 1, n, , , log , = λt + c,  1000 − n , 1000, , ⇒, , 53. (a) : When, t = 0, n = 1, This condition is satisfied by option (a) only., 54. (b) : Let x2 + y2 + 2gx + 2fy + c = 0, Here, in this equation, there are three constants., \ Order = 3, Reason is also correct., 55. (d) : Q y = (c1ec2 + c3ec4) ex = cex...(i), dy, dy, \, = ce x ⇒, = y (Using (i)), dx, dx, \ Order is 1., 56. (b) : Q y = a sin x + b cos x...(i), \ y′ = a cos x – b sin x, ⇒ y′′ = – a sin x – b cos x = – y, , [Using (i)], ⇒ y′′ + y = 0, 57. (a) : Let c1 + c2 + c3ec4 = A (constant), Then, y = Ax, …(i), dy y, dy, = [Using(i)], =A ⇒, ⇒, dx, dx x, ⇒, , x, , dy, =y, dx, , 58. (b) :, 2, , dy 2, ex, + y=, dx x, x, , ∫ dx, log x 2, I.F. = e x = e2logx = e, = x2, ⇒ Assertion is correct., , Now,, ⇒, , dy, d,  dy, , + 2y , (x2 y) = x2, + y⋅2x = x  x, , dx, dx, dx, , Reason is correct., , y, 1, 59. (a) : dy +  1 + cot x + 1  =, x, dx  sin x, x, , x, , , 1,  1, + cot x +  dx = exp ln  x tan sin x , I.F.= exp ∫ , 2, , , x, x, sin, , , x, x, x, = x tan × 2 sin cos = x(1 – cos x), 2, 2, 2, 1, \ Solution is, yx(1 – cos x) = ∫ . x(1 − cos x) dx, x, , = x – sin x + c, 2,  π, ⇒ c=0, y  = 1−, π,  2, x − sin x, \ y=, x(1 − cos x), , ⇒, , , x3 , , x2  x2, x −  x − ...,  1 − + ...., 6 , , 6, 20, y=, = 2,   x2  , , x  x2,  1 − + ...., x  1 −  1 − ... , 2, 12, 2, ,  , , As x → 0, y →, , 1, 3, , 1 dy x, 3, + 2 =x, 3 dx, y, y, 1, 2, Put 2 = z ⇒ − 3 dy = dz, y, y, dz, \, – 2xz = –2x3,, dx, 2, which is a linear differential equation with I.F. = e − x, 60. (c) :, , 2, , 2, , \, , Solution, ze − x = −∫ e − x 2x 3dx, , ⇒, , ze–x = (x2 + 1)e–x + C ⇒ z = x2 + 1 + Cex, 2, 1, = x2 + 1 + Ce x, 2, y, y(0) = 1 ⇒ C = 0, 1, 1, 1, y2 =, ⇒y=, ⇒ y(1) =, ., 2, 2, 2, x +1, x +1, , \, Q, \, , 2, , 2, , 2, , SUBJECTIVE TYPE QUESTIONS, 1., x2, , The given differential equation is,   dy  2 , = 1 +   , dx 2   dx  , d2y, , 4, , \, , Its order is 2 and degree is 1., , 2., \, 3., , Required Sum = 2 + 2 = 4, Degree of the given differential equation is 2., , 4., , v=, , Order = 2, Degree = 2., , dv, A, A, ⇒, =−, +B ⇒, dr, r, r2, , d2v, dr, , 2, , =, , 2A, r3

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53, , Differential Equations, \, , (i) becomes, t + x, , dt, x, =t+ t, dx, e, , dt, dt, = xe −t ⇒, = e −t ⇒ dx = etdt, dx, dx, Integrating both sides, we get, ⇒, , x, , x = et + C ⇒ x = ey/x + C, dy, 15. We have,, = 1 + x2 + y 2 + x2y 2, dx, dy, ∴, = 1 + x 2 + y 2 (1 + x 2 ) = (1 + x2) (1 + y2), dx, dy, ⇒, = (1 + x 2 ) dx, 2, 1+ y, x3, −1, +C, Integrating both sides, we get tan y = x +, 3, when x = 0, y = 1, , \, , tan −1 1 = 0 + 0 + C ⇒ C =, , π, 4, , 1, π, ∴ tan −1 y = x + x 3 + is the required solution., 3, 4, 16. We have, (1 – y2)(1 + log x) dx + 2xy dy = 0, ⇒, , (1 – y2)(1 + log x) dx = – 2xy dy, , ⇒, , (1 + log x ), 2y, dy, dx = −, x, 1 − y2, , On integrating both sides, we get, (1 + log x )2, = log|1 − y 2 |+ C, 2, When x = 1, y = 0, ∴, , (1 + log 1)2, 1, = log(1) + C ⇒ C =, 2, 2, , (1 + log x )2, 1, = log|1 − y 2 |+, 2, 2, ⇒ (1 + log x)2 = 2 log |1 – y2| + 1 is the required, particular solution., ⇒, , 17. We have,, , dy x( 2 log x + 1), =, dx sin y + y cos y, , ⇒ (siny + y cosy)dy = x(2 logx + 1)dx, On integrating both sides, we get, – cosy + y siny – (– cosy), , 1 x2  x2, x2, = 2 log x ×, −∫ ×, dx  +, +C, 2, x 2, ,  2, , ⇒, , y siny = x 2 log x −, , x2 x2, +, +C, 2, 2, , y siny = x2 logx + C, π, When x = 1, y =, 2, ⇒, , ∴, , π, π, π, sin = 1 ⋅ log(1) + C ⇒ = C, 2, 2, 2, , \ y siny = x2 logx + p/2 is the required particular, solution., 18. We have, x(1 + y2) dx – y(1 + x2) dy = 0, y, x, dx −, dy = 0, ⇒, 2, 1+x, 1 + y2, ⇒, , 2x, , 1 + x2, , dx =, , 2y, , 1 + y2, , dy, , Integrating both sides, we get, log(1 + y2) = log(1+ x2) + log C, ⇒ 1 + y2 = C (1+ x2), When x = 0, y = 1, \ 1 + 1 = C(1 + 0) ⇒ C = 2, \ 1 + y 2 = 2(1 + x 2 ) is the required particular, solution., dy, = (x + 2) (y + 2), dx, y dy, x+2, ⇒, =,  dx, (y + 2)  x , , 19. We have, xy, , ⇒ dy −, , 2, 2, dy = dx + dx, x, (y + 2), , Integrating both sides, we get, y – 2log(y + 2) = x + 2log x + C, when x = 1, y = – 1, So, – 1 – 2log (– 1 + 2) = 1 + 2 log 1 + C, ⇒ C=–1–1=–2, So, we have y – 2log(y + 2) = x + 2log x – 2, ⇒ y – x + 2 = 2log (x(y + 2))., dy, = 2 e − y − 1 ...(i), dx, dy, dx, ey, dx, ⇒, =, ⇒, dy =, −y, y, +, x, 1, x+1, 2e − 1, 2−e, , 20. We have, ( x + 1), , Integrating both sides, we get, –log (2 –ey) = log (x + 1) + C...(ii), When x = 0, y = 0, \ – log (2 – 1) = log (0 + 1) + C ⇒ C = 0, \ (ii) becomes, – log (2 – ey) = log (x + 1), ⇒ log (x + 1) + log (2 – ey) = 0, ⇒ log[(x + 1) (2 – ey)] = 0, ⇒ (x + 1) (2 – ey) = 1 is the required particular solution., 21. We have (x + a)2 + (y – a)2 = a2, ...(i), which has only one arbitrary constant a., Differentiating (i) w.r.t. x, we get, dy, x + yy ′, 2 (x + a) + 2 ( y − a) = 0 ⇒ a =, ...(ii), dx, y′ − 1, Substituting value of a from (ii) in (i), we get

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CBSE Board Term-II Mathematics Class-12, , 54, 2, , 2, , x + y′ y , x + y′ y , , ,  x + y′ y ,  x + y ′ − 1  +  y − y ′ − 1  =  y ′ − 1 , , 2, , π, 4, , When x = 1, y =, , ⇒ [x(y′ – 1) + x + y′y]2 + [y(y′ – 1) – x – y′y]2 = (x + y′y)2, ⇒ (x + y)2 (y′)2 + (x + y)2 = (x + y′y)2, , 1, π, 1 ⋅ sin   = C ⇒ C =, 4, 2, 1, y, So, x sin   =, is the required particular solution., x, 2, , 2,   dy  2  , dy , ⇒ ( x + y )2    + 1 =  x + y  , is the required, dx ,  ,   dx , , 25. We have, x, , differential equation., , 22. We have x2 = 4ay, where a is the constant. ...(i), Differentiating (i) w.r.t. x, we get, 2x, = 4a, 2 x = 4ay1 ⇒, ...(ii), y1, Substituting the value of 4a from (ii) in (i), we get, 2x, x2 =, y ⇒ x 2 y1 − 2 xy = 0 ⇒ xy1 − 2 y = 0, y1, , dy, − y = x2 + y 2, dx, , dy y, y, = + 1+  , x, dx x, , ⇒, , 2, , , , ...(i), , This is a homogeneous differential equation., dy, dv, Put y = vx ⇒, = v+x, dx, dx, \ (i) becomes, , dy, ⇒ x, − 2 y = 0, is the required differential equation., dx, , v+x, , dv, dv, = v + 1 + v2 ⇒ x, = 1 + v2, dx, dx, , 23. We have, xexdy = (x3 + 2yex)dx, , ⇒, , dx, dv, =, x, 1 + v2, , ⇒, , log x + log C 1 = log|v + 1 + v 2 |, , dy x 3 + 2 ye x, dy 2, =, ⇒, − y = x 2 e− x , x, dx, dx x, xe, dy, This is a linear D.E. of the form, + Py = Q, dx, ⇒, , \, , I.F. = e, , 2, − ∫ dx, x, , = e −2 log x = elog x, , −2, , =, , 1, , x2, So, the solution of (i) is, 1, 1, y ⋅ 2 = ∫ 2 ⋅ x 2 e − x dx, x, x, y, −x, ⇒, = − e + C ⇒ y = –x2e–x + Cx2, x2, which is the required solution., 24. We have, x, , dy, y, = y − x tan  , x, dx, , dy y, y, = − tan   ,, x, dx x, which is a homogeneous differential equation., dy, dv, Now, put y = vx ⇒, = v+x, dx, dx, dv, \ v+x, = v − tan v, dx, dv, dv, dx, ⇒ x, = − tan v ⇒, =−, dx, tan v, x, dx, ⇒ cot v dv +, =0, x, Integrating both sides,we get, log|sin v| + log x = log C, y, ⇒ x sin v = C ⇒ x sin   = C, x, ⇒, , ...(i), , ⇒, , ∫, , ⇒ log x + log C 1 = log, , dx, dv, =, x ∫ 1 + v2, , y, y2, + 1+ 2, x, x, , ⇒ log C 1x = log|y + x 2 + y 2 |− log x, ⇒ ±C 1x 2 = y + x 2 + y 2, ⇒ Cx 2 = y + x 2 + y 2, , , , When x = 1, y = 0, , [where C = ±C1], , \, , C = 0+ 1+0 ⇒C = 1, , \, , Required particular solution is x 2 = y + x 2 + y 2 ., , 26. , We have (1 + x 2 ), , dy, + 2 xy = 4 x 2, dx, , dy, 2x, 4x 2, +, y, =, dx 1 + x 2, 1 + x2, This is a linear differential equation of the form, ⇒, , 2x, 4x 2, dy, and Q =, + Py = Q where P =, 2, dx, 1+x, 1 + x2, Pdx, =e, ∴ I.F. = e ∫, , ∫, , 2x, , 1+ x 2, , dx, , 2, , = elog(1 + x ) = 1 + x2, Hence, the required solution is, y( 1 + x 2 ) = ∫, , 4x 2, , 1 + x2, , (1 + x 2 ) dx + C, , ⇒ y(1 + x 2 ) = 4 ∫ x 2 dx + C

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CBSE Board Term-II Mathematics Class-12, , 56, 31. We have,, , dy, = 1 + x + y + xy, dx, , dy, = (1 + x ) + (1 + x ) y = (1 + x) (1+ y), dx, dy, ⇒, = (1 + x ) dx, 1+ y, ⇒, , Integrating both sides, we get, dy, ∫ 1 + y = ∫ (1 + x ) dx + C, ⇒ log (1 + y ) = x +, , x2, + C ...(i), 2, , When x = 1, y = 0, 1, 3, ∴ log 1 = 1 + + C ⇒ C = −, 2, 2, \, , The particular solution is, , log (1 + y ) = x +, , x2 3, − ., 2 2, , 32. We have,, , dy, + y cot x = 2 cos x, dx, , This is a linear differential equation of the form, dy, + Py = Q , where P = cot x, Q = 2cos x, dx, ∴, , cot x dx, I.F. = e ∫, = elog|sin x| =|sin x|, , \, , y|sin x| = ∫|sin x|(2 cos x ) dx, , ⇒ y|sin x|= ∫ sin 2 xdx, , 1, ⇒ y (sin x ) = − cos 2 x + C, 2, π, When x = , y = 0, 2, 1, 1, π, π, 0(sin ) = − cos 2   + C ⇒ C = −, 2, 2, 2, 2, 1, 1, ∴ y (sin x ) = − cos 2 x −, 2, 2, i.e., 2y sin x + cos 2x + 1 = 0 is the required solution.,  2 + sin x  dy, 33. , We have, , = − cos x,  1 + y  dx, dy, cos x, ⇒, =−, dx, 1+ y, 2 + sin x, Integrating both sides, we get, log(y + 1) = –log|2 + sinx| + log C,  C, , ⇒ log( y + 1) = log ,  2 + sin x , C, ⇒ y+1=, ⇒ (y + 1) (2 + sin x) = C, 2 + sin x, Given : y(0) = 1 ⇒ x = 0, y = 1, (1 + 1)(2 +sin 0) = C ⇒ C = 4, , (y + 1)(2 + sin x) = 4, 4, ⇒ y=, − 1 ...(i), 2 + sin x, Put x =, , π, in (i),, 2, , 4, 1, π, y  =, −1= ., 2 2+1, 3, , dy, y, − y + x sin   = 0, x, dx, dy y, y,  , ⇒, − + sin   = 0, x, dx x, , 34. We have x, , This is a linear homogeneous differential equation., dy, dv, Put y = vx ⇒, = v⋅1+ x, dx, dx, dv, ∴ v + x − v + sin v = 0, dx, dv, dx, ⇒ x + sin v = 0 ⇒ cosec v dv +, =0, dx, x, Integrating both sides, we get, log |cosec v – cot v| + log x = log C, ⇒ x (cosec v – cot v) = C, , y,  y , ⇒ x  cosec   − cot    = C, , ,  x , x, , when x = 2, y = p, π, π, , ∴ 2  cosec − cot  = C ⇒ C = 2, 2, 2, , , y,  y , ⇒ x  cosec   − cot    = 2 is the required particular, , ,  x , x, , solution., 35. We have,, ⇒, ⇒, \, \, , dy, y2, y 2 / x2, =, =, , 2, dx xy − x, ( xy − x 2 ) / x 2, , ... (i), , dy y 2 /x 2, =, y, dx, −1, x, It is a homogeneous differential equation, dy, dv, Put y = vx ⇒, = v⋅1+ x, dx, dx, (i) becomes, , v+x, , v2, v2, v, dv, dv, dv, =, ⇒ x, =, −v⇒x, =, dx v − 1, dx v − 1, dx v − 1, , v−1, dx, 1, dx, , dv =, ⇒  1 −  dv =, , , v, x, v, x, Integrating both sides, we get, v – log v = log x + C ⇒ v = log vx + C, y, ⇒, = log y + C, x, ⇒ y = x(log y + C) is the required solution., dy, 36. We have, ( x − y ) = x + 2 y, dx, dy x + 2 y, =, ...(i), dx x − y, ⇒

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CHAPTER, , 4, , Vector Algebra, Recap Notes, , VECTOR, h, , h, , h, , A physical quantity having magnitude as well as, direction is called a vector. A vector is represented, , , by a line segment, denoted as AB or a . Here, point, A is the initial point and B is the terminal point of, , the vector AB ., Magnitude : The distance between the points A, and B is called the magnitude of the directed line, , , segment AB . It is denoted by | AB|., Position Vector : Let P be any point in space, having, coordinates (x, y, z) with respect to some fixed point, , O (0, 0, 0) as origin, then the vector OP having O as, its initial point and P as its terminal point is called, the position vector of the point P with respect to O., , , The vector OP is usually denoted by r ., , n=, , TYPES OF VECTORS, h, , h, , h, , h, , x + y 2 + z2, , The numbers lr , mr and nr, proportional to the, , direction cosines of vector r are called direction, , ratios of the vector r and denoted as a, b and c, respectively., i.e., a = lr, b = mr and c = nr, Note : l2 + m2 + n2 = 1 and a2 + b2 + c2 ≠ 1, (in general)., , h, , , , Magnitude of OP is , OP = x 2 + y 2 + z2, , i.e.,| r | = x 2 + y 2 + z2 ., In general, the position vectors of points A, B, C, etc.,   , with respect to the origin O are denoted by a , b , c ,, etc. respectively., Direction Cosines and Direction Ratios :, , The angles a, b, g made by the vector r with the, positive directions of x, y and z-axes respectively are, called its direction angles. The cosine values of these, angles, i.e., cosa, cosb and cosg are called direction, , cosines of the vector r , and usually denoted by l,, m and n respectively., , Direction cosines of r are given as, y, x, and, l=, ,m =, 2, 2, 2, 2, x +y +z, x + y 2 + z2, , z, 2, , h, , h, , Zero vector : A vector whose initial and terminal, points coincide is called a zero (or null) vector. It, cannot be assigned a definite direction as it has zero, , magnitude and it is denoted by the 0 ., Unit Vector : A vector whose magnitude is unity, , i.e., | a| = 1 . It is denoted by a ., , , Equal Vectors : Two vectors a and b are said to, , , be equal, written as a = b , iff they have equal, magnitudes and direction regardless of the positions, of their initial points., Coinitial Vectors : Vectors having same initial point, are called co-initial vectors., Collinear Vectors : Two or more vectors are called, collinear if they have same or parallel supports,, irrespective of their magnitudes and directions., Negative of a Vector : A vector having the same, magnitude as that of a given vector but directed in, the opposite sense is called negative of the given, , , vector i.e., BA = − AB ., , ADDITION OF VECTORS, h, , Triangle law : Let the vectors, , , be a and b so positioned such, that initial point of one, coincides with terminal point,    , of the other. If a = AB, b = BC ,,  , then the vector a + b is represented by the third side,   , of DABC i.e., AB + BC = AC

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CBSE Board Term-II Mathematics Class-12, , 60, h, , Parallelogram law : If the two, , , vectors a and b are, represented by the two, adjacent sides OA and OB of, a parallelogram OACB, then,  , their sum a + b is represented, in magnitude and direction by the diagonal OC of, parallelogram OACB through their common point,   , O i.e., OA + OB = OC, ,  , (iv) a = b ⇔ a1 = b1 , a2 = b2 and a3 = b3, , , (v) a and b are collinear iff, b1 b2 b3, a = a = a = λ., 1, 2, 3, , VECTOR JOINING TWO POINTS, h, , joining P 1 and P 2 is the, , vector P1P2 ., Applying triangle law in, DOP1P2, we get,   , OP1 + P1P2 = OP2,   , ⇒ P1P2 = OP2 − OP1, = ( x2 i + y 2 j + z2 k ) − ( x1 i + y1 j + z1 k ), = ( x − x )i + ( y − y )j + ( z − z )k, , Properties of Vector Addition, h, , h, , h, , h, , Vector addition is commutative i.e.,,    , a + b = b + a., Vector addition is associative i.e.,,   ,   , a + ( b + c ) = ( a + b ) + c., Existence of additive identity : The zero vector acts, as additive identity i.e.,, ,     , a + 0 = a = 0 + a for any vector a ., Existence of additive inverse : The negative, , , o f a i.e., − a a c t s a s a d d i t i v e i n v e r s e i . e . ,, , , ,  , , a + ( − a) = 0 = ( − a) + a for any vector a., , MULTIPLICATION OF A VECTOR BY A SCALAR, h, , , Let a be a given vector and l be a given, scalar (a real number), then λ a is defined as, , the multiplication of vector a by the scalar l., Its magnitude is λ times the modulus of, , , , a i.e., λ a = λ a ., , , Direction of λa is same as that of a if λ > 0 and, opposite to that of a if λ < 0., 1, , , Note : If l =  , provided that a ≠ 0, then λa, | a|, represents the unit vector in the direction of, , a, , a i.e. a = , | a|, , COMPONENTS OF A VECTOR, h, , h, , Let O be the origin and P ( x, y, z ) be any point, in space. Let i , j and k be unit vectors along, the X -axis, Y -axis and Z -axis respectively. Then, , OP = xi + y j + zk , which is called the component, , form of OP. Here x, y and z are scalar components, , of OP and xi, y j and zk are vector components, , of OP., , , , a = a1i + a2 j + a3 k, If a and,  b are two given vectors as, and b = b1i + b2 j + b3 k and l be any scalar, then,  , (i) a + b = ( a1 + b1 )i + ( a2 + b2 )j + ( a3 + b3 )k,  , (ii) a − b = ( a1 − b1 )i + ( a2 − b2 )j + ( a3 − b3 )k, , (iii) λa = ( λa )i + ( λa )j + ( λa )k, 1, , 2, , 3, , If P1(x1, y1, z1) andP2 (x2, y2, z2), are any two points in the, space, then the vector, , 2, , 1, , 2, , 1, , 2, , 1, , , ∴ | P1P2 |= ( x2 − x1 )2 + ( y 2 − y1 )2 + ( z2 − z1 )2, , SECTION FORMULA, h, , h, , h, , h, , Let A, B be two points such that,  ,  , OA = a and OB = b., , The position vector r of the point P which divides, the line segment AB internally in the ratio m : n is, , ,  mb + na, given by r =, ., m+n, , The position vector r of the point P which divides, the line segment AB externally in the ratio m : n is, , ,  mb − na, given by r =, ., m−n, , The position vector r of the mid-point of the line,  ,  a+b, segment AB is given by r =, ., 2, , PRODUCT OF TWO VECTORS, h, , Scalar (or dot) product : The scalar (or dot), , , product of two (non-zero) vectors a and b ,,  , , , denoted by a ⋅ b (read as a dot b ) , is defined as,    , , , a ⋅ b = a b cos θ = ab cos θ, where, a = a , b = b and, , , q(0 ≤ q ≤ p) is the angle between a and b., , h, , Properties of Scalar Product :, ,    , (i) Scalar product is commutative : a ⋅ b = b ⋅ a,  , (ii) a ⋅ 0 = 0, (iii) Scalar product is distributive over addition :,   ,    , a ⋅ (b + c ) = a ⋅ b + a ⋅ c,       , ( a + b ) ⋅ c = a ⋅ c + b ⋅ c, ,  ,   , (iv) λ( a ⋅ b ) = ( λ a ) ⋅ b = a ⋅ ( λ b ), λ be any scalar.

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61, , Vector Algebra, (v) If i , j and k are three unit vectors along three, mutually perpendicular lines, then, i ⋅ i = j ⋅ j = k ⋅ k = 1 and i ⋅ j = j ⋅ k = k ⋅ i = 0, , , , , , (vi) Angle between two non-zero vectors a and b is,  , a⋅b, given by cosθ =   ., a b,  , , a⋅b , i.e., q = cos−1    , | a || b |, , , (vii) T wo non-zero vectors a and b are mutually, , , perpendicular if and only if a ⋅ b = 0,    , (viii) If q = 0, then a ⋅ b = a b,  ,  , If q = p, then a ⋅ b = − a b, h, , Projection of a vector on a line :, , Let the vector AB makes an, angle q with directed line  ., A, , , Projection of AB on l, , h, , (v) i × j = k , j × k = i , k × i = j,  , (vi) Two non-zero vectors a , b are collinear if and,   , only if a × b = 0,   , , , , , S imilarly, a × a = 0 and a × ( − a) = 0 , since in the, first situation q = 0 and in the second one,, q = p, making the value of sin q to be 0., , , (vii) If a and b represent the adjacent sides of a, triangle as given in the figure. Then,, C, , , a, A, , B, , , , p, , C, , , , Vector (or Cross) Product : The vector (or cross), , , product of two (non-zero) vectors a and b (in an, , ,  , assigned order), denoted by a × b (read as a cross b ),,    , is defined as a × b = a b sin θ n where q(0 ≤ q ≤ p), , , is the angle between a and b and n is a unit vector, , , perpendicular to both a and b ., Properties of Vector Product :,  ,  , (i) Non-commutative : a × b = −b × a, , (ii) Vector product is distributive over addition :,   ,    , a × (b + c ) = a × b + a × c, , , , , b, , D, , B, , 1, AB ⋅ CD, 2, 1  , 1  , = | b || a |sin θ = | a × b |, 2, 2, , , D, (viii) If a and b represent the, , adjacent sides of a, a, , parallelogram as given in, , A E, b, the figure., , , ,  , = AB cos θ = AC = p., , The vector p is called the projection vector. Its, , magnitude is p , which is known as projection of, , vector AB ., , , , Projection of a vector a on b , is given as a ⋅ b, 1  , i.e.,  ( a ⋅ b ) ., |b |, h, , ,  ,   , (iii) λ( a × b ) = ( λa ) × b = a × ( λb ) , l be any scalar., , ,  , (iv) ( λ 1a ) × ( λ 2 b ) = λ 1λ 2 ( a × b ), , Area of triangle ABC =, , C, , B, , Then, area of parallelogram ABCD = AB⋅DE,  ,  , = | b | | a | sin θ = | a × b |, , , (ix) If a = a1i + a2 j + a3 k , b = b1i + b2 j + b3 k ,, , , i,  , Then, a × b = a1, b1, , j, a2, b2, , k, a3, b3, , = ( a2 b3 − a3b2 )i + ( a3b1 − a1b3 )j + ( a1b2 − a2 b1 )k, , , (x) Angle between two vectors a and b is given by,  , |a × b|, sinq =  , | a || b |,  , |a × b| , i.e., q = sin–1    , | a || b |

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CBSE Board Term-II Mathematics Class-12, , 62, , Practice Time, , OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), , ,, 1. Find the sum of the vectors a = i − 2 j + k, , ,  and c = i − 6 j − 7 k, ., b = −2i + 4 j + 5k, ,  , (a) − 4 j − k, (b) −i − 4 j − k, , ,  , (c) 4 j + k, (d) i − 4 j, , , 2. The magnitude of the vector a = 3i − 2 j + 6 k, is equal to, (a) 6, (b) 7, (c) 7.5, (d) 8.5, , ,   2, 3. (a ⋅ i)2 + (a ⋅ j)2 + (a ⋅ k, ) is equal to, , , (a) 1, (b) | a |, (c) − a, (d), , 2, a, 4. If ABCD is a rhombus,, whose,   ,   diagonals, intersect at E, then, EA, +, EB, +, EC, , + ED equals, , , (a) 0, (b) AD, (c) 2BC, (d) 2AD, , , , , 5. If a = 2i^− j^+ 2k^ and b = 4i^+ 4 j^ − 2k^ then find, , , the angle between the vectors a and b., π, π, π, (a) 0, (b), (c), (d), 3, 2, 4, , ^, 6. The projection of the vector a = 2 i^ + 3 ^j + 2 k, , ^, on the vector b = i^ + 2 ^j + k, is, 5, 10, 10, 5, (a), (b), (c), (d), 3, 6, 3, 6, 7. If A and B are the points (– 3, 4, – 8) and, (5, – 6, 4) respectively, then find the ratio in, which yz-plane divides AB., (a) 5 : 2 (b) 7 : 5 (c) 3 : 5, (d) 5 : 3, 8. The vector in the direction of the vector, i − 2 j + 2k,  that has magnitude 9 is, i − 2 j + 2k, ,  , (a) i − 2 j + 2k, (b), 3, ), ), (c) 3(i − 2 j + 2k, (d) 9(i − 2 j + 2k, 9. If the angle between i + k and i + j + ak is, π, then the value of a is, 3, (a) 0 or 2 , (b) –4 or 0, (c) 0 or –2 , (d) 2 or –2, , ,  , , a − b = a = b = 1, t h e n, , between a and b is, π, 3π, π, (a), (b), (c), 3, 4, 2,        , 11. If a = 2i + j + 3k,b = − i + 2 j + k and,   , then the value of a ⋅ (b × c ) is, (a) – 20 (b) – 10, (c) 10, 10. I f, , the angle, π, 6,    , c = 3i + j + 2k,, (d), , (d) 20, , 12. The magnitude of each of the two vectors, , , a and b , having the same magnitude such that, the angle between them is 60° and their scalar, 9, product is , is, 2, (a) 2, (b) 3, (c) 4, (d) 5,  , , , 13. If u = i + 2 j, v = − 2i + j and w = 4i + 3 j,, , , , then find scalars x and such that w = xu + yv ., (a) x = 4, y = –2, (b) x = 2, y = –1, (c) x = 3, y = 5 , (d) x = –5, y = 2, 14. Write the direction cosines of the vector, −2iˆ + jˆ − 5kˆ ., 1, 5 ,  2, ,, ,, (a) ,  30 30 30 , 1, −5 ,  −2, ,, ,, (b) ,  30 30 30 , 2, 1, −5 , , ,−, ,, (c)  −, , , 30, 30 30 , (d) None of these, , ,  , , 15. Let a and b are non-collinear. If c = ( x − 2)a + b, ,  , and d = (2x + 1)a − b are collinear, then find the, value of x., 1, 2, −1, −2, (a), (b), (c), (d), 3, 3, 3, 3, , , , , , , , , 16. If a = 2i + j + 3k and b = 3i + 5 j − 2k , then,  , | a × b | is equal to, (a), , 507 (b), , 506, , (c), , 508, , (d), , 509

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63, , Vector Algebra, ,  ) ⋅ (k,  + i) is equal to, 17. (i + j ) × ( j + k, (a) 0, , (b) 1, , (c) 2, , (d) –1, , , , 18. If a and b are two unit vectors inclined to, , x-axis, at angles 30° and 120° respectively, then,  , a + b equals, , 2, (b), 2, (c), (d) 2, 3, 3, , 19. If b and c are any two non-collinear unit, , vectors and a is any vector, then find the value,   ,   ,    a ⋅ (b × c )  , of (a ⋅ b)b + (a ⋅ c )c +   ⋅ (b × c ) ., b×c, ,   , (a) a + b + c , (b) c, , , (c) a (d) b, , , 20. If a and b are unit vectors enclosing an,  , angle q and a + b < 1, then, (a) θ = π , (b) θ < π, 2, 3, 2π, π, 2π, (c) π ≥ θ >, , (d), <θ<, 3, 3, 3, , ,  , 21. If a = 10, b = 2 and a ⋅ b = 12, then the,  , value of a × b is, (a) 5, (b) 10, (c) 14, (d) 16, ,  ,  and b = 3i − j + 2k,  , then, 22. If a = i + 2 j − 3k, , , , , a + b and a − b are, (a) parallel , (b) perpendicular, (c) skew , (d) None of these, (a), , 23. Area of a parallelogram whose adjacent, sides are represented by the vectors,  and 4 j + 2k,  is, 2i − 3k, (a), , 4 14 sq. units, , (b) 2 7 sq. units, , (c), , 4 7 sq. units, , (d) 4 19 sq. units, , , , , , 24. The direction ratios of the vector 3a + 2b ,, , , where a = i + j − 2k and b = 2i − 4 j + 5k are, (a) 7, 5, 4 , (b) 7, –5, 4, (c) – 7, 5, 4 , (d) 7, 5, – 4, , , 25. The angle between two vectors a and b, with magnitudes 3 and 4, respectively and,  , a ⋅ b = 2 3 is, π, 5π, π, π, (a), (b), (c), (d), 3, 2, 6, 2, 26. The position vector of the point which, , ,  , divides the joining of points 2a − 3b and a + b, in the ratio 3 : 1 is, , , , , , 3 a − 2b, 7 a − 8b, (a), , (b), 2, 4, , , 5a, 3a, (c), , (d), 4, 4, , 27. If a = 4 and –3 ≤ l ≤ 3, then the range of, , λa is, (a) [0, 8] , (b) [–12, 8], (c) [0, 12] , (d) [8, 12], , ^, ^, ^, ^, ^, ^, 28. If (2 i + 6 j + 27 k) × (i + p j + q k) = 0 , then, the values of p and q are, 27, (a) p = 6, q = 27, (b) p = 3, q =, 2, 27, (c) p = 6, q =, (d) p = 3, q = 27, 2, , , , 29. If a, b, c are unit vectors such that,    , a +b +c = 0, t h e n w r i t e t h e v a l u e o f, ,   , a ⋅ b + b ⋅ c + c ⋅ a., 3, −3, (a), (b) 3, (c), (d) – 3, 2, 2, 30. Find the value of l for which the vectors,  and 2i − 4 j + λ k,  are parallel., 3i − 6 j + k, −3, 3, 2, −2, (a), (b), (c), (d), 2, 2, 3, 3, 31. Find the value of l so that the vectors,  and 4i − 8 j + λ k,  are perpendicular., 2i − 4 j + k, (a) 20, (b) – 40, (c) 40, (d) – 20, , 32. Find the projection of the vector i + 3 j + 7 k, ., on the vector 2i − 3 j + 6 k, (a) 5, (b) 6, (c) 7, (d) 8, 33. The vector having initial and terminal, points as (2, 5, 0) and (–3, 7, 4) respectively is, , , (a) −i + 12 j + 4 k, (b) 5i + 2 j − 4 k, (c), , , −5i + 2 j + 4 k, , , (d) i + j + k, , 34. The vectors from origin to the points A, , ,  and b = 2i + 3 j + k, ,, and B are a = 2i − 3 j + 2k, respectively, then the area of triangle OAB, (in sq. units) is, (a), 340 , (b) 325, 1, 229 , (c), (d), 229, 2, , , 35. If a = 3, b = 4, then the value of l for, , , , , which a + λ b is perpendicular to a − λ b, is, 3, 3, 4, 9, (a), (b), (c), (d), 4, 2, 3, 16

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CBSE Board Term-II Mathematics Class-12, , 64, , Case Based MCQs, Case I : Read the following passage and answer, the questions from 36 to 40., Ginni purchased an air plant holder which is in, the shape of a tetrahedron., Let A, B, C and D are the coordinates of the air plant, holder where A ≡ (1, 1, 1), B ≡ (2, 1, 3), C ≡ (3, 2, 2) and, D ≡ (3, 3, 4)., , C displaying A (Hub of Learning), B (Creating, a better world for tomorrow) and C (Education, comes first). The coordinates of points A, B and C, are (1, 4, 2), (3, –3, –2) and (–2, 2, 6) respectively., , A Hub of Learning, , D, B, , C, , Creating a, better world, for tomorrow, , A, , Education, comes first, , C, B, , , 36. Find the position vector of AB ., ,  (b) 2i + k, (a) −i − 2k, ,  , (c) i + 2k, (d) −2i − k, , , 37. Find the position vector of AC ., ,  , (a) 2i − j − k, (b) 2i + j + k,  , , (c) −2i − j + k, (d) i + 2 j + k, , 38. Find the position vector of AD .,  , , (a) 2i − 2 j − 3k, (b) i + j − 3k, (c), ,  , 3i + 2 j + 2k, , , (d) 2i + 2 j + 3k, , 39. Area of DABC =, (a), , 11, sq. units, 2, , (b), , 14, sq. units, 2, , 17, sq. units, 2, , 40. Find the unit vector along AD ., 1, 1,  ) (b), ), (2i + 2 j + 3k, (3i + 3 j + 2k, (a), 17, 17, 13, , 2, , (d), , (c), , 1, , ), (d) (2i + 2 j + 3k, , ), (2i + 2 j + 3k, , (c), ,  , 2i + 8 j + 3k, , Case II : Read the following passage and answer, the questions from 41 to 45., Three slogans on chart papers are to be placed, on a school bulletin board at the points A, B and, , ) , (d) 2(7i + 8 j + 3k, , 42 Which of the following,    , (a) AB + BC + CA = 0 (b),    , (c) AB + BC − CA = 0 (d), , is not true?,    , AB + BC − AC = 0,    , AB − CB + CA = 0, , 43 Area of DABC is, (a) 19 sq. units , , (c), , 11, ,  , , 41. Let a, b and c be the position vectors of,   , points A, B and C respectively, then a + b + c, is equal to,  , , (a) 2i + 3 j + 6 k, (b) 2i − 3 j − 6 k, , (b), , 1937 sq. units, , 1, 1937 sq. units (d) 1837 sq. units, 2, 44. Suppose, if the given slogans are to be, placed on a straight line, then the value of,      , a × b + b × c + c × a will be equal to, (a) –1 , (b) –2, (c) 2 , (d) 0, ,  , then unit vector in the, 45. If a = 2i + 3 j + 6 k, , direction of vector a is, 2 3  6 , 2 3  6 , i+ j+ k, (a), (b), i− j− k, 7, 7, 7, 7, 7, 7, (c), , (c), , 3 2  6 , i+ j+ k, 7, 7, 7, , (d) None of these

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65, , Vector Algebra, , Case III : Read the following passage and answer, the questions from 46 to 50., A barge is pulled into harbour by two tug boats, as shown in the figure., , x', , y, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, , O, y', , A, , 47. Position vector of B is, (a) 4i + 4 j , (b) 6i + 6 j, (c) 9i + 7 j , (d) 3i + 3 j, , 48. Find the vector AC ., (a) 8 j (b) −8 j, 8i (d) None of these, ,  , then its unit vector is, 49. If A = i + 2 j + 3k, (c), , B, , (a), C, 1 2 3 4 5 6 7 8 9 10 11 12, , x, , 46. Position vector of A is, (a) 4i + 2 j , (b) 4i + 10 j, (c) 4i − 10 j , (d) 4i − 2 j, , (c), , i, 14, 2i, 14, , +, +, , 2 j, 14, 3 j, 14, , +, , , 3k, , +, , 14, , k, 14, , , 50. I f A = 4i + 3 j, , , | A| + | B| =, (a) 12, , (b) 13, , (b), , 3i, 14, , +, , 2 j, 14, , +, , , k, 14, , (d) None of these, and, , , B = 3i + 4 j , t h e n, , (c) 14, , (d) 10, , Assertion & Reasoning Based MCQs, Directions (Q.-51 to 60) : In these questions, a statement of Assertion is followed by a statement, of Reason is given. Choose the correct answer out of the following choices :, (a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion., (b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion., (c) Assertion is correct statement but Reason is wrong statement., (d) Assertion is wrong statement but Reason is correct statement., 51. Assertion : The magnitude of the resultant, , ,  and b = i + 2 j + 3k,  is, of vectors a = 2i + j + k, 34 ., Reason : The magnitude of a vector can never, be negative., 52. Assertion : The unit vector in the direction of, sum of the vectors i + j + k , 2i − j − k and 2 j + 6 k, 1,  )., is − (3i + 2 j + 6 k, 7, , , a, Reason : Let a be a non-zero vector, then , a, , is a unit vector parallel to a ., , ,  and b = 2i + j + k, ., 53. Let a = i + j − 3k, , , Assertion : Vectors a and b are perpendicular, to each other.,  , Reason : a ⋅ b = 0, 54. Assertion : The adjacent sides of a, , , parallelogram are along a = i + 2 j and b = 2 i + j., , The angle between the diagonals is 150°., Reason : Two vectors are perpendicular to each, other if their dot product is zero.,     , 55. Assertion : If a + b + c = 0, a = 3 ,, ,      , , b = 4, c = 5, then a ⋅ b + b ⋅ c + c ⋅ a is equal, to –25., ,    , Reason : If a + b + c = 0, then the angle q between, , , , , , a2 − b2 − c 2, b and c is given by cosθ =, , ., 2b c, 56. Assertion : The length of projection of the,  on the vector i + 2 j − 3k,  is, vector 3i − j − 2k, 7, , ., 14, , Reason : The projection of a vector a on another,  , , (a ⋅ b), vector b is,  ., b

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CBSE Board Term-II Mathematics Class-12, , 66, , , 57. Let a and b be proper vectors and q be the, angle between them., ,  ,  , , Assertion : (a × b)2 + (a ⋅ b)2 ≠ (a )2 ( b)2, Reason : sin2q + cos2q = 1,  ,  , 58. Assertion : If (a × b)2 + (a ⋅ b)2 = 400 and, , , a = 4, then b = 9 ., , , Reason : If a and b are any two vectors, then, ,  , ,  , (a × b)2 is equal to (a )2 (b)2 − (a ⋅ b)2 ., , , , 59. Assertion : If a = 2i + 3 j − k , b = −i + 3 j + 4k, then projection of on ., , , Reason : Projection of a on b =, , 3, 26, , ., , 60. Assertion : Three points with position vectors,  , ,       , a, b and c are collinear if a × b + b × c + c × a = 0,  ,  , Reason : If AB ⋅ AC = 0, then AB ⊥ AC ., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), , π, 1. If a unit vector a makes angles, with i,, 3, π, with j and an acute angle q with k , then find, 4, the value of q., 2. Find the sum of the following vectors., , , , a = i − 3k , b = 2 j − k , c = 2i − 3 j + 2k, 3. Find a unit vector in the direction of the sum, , , of the vectors a = 2i + 2 j − 5k and b = 4i − 3 j + 2k ., 4. L and M are two points with position, ,  , , vectors 2a − b and a + 2b respectively. What is, the position vector of a point N which divides the, line segment LM in the ratio 2 : 1 externally ?,  2  2, , 5. Find the value of a × b + a ⋅ b if a = 5, , and |b| = 4. , , 6. Find the area of a parallelogram whose, adjacent sides are represented by the vectors, i − 3 k and 2 j + k . , 7. F i n d t h e p r o j e c t i o n o f t h e v e c t o r, , , a = 2i + 3 j + 2k on the vector b = 2 i + 2 j + k . , , , 8. If a and b are unit vectors, then find the, ,  , , angle between a and b , given that 2 a − b is a, unit vector., , (, , ), , Find the angle between x -axis and the, vector i + j + k . , ,  , , , 10. If a and b are two vectors such that a + b = a ,, , , then prove that vector 2a + b is perpendicular to, , vector b ., 9., , Short Answer Type Questions (SA-I), 11. X and Y are two points with position, , ,  , vectors 3a + b and a − 3b respectively. Write, the position vector of a point Z which divides the, line segment XY in the ratio 2 : 1 externally., 12. Find a unit vector perpendicular to each of, , , , , the vectors a and b and where a = 5i + 6 j − 2k, , . , and b = 7i + 6 j + 2k, 13. Show that for any two non-zero vectors, ,   ,  , , , a and b, a + b = a − b iff a and b are, perpendicular vectors.,  , 3i + 7 j + k, , 14. Show that the vectors 2 i − j + k, ,  form the sides of a right-angled, and 5 i + 6 j + 2k, triangle., , , 15. If two vectors, a and b are such that, , , , | a | = 2, | b | = 1 and a ⋅ b = 1, then find the value of,  ,  , (3a − 5b ) ⋅ (2a + 7b ). , , ,  and b = 3 i − j + 2k,  be, 16. Let a = i + 2 j − 3k,  , two vectors. Show that the vectors (a + b) and,  , (a − b) are perpendicular to each other., 17. If q is the angle between two vectors, i − 2 j + 3k,  and 3 i − 2 j + k,  , then find sin q.

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67, , Vector Algebra, , , , 18. Find | 2a ⋅ ( −b × 3c ) | , where, , ,  , b = 3i + 4 j − 5k,  and c = 2i − j + 3k, . , a = i − j + 2k, , , ^, ^ ^  ^, ^ ^, 19. If a = 2 i + 3 j + k, b = i − 2 j + k and, ,   , c = −3i + j + 2k , then find a ⋅ (b × c ). , , , , , , , 20. If r = xiˆ + yjˆ + zkˆ , find (r × iˆ ) ⋅(r × jˆ ) + xy ., , Short Answer Type Questions (SA-II),  and 3i − j + 4 k, , 21. The two vectors j + k, represent the two sides AB and AC, respectively, of a DABC. Find the length of the median through A. , 22. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors, , ,  and b = i − 2 j + k, ., a = 2i + 3 j − k, , ,  , b = i − 4 j + 5k,  and, 23. Let a = 4i + 5 j − k, , ,  . Find a vector d which is, c = 3i + j − k, ,  , , perpendicular to both c and b and d ⋅ a = 21 ., , , 24. The scalar product of the vector a = i + j + k, with a unit vector along the sum of vectors, ,  is equal to,  and c = λi + 2 j + 3k, b = 2i + 4 j − 5k, one. Find the value of l and hence find the unit,  , vector along b + c . ,    ,    , 25. If a × b = c × d and a × c = b × d , show that,  ,  ,  ,  , a − d is parallel to b − c , where a ≠ d and b ≠ c . , , 26. Dot product of a vector with vectors,  , 2i + j − 3k,  and i + j + k,  are respectively, i − j + k, 4, 0 and 2. Find the vector., 27. Find the values of l for which the angle, , b e t w e e n t h e v e c t o r s a = 2 λ2 iˆ + 4 λjˆ + kˆ a n d, , b = 7iˆ − 2 jˆ + λkˆ is obtuse., , 28. Using vectors, find the area of the triangle, with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).,  , , 29. If a , b and c are three vectors such that, , , , a = 3, b = 4 and c = 5 and each one of them, is perpendicular to the sum of the other two,,   , then find | a + b + c|., , , 30. If a = iˆ + 2 jˆ + 3kˆ and b = 2iˆ + 4 jˆ − 5kˆ represent, two adjacent sides of a parallelogram, find, unit vectors parallel to the diagonals of the, parallelogram., 31. Using vectors, find the area of the triangle, ABC with vertices A (1, 2, 3), B (2, –1, 4) and, C(4, 5, –1), , , , 32. If a = iˆ + 2 jˆ + kˆ , b = 2iˆ + jˆ and c = 3iˆ − 4 jˆ − 5kˆ , then, find a unit vector perpendicular to both of the,  ,  , vectors (a − b ) and (c − b ) .], 33. Find a unit vector perpendicular to,  ,  , both of the vectors a + b and a − b where, , , a = iˆ + jˆ + kˆ ,b = iˆ + 2 jˆ + 3kˆ ., 34. If the sum of two unit vectors is a unit vector,, prove that the magnitude of their difference is, 3., ,  ^ ^, ,  ^ ^ ^, 35. If a = i + j + k and b = j − k, find a vector c,, ,  ,  , such that a × c = b and a ⋅ c = 3., , Long Answer Type Questions (LA),   , 36. If a, b, c are mutually perpendicular, vectors of equal magnitudes, show that the vector,   ,  , , a + b + c is equally inclined to a, b and c. Also,, , , , , find the angle which a + b + c makes with a or, , , b or c ., 37. Show that the points A, B, C with position, vectors 2i − j + k , i − 3 j − 5k and 3i − 4 j − 4 k, respectively, are the vertices of a right-angled, triangle. Hence find the area of the triangle., 38. The two adjacent sides of a parallelogram,  and 2i + 2 j + 3k,  . Find the two, are 2i − 4 j − 5k, , unit vectors parallel to its diagonals. Using, the diagonal vectors, find the area of the, parallelogram., , , 39. If a = 3iˆ − jˆ and b = 2iˆ + jˆ − 3kˆ then express, ,   , , , , , b in the form b = b1 + b2 where b1 | | a and b2 ⊥ a ., ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , 40. If i + j + k, 2 i + 5 j , 3 i + 2 j − 3 k and i − 6 j − k, respectively are the position vectors of points, A , B , C and D , then find the angle between, the straight lines AB and CD . Find whether, , , AB and CD are collinear or not.

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CBSE Board Term-II Mathematics Class-12, , 68, , OBJECTIVE TYPE QUESTIONS, 1. (a) : The given vectors are, , , , a = i − 2 j + k , b = −2i + 4 j + 5k , c = i − 6 j − 7 k,   , \ Required sum = a + b + c, = (i − 2 j + k ) + ( −2i + 4 j + 5k ) + (i − 6 j − 7 k ), , (b) : Here, a = 3i − 2 j + 6 k, , Its magnitude = a, 2, , 2, , ∴, , , , , 2, ( a ⋅ i )2 + ( a ⋅ j )2 + ( a ⋅ k )2 = x 2 + y 2 + z2 = a, , 4., , (a) : EA + EB + EC + ED, ,    , ,    , , = EA + EB − EA − EB, [As diagonals of a rhombus bisect each other], , , , = 0, , , 5. (c) : We have, a = 2 ^i − ^j + 2 ^k and b = 4 ^i + 4 j^ − 2 k^,  , Now, a ⋅ b = ( 2 i^ − ^j + 2 k^) ⋅ ( 4 i^ + 4 j^ − 2 k^),  , = 8 – 4 – 4 = 0. Therefore, a ⋅ b = 0 ⇒ cosq = 0, , π, , So, angle between a and b is ., 2, , 6. (a) : We have, a = 2 i^+ 3 ^j + 2 k^ and b = i^ + 2 ^j + k^,  , ∴ a ⋅ b = ( 2 i^ + 3 ^j + 2 k^) ⋅ ( i^ + 2 ^j + k^) = 2 + 6 + 2 = 10, , and| b | = 12 + 2 2 + 12 = 6, ,  a ⋅ b 10, , Hence, projection of a on b is  =, ., |b |, 6, , , 7. (c) : Let a = − 3 i + 4 j − 8k , b = 5 i − 6 j + 4 k, , , Let C (c ) be the point in yz-plane which divides AB, in the ratio r : 1., 5r − 3, Then, 0 =, (, In yz-plane, x = 0), r+1, 3, 5, Thus required ratio is 3 : 5, , 8. (c) : Let a = i − 2 j + 2 k, ⇒, , \, \, 9., , 5r − 3 = 0 ⇒ r =, , , a =, , 10., , ⇒, , 2, , 3 + ( −2 ) + 6 = 9 + 4 + 36 = 49 = 7.,  2, , 2, 3. (d) : Let a = xi + y j + zk ⇒ ( a ⋅ i ) = x, , , Similarly, ( a ⋅ j )2 = y 2 and ( a ⋅ k )2 = z2, =, , ⇒, , ⇒, , = −4 j − k ., 2., \, , ⇒, , 1+ 4+ 4 =, , Required vector =, (b) : We have, cos, , 1, =, 2, , 1+ a, 2 2 + a2, , ⇒, , 2 + a2 = 2(1 + a2 + 2a) ⇒ a2 + 4a = 0 ⇒ a = 0, –4, , , (a) : Given, a − b = a = b = 1,  2, 2 2,  ,  , a − b = a + b − 2 a ⋅ b ⇒ 1 = 1 + 1 − 2 a b cosθ, , , (Here q is angle between a and b ), 1, π, cosθ = ⇒ θ =, 2, 3, , , (b) : Here, a = 2i + j + 3k , b b = −i + 2 j + k and, , 11., , c = 3i + j + 2 k, , i j k,  , Now, b × c = −1 2 1 = 3i + 5 j − 7 k, 3 1 2, , , , ∴ a ⋅ (b × c ) = ( 2i + j + 3 k ) ⋅ ( 3i + 5 j − 7 k ), = 2 × 3 + 1 × 5 + 3 × (–7), = 6 + 5 – 21 = –10, ,   9, , 12. (b) : Given, | a | = | b |, θ = 60° and a ⋅ b =, 2,  , a⋅b, Now, cos θ =  , | a || b |, 9 /2, 1 9 /2, cos 60° =  2 ⇒, = , 2 | a |2, | a|, , , , , ⇒ | a |2 = 9 ⇒ | a | = 3 ∴ | a | = | b | = 3, ⇒, , , , , 13. (b) : We have, w = xu + yv, ⇒, ⇒, ⇒, ⇒, , 4i + 3 j = x(i + 2 j ) + y( − 2i + j ), , ( x − 2 y − 4 )i + ( 2 x + y − 3)j = 0, x – 2y – 4 = 0 and 2x + y – 3 = 0, x = 2 and y = – 1, , , , 14. (b) : We have, a = −2iˆ + jˆ − 5kˆ, Direction cosines of the given vector are, , −2, , ,, , 2,  (−2) + (1)2 + (−5)2, , 9=3, 9(i − 2 j + 2 k ), 3, , (1 + a)2, 1, =, 4 2( 2 + a2 ), , 1, 2, , (−2) + (1)2 + (−5)2, , π (i + k ) ⋅ (i + j + ak ), =, 3, 2 1 + 1 + a2, , −2, , =, ,,  4 + 1 + 25, , 1, 4 + 1 + 25, , −5, , , , (−2) + (1) + (−5) , 2, , = 3(i − 2 j + 2 k ), , ,, , ,, , −5, , , 4 + 1 + 25 , ,  −2, 1, −5 , ,, ,, ∴ Direction cosines are , ,  30, 30, 30 , , 2, , 2

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69, , Vector Algebra, ,   ,  , 15. (d) : W e have, c = ( x − 2 ) a + b , d = ( 2 x + 1) a − b, , , are collinear, then c = md,  ,  , ⇒ ( x − 2 ) a + b = m ( 2 x + 1) a − b, ⇒ –m = 1 ⇒ m = –1, 1, and m (2x + 1) = x – 2 ⇒ – 2x – 1 = x – 2 ⇒ x =, 3, , (, , ), , ^, , i, , ^, , ^, , j, , k, ,  , 16. (a) : We have, a × b = 2 1, , 3, , 3 5 −2, ^, , ^, , ^, ,    ,    ,    , 22. (b) : ( a + b ) ⋅ ( a − b ) = a ⋅ a − a ⋅ b + b ⋅ a − b ⋅ b,    ,    , [∵ a⋅b = b ⋅ a], =, a, ⋅, a, −, b, ⋅, b, , , 2 2, 2, 2, 2, 2, = a − b = [1 + 2 + ( −3) ] − [ 3 + ( −1)2 + 2 2 ] = 0,  ,  , ⇒ (a + b ) ⊥ (a − b ), , , 23. (a) : Let a = 2i − 3 k and b = 4 j + 2 k, , , The area of a parallelogram with a and b as its adjacent,  , sides is given by | a × b |., i j k,  , Now, a × b = 2 0 −3 = 12i − 4 j + 8k, , = i ( −2 − 15) − ( −4 − 9 ) ^j + (10 − 3) k = −17 ^i + 13 ^j + 7 k,  , Hence,| a × b | =, , ( −17 )2 + (13)2 + (7 )2 =, , 507, , 17. (c) : (i + j ) × ( j + k ) ⋅ ( k + i ) = (i × j + i × k + j × k ) ⋅ ( k + i ), (∵ i ⋅ j = j ⋅ k = k ⋅ i = 0 ), = ( k − j + i ) ⋅ ( k + i ) = k ⋅ k + i ⋅ i , 2, 2, = k + i = 1 + 1 = 2, , , π, , 18. (b) : Clearly, angle between a and b is ., 2,  , ⇒ a⋅b = 0,  2, 2 2,  , ∴ a + b = a + b + 2 a ⋅ b = 1+ 1 + 0 = 2,  , ⇒ a+b = 2, , , , 19. (c) : Let b = i and c = j and a = a1 i + a2 j + a3 k,  ,  b ×c, ,  , Now, a ⋅ b = a1 , a ⋅ c = a2 and a ⋅   = a ⋅ k = a3, b ×c,    , , , ,    a ⋅ (b × c ), , ∴ ( a ⋅ b )b + ( a ⋅ c )c +   (b × c ), b ×c, , , , = a1b + a2c + a3 k = a1 i + a2 j + a3 k = a,  ,  2, 20. (c) : a + b < 1 ⇒ a + b < 1, , , 2 2,  ,  , ⇒ a + b + 2 a ⋅ b < 1 ⇒ 1 + 1 + 2 a ⋅ b < 1 [∵ a = b = 1], 1, 1,  ,  , ⇒ a ⋅ b < − ⇒ a b cos θ < −, 2, 2, 1, 1, ⇒ 1 × 1 × cos θ < − ⇒ cos θ < −, 2, 2, 1, 2π, ⇒ − 1 ≤ cos θ < − ⇒ π ≥ θ >, 2, 3, , ,  , 21. (d) : a = 10 , b = 2 , a ⋅ b = 12,    , We know, a ⋅ b = a b cos θ, ⇒, , 12 = 10 × 2cosq ⇒ cos θ =, , 3, 5, , 4, ∴ sin θ =, 5, , , , Now, a × b = a b sin θ = 10 × 2 × 4 = 16, 5, , 0 4, \, , 2, ,  , | a × b| = (12 )2 + ( −4 )2 + (8)2 = 144 + 16 + 64, , , , = 224 = 4 14 sq. units., ,   , 24. (b) : We have, a = i + j − 2 k , b = 2i − 4 j + 5k, , , ∴ 3a + 2 b = 3(i + j − 2 k ) + 2( 2i − 4 j + 5k ), = ( 3i + 3 j − 6 k ) + ( 4i − 8 j + 10 k ) = 7i − 5 j + 4 k, , , \ The direction ratios of the vector 3a + 2 b are 7 , − 5, 4., ,  , , 25. (b) : We have a ⋅ b = 2 3 , a = 3 , b = 4, , , Let q be the angle between a and b,    , ∴ a ⋅ b = a b cos θ, ⇒ 2 3 = 3 ⋅ 4 ⋅ cos θ ⇒ cos θ = 1 ⇒ θ = π, 2, 3, , ,  , 26. (d) : Given points are 2 a − 3b and a + b and, Given ratio = 3 : 1, , ,  , ( 2 a − 3b ) × 1 + ( a + b ) × 3, \ Required vector =, 3+1, , , , , 2 a − 3b + 3a + 3b 5 , =, = a, , 4, 4, 27. (c) : We have, –3 ≤ l ≤ 3 ⇒ |l| ≤ 3, , , , Now, λ a ≤ 3 a ⇒ λa ≤ 12, , \ Range of λa is [0, 12], i, , j, , k, , 28. (b) : Given, 2 6 27 = 0, 1, , p q, , ⇒, , i (6q– 27p) − j (2q – 27) + k (2p – 6) = 0, 6q – 27p = 0, 2q – 27 = 0 and 2p – 6 = 0, , ⇒, , q=, , ⇒, , 27, and p = 3., 2, ,   , 29. (c) : We have a , b , c are unit vectors., , Therefore, a = 1, b = 1 and c = 1

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CBSE Board Term-II Mathematics Class-12, , 70,    , Also, a + b + c = 0 (given), , ⇒ a + b + c 2 = 0, 2 , ,      , ⇒ a 2 + b + c 2 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0,      , ⇒ 1 + 1 + 1 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0,      , ⇒ 3 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0,  , ⇒ ( a ⋅ b + b ⋅ c + c ⋅ a ) = − 3, 2, , , , , , 30. (a) : a = 3i − 6 j + k and b = 2i − 4 j + λ k, ,   , , Since, a and b are parallel ∴ a × b = 0, i, ⇒, , j, , 3 −6, , k, , 2, , 2, a, 3, a, 2, a − λ2 b = 0 ⇒ λ2 =  2 ⇒ λ =  =, 4, b, b, , 36. (c) : Position vector of AB, = ( 2 − 1)i + (1 − 1)j + ( 3 − 1)k = i + 2 k, , 37 (b) : Position vector of AC, = ( 3 − 1)i + ( 2 − 1)j + ( 2 − 1)k = 2i + j + k, , 38. (d) : Position vector of AD, = ( 3 − 1)i + ( 3 − 1)j + ( 4 − 1)k = 2i + 2 j + 3k, ⇒, , 39. (b) : Area of DABC =, , i j k,  , Now, AB × AC = 1 0 2 = i(0 − 2 ) − j(1 − 4 ) + k (1 − 0 ), , , 1 =0, , 2 −4 λ, , 2 1 1, , , ⇒ (− 6 λ + 4 )i − ( 3λ − 2 )j + (− 12 + 12 )k = 0, ⇒ (− 6 λ + 4 )i + ( 2 − 3λ )j = 0i + 0 j, Comparing coefficients of i and j , we get, –6l + 4 = 0 and 2 – 3l = 0 ⇒ l = 2/3, 31. (b) : The given vectors will be at right angles if, their dot product vanishes, i.e.,, ( 2i − 4 j + k ) ⋅ ( 4i − 8 j + λ k ) = 0, , ⇒ 8 + 32 + l = 0 ⇒ l = –40, , , 32. (a) : Let a = i + 3 j + 7 k and b = 2i − 3 j + 6 k ., ,  a ⋅ b, , Now, projection of a on b = , |b |, =, , (i + 3 j + 7 k ) ⋅ ( 2i − 3 j + 6 k ), 2, , 2, , 2 + ( −3) + 6, , 2, , =, , 35, 2 − 9 + 42, =, =5, 7, 4 + 9 + 36, , 33. (c) : Let A(2, 5, 0) and B(–3, 7, 4), \, , Required vector = (− 3 − 2 )i + (7 − 5)j + ( 4 − 0 )k, , = −5i + 2 j + 4 k, , , 34. (d) : a = 2i − 3 j + 2 k and b = 2i + 3 j + k, i j k,  , a × b = 2 −3 2 = −9i + 2 j + 12 k, 2, , 3, , 1, , 1, 1, 1  , a×b =, 81 + 4 + 144 =, 229, 2, 2, 2, , , , , 35. (b) : Given that, a = 3, b = 4 and a + λb is, , , perpendicular to a − λb .,  , , , ∴ ( a + λb ) ⋅ ( a − λb ) = 0,  ,  ,    , ⇒ a ⋅ a − a ⋅ b λ + λb ⋅ a − λ 2 b ⋅ b = 0, Area of DOAB =, , 1  , | AB × AC |, 2, , ⇒, , = −2i + 3 j + k,  , | AB × AC | = ( −2 )2 + 32 + 12, =, , 4+9+1 =, , 14, , 1, 14 sq. units, 2, , , 40. (a) : Unit vector along AD = AD, , | AD |, 2i + 2 j + 3k, 2i + 2 j + 3k, 1, =, =, =, ( 2i + 2 j + 3k ), 2, 2, 2, 4+4+9, 17, 2 +2 +3, , , 41. (a) : a = i + 4 j + 2 k , b = 3i − 3 j − 2 k, and c = −2i + 2 j + 6 k,   , \ a + b + c = 2i + 3 j + 6 k, \, , Area of DABC =, , 42. (c) : Using triangle law of addition in DABC, we get,    , AB + BC + CA = 0, which can be rewritten as,    ,    , AB + BC − AC = 0 or AB − CB + CA = 0, 43. (c) : We have, A(1,4, 2), B(3, –3, –2) and C(–2, 2, 6),   , Now, AB = b − a = 2i − 7 j − 4 k,   , and AC = c − a = −3i − 2 j + 4 k, i, , \, ,  , AB × AC = 2, , j, , k, , −7 −4, , −3 −2, , 4, , = i( −28 − 8) − j(8 − 12 ) + k ( −4 − 21) = −36i + 4 j − 25 k, ,  , Now, | AB × AC | = ( −36 )2 + 4 2 + ( −25)2, 1296 + 16 + 625 = 1937, 1  , 1, \ Area of DABC = | AB × AC | =, 1937 sq. units, 2, 2, =

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71, , Vector Algebra, 44. (d) : If the given points lie on the straight line,, then the points will be collinear and so area of, DABC = 0.,      , ⇒, a×b +b ×c +c ×a = 0,   , [ If a , b , c are the position vectors of the three, vertices A, B and C of DABC, then area of triangle, 1      , =, a×b +b ×c +c ×a], 2, , 45. (b) : Here, | a | = 2 2 + 32 + 6 2 = 4 + 9 + 36, , = 49 = 7, , \ Unit vector in the direction of vector a is, 2i + 3 j + 6 k 2  3  6 , a =, = i+ j+ k, 7, 7, 7, 7, 46. (b) : Here, (4, 10) are the coordinates of A., \ P.V. of A = 4i + 10 j, 47. (c) : Here, (9, 7) are the coordinates of B., \ P.V. of B = 9i + 7 j, 48. (b) : Here, P.V. of A = 4i + 10 j and P.V. of, C = 4i + 2 j, , , AC = (4 – 4) i + (2 – 10) j = – 8 j, , 49. (a) : Here A = i + 2 j + 3k, , \, A = 12 + 2 2 + 32 = 1 + 4 + 9 = 14, , A i + 2 j + 3k, ^, 1 , 2 , 3 , \ A, =  =, =, i+, j+, k, 14, A, 14, 14, 14, , , 50. (d) : We have, A = 4i + 3 j and B = 3i + 4 j, , \, A = 4 2 + 32 = 16 + 9 = 25 = 5, , and B = 32 + 4 2 = 9 + 16 = 25 = 5, , , Thus, A + B = 5 + 5 = 10, , , 51. (b) : a = 2i + j + k , b = i + 2 j + 3k,   , , Resultant of a and b is a + b, \, , = ( 2i + j + k ) + (i + 2 j + 3 k ) = 3i + 3 j + 4 k, ∴, ,  , a + b = 32 + 32 + 4 2 = 9 + 9 + 16 = 34, , Also, the magnitude of a vector can never be negative., Hence, both Assertion and Reason are correct but, Reason is not the correct explanation of Assertion., 52. (d) : Sum of the given vectors, = (i + j + k ) + ( 2i − j − k ) + ( 2 j + 6 k ) = 3i + 2 j + 6 k, \ The unit vector in the direction of the sum of the, given vectors, 3i + 2 j + 6 k, 3i + 2 j + 6 k, 1, = ( 3i + 2 j + 6 k ), =, =, 2, 2, 2, 7, 9 + 4 + 36, 3 +2 +6, Hence, Assertion is wrong., , , a, , Also,  is a unit vector which is parallel to a., a, Hence, Reason is correct., , , 53. (a) : a = i + j − 3k , b = 2i + j + k,  , a ⋅ b = (i + j − 3 k ) ⋅ ( 2i + j + k ), = 1 ⋅ 2 + 1 ⋅ 1 + (–3) ⋅ 1 = 2 + 1 – 3 = 0, π, cosq = 0 ⇒ θ =, 2, , , Hence, a and b are perpendicular to each other., Hence, both Assertion and Reason are correct and Reason, is the correct explanation of Assertion., , , 54. (d) : a = i + 2 j , b = 2i + j, , , Diagonals of the parallelogram are along a + b and a − b .,  , Now, a + b = (i + 2 j ) + ( 2i + j ) = 3i + 3 j,  , and a − b = (i + 2 j ) − ( 2i + j ) = −i + j, ⇒, , Let q be the angle between these vectors, then, cosθ =, , ( 3i + 3 j ) ⋅ ( − i + j ), −3 + 3, =, =0, 9+9 1+1, 18 2, , ⇒ q = 90°, Hence, Assertion is wrong and Reason is correct., , , , 55. (b) : We have, a = 3 , b = 4 , c = 5 and, , , 2,    , a + b + c = 0 ⇒ (a + b + c ) = 0, 2 2 2,      , ⇒ a + b + c + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0,      , ⇒ ( 3)2 + ( 4 )2 + ( 5)2 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) = 0, 1, 1,      , ⇒ a ⋅ b + b ⋅ c + c ⋅ a = − [9 + 16 + 25] = − ( 50 ) = −25, 2, 2,  , ,    , Now, a + b + c = 0 ⇒ b + c = − a,  2, ,   , 2, , ⇒ b + c = −a, ⇒ b 2 + c 2 + 2b ⋅ c = a 2, , , , , ⇒ b 2 + c 2 + 2 bc cos θ = a 2, , , , a2 − b 2 − c 2, , ⇒ cos θ =, 2 bc, Hence, both Assertion and Reason are correct but Reason, is not the correct explanation of Assertion., 56. (a) : Required length =, 3−2+6, =, 1+ 4+9, , ( 3i − j − 2 k ) ⋅ (i + 2 j − 3k ), 12 + 2 2 + (− 3)2, , 7, 14, ,  , ,  a⋅b, ,  , Also, vector projection of a on b = ( a ⋅ b ) =   ,  b , Hence, both Assertion and Reason are correct and Reason, is the correct explanation of Assertion.

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CBSE Board Term-II Mathematics Class-12, , 72,  ,  ,  2,  , 57. (d) : ( a × b )2 + ( a ⋅ b )2 = a × b + ( a ⋅ b )2, , ,  , = ( ab sin θ)2 + ( ab cos θ)2 = a 2 b 2, Hence, Assertion is wrong., But sin2q + cos2q = 1, Hence, Reason is correct.,  ,  , , 58. (d) : ( a × b )2 + ( a ⋅ b )2 = 400 , a = 4, We know that,,  , 2 2,  , ( a × b )2 + ( a ⋅ b )2 = a b, 2, 2, ⇒ 400 = ( 4 )2 b ⇒ 16 b = 400, 2, , ⇒ b = 25 ⇒ b = 5, Hence, Assertion is wrong.,  ,  ,  2,  , ( a × b )2 + ( a ⋅ b )2 = a × b + ( a ⋅ b )2, , ,  , = ( ab sin θ)2 + ( ab cos θ)2 = a 2 b 2,  ,  ,  , ⇒ ( a × b )2 = a 2 b 2 − ( a ⋅ b )2, Hence, Reason is correct.,  ,  a⋅b, , 59. (a) : Projection of a on b = , b, 2i + 3 j − k ⋅ −i + 3 j + 4 k, −2 + 9 − 4, =, =, =, 2, 2, 2, 26, ( −1) + ( 3) + ( 4 ), , (, , )(, , ), , 3., , the sum of the given vectors is,   , c = a + b = ( 2 + 4 )i + ( 3 − 3) j + ( −1 + 2 ) k = 6i + k, ,  , and | c |=| a + b |= 6 2 + 12 = 36 + 1 = 37, , 6 , 1 , c, 6i + k, \ Unit vector, c =  =, =, i+, k, |c |, 37, 37, 37, 4., , 2, 2,  ,  , = { a b sin θ} + { a b cos θ}, , 2 2, 2 2, = a b sin 2 θ + a b cos 2 θ, 2 2, = a b = 25 × 16 , , 2, , , , [∵ | a | = 5 and |b | = 4 ], , = 400, 3, 26, , π 1, π 1, 1. We have, l = cos = , m = cos =, and n = cos θ, 3 2, 4, 2, Now, l 2 + m2 + n2 = 1, 2, ,  2  2, a × b + a⋅b, , 5., , , , Let a = i − 3k and b = 2 j + k, , , The area of a parallelogram with a and b as its adjacent,  , sides is given by | a × b |., 6., , i j k,  , Now, a × b = 1 0 −3 = 6i − j + k, 0 2 1, , \, ,  , | a × b |= (6 )2 + ( −1)2 + ( 2 )2 = 36 + 1 + 4, , =, , 41 sq. units., , 7., , SUBJECTIVE TYPE QUESTIONS, ,  1,  1 , + n2 = 1,   + ,  2 , 2, 1 1, 1, 1, ⇒, + + n2 = 1 ⇒ n2 =, ⇒ n=±, 4 2, 4, 2, 1, ⇒ cos θ = ±, 2, But q is an acute angle (given).,  1 π, ∴ θ = cos −1   =,  2 3,   , 2. Required sum = a + b + c, , Required position vector, , , ,  , , 2( a + 2 b ) − 1( 2 a − b ), =, = 5b, 2−1, , \ Assertion and Reason are correct and Reason is the, correct explanation of Assertion., , , 60. (b) : If A, B, C are collinear, then AB = k AC,   ,  ,  , ∴ AB × AC = 0 ⇒ ( b − a ) × ( c − a ) = 0,       ,       , ⇒ b × c + a × b + c × a = 0 i.e., a × b + b × c + c × a = 0, Hence, both Assertion and Reason are correct but, Reason is not the correct explanation of Assertion., , ⇒, , , , Let a = 2 iˆ + 3 jˆ − kˆ and b = 4iˆ − 3 jˆ + 2 kˆ . Then,, , =, , ,  a ⋅ b, , Projection of a on b = , |b |, , ( 2i + 3 j + 2 k ) ⋅ ( 2i + 2 j + k ), 2, , 2, , ( 2 ) + ( 2 ) + ( 1), , 2, , =, , 4 + 6 + 2 12, =, =4, 3, 3, , , , Let q be the angle between the unit vectors a and b.,  , , , a⋅b,  , (∵ | a |= 1 =| b |) ...(1), ∴ cos θ =   = a ⋅ b , | a || b |,  , Now, 1 = 2 a − b, 8., ,  2,  ,  , ⇒ 1= 2a−b = 2a−b ⋅ 2a−b, ,   ,  ,  , = 2 | a |2 − 2 a ⋅ b − b ⋅ 2 a + | b |2 = 2 − 2 2 a ⋅ b + 1,    , (∵ a ⋅ b = b ⋅ a ),  , = 3 − 2 2 a⋅b, , (, , )(, , = (i − 3k ) + ( 2 j − k ) + ( 2i − 3j + 2k ), , 1, 1,  , ⇒ a⋅b =, ⇒ cos θ =, , 2, 2, , = 3i − j − 2k ., , \, , q = p/4, , ), , [By using (1)]

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73, , Vector Algebra, , Here, a = iˆ + jˆ + k̂ and vector along x-axis is î ., , \ Angle between a and iˆ is given by, , iˆ + jˆ + kˆ ⋅ iˆ, a . i, 1, cos θ =, =, =, 2, 2, 2, 2, 3, ⋅, 1, 3, 1 +1 +1 . 1, 9., , (, , ), ,  1 , ⇒ θ = cos −1 ,  3 ,  , , 10. Here a + b = a,    ,  ,  2,  2, ⇒ (a + b ) ⋅ (a + b ) = a ⋅ a, ⇒ a+b = a,          , ⇒ a⋅a + a⋅b + b ⋅a + b ⋅b = a⋅a,    ,    , ∵ b ⋅ a = a ⋅ b , ⇒ 2a ⋅ b + b ⋅ b = 0, ,  ,   , ⇒ (2a + b ) ⋅ b = 0 ⇒ (2a + b ) ⊥ b, 11. Position vector which divides the line segment, ,  , , joining points with position vectors 3a + b and a − 3b, in the ratio 2 : 1 externally is given by, , , ,  , ,  , 2( a − 3b ) − 1( 3a + b ) 2 a − 6b − 3a − b, =, 2−1, 1, , , = −a − 7b, , , 12. Here, a = 5iˆ + 6 jˆ − 2 kˆ and b = 7 iˆ + 6 jˆ + 2 kˆ, , , Vector perpendicular to both a and b is, i j k,  , a × b = 5 6 −2, 7 6 2, = i( 12 + 12 ) − j(10 + 14 ) + k ( 30 − 42 ), = 24iˆ − 24 jˆ − 12 kˆ = 12( 2 iˆ − 2 jˆ − kˆ ), , , , Unit vector perpendicular to both a and b,  , 24i − 24 j − 12 k, 12( 2i − 2 j − k ), a×b, =   =, =, 576 + 576 + 144, 1296, a×b, 12( 2 iˆ − 2 jˆ − kˆ ) 1 ˆ, =, = ( 2 i − 2 jˆ − kˆ ), 36, 3, , , 13. For any two non-zero vectors a and b , we have,  ,  ,  2,  2, a+b = a−b ⇒ a+b = a−b, , , ,   ,  , ⇒ a2 + b 2 + 2a ⋅ b = a2 + b 2 − 2a ⋅ b, , , , , ⇒ 4a ⋅ b = 0 ⇒ a ⋅ b = 0, , , So, a and b are perpendicular vectors., 14. Let A( 2 iˆ − jˆ + kˆ ), B( 3iˆ + 7 jˆ + kˆ ) and C ( 5iˆ + 6 jˆ + 2 kˆ ), , Then, AB = ( 3 − 2 )iˆ + (7 + 1)jˆ + (1 − 1)kˆ = iˆ + 8 jˆ, , AC = ( 5 − 2 )i + ( 6 + 1)j + ( 2 − 1)k = 3i + 7 j + k, , BC = ( 5 − 3)i + (6 − 7 )j + ( 2 − 1)k = 2i − j + k, , \, , , , , , Now, angle between AC and BC is given by,  , AC ⋅ BC, 6−7+1, ⇒ cos q =   =, 9 + 49 + 1 4 + 1 + 1, AC BC, , ⇒ cos q = 0 ⇒ AC ⊥ BC, So, A, B, C are the vertices of right angled triangle., , ,  , We have a = 2 , b = 1 and a ⋅ b = 1, , , , , Now, ( 3a − 5b ) ⋅ ( 2 a + 7 b ), 2, 2,  ,  , = 6 a + 21 a ⋅ b − 10 a ⋅ b − 35 b, , ,  , = 6| a |2 + 11a ⋅ b − 35| b |2, = 6(2)2 + 11(1) – 35(1)2 = 24 + 11 – 35 = 0, , , 16. Given , a = ^i + 2 ^j − 3 ^k and b = 3 ^i − ^j + 2 ^k,  , ^ ^ ^, Now, a + b = 4 i + j − k,  , ^, ^, ^, Also, a − b = −2 i + 3 j − 5 k,    , ^ ^ ^, ^, ^, ^, Now, ( a + b ) ⋅ ( a − b ) = ( 4 i + j − k ) ⋅ ( −2 i + 3 j − 5 k ), 15., , = (4)(–2) + (1)(3) + (–1)(–5) = –8 + 3 + 5 = 0,  ,  , Hence, ( a + b ) and ( a − b ) are perpendicular to each other., , , 17. Let a = i − 2 j + 3k , b = 3i − 2 j + k,  ,  , Now, a ⋅ b = | a || b |cos θ, ⇒ (i − 2 j + 3k ) ⋅ ( 3i − 2 j + k ) = (1)2 + ( −2 )2 + ( 3)2, × ( 3)2 + ( −2 )2 + (1)2 cos θ, , , ⇒, , 3+4+3 =, , ⇒ cos θ =, , 14 × 14 cos θ, , 10 5, =, 14 7, , ∴ sin θ = 1 − cos 2 θ = 1 −, 2 6, 7,  ˆ ˆ, 18. Given, a = i − j + 2 kˆ ,, , c = 2 iˆ − jˆ + 3kˆ, , \ 2 a = 2 iˆ − 2 jˆ + 4 kˆ, , −b = −3iˆ − 4 jˆ + 5kˆ, , 3c = 6iˆ − 3 jˆ + 9 kˆ, 2, , , , Now, 2 a ⋅ ( − b × 3c ) = −3, 6, , 25, =, 49, , 24, 49, , ⇒ sin θ =, , , b = 3iˆ + 4 jˆ − 5kˆ and, , −2 4, −4 5, −3 9, , = 2(–36 + 15) + 2(–27 – 30) + 4(9 + 24), = 2(–21) – 2(57) + 4(33), = –42 – 114 + 132 = – 24,  , , \ | 2 a .( −b × 3c )| = |–24| = 24, , ^, ^ ^ , ^, ^ ^, 19. Given, a = 2 i + 3 j + k , b = i − 2 j + k and, , ^ ^, ^, c = −3 i + j + 2 k, ^, , ^, , ^, , i, j k,  , Now, b × c = 1 −2 1, −3 1 2

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CBSE Board Term-II Mathematics Class-12, , 74, ^, , ^, , Solving (iv) and (v), we get, , ^, , = i ( −4 − 1) − j ( 2 + 3) + k (1 − 6 ), ^, , ^, , ^, , = −5 i − 5 j − 5 k, , −1, 16, ,y=, 3, 3, 13, Putting the values of x and y in (i), we get z =, 3,  −1, i + 16 j + 13 k is the required vector., ∴ d=, 3, 3, 3,  ˆ ˆ ˆ , 24. Here, a = i + j + k ; b = 2 iˆ + 4 jˆ − 5 kˆ and, , c = λiˆ + 2 jˆ + 3kˆ,  , ⇒ b + c = ( 2 + λ ) iˆ + 6 jˆ − 2 kˆ,  ,   , b+c, The unit vector along b + c is p =  , b+c, x=, ,   , ^, ^ ^, ^, ^, ^, ∴ a ⋅ (b × c ) = ( 2 i + 3 j + k ) ⋅ ( −5 i − 5 j − 5 k ), , = –10 – 15 – 5 = –30,    , 20. (r × i ) ⋅ (r × j ) + xy, = [( xi + y j + zk ) × i ] .[( xi + y j + zk ) × j )] + xy, = ( − yk + z j ) ⋅ ( xk − zi) + xy = –xy + xy = 0, 21. Take A to be as origin (0, 0, 0)., \ Coordinates of B are (0, 1, 1) and coordinates of C, are (3, –1, 4)., , =, , ( 2 + λ ) iˆ + 6 jˆ − 2 kˆ, , =, , ( 2 + λ )2 + 6 2 + ( −2 )2, ,  , Also, a . p = 1 (Given), ⇒, Let D be the mid point of BC and AD is a median of, DABC., , \, , 5, 3, Coordinates of D are  , 0 , , 2, 2, , So, length of AD =, , 2, , 3, , , 2 5,  − 0 + (0 ) +  − 0, 2, 2, , 2, , ), , \, , A vector of magnitude 5 units in the direction of,  , , 5( a + b ) 5( 3i + j ), , a + b is   =, 10, a+b, , 23. Let d = xi + y j + zk, , Now, it is given that, d is perpendicular to, , , b = i − 4 j + 5k and c = 3i + j − k,  ,  , ∴ d ⋅ b = 0 and d ⋅ c = 0, , ⇒ x – 4y + 5z = 0, and 3x + y – z = 0 ,  , , Also, d ⋅ a = 21, where a = 4iˆ + 5 jˆ − kˆ, , ⇒ 4x + 5y – z = 21, Eliminating z from (i) and (ii), we get, 16x + y = 0, Eliminating z from (ii) and (iii), we get, x + 4y = 21, , =1, , λ2 + 4 λ + 44 = λ + 6, , 1 + 4 + 44, , 9 25, 34, +, =, units, 4 4, 2, , , 22. a = 2 iˆ + 3 jˆ − kˆ , b = iˆ − 2 jˆ + kˆ,  , ∴ a + b = 2 iˆ + 3 jˆ − kˆ + iˆ − 2 jˆ + kˆ = 3iˆ + jˆ,  , a + b = 32 + 12 = 10, , ) (, , λ2 + 4 λ + 44, , λ 2 + 4 λ + 44, , ⇒ l2 + 4l + 44 = l2 + 12l +36, ⇒ 8l = 8 ⇒ l = 1, \ The required unit vector, 1,  ( 2 + 1 ) iˆ + 6 jˆ − 2 kˆ, = ( 3iˆ + 6 jˆ − 2 kˆ ) ., p=, , =, , (, , ⇒, , (2 + λ) + 6 − 2, , ( 2 + λ ) iˆ + 6 jˆ − 2 kˆ, , ...(i), ...(ii), ...(iii), ...(iv), ...(v), , 7, , 25. Two non zero vectors are parallel if and only if their, cross product is zero vector.,  , So, we have to prove that cross product of a − d and,  , b − c is zero vector., , , ,  , , Now, ( a − d ) × ( b − c ) = ( a × b ) − ( a × c ) − ( d × b ) + ( d × c ),    ,    , Since, it is given that a × b = c × d and a × c = b × d,  ,    ,  , And, d × b = − b × d , d × c = − c × d,  ,  ,  ,  ,  ,   , \ ( a − d ) × ( b − c ) = (c × d ) − ( b × d ) + ( b × d ) − (c × d ) = 0,  ,  , Hence, a − d is parallel to b − c , where,  ,  , a ≠ d and b ≠ c ., , 26. Let the required vector be r = x iˆ + y jˆ + zkˆ ., , , , Also let, a = iˆ − jˆ + kˆ , b = 2 iˆ + jˆ − 3kˆ and c = iˆ + jˆ + kˆ, ,  , ,  , r ⋅ a = 4 , r ⋅ b = 0 , r ⋅ c = 2 (Given), , ⇒ x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2, Now (iii) – (i) ⇒ 2y = –2 ⇒ y = –1, From (ii) and (iii), 2x – 3z – 1 = 0, x + z – 3 = 0 ⇒ x = 2, z = 1, , \ The required vector is r = 2 iˆ − jˆ + kˆ ., , ...(i), ...(ii), ...(iii)

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CBSE Board Term-II Mathematics Class-12, , 76, , , 33. We have a = ˆi + ˆj + kˆ , b = ˆi + 2ˆj + 3kˆ,   , Let r = a + b = 2ˆi + 3ˆj + 4kˆ,   , and p = a − b = −ˆj − 2 kˆ, , , , , , A unit vector perpendicular to both r and p is given,  , r×p, as ±   ., r×p, ˆi ˆj, kˆ,  , Now, r × p = 2 3, 4 = −2ˆi + 4ˆj − 2 kˆ, 0 −1 −2, ,  2, a +0+0, =    , [ Using (ii )], a+b+c a, , a, =    ...(iii), a+b+c, , b, Similarly, cosβ =    ...(iv), a+b+c, , c, and cos γ =    ...(v), a+b+c, , So, the required unit vector is, =±, , ( −2ˆi + 4ˆj − 2kˆ), , ( −2 )2 + 4 2 + ( −2 )2, , =+, , (ˆi − 2ˆj + kˆ) ., 6, , 34. Given, aˆ + bˆ = cˆ, , ⇒ ( aˆ + bˆ ) ⋅ ( aˆ + bˆ ) = cˆ ⋅ cˆ, ⇒ aˆ ⋅ aˆ + aˆ ⋅ bˆ + bˆ ⋅ bˆ + bˆ ⋅ aˆ = cˆ ⋅ cˆ, ⇒ 1 + aˆ ⋅ bˆ + 1 + aˆ ⋅ bˆ = 1, , From (i), (iii), (iv) and (v), we get, , ⇒ 2 aˆ ⋅ bˆ = − 1 ...(i), 2, Now, ( aˆ − bˆ ) = ( aˆ − bˆ ) ⋅ ( aˆ − bˆ ), , = aˆ ⋅ aˆ − aˆ ⋅ bˆ − bˆ ⋅ aˆ + bˆ ⋅ bˆ = 1 − aˆ ⋅ bˆ − aˆ ⋅ bˆ + 1, = 2 − 2 aˆ ⋅ bˆ = 2 − ( −1), , =3, , , , , a = b = c, (Given), ...(i), , , , ,  , and a ⋅ b = 0 , b ⋅ c = 0 , c ⋅ a = 0 ...(ii),   ,   , Let ( a + b + c ) be inclined to vectors a , b , c by angles, a, b and g respectively. Then, , , ( a + b + c ) ⋅ a a ⋅ a + b ⋅ a + c ⋅ a, cos α =     =    , a+b+c a, a+b+c a, 36., , cos a = cos b = cos g ⇒ a = b = g,   , Hence, the vector a + b + c is equally inclined to the,  , , vector a , b and c ., Also the angle between them is given as, , , |b | , | a| , −1 , −1 , α = cos      , β = cos      ,, | a + b + c |, | a + b + c |, , [ Using(i)], , ∴ aˆ − bˆ =, , 3, ,  ˆ ˆ ˆ, 35. Given a = i + j + k and b = ˆj − kˆ, , , |c | , , γ = cos −1     , | a + b + c |, , , Let c = xiˆ + yjˆ + zkˆ,   , Now we have, a × c = b, , 37. W e h a v e , A( 2i − j + k ) , B (i − 3 j − 5k ), C ( 3 i − 4 j − 4 k ), , Then, AB = (1 − 2 ) i + ( −3 + 1)j + ( −5 − 1)k, , = − i − 2 j − 6 k, , AC = ( 3 − 2 ) i + ( − 4 + 1) j + ( −4 − 1)k = i − 3 j − 5k, , ⇒ (i + j + k ) × (xi + y j + zk ) = j − k, ⇒, , ˆi ˆj kˆ, 1 1 1 = ˆj − kˆ, , x y, , , and BC = ( 3 − 1) i + ( −4 + 3)j + ( −4 + 5)k = 2i − j + k, , z, , ⇒ ˆi (z − y) − ˆj(z − x) + kˆ(y − x) = ˆj − kˆ, ⇒ z – y = 0, x – z = 1 and y – x = –1, ⇒ y = z, x – z = 1, x – y = 1,  , Also, we have a ⋅ c = 3, ⇒ (i + j + k ) ⋅ (xi + y j + zk ) = 3, , ⇒ x+y+z=3, ⇒ x + x – 1 + x – 1 = 3, , 5, 2, 2, ,y= ,z=, 3, 3, 3,  5ˆ 2 ˆ 2 ˆ, Hence, c = i + j + k, 3, 3, 3, , ⇒ 3x – 2 = 3 ⇒ x =, , .....(i), , , , Now angle between AC and BC is given by,  , 2+3−5, ( AC ) ⋅ ( BC ), cos θ =   =, | AC || BC |, 1 + 9 + 25 . 4 + 1 + 1, , ⇒ cosq = 0 ⇒ BC ^ AC, , [Using (i)], , So, A, B, C are vertices of right angled triangle., 1  , Now area of DABC = | AC × BC |, 2, i j, k, 1, 1, =, 1 −3 −5 = |( −3 − 5) i − ( 1 + 10 )j + ( −1 + 6 )k |, 2, 2, 2 −1 1, , and

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77, , Vector Algebra, =, , (, , 1, | − 8i − 11j + 5k |, 2, , 1, =, 64 + 121 + 25 =, 2, , 38. Let a = 2i − 4 j −, , 210, sq . units., 2, , 5 k and b = 2i + 2 j + 3k, , )(, , ), , ⇒ xiˆ + yjˆ + zkˆ ⋅ 3iˆ − jˆ = 0, ⇒ 3x – y = 0,   , Now, b = b1 + b2, , ...(i), , ) (, , (, , ⇒ 2 iˆ + jˆ − 3kˆ = λ 3iˆ − jˆ + xiˆ + yjˆ + zkˆ, On comparing, we get, , ), , 2 = 3λ + x , ⇒ x + 3y = 5 ...(ii), 1 = − λ + y , , and –3 = z ⇒ z = –3, , Then diagonal AC of the parallelogram is,   , p=a+b, = 2i − 4 j − 5k + 2i + 2 j + 3k = 4i − 2 j − 2 k, Therefore, unit vector parallel to it is, , p, 4i − 2 j − 2 k 2i − j − k, =,  =, | p|, 16 + 4 + 4, 6, , Now, diagonal BD of the parallelogram is,  , , p′ = b − a = 2i + 2 j + 3k − 2i + 4 j + 5k = 6 j + 8k, Therefore, unit vector parallel to it is, , p′, 6 j + 8 k, 6 j + 8 k 3 j + 4 k, =, =,  =, | p ′|, 10, 5, 36 + 64, i j, k,  , Now, p × p ′ = 4 −2 −2, 0 6, 8, = i ( −16 + 12 ) − j ( 32 − 0 ) + k ( 24 − 0 ), = − 4i − 32 j + 24 k, , \, , Area of parallelogram =, ,  , | p × p ′|, 2, , 16 + 1024 + 576, = 2 101 sq . units., 2, , , 39. Here a = 3iˆ − jˆ , b = 2 iˆ + jˆ − 3kˆ,   , We have to express : b = b1 + b2 , where, , , , , b1 || a and b2 ⊥ a, , , , Let b1 = λ a = λ 3iˆ − jˆ and b2 = xiˆ + yjˆ + zkˆ, ,  , , Now b2 ⊥ a ⇒ b2 ⋅ a = 0, =, , (, , ), , Solving (i) and (ii), we get x =, , 1, 3, ,y =, 2, 2, , 3, 1, ⇒λ=, 2, 2, , 3 1 , , , Hence, b1 = λ 3i − j = i − j, 2, 2, , 1ˆ 3ˆ, and b2 = i + j − 3k, 2, 2, , ∴ 1 = −λ + y ⇒ 1 = −λ +, , (, , ), , ^, , ^, , ^, , 40. Given, position vector of A = i + j + k, ^, , Position vector of B = 2 i + 5 ^j, ^, , ^, , ^, , Position vector of C = 3 i + 2 j − 3 k, ^, , ^, , ^, , Position vector of D = i − 6 j − k, , ^, ^ ^ ^, ^, ^, ^ ^, ∴ AB = ( 2 i + 5 j ) − ( i + j + k ) = i + 4 j − k and,  ^, ^, ^, ^, ^, ^ ^, ^, ^, CD = ( i − 6 j − k ) − ( 3 i + 2 j − 3 k ) = − 2 i − 8 j + 2 k, , 2, 2, 2, Now AB = (1) + ( 4 ) + (1) = 18, , CD = ( −2 )2 + ( −8)2 + ( 2 )2 = 4 + 64 + 4, = 72 = 2 18, , , , Let q be the angle between AB and CD.,  , ^, ^ ^, ^, ^, ^, ( i + 4 j − k ) ⋅ ( −2 i − 8 j + 2 k ), AB ⋅ CD, ∴ cos θ =   =, | AB||CD|, ( 18 )( 2 18 ), =, , −2 − 32 − 2 −36, =, = −1, 36, 36, , ⇒ cosq = –1 ⇒, , q=p, , , , Since, angle between AB and CD is 180°., , , \ AB and CD are collinear., , 

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Three Dimensional, Geometry, , CHAPTER, , 5, , Recap Notes, Note : Equation of x -axis, y -axis and z -axis are, respectively y = 0 = z, z = 0 = x and x = 0 = y., , INTRODUCTION, Everything in the real world is in a three dimensional, shape. The three dimensional geometry is used in, various fields such as art and architecture, space study, and astronomy, geographic information systems etc., In this chapter we shall study the direction cosines, and direction ratios of line m and also study about the, equations of lines and planes etc., , Direction Cosines and Direction Ratios of a Line, h, , , , If a non-zero vector r (or any line along which r lies), makes angles a, b, g with positive directions of axes,, , then a, b, g are called direction angles of r (or of the, line). cosa, cosb, cosg are called the direction cosines, , of r (or line). Any three numbers proportional, to the direction cosines (d.c.’s) of a vector are, , called its direction ratios. If r = xi + y j + zk , then, x y z, x, y, z are direction ratios (numbers) and  ,  , , |r | |r | |r |, , are direction cosines of r (or any lines along which, , r lies)., , Relation Between the Direction Cosines of a Line, 2, 2, 2, h If l, m, n are d.c.’s of a line, then l + m + n = 1., Direction Cosines of Line Passing Through Two, Points, h Direction ratios of the line joining (x1, y1, z1) and, (x2, y2, z2) are x2 – x1, y2 – y1, z2 – z1 and d.c.’s are, ±, , x 2 − x1, , ∑ ( x 2 − x1 ), , 2, , ,±, , y 2 − y1, 2, , ∑ ( x 2 − x1 ), , , ±, , z2 − z1, , ∑ (x2 − x1 )2, , EQUATION OF A LINE IN SPACE, Equation of a Line Through a Given Point and, Parallel to a Given Vector, h, , Cartesian Equation : Cartesian equation of a line, passing through (x1, y1, z1) and having direction, x − x1 y − y1 z − z1, ratios a, b, c is, =, =, a, b, c, , Vector Equation : Vector equation of line passing, , , through the point A(a) and parallel to vector b is, ,  , r = a + λb , where l ∈ R is a parameter., , Equation of a Line Passing Through Two Given, Points, h, , Cartesian Equation : Cartesian equations of a line, passing through two points having coordinates, y − y1, x − x1, z − z1, (x1, y1, z1) and (x2, y2, z2) is, =, =, x2 − x1 y 2 − y1 z2 − z1, , h, , Vector Equation : Vector equation of a line passing, , through two points having position vector a and, , ,  , , b is r = a + λ(b − a) , where l ∈ R is a parameter., , h, , Collinearity of Three Points : The points A(x1, y1, z1),, B ( x 2 , y 2 , z 2 ) and C ( x 3 , y 3 , z 3 ) are collinear, if, x1 − x2 y1 − y 2 z1 − z2, =, =, x2 − x3 y 2 − y 3 z2 − z3, , CONDITION OF PARALLELISM AND, PERPENDICULARITY OF TWO LINES, h Two lines with direction ratios a1, b1, c1 and a2, b2,, c2 are, (i) Perpendicular, i.e., if a1 a2 + b1b2 + c1c2 = 0, (ii) Parallel, if a1 = b1 = c1, a2 b2 c 2, , SHORTEST DISTANCE BETWEEN TWO LINES, Distance between Skew Lines, , (The lines which are neither parallel nor intersecting.)

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79, , Three Dimensional Geometry, h, h, , Shortest distance between two skew lines is the line, segment perpendicular to both the lines., Let l1 and l2 be two skew lines with equations, , ,  ,  , r = a1 + λb1 and r = a2 + µb2 , then shortest distance, in vector form is,  ,  , (b1 × b2 ) ⋅ (a2 − a1 ),  , d=, b1 × b2, , If the equation of lines are given by, x − x1 y − y1 z − z1, and, l1 :, =, =, a1, b1, c1, x − x2 y − y 2 z − z2, l2 :, =, =, a2, b2, c2, then shortest distance is, x2 − x1 y 2 − y1 z2 − z1, a1, b1, c1, a2, b2, c2, , h, , Cartesian Equation : Let P(x, y, z) be any point on, the plane. Then,  , OP = r = xi + y j + zk., Let l , m , n be the direction cosines of n . Then, n = li + mj + n k, , Now, r ⋅ n = d ⇒ (xi + y j + zk ) ⋅ (li + mj + nk ) = d, ⇒ lx + my + nz = d, which is cartesian equation of, the plane in the normal form., , Equation of a Plane Perpendicular to a Given, Vector and Passing through a Given Point, h, , Vector Equation : Let a plane,  passing through a point, a and perpendicular to the, A with position, vector, , , vector N . Also, let r be the position vector of any, point P in the plane. Then,,   , (r − a) ⋅ N = 0 , which is the equation of plane in, vector form., , h, , Cartesian Equation : Let the coordinates of given, point A be (x1, y1, z1), arbitrary, point P be (x, y, z), , and direction ratios of N are A, B, and C. Then,, , , a = x1i + y1 j + z1 k , r = xi + y j + zk, , and N = Ai + , Bj + Ck,   , Now, (r − a) ⋅ N = 0, ⇒ A(x – x1) + B(y – y1) + C(z – z1) = 0,, which is equation of plane in cartesian form., , (b1c 2 − b2c1 )2 + (c1 a2 − c 2 a1 )2 + (a1b2 − a2 b1 )2, , Distance between Parallel Lines, , h, , Let two lines l1 and l2 are parallel, given by, , ,  , ,  , r = a1 + λb and r = a2 + µb respectively, where, a1, , is the position vector of a point S on l1 and a2 is the, , position vector of a point T on l2. Then the shortest, distance ,  , , b × (a2 − a1 ), , d =|PT | =, |b|, Note : If two lines are parallel, then they are coplanar also., , PLANE, Equation of a Plane in Normal Form, h, , , Vector Equation : Let r be the position vector of, any point P on the plane. Then equation of plane, , is r ⋅ n = d , where n is the normal unit vector to, the plane and d is perpendicular distance of plane, from the origin., , Equation of a Plane Passing Through Three, Non Collinear Points, h Vector Equation : Let R , S and T be three non, collinear, points on the plane with position vectors,  , , , a , b and c respectively. Also, let r be the position, vector of any point, plane. Then equation of,  P in the,  ,  , plane is (r − a) ⋅ [(b − a) × (c − a)] = 0, Z, , P, , O, , R, ,  , r a S, b, ,  , (RS × RT), , c, , T, Y, , X, h, X, , Cartesian Equation : Let (x1, y1, z1), (x2, y2, z2) and, (x3, y3, z3) be the coordinates of the points R, S and

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81, , Three Dimensional Geometry, , Practice Time, OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1. If a line makes an angle q 1 , q 2 , q 3 with, the axes respectively, then the value of, cos 2q1 + cos 2q2 + cos 2q3 is, (a) 1, (b) –1, (c) 4, (d) 3, x −3, y+2, If the equation of a line AB is, =, −2, 1, z −5, =, , find the direction ratios of a line parallel, 4, to AB., (a) 1, 2, 4, (b) 1, 2, –4, (c) 1, –2, –4, (d) 1, –2, 4, 2., , 3. If a, b, g are the angles made by a line with the, co-ordinate axes. Then sin2a + sin2b + sin2g is, (a) 0, (b) –1, (c) 2, (d) 1, If a, b, g are the direction angles of a vector, 14, 1, and cos a =, , cos b = , then cos g =, 15, 3, 1, 4, 2, 1, (a) ±, (b) ±, (c) ±, (d) ±, 5, 15, 15, 15, 4., , 5. If a line makes angles 90°, 60° and 30°, with the positive directions of x, y and z-axis, respectively, then its direction cosines are, (a), , 1, 3, , 0,, 2, 2, , (b), , 3 1, , ,0, 2 2, , (c), , 3, 1, , 0,, 2, 2, , (d), , 1 3, 0, ,, 2 2, , The direction cosines of the line passing through, two points (2, 1, 0) and (1, –2, 3) are, 6., , (a), (c), , 1, 19, −1, 19, , ,, ,, , 3, 19, −3, 19, , ,, ,, , 3, 19, 3, 19, , (b), (d), , −3, 19, 1, 19, , ,, ,, , 3, 19, 3, 19, , ,, ,, , −1, 19, −3, 19, , 7. Find the direction cosines of the line that, makes equal angles with the three axes in space., , (a) ±, , 1, , (b) ±1, , 1, , (d) 3, 3, 2, 8. Find the direction cosines of the line joining, A(0, 7, 10) and B(–1, 6, 6)., 1, 4 ,  1, (a) , (b)  1 , 4 , 1 , ,, ,, , , , 3 2 3 2 3 2, 3 2 3 2 3 2, 1, 1 ,  1, ,, ,, (c) ,  3 2 3 2 3 2 , , (c) ±, , 1, 1 ,  4, ,, ,, (d) ,  3 2 2 3 2 , , 9. Find the equation of a line passing through, a point (2, –1, 3) and parallel to the line, , r = (i + j ) + λ(2i + j − 2k )., , (a) r = (i + j ) + µ (2i − j + 3k ), , (b) r = (2i − j + 3k ) + µ (2i + j − 2k ), , (c) r = (i − j ) + µ (2i − j + 3k ), , (d) r = (2i + j + 3k ) + µ (2i + j − 2k ), 1 1, 10. If  , , n are the direction cosines of a, 2 3 , line, then the value of n is, (a), , 23, 6, , (b), , 23, 6, , (c), , 2, 3, , (d), , 3, 2, , 11. The distance of the plane 3x – 6y + 2z + 11 = 0, from the origin is, (a) 11 units, (b) 1 unit, 7, 7, 7, 13, (c), (d), unit, units, 11, 7, 12. Find the distance of the point (2, 3, 4) from, , the plane r ⋅ (3i − 6 j + 2k ) + 11 = 0., (a) 1 unit, (c) 2 units, , (b) 4 units, (d) 3 units, , 13. Write the direction cosines of a line parallel, 3− x, y + 2, z + 2, =, ., =, to the line, −2, 3, 6

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CBSE Board Term-II Mathematics Class-12, , 82, 1 2 3, , ,, 7 7 7, 3 2 6, , ,, (c), 7 7 7, , −3 −2 6, ,, ,, 7 7 7, 3 −2 6, (d), ,, ,, 7 7 7, x −1 y − 2 z − 3, x −1 y − 5, 14. If lines, =, =, and, =, −3, 2k, 2, 3k, 1, z−6, are mutually perpendicular, then k is, =, −5, equal to, (a), , (a) − 10, 7, , (b), , (b) −, , 7, 10, , (c) –10, , (d) –7, , 15. The equation of a plane with intercepts 2, 3, and 4 on the X, Y and Z-axes respectively is A., Here, A refers to, (a) 2x + 3y + 4z = 12, (b) 6x + 4y + 3z = 12, (c) 2x + 3y + 4z = 1, (d) 6x + 4y + 3z = 1, 16. What is the distance between the planes, 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0 ?, (a) 2/6 unit, (b) 3/2 units, (c) 1/6 unit, (d) 1/4 unit, 17. Th e e q ua t io n o f a line is gi ve n by, 4 − x, y + 3, z + 2, , the direction cosines, =, =, 2, 3, 6, of line parallel to the given line is, (a) −2 , −3 , −6, (b) 2 , −3 , −6, 7 7 7, 7 7 7, −2 3 6, 2, 3, 6, (c), (d), , ,, , ,, 7 7 7, 7 7 7, , 18. A line makes angles a, b and g with the, co-ordinate axes. If a + b = 90°, then the value, of angle g is, (a) 60°, (b) 90°, (c) 45°, (d) 30°, 19. Find the equation of a line passing, through (1, 2, –3) and parallel to the line, x − 2 y +1 z −1, =, =, ., 1, 3, 4, (a), , x − 2 y +1 z −1, x −1 y − 2 z + 3, =, =, =, =, (b), −1, 1, 1, 1, 3, 4, , x − 2 y +1 z −1, x +1 y − 2 z + 3, =, =, =, =, (d), −1, 2, 1, 1, 3, 4, 20. The vector equation of the plane passing, through a point having position vector, 2i + 3 j + 4k and perpendicular to the vector, , (c), , 2i + j − 2k is, ^, ^, ^, , (a) r . (2 i + j − 2 k ) = −1, , ^, ^, ^, , (b) r . (2 i + j − 2 k ) = 8, ^, ^, ^, , (c) r . (2 i + j − 2 k ) = 9, ^, ^, ^, , (d) r . (2 i + j − 2 k ) = 15, , 21. The cartesian equation of a line is, x +3 y−5 z+6, . The vector equation for, =, =, 2, 4, 2, the line is, ^, , ^, , ^, , ^, , ^, , ^, , (a), , 2 i + 3 j − 6 k + λ(2 i − 3 j + 2 k ), , (b), , −3 i + 5 j − 6 k + λ(2 i + 4 j + 2 k ), , (c), , −3 i − 5 j + 6 k + λ(2 i − 3 j − 2 k ), ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , ^, , (d) 3 i + 5 j + 6 k + λ( 2 i − 4 j − 2 k ), , Find the equation of plane passing through, point (1, 2, 3) and the direction cosines of, normal as l, m, n., lx + my + nz = l + 2m + 3n, lx + my + nz + (l + 2m + 3n) = 0, 1, (c) lx + my + nz = (l + 2m + 3n), 2, (d) None of these, 22., the, the, (a), (b), , x − 1 y + 1 z + 10, and, =, =, 2, 8, −3, x − 4 y + 3 z +1, are coplanar if k =, =, =, 1, k, 7, (a) 4, (b) –4, (c) 2, (d) –2, 23. The lines, , 24. Find the equation of the plane passing, through (2, 3, –1) and is perpendicular to the, vector 3i − 4 j + 7k ., (a) 3x – 4y + 7z + 13 = 0, (b) 3x + 4y – 7z – 13 = 0, (c) 3x + 4y + 7z – 13 = 0, (d) 3x – 4y – 7z + 13 = 0, 25. The equation of a line passing through the point, (– 3, 2, –4) and equally inclined to the axes are, (a) x – 3 = y + 2 = z – 4, (b) x + 3 = y – 2 = z + 4, x +3 y−2 z+4, (c), =, =, 1, 2, 3, (d) None of these, 26. Find the direction cosines of the line, x − 2 2y − 5 z + 1, =, =, ., 2, −3, 0, (a) –2, –5, 1, 3, (c) 2, − , 0, 2, , (b) 2, –3, 0, 4 3, , − ,0, (d), 5 5

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83, , Three Dimensional Geometry, , 27. The distance of the plane, 3, 6 ,  2, r ⋅  i + j − k  = 1 from the origin is, 7, 7, 7 , (a) 1 unit, (b) 7 units, 1, (c), unit, (d) 2 units, 7, , , (a) r, , (b) r, , (c) r, , (d) r, , 28. An equation of the plane passing through, the points (3, 2, –1), (3, 4, 2) and (7, 0, 6) is, 5x + 3y – 2z = l, where l is, (a) 23, , (b) 21, , (c) 19, , (d) 27, , 29. The distance of the plane 2x – 3y + 4z – 6 = 0, from the origin is A. Here, A refers to, (a) 6, , (b) –6, , 6, , (c) −, , 29, , (d), , 6, 29, , 30. What is the distance (in units) between the, two planes 3x + 5y + 7z = 3 and 9x + 15y + 21z = 9?, (a) 0, , (b) 3, , (c), , 6, 83, , (d) 6, , 31. Distance of the point (a, b, g) from y-axis is, (a) b, , (b) |b|, , (c) |b| + |g|, , (d), , α2 + γ 2, , 32. The reflection of the point (a, b, g) in the xyplane is, (a) (a, b, 0), , (b) (0, 0, g), , (c) (–a, –b, g), , (d) (a, b, –g), , 33. P is a point on the line segment joining the, points (3, 2, –1) and (6, 2, –2). If x co-ordinate of, P is 5, then its y co-ordinate is, (a) 2, , (b) 1, , (c) –1, , (d) –2, , 34. The equation of the line joining the points, (–3, 4, 11) and (1, –2, 7) is, x + 3 y − 4 z − 11, (a), =, =, 2, 3, 4, x + 3 y − 4 z − 11, (b), =, =, −2, 3, 2, x + 3 y + 4 z + 11, (c), =, =, −2, 3, 4, x + 3 y + 4 z + 11, =, =, (d), −3, 2, 2, 35. The vector equation of the line through the, points A(3, 4, –7) and B(1, –1, 6) is, , = (3i − 4 j − 7k ) + λ(i − j + 6k ), = (i − j + 6k ) + λ(3i − 4 j − 7k ), = (3i + 4 j − 7k ) + λ( −2i − 5 j + 13k ), = (i − j + 6k ) + λ(4i + 3 j − k ), , 36. If the line joining (2, 3, –1) and (3, 5, –3), is perpendicular to the line joining (1, 2, 3) and, (3, 5, l), then l =, (a) –3, , (b) 2, , (c) 5, , (d) 7, , 1 − x 7 y − 14, =, 3, 2p, z −3, 7 − 7x, y−5 6−z, =, and, intersect at, =, =, 2, 3p, 1, 5, right angle, is, 37. The value of p, so that the lines, , (a), , 10, , 11, , (b), , 70, 11, , 10, 70, , (d), 7, 9, , ,  ,  , 38. Two lines r = a1 + λb1 and r = a2 + µb2 are, said to be coplanar, if,  , (a) ( a − a ) ⋅ (b × b ) = 0, 2, 1, 1, 2, (c), , x1 y1, (b) a1 b1, a2 b2, , z1, c1 = 0, where (x 1, y 1 , z 1 ) are the, c2, , coordinates of a point on any of the line, and, a1, b1, c1 and a2, b2, c2 are the direction ratios, , , of b1 and b2, (c) both (a) and (b), (d) none of these, x + 3 y −1 z − 5, =, =, and, −3, 1, 5, x +1 y − 2 z − 5, are, =, =, −1, 2, 5, (a) coplanar, (b) non-coplanar, 39. The lines, , (c) perpendicular, , (d) none of these, , x −2, y − 9 z − 13, and, =, =, 1, 2, 3, x −a, y − 1 z +2, are coplanar, then a =, =, =, 1, −2, 3, (a) 2 , (b) –2, 40. If the lines, , (c) 3 , , (d) –3

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CBSE Board Term-II Mathematics Class-12, , 84, , Case Based MCQs, Case-I : Read the following passage and answer, the questions from 41 to 45., In a diamond exhibition, a diamond is covered, in cubical glass box having coordinates, O(0, 0, 0), A(1, 0, 0), B(1, 2, 0), C(0, 2, 0), O′(0, 0, 3),, A′(1, 0, 3), B′(1, 2, 3) and C′(0, 2, 3)., , A and B are respectively sitting on the plane, , represented by the equation r ⋅ (i + j + 2k ) = 5, , and r ⋅ (2i − j + k ) = 6, to cheer up the team of, their respective schools., , O′(0, 0, 3), , A′(1, 0, 3), , O(0, 0, 0), , B′(1, , ), , 2, 3, , A(1, 0, 0), , C′(0, 2, 3), , C (0, 2, 0), B(1, 2, 0), , 41. Direction ratios of OA are, (a) < 0, 1, 0 >, (b) < 1, 0, 0 >, (c) < 0, 0, 1 >, (d) none of these, 42. Equation of diagonal OB′ is, x, y z, x, y z, (a), (b), = =, = =, 1 2 3, 0 1 2, x, y z, (c), (d) none of these, = =, 1 0 2, 43. Equation of plane OABC is, (a) x = 0, (b) y = 0, (c) z = 0, (d) none of these, 44. Equation of plane O′A′B′C′ is, (a) x = 3, (b) y = 3 (c) z = 3, , (d) z = 2, , 45. Equation of plane ABB′A′ is, (a) x = 1, (b) y = 1 (c) z = 2, , (d) x = 3, , Case-II : Read the following passage and answer, the questions from 46 to 50., A football match is organised between students, of class XII of two schools, say school A and, school B. For which a team from each school is, chosen. Remaining students of class XII of school, , 46. The cartesian equation of the plane on which, students of school A are seated is, (a) 2x – y + z = 8, (b) 2x + y + z = 8, (c) x + y + 2z = 5, (d) x + y + z = 5, 47. The magnitude of the normal to the plane, on which students of school B are seated, is, (a) 5, (b) 6, (c), (d) 2, 3, 48. The intercept form of the equation of the, plane on which students of school B are seated, is, x y z, x, y, z, + + =1, (a), (b), +, + =1, 6 6 6, 3 ( −6) 6, x y z, x y z, (c), (d), + + =1, + + =1, 3 6 3, 3 6 6, 49. Which of the following is a student of school B?, (a) Mohit sitting at (1, 2, 1), (b) Ravi sitting at (0, 1, 2), (c) Khushi sitting at (3, 1, 1), (d) Shewta sitting at (2, –1, 2), , 50. The distance of the plane, on which students, of school B are seated, from the origin is, 1, (a) 6 units, (b), units, 6, 5, units, (c), (d) 6 units, 6, , Assertion & Reasoning Based MCQs, Directions (Q.-51 to 60) : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose, the correct answer out of the following choices :, (a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion., (b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion., (c) Assertion is correct statement but Reason is wrong statement., (d) Assertion is wrong statement but Reason is correct statement.

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85, , Three Dimensional Geometry, , 51. Assertion : The points (1, 2, 3), (–2, 3, 4) and, (7, 0, 1) are collinear., π 3π, Reason : If a line makes angles ,, and, 2, 4, π, with X, Y, and Z-axes respectively, then its, 4, −1, 1, direction cosines are 0,, and, ., 2, 2, 52. Assertion : If the cartesian equation of a, x −5 y+4 z −6, line is, , then its vector form is, =, =, 3, 7, 2, , r = 5i − 4 j + 6k + λ(3i + 7 j + 2k ) ., Reason : The cartesian equation of the line, which passes through the point (–2, 4, –5) and, x +3 y−4 z+8, parallel to the line given by, =, =, 3, 5, 6, x +3 y−4 z+8, is, ., =, =, −2, 4, −5, 53. Assertion : The three lines with direction, 12 −3 −4 4 12 3 3 −4 12, ,, ,, ; , , ; ,, ,, cosines, are, 13 13 13 13 13 13 13 13 13, mutually perpendicular., Reason : The line through the points (1, –1, 2) and, (3, 4, –2) is perpendicular to the line through the, points (0, 3, 2) and (3, 5, 6)., 54. Assertion : The pair of lines given by, , , r = i − j + λ(2i + k ) and r = 2i − k + µ(i + j − k ), intersect., Reason : Two lines intersect each other, if they, are not parallel and shortest distance = 0., 55. Assertion : There exists only one plane that, is perpendicular to the given vector., Reason : Through a given point perpendicular to, the given vector only one plane exists., 56. Assertion : If a variable line in two adjacent, positions has direction cosines l, m, n and l + dl,, m + dm, n + dn, then the small angle dq between, the two positions is given by dq2 = dl2 + dm2 + dn2., , Reason : If O is the origin and A is (a, b, c), then, the equation of plane through A at right angle to, OA is given by ax + by + cz = a2 + b2 + c2., x +1 y + 2 z +1, =, =, ,, 57. Consider the lines L1 :, 3, 1, 2, x −2 y+2 z −3, L2 :, =, =, ., 1, 2, 3, Assertion : The distance of point (1, 1, 1) from, the plane passing through the point (–1, –2, –1), and whose normal is perpendicular to both the, 13, lines L1 and L2 is, ., 5 3, Reason : The unit vector perpendicular to both, −i − 7 j + 5k, ., the lines L1 and L2 is, 5 3, 58. Assertion : The equation of a plane which, passes through (2, –3, 1) and normal to the line, joining the points (3, 4, –1) and (2, –1, 5) is given, by x + 5y – 6z + 19 = 0., Reason : The length of perpendicular from the, point (7, 14, 5) to the plane 2x + 4y – z = 2 is 2 21., 59. Assertion : Two systems of rectangular axis, have the same origin. If a plane cuts them at, distances a, b, c and a′, b′, c′ respectively from, 1, 1, 1, 1, 1, 1, the origin, then, ., +, +, =, +, +, 2, 2, 2, 2, 2, a, b, c, a′, b′, c ′2, Reason : The points (i − j + 3k ) and 3(i + j + k ) are, , equidistant from the plane r ⋅ (5i + 2 j − 7k ) + 9 = 0., 60. Assertion : The straight line, , x −3 y−4, =, −4, −7, , z+3, lies in the plane 5x – y + z = 8., 13, x − x1 y − y1 z − z1, =, =, Reason : The straight line, l, m, c, lies in the plane ax + by + cz + d = 0 iff normal, to the plane is perpendicular to the line & every, point of the line satisfies the equation of the, plane., =, , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), Find the direction cosines of the line, 4− x y 1− z, ., = =, 2, 6, 3, 2. Find the vector equation of a line which, passes through the points (3, 4, –7) and (1, –1, 6)., 1., , 3. The equation of a line are 5x – 3 = 15y + 7 =, 3 –10z. Write the direction cosines of the line., 4. Find the length of the intercept, cut off by, the plane 2x + y – z = 5 on the x-axis.

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CBSE Board Term-II Mathematics Class-12, , 86, 5. Write the vector equation of a line passing, through the point (1, –1, 2) and parallel to the, x − 3 y −1 z +1, line whose equation is, ., =, =, 1, 2, −2, 6. Find the vector equation of a plane which is, at a distance of 5 units from the origin and its, normal vector is 2i − 3j + 6k., 7. Write the equation of a plane which is at a, distance of 5 3 units from origin and the normal, , to which is equally inclined to coordinate axes., 8. Find the distance between the planes, 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20., 9. Find the distance of the plane 3x – 4y + 12z = 3, from the origin., 10. Write the vector equation of the line passing, through (1, 2, 3) and perpendicular to the plane, , r . (i + 2j − 5k ) + 9 = 0., , Short Answer Type Questions (SA-I), 11. The x-coordinate of a point on the line joining, the points P(2, 2, 1) and Q(5, 1, –2) is 4. Find its, z-coordinate., 12. Find the value of k so that the lines x = –  y = kz, and x – 2 = 2y + 1 = –z + 1 are perpendicular to, each other., 13. Find the vector equation of the line passing, through the point A(1, 2, –1) and parallel to the, line 5x – 25 = 14 – 7y = 35z., 14. Find the coordinates of the point where the, line through (–1, 1, –8) and (5, –2, 10) crosses, the ZX-plane., 15. Show that the lines, , x +1 y + 3 z + 5, and, =, =, 3, 5, 7, , x −2 y−4 z−6, intersect. Also find their, =, =, 1, 3, 5, point of intersection., , 16. Find the perpendicular distance of the point, x − 1 y + 1 z + 10, (1, 0, 0) from the line, =, =, . Also, 2, −3, 8, , find the coordinates of the foot of the perpendicular, and the equation of the perpendicular., 17. Find the value of l , so that the lines, 1 − x 7 y − 14 z − 3, 7 − 7x y − 5 6 − z, and, =, =, are, =, =, 3, λ, 2, 3λ, 1, 5, at right angles. Also, find whether the lines are, intersecting or not., 18. Find the equation of the line passing through, the point (–1, 3, –2) and perpendicular to the, x y z, x + 2 y −1 z +1, lines = = and, ., =, =, 1 2 3, −3, 2, 5, 19. Find the distance between the point, (–1, –5, –10) and the point of intersection of the, x − 2 y +1 z − 2, line, and the plane x – y + z = 5., =, =, 3, 4, 12, 20. A plane makes intercepts –6, 3, 4 respectively, on the coordinate axes. Find the length of the, perpendicular from the origin on it., , Short Answer Type Questions (SA-II), 21. Prove that the line through A(0, –1, –1) and, B(4, 5,1) intersects the line through C(3, 9, 4), and D(–4, 4, 4)., 22. Show that the lines, , r = 3i + 2j − 4k + λ(i + 2j + 2k ) ;, , r = 5 i − 2j + µ(3 i + 2j + 6k ) are intersecting., Hence find their point of intersection., 23. Using vectors, show that the points A(–2, 3, 5),, B(7, 0, –1), C(–3, –2, –5) and D(3, 4, 7) are such, that AB and CD intersect at the point P(1, 2, 3)., 24. Find the vector and cartesian equations, of the line through the point (1, 2, –4) and, perpendicular to the two lines, , , r = (8i − 19 j + 10k ) + λ(3i − 16 j + 7k ) and, , r = (15i + 29 j + 5k ) + µ(3i + 8 j − 5k )., 25. Find the shortest distance between the two, lines whose vector equations are, , r = ( i + 2j + 3k ) + λ (i − 3j + 2k )and, , r = (4 i + 5j + 6k ) + µ (2i + 3j + k )., 26. Find the direction cosines of the line, x + 2 2y − 7 5 − z, =, =, . Also, find the vector, 2, 6, 6, equation of the line through the point A(–1, 2, 3), and parallel to the given line.

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87, , Three Dimensional Geometry, , 27. Find the shortest distance between the, following lines :, , r = 2i − 5j + k + λ(3 i + 2j + 6k ) and, , r = 7i − 6k + µ( i + 2j + 2k ), 28. Find the shortest distance between the, following lines:, x +1 y +1 z +1 x − 3 y − 5 z − 7, =, =, ,, =, =, 7, −6, 1, 1, −2, 1, 29. Find the unit vector perpendicular to the, plane ABC where the position vectors of A, B and, C are 2 i − j + k , i + j + 2 k and 2 i + 3 k respectively., 30. Find the vector equation of the plane, through the points (2, 1, –1) and (–1, 3, 4) and, perpendicular to the plane x – 2y + 4z = 10., 31. Show that the lines, , 5− x y −7 z +3, =, =, and, −4, −5, 4, , x − 8 2y − 8 z − 5, are coplanar., =, =, 7, 2, 3, 32. Find the image of the point having position, vector i + 3j + 4k in the plane, , r . (2 i − j + k ) + 3 = 0 ., , 33. Find the equation of the plane passing, through the points (–1, 2, 0), (2, 2, –1) and, x − 1 2y + 1 z + 1, parallel to the line, ., =, =, 1, 2, −1, 34. Find the equation of a plane which passes, through the point (3, 2, 0) and contains the line, x −3 y−6 z −4, ., =, =, 1, 5, 4, 35. Find the coordinates of the point where the, line through the points A(3, 4, 1) and B(5, 1, 6), crosses the XY-plane., , Long Answer Type Questions (LA), 36. F i n d t h e c o o r d i n a t e s o f t h e f o o t o f, perpendicular and the length of the perpendicular, drawn from the point P(5, 4, 2) to the line,, , r = − i + 3j + k + λ(2 i + 3j − k )., Also find the image of P in this line., 37. Find the distance of the point (2, 12, 5), from the point of intersection of the line, , r = 2 i − 4 j + 2k + λ(3 i + 4 j + 2k ) and the plane, , r . ( i − 2j + k ) = 0., 38. Find the vector equation of a line passing, through the point (2, 3, 2) and parallel to the, , line r = ( − 2i + 3j) + λ(2i − 3j + 6k ). Also, find the, distance between these two lines., , OBJECTIVE TYPE QUESTIONS, 1. (b) : Consider, cos 2q1 + cos 2q2 + cos 2q3, = 2(cos2q1+ cos2q2 + cos2q3) – 3 (Q cos2x = 2cos2 x – 1), = 2(1) – 3 = –1, 2. (d) : The direction ratios of line parallel to AB is, 1, –2 and 4., 3. (c) : a, b and γ are the angles made by line with, the co-ordinate axes., \ cos2 a + cos2 b + cos2 g = 1, ⇒ 1– sin2 a + 1 – sin2 b + 1 – sin2 g = 1, ⇒ sin2 a + sin2 b + sin2 g = 2, 4. (a) : cos2 a + cos2 b + cos2 g = 1, ( a, b, γ are direction angles), , 39. Show that the lines, , x −2 y−2 z −3, =, =, and, 1, 3, 1, , x −2 y−3 z −4, =, =, intersect., 1, 4, 2, , Also, find the coordinates of the point of intersection., Find the equation of the plane containing the two, lines., , 40. Find the vector and cartesian equations of, a plane containing the two lines, , r = 2i + j − 3k + λ(i + 2j + 5k ) and, , r = 3 i + 3j + 2k + µ(3i − 2j + 5k ), Also show that the line, , , r = (2 i + 5j + 2k ) + p (3i − 2j + 5k ) lies in the plane., , 196 1, + + cos 2 γ = 1, 225 9, 221, 4, 2, ⇒ cos 2 γ = 1 −, =, ⇒ cos γ = ±, 225 225, 15, 5. (d) : Let the direction cosines of the line be l, m, n., 1, 3, Then, l = cos 90° = 0, m = cos60° = and n = cos 30° =, ., 2, 2, ⇒, , So, direction cosines are 0 ,, 6., , 1, 3, ,, ., 2 2, , (c) : Here, P(2, 1, 0) and Q(1, –2, 3), , So, PQ = (1 − 2 )2 + ( −2 − 1)2 + ( 3 − 0 )2, = 1 + 9 + 9 = 19

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CBSE Board Term-II Mathematics Class-12, , 88, Thus, the direction cosines of the line joining two points, 1 − 2 −2 − 1 3 − 0, ,, ,, =, 19, 19, 19, , are, , −1 − 3, 3, ,, ,, 19, 19, 19, , (c) : Since l = m = n and l2 + m2 + n2 = 1, 1, ⇒ l=m=n=±, ., 3, , 7., , 16. (c) : Given planes are 2x + 2y – z + 2 = 0 and, 4x + 4y – 2z + 5 = 0., Let point P(x1, y1, z1) lie on plane 2x + 2y – z + 2 = 0., ⇒ 2x1 + 2y1 – z1 = –2, ∴ d=, , 8. (a) : Direction ratios of AB are, (–1 – 0, 6 – 7, 6 – 10) or (–1, –1, – 4), Also,, \, or, , =, , ( −1)2 + ( −1)2 + ( −4 )2 = 3 2, , ^, , ^, , ^, , 2 i + j − 2 k and the required line is parallel to the given, ^, , ^, , ^, , line. So, required line is parallel to the vector 2 i + j − 2 k ., Thus, the equation of the required line passing through, (2, –1, 3) is, , ^ ^, ^, ^ ^, ^, r = ( 2 i − j + 3 k ) + µ( 2 i + j − 2 k ), 1 1 , 10. (a) : Q  , , n are the direction cosines of a line, 2 3 , 2, , 2, , 23, ± 23,  1,  1, ∴   +   + n2 = 1 ⇒ n2 =, ⇒n=,  2,  3, 36, 6, 11. (a) : We have, equation of plane is 3x – 6y + 2z + 11 = 0., Its distance from origin (0, 0, 0) is, 3 × 0 − 6 × 0 + 2 × 0 + 11, 2, , 2, , 2, , =, , 11, 11, =, units., 9 + 36 + 4 7, , 3 + ( −6 ) + ( 2 ), , 12. (a) : Here, a = 2i + 3 j + 4 k, The distance of the point ( 2i + 3 j + 4 k ) is, ( 2i + 3 j + 4 k ) ⋅ ( 3i − 6 j + 2 k ) + 11 6 − 18 + 8 + 11, =, = 1 unit, 7, 9 + 36 + 4, x−3 y+2 z+2, 13. (b) : We have,, =, =, −2, −3, 6, ⇒ Direction ratios are – 3, – 2, 6., −3 −2 6, ,, , ., \ Direction cosines are, 7 7 7, These are direction cosines of a line parallel to given line., 14. (a) : Lines, , 2, , 2, , 2, , 4 + 4 + ( −2 ), , =, , 2( 2 x1 + 2 y1 − z1 ) + 5, 16 + 16 + 4, , 2( −2 ) + 5 1, = unit, 6, 36, , 4−x y+3 z+2, =, =, ., 2, 3, 6, The direction ratios of the given line are –2, 3, 6., \ The direction cosines of the given line are, 3, 6, −2, , , ,, ,,  4 + 9 + 36, 4 + 9 + 36, 4 + 9 + 36 , 17. (d) : Equation of given line is, , −4 , 1, 1, , ,−, ,, Direction cosines are  −, ,  3 2, 3 2 3 2, 1, 1, 4, , , ,, ,, , , 3 2 3 2 3 2, (b) : The given line is parallel to the vector, , 9., , 4 x1 + 4 y1 − 2 z1 + 5, , x−1 y−2 z−3, x−1 y−5 z−6, and, =, =, =, =, −3, 2k, 2, 3k, 1, −5, , are perpendicular if a1a2 + b1b2 + c1c2 = 0., 10, ⇒ –3(3k) + 2k + 2(–5) = 0 ⇒ k = −, 7, 15. (b) : As the plane has intercepts 2, 3 and 4 on X, Y, and Z axes respectively., \ The required equation of the plane is, x y z, + + = 1 ⇒ 6x + 4y + 3z = 12, 2 3 4, ,  −2, =, ,,  49, , 3, ,, 49, , 6   −2 3 6 , , , ,  =, 49   7 7 7 , , 18., ⇒, ⇒, ⇒, , (b) : We know that cos2a + cos2b + cos2g = 1, cos2a + cos2(90° – a) + cos2g = 1 [Q a + b = 90°], cos2a + sin2a + cos2g = 1 ⇒ 1 + cos2g = 1, cos2g = 0, , ⇒, , cosg = 0 ⇒ g =, , π, = 90°., 2, , 19. (b) : Since, the line is parallel to the line, x−2 y+1 z−1, =, =, ., 1, 3, 4, \ D.r.’s of the required line are < 1, 3, 4 >, Hence, equation of the line passing through (1, 2, –3), with d.r.’s <1, 3, 4> is, , x−1 y−2 z+3, =, =, 1, 3, 4, , 20. (a) : Vector equation of plane passing through a, , point having position vector a and perpendicular to, ,    , n is given by r . n = a . n, , , Here a = 2 i + 3 j + 4 k , n = 2 i + j − 2 k, ^,  ^ ^, \ Required equation = r .( 2 i + j − 2 k ) = 4 + 3 − 8 = −1, , 21. (b) : The given cartesian equation is, x+3 y−5 z+6, =, =, ., 2, 4, 2, The line passes through the point (–3, 5, –6) and is parallel, to vector 2i + 4 j + 2 k ., Hence, the vector equation of the line is, , r = − 3i + 5 j − 6 k + λ ( 2i + 4 j + 2 k )., 22. (a) : Equation of plane passing through (1, 2, 3), having direction cosines of normal as l, m, n is, l(x – 1) + m(y – 2) + n(z – 3) = 0, ⇒ lx + my + nz = l + 2m + 3n

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89, , Three Dimensional Geometry, 23. (b) : Given lines are, , x − 1 y − ( −1) z − ( −10 ), =, =, −3, 2, 8, , x − 4 y − ( −3) z − ( −1), and, =, =, k, 1, 7, Since the lines are coplanar., \, , 4 − 1 − 3 − ( − 1) − 1 − ( − 10 ), −3, =0, 2, 8, k, 1, 7, , ⇒ 3(– 21 – 8k) + 2(14 – 8) + 9(2k + 3) = 0, ⇒ 6k = –24 ⇒ k = – 4, 24. (a) : The equation of the plane passing through, is, (2, 3, – 1) and perpendicular to the vector 3i − 4 j + 7k, 3(x – 2) + (– 4) (y – 3) + 7(z – (– 1)) = 0, ⇒ 3x – 4y + 7z + 13 = 0, 25. (b) : Since, line equally inclined to the axes., \ l = m = n, …(i), The required equation of line is, x+3 y−2 z+4, , [using (i)], =, =, l, l, l, x+3 y−2 z+4, ⇒ x+3=y–2=z+4, ⇒, =, =, 1, 1, 1, 26. (d) : The given line is, x − 2 2y − 5 z + 1, =, =, −3, 2, 0, x − 2 y − 5 /2 z + 1, ⇒, =, =, 2, −3 / 2, 0, Direction cosines are, 2, −3 / 2, 0, ,, ,, 2, 2, 2,  −3 ,  −3 ,  −3 , 2 2 +   + 02 2 2 +   + 02 2 2 +   + 02,  2,  2,  2, 4 −3, 2 −3 / 2, ,, ,0, , 0 i.e., ,, 5 /2 5 /2, 5 5, 3, 6,  2, 27. (a) : We have, r ⋅  i + j − k  = 1, 7, 7, 7 , 3, 6,  2, Here n = i + j − k, 7, 7, 7, i.e.,, , 2, , 2, , \, , 2, , 49, =1, 49, |1| 1, Required distance =  = = 1 unit, |n| 1, , ,  2,  3,  −6 , n =   +  + ,  7,  7,  7, , =, , 28. (a) : Equation of the plane through (3, 2, –1), (3, 4, 2), and (7, 0, 6) is, x−3 y−2 z+1, 0, 2, 3 =0, −2, 4, 7, i.e., 5x + 3y – 2z = 23, \ l = 23, 29. (d) : Distance of plane 2x – 3y + 4z – 6 = 0 from the, origin is given by,, , −6, 2, , 2, , 2, , ( 2 ) + ( −3) + ( 4 ), , =, , 6, 29, , 30. (a) : Given planes are 3x + 5y + 7z = 3 and 9x + 15y, + 21z = 9, ⇒ 3x + 5y + 7z = 3, Both the planes are coincident planes, so distance, between them is zero., 31. (d) : Foot of perpendicular from (a, b, g) on the, y-axis is (0, b, 0), \ Distance of (a, b, g) from y-axis = distance of, (a, b, g) from (0, b, 0), = (0 − α )2 + (β − β )2 + (0 − γ )2 = α 2 + γ 2, 32. (d) : Projection of P(a, b, g) on xy-plane is Q(a, b, 0)., If P′ (a′, b′, g′) is the reflection of P in xy-plane, then, Q is the mid-point of PP′.,  α + α ′ β + β′ γ + γ ′ , ⇒ (α , β , 0) = , ,, ,, ,  2, 2, 2 , ⇒, ⇒, \, , α + α′, β + β′, γ + γ′, = α,, = β,, =0, 2, 2, 2, a′ = a, b′ = b, g′ = –g, Required reflection is (a, b, –g)., , 33. (a) : Equation of line joining the points (3, 2, –1) and, (6, 2, –2) is,, x−3 y−2, z+1, x−3 y−2 z+1, =, =, =, i.e.,, =, = λ (say), −1, 6 − 3 2 − 2 −2 + 1, 3, 0, ⇒ x = 3l + 3, y = 2, z = –l – 1, So, y-coordinate of P is 2., 34. (b) : DR’s of the line joining the given points are, {1 – (– 3), – 2 – 4, 7 – 11}, i.e., (4, – 6, – 4) or (– 2, 3, 2), Now, Equation of line passing through (–3, 4, 11) and, x + 3 y − 4 z − 11, having direction ratios –2, 3, 2 is, =, =, −2, 3, 2, , 35. (c) : If a = P.V. of A = 3i + 4 j − 7 k, , and b = P.V. of B = i − j + 6 k , then the equation of line,  ,  , AB is r = a + λ(b − a ), , \ r = ( 3i + 4 j − 7 k ) + λ( −2i − 5 j + 13 k ), 36. (d) : DR’s of the given lines are 1, 2, –2 and 2, 3,, l – 3., Since, lines are perpendicular., \ a 1a 2 + b 1b 2 + c 1c 2 = 0, ⇒ 1 × 2 + 2 × 3 – 2 (l – 3) = 0 ⇒ l = 7, 37. (b) : Equation of the given lines can be written in, the standard form as, x−1 y−2 z−3, x−1 y−5 z−6, and, =, =, =, =, 2p, 3p, −3, 2, 1, −5, −, 7, 7

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CBSE Board Term-II Mathematics Class-12, , 90, \, , Lines are perpendicular to each other., a1a2 + b1b2 + c1c2 = 0., , 70,  −3 p   2 p , ⇒ ( −3) , +, (1) + ( 2 )( −5) = 0 ⇒ p =,  7   7 , 11, , ,  ,  , 38. (a) : Two lines r = a1 + λb1 and r = a2 + µb2,  ,  , are coplanar if ( a2 − a1 ) ⋅(b1 × b2 ) = 0., 39. (a) : Here, x1 = –3, y1 = 1, z1 = 5;, a1 = – 3, b1 = 1, c1 = 5;, x2 = –1, y2 = 2, z2 = 5 and a2 = – 1, b2 = 2, c2 = 5, x 2 − x1, a1, Now,, a2, , y 2 − y1, b1, b2, , z2 − z1, 2 1 0, c1 = −3 1 5, c2, −1 2 5, , = 2(5 – 10) – 1(–15 + 5) + 0 = –10 + 10 = 0, Therefore, lines are coplanar., 40. (d) :, , The lines, , x − 2 y − 9 z − 13, and, =, =, 1, 2, 3, , 49. (c) : Since, only the point (3, 1, 1) satisfy the equation, of plane representing seating position of students of, school B, therefore Khushi is the student of school B., 50. (d) : Equation of plane representing students of, school B is r ⋅ ( 2i − j + k ) = 6 , which is not in normal, , form, as | n | ≠ 1 ., On dividing both sides by, 6,   2  1  1 , r ⋅, i−, j+, k =, ,,  6, , 6, 6, 6, , which is of the form r ⋅ n = d, Thus, the required distance is, , Now,, , x2 − x1 y 2 − y1 z2 − z1, =, =, x3 − x2 y 3 − y 2 z3 − z2, , −2 − 1, 3−2 4−3, =, =, 7 − ( −2 ) 0 − 3 1 − 4, , ⇒, , 41. (b) : D.R.’s of OA are < 1 – 0, 0 – 0, 0 – 0 >,, i.e., < 1, 0, 0 >., , Here, l = cos, , 44. (c) : Plane O′A′B′C′ is parallel to xy-plane passing, through (0, 0, 3), therefore its equation is z = 3., 45. (a) : Plane ABB′A′ is parallel to yz-plane passing, through (1, 0, 0), therefore its equation is x = 1., 46. (c) : Clearly, the plane for students of school A is, , r ⋅ (i + j + 2 k ) = 5, which can be rewritten as, ( xi + y j + zk ) ⋅ (i + j + 2 k ) = 5, ⇒ x + y + 2z = 5, which is the required cartesian, equation., 47. (b) : Clearly, the equation of plane for students, , of school B is r ⋅ ( 2i − j + k ) = 6 , which is of the form,  , r ⋅n = d, , \ Normal vector to the plane is, n = 2i − j + k and its, , magnitude is | n | = 2 2 + ( −1)2 + 12 = 6, 48. (b) : The cartesian form is 2x – y + z = 6,, which can be rewritten as, y, x, z, 2x y z, + =1, +, − + =1 ⇒, 3 ( −6 ) 6, 6 6 6, , 6 units., , 51. (b) : We have, x1 = 1, y1 = 2, z1 = 3;, x2 = –2, y2 = 3, z2 = 4 and x3 = 7, y3 = 0, z3 = 1, , x − a y −1 z+ 2, are coplanar., =, =, 1, −2, 3, 2 − a 8 15, 1, 2 3 = 0⇒ a= −3, ∴, 1, −2 3, , 42. (a) : Equation of diagonal OB′ is, x−0 y−0 z−0, x y z, i.e., = =, =, =, 1, 2, 3, 1 2 3, 43. (c) : OABC is xy-plane, therefore its equation is z = 0., , 2 2 + ( −1)2 + 12 = 6 , we get, , ⇒, \, , 1, −3 1, −1 −1 −1, ⇒, =, =, =, =, 9 −3 −3, 3, 3, 3, Given points are collinear., , m = cos, , −1, 3π, π, π, , = cos  π −  = – cos, =, , 2, 4, 4, 4, , and n = cos, \, , π, =0, 2, , π, 1, =, 2, 4, , Direction cosines are 0,, , −1 1, ., ,, 2, 2, , Hence, both Assertion and Reason correct but Reason is, not the correct explanation of Assertion., 52. (c) : In assertion the given cartesian equation is, x−5 y+4 z−6, ., =, =, 3, 7, 2, , , ⇒ a = 5i − 4 j + 6 k and b = 3i + 7 j + 2 k ., ,  , The vector equation of the line is given by r = a + λb , λ ∈ R., ⇒ r = 5i − 4 j + 6 k + λ( 3i + 7 j + 2 k ), Thus Assertion is correct., In reason it is given that the line passes through the, point (–2, 4, –5) and is parallel to, x+3 y −4 z+8, =, =, ., 3, 5, 6, Clearly, the direction ratios of line are (3, 5, 6)., Now the equation of the line (in cartesian form) is, x − ( −2 ) y − 4 z − ( −5), x+2 y−4 z+5, =, =, =, =, ⇒, 3, 5, 6, 3, 5, 6, Hence, Reason is wrong.

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91, , Three Dimensional Geometry, 53. (b) : For first two lines, having direction cosines, 12 −3 −4, 4 12 3, ,, ,, ,, ,, and, , we obtain, 13 13 13, 13 13 13, l 1l 2 + m 1m 2 + n 1n 2 =, , 12 4  −3  12  −4  3, =0, ×, +  ×, +  ×, 13 13  13  13  13  13, , Therefore, the lines are perpendicular., For second and third lines having direction cosines, 4 12 3, 3 −4 12, ,, ,, and, , we obtain, ,, ,, 13 13 13, 13 13 13, 4, 3 12  −4  3 12, ×, +, ×  +, ×, l 1l 2 + m 1m 2 + n 1n 2 =, =0, 13 13 13  13  13 13, Therefore, the lines are perpendicular., For third and first lines, having directions cosines, 3 −4 12, 12 −3 −4, , ,, and, , we obtain, , ,, 13 13 13, 13 13 13, 3 12  −4   −3  12  −4 , l 1l 2 + m 1m 2 + n 1n 2 =, × +  ×  + ×  = 0, 13 13  13   13  13  13 , Therefore, the lines are perpendicular., Hence, all the lines are mutually perpendicular., Let the given points are A(1, –1, 2), B(3, 4, –2),, C(0, 3, 2) and D(3, 5, 6), Direction ratios of AB are (3 – 1, 4 – (–1), –2 – 2) = (2, 5, –4), and direction ratios of CD are (3 – 0, 5 – 3, 6 – 2) = (3, 2, 4), We know that, two lines AB and CD with direction, ratios, a1, b1, c1 and a2, b2, c2 are perpendicular, if, a 1a 2 + b 1b 2 + c 1c 2 = 0, Here, 2 × 3 + 5 × 2 + (–4) × 4 = 6 + 10 – 16 = 0, Therefore, the lines AB and CD are perpendicular., Hence, both Assertion and Reason are correct but, Reason is not the correct explanation of Assertion., , , 54. (a) : Here, a = i − j , b = 2i + k, 1, , 1, , , , a2 = 2i − k and b2 = i + j − k, , , b1 ≠ kb2 , for any scalar k, Given lines are not parallel.,  , Now, a2 − a1 = ( 2i − k ) − (i − j ) = i + j − k,  , and b1 × b2 = −i + 3 j + 2 k,  , ⇒ b1 × b2 = ( −1)2 + ( 3)2 + ( 2 )2 = 1 + 9 + 4 = 14,  ,  , (i + j − k ) ⋅ ( −i + 3 j + 2 k ), ( a2 − a1 ) ⋅ (b1 × b2 ), \ S.D. =, =, =0,  , 14, b1 × b2, \, , Hence, two lines intersect each other., Two lines intersect each other, if they are not parallel, and shortest distance = 0., 55. (d) : In the space, there, can be many planes that, are perpendicular to the, given vector, but if it also, passes through a given, O, point then only one such, plane exist., , 56. (b) : Given dq be the angle between the two adjacent, positions., \ cosdq = l(l + dl) + m(m + dm) + n(n + dn), ⇒ cosdq = (l2 + m2 + n2) + ldl + mdm + ndn, ⇒ cosdq = 1 + ldl + mdm + ndn, ...(i), Also, we have, (l + dl)2 + (m + dm)2 + (n + dn)2 = 1, ⇒ (l2 + m2 + n2) + ((dl)2 + (dm)2 + (dn)2,, , + 2 (l.dl + m.dm + n.dn) = 1, ⇒ 1 + ((dl)2 + (dm)2 + (dn)2) + 2 (cosdq – 1) = 1, , [using (i)], 2, 2, 2, ⇒ (dl) + (dm) + (dn) = 2 (1 – cosdq), , ,  δθ  ,  δθ  , = 2  2.sin 2    = 4  sin   ,  2 ,  2 , , , , 2, , 2, ,  δθ , = 4   [,  2, , For very small angle q, sin q = q], , = (dq)2, ⇒ dq2 = dl2 + dm2 + dn2,  , In reason, we have n = OA = ai + b j + ck, The required equation of plane is given by, , , [r − ( ai + b j + ck )] ⋅ n = 0,   , ⇒ r ⋅ ( ai + b j + ck ) − ( ai + b j + ck ) ⋅ ( ai + b j + ck ) = 0, ⇒, , ( xi + y j + zk ) ⋅ ( ai + b j + ck ) − ( a2 + b 2 + c 2 ) = 0, , ⇒ ax + by +cz = a2 + b2 + c2, 57. (a) : Lines L 1 and L 2 are parallel to the vectors, , , a = 3i + j + 2 k and b = i + 2 j + 3k respectively. The unit, vector perpendicular to both L1 and L2 is,  , −i − 7 j + 5k, −i − 7 j + 5 k, a×b, =,   =, |a × b |, 5 3, 1 + 49 + 25, Now, equation of plane through (–1, –2, –1) is, –(x + 1) – 7(y + 2) + 5(z + 1) = 0 whose distance from, 13, (1, 1, 1) is, ., 5 3, , , 58. (c) : Here, N = ( 2 − 3)i + ( −1 − 4 )j + ( 5 + 1)k, = −i − 5 j + 6 k, , Now, equation of the plane passing through (2, –3, 1), perpendicular to N is, –1(x – 2) – 5(y + 3) + 6(z – 1) = 0, ⇒ –x + 2 – 5y – 15 + 6z – 6 = 0 ⇒ x + 5y – 6z + 19 = 0, which is the required equation., 2(7 ) + 4(14 ) − ( 5) − 2, Length of perpendicular =, = 3 21, 2 2 + 4 2 + ( −1)2, 59. (b) : Equation of the plane which cuts the axes at, distances a, b, c is given by, x y z, + + =1, a b c, Since, the same plane passes through (a′, 0, 0), (0, b′, 0), and (0, 0, c′) in other system.

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CBSE Board Term-II Mathematics Class-12, , 92, Therefore,, , 1 1, c′, b′, 1 1, =, =1 ⇒, and, =1 ⇒, =, b, b b′, c, c c′, 1, 1, 1, 1, 1, 1, + 2+ 2 =, +, +, ⇒, 2, 2, 2, ( a′ ), (b′ ), a, b, c, (c ′ )2, Let d1 and d2 be the distances of the point (i − j + 3 k ) and, ( 3i + 3 j + 3k ) from the plane., Then, d1 =, d2 =, , (i − j + 3 k ) ⋅ ( 5i + 2 j − 7 k ) + 9, 52 + 2 2 + ( −7 )2, , ( 3i + 3 j + 3k ) ⋅ ( 5i + 2 j − 7 k ) + 9, 2, , 2, , 2, , 5 + 2 + ( −7 ), , =, , =, , and parallel to the line,, , 9, 78, , 6., , x−3 y−4 z+3, is, =, =, −7, −4, 13, (3, 4, –3), which satisfies the equation of the plane, 5x – y + z = 8., Also, direction cosines of the line are, −4, −7, 13, <, ,, ,, > and direction cosines of the, 234, 234, 234, normal to the plane 5x – y + z = 8 are < 5, –1, 1 >., −20, 7, 13, \ al + bm + cn =, +, +, =0., 234, 234, 234, 60. (a) : Any point on the line, , SUBJECTIVE TYPE QUESTIONS, 4 − x y 1− z, = =, The eq. of given line is, 2, 6, 3, , x−4 y z−1, ⇒, = =, −2, 6, −3, 2, , As, , 2, , , , ...(i), , 2, , 2 + (−6) + 3 = 7, , 2 −6 3, , ., \ D.c’s. of (i) are ,, 7 7 7, 2. , Vector equation of a line passes through the points, (3, 4, –7) and (1, –1, 6) is given by, , r = ( 3i + 4 j − 7 k ) + λ[(i − j + 6 k ) − ( 3i + 4 j − 7 k )], , \ r = 3i + 4 j − 7 k + λ( − 2i − 5 j + 13k ), 3. , The given line is 5x – 3 = 15y + 7 = 3 –10z, , x−, , i.e., Its direction ratios are proportional to 6, 2, –3., Now,, \, , 62 + 22 + (−3)2 = 7, , Its direction cosines are, , 6 2 3, , ,− ., 7 7 7, , ∴, , , Let n = 2i − 3j + 6k, , 2i − 3j + 6k 2i − 3j + 6k, n, n =  =, =, |n|, 7, 4 + 9 + 36, , So, the required equation of the plane is, ,  2, 3, 6, r ⋅  i − j + k  = 5 ⇒ r ⋅ (2i − 3j + 6k ) = 35, 7, 7, 7 , , , , 7. , Let a, b and g be the angles made by n with x, y, and z-axis, respectively., Given a = b = g ⇒ cos a = cos b = cos g, , ⇒ l = m = n, where l, m, n are direction cosines of n ., But l2 + m2 + n2 = 1 ⇒ l2 + l2 + l2 = 1 ⇒ l = ±, So, l = m = n = ±, , y, x−4, z−1, ⇒, =, =, −6, 2, 3, , 3, 7, 3, y+, z−, 5, 15, 10, ⇒, =, =, 1, 1, 1, −, 5, 15, 10, 1 1, 1, Its direction ratios are , , −, 5 15 10, , 5. , V ector eq. of the line passing through (1, –1, 2), , x − 3 y −1 z +1, is, =, =, 1, 2, −2, , r = (i − j + 2k ) + λ (i + 2 j − 2k ) ., , 9, 78, , Clearly, d1 = d2, , 1., , We have, 2x + y – z = 5, y z, x, ⇒, + +, =1, 5 / 2 5 −5, which is the equation of plane in intercept form., \ Intercepts on x, y and z-axis respectively are, 5, , 5, − 5., 2, 5, \ Required length of intercept =, 2, 4., , a′, 1 1, =1 ⇒, =, a, a a′, , \, , 3, , 3, , The normal form of the plane is lx + my + nz = d, , ⇒ ±, ⇒, , 1, , 1, , 1, 3, , x±, , 1, 3, , y±, , 1, 3, , z =5 3, , ± x ± y ± z = 15, , 8. , E quation of both the planes can be written as, 2x – y + 2z = 5 and 2x – y + 2z = 8., Distance between both the planes, , =, , 8−5, 4 +1+ 4, , =, , 3, 9, , = 1 unit, , 9. , Perpendicular distance from the origin (0, 0, 0) to, the plane 3x – 4y + 12z – 3 = 0 is, , 3 × 0 − 4 × 0 + 12 × 0 − 3, 32 + (−4)2 + 122, , =, , 3, unit ., 13, , , , 10. , The given plane is r ⋅ (i + 2 j − 5k ) + 9 = 0, , , ⇒ n = i + 2 j − 5k, , Q D.r’s of ^ to this plane are 1, 2, –5., So, the line has direction ratios proportional to, 1, 2, –5., \ Eq. of line through (1, 2, 3) and ^ to the plane is, , , r = ( i + 2 j + 3k ) + λ ( i + 2 j − 5k ).

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93, , Three Dimensional Geometry, 11. , Given that P(2, 2, 1) and Q(5, 1, –2), k, , Let the point R on the line PQ, divides the line in the, ratio k : 1 and x-coordinate of point R on the line is 4., So, by section formula, , 4=, , 5k + 2, ⇒ k=2, k +1, , Now, z-coordinate of point R,, , z=, , −2k + 1 −2 × 2 + 1, =, = −1, k +1, 2 +1, , ⇒ z-coordinate of point R = –1, 12. The given lines are, , x −0 y −0 z −0, =, =, 1, 1, −1, k, 1, y, +, x −2, 2 = z −1, =, l2 :, 1, 1, −1, 2, l1 :, , Q, \, , l1 is perpendicular to l2., , 1 1, 1(1) + (−1)   +   (−1) = 0, 2 k, , 1 1, 1 1, − =0 ⇒, =, ⇒ k=2, 2 k, 2 k, 13. , V ector equation of the line passing through, (1, 2, –1) and parallel to the line, 5x – 25 = 14 – 7y = 35z, y−2, x−5, z, x−5 y−2 z, i.e.,, =, =, =, or, =, 1, 1 / 5 −1 / 7 1 / 35, 7, −5, , is r = (i + 2 j − k ) + λ(7i − 5 j + k ), 14. , T he equation of line through A(–1, 1, –8) and, y−1, x+1, z+8, B(5, –2, 10) is, =, =, 5 + 1 −2 − 1 10 + 8, x+1 y−1 z+8, i.e.,, =, =, = k(say) ...(i), −3, 6, 18, Any point on (i) is given by (6k – 1, –3k + 1, 18k – 8)., We know that the coordinates of any point on the, ZX-plane is given by (x, 0, z)., 1, \ –3k + 1 = 0 ⇒ k =, 3, Thus the coordinates of the point where the line joining, A and B crosses the ZX-plane are, 1, 1, , ,  6 × − 1, 0 , 18 × − 8 = (1, 0 , − 2 ), 3, 3, 15. , Any point on the line, x+1 y+3 z+5, =, =, = r (say) ...(i), 3, 5, 7, is (3r – 1, 5r – 3, 7r – 5)., Any point on the line, x−2 y−4 z−6, =, =, = k (say) ...(ii), 1, 3, 5, ⇒, , 1−, , is (k + 2, 3k + 4, 5k + 6), For lines (i) and (ii) to intersect, we must have, 3r – 1 = k + 2, 5r – 3 = 3k + 4, 7r – 5 = 5k + 6, 1, 3, On solving these, we get r = , k = −, 2, 2, \ Lines (i) and (ii) intersect and their point of, 1, 1, 3, intersection is  , − , − , 2, 2, 2, 16. , Any point on the given line,, x − 1 y + 1 z + 10, =, =, = k (say) ...(i), −3, 2, 8, is R(2k + 1, –3k – 1, 8k –10), If this is the foot of the ^ from P(1, 0, 0) on (i), then, (2k + 1 –1) · 2 + (–3k – 1 – 0) · (–3) + (8k –10 – 0) · 8 = 0, ⇒ 4k + 9k + 3 + 64k – 80 = 0, ⇒ 77k = 77 ⇒ k = 1, \ R is (3, –4, –2)., This is the required foot of perpendicular., Also, perpendicular distance = PR, = ( 3 − 1)2 + ( − 4 − 0 )2 + ( − 2 − 0 )2 = 24 = 2 6 units., , Also eq. of PR is, , y, x−1, z, =, =, 2, −4 −2, , 17. , The given lines are, x−1 y−2 z−3, l1 :, =, =, −3, λ /7, 2, y−5 z−6, x−1, =, =, and l2 :, −3λ / 7, 1, −5, Now, l1 ^ l2 , , [Given], ,  3λ  λ, ∴ ( −3)  −  + − 10 = 0,  7  7, 10 λ, 9λ λ, ⇒, + − 10 = 0 ⇒, = 10 ⇒ λ = 7, 7, 7, 7, Since for l = 7, given lines are at right angles., \ Lines are intersecting., 18. , Let l, m, n be the direction ratios of the line which, is perpendicular to the lines, x y z, x+2 y−1 z+1, = = and, =, =, ., 1 2 3, −3, 2, 5, Then, l + 2m + 3n = 0 and –3l + 2m + 5n = 0, l, m, n, l, m n, ⇒, =, =, ⇒, =, =, 10 − 6 −9 − 5 2 + 6, 2 −7 4, \ Eq. of the required line through (–1, 3, –2) having, d.r’s proportional to 2, –7, 4 is, x+1 y−3 z+2, =, =, ., −7, 2, 4, 19. Equation of given line is, x−2 y+1 z−2, =, =, = k (say), 3, 4, 12, ⇒ x = 3k + 2, y = 4k –1, z = 12k + 2, Since point (3k + 2, 4k –1, 12k + 2) lie on plane, x–y+z=5

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CBSE Board Term-II Mathematics Class-12, , 94, \, ⇒, \, , 3k + 2 – 4k + 1 + 12 k + 2 = 5, 11k = 0 ⇒ k = 0, Point is (2, –1, 2), , Required distance = ( 2 + 1)2 + ( −1 + 5)2 + ( 2 + 10 )2, , , = 9 + 16 + 144 = 169 = 13 units, 20. , G iven that –6, 3, 4 are intercepts on x, y and, z-axes respectively., y z, x, \ Eq. of the plane is, + + =1, −6 3 4, ⇒ –2x + 4y + 3z –12 = 0, \ Length of the perpendicular from origin to the, plane – 2x + 4y + 3z – 12 = 0 is, −2 × 0 + 4 × 0 + 3 × 0 − 12, 2, , 2, , ( −2 ) + 4 + 3, , 2, , =, , 12, unit, 29, , \ The given lines intersect and their point of, intersection is (–1, –6, –12) ., , , 23. , Here a = −2i + 3 j + 5k , b = 7i − k, \ Equation of line joining A and B is,,  ,  , r = a + λ(b − a ), , ⇒ r = −2i + 3 j + 5k + λ(9i − 3 j − 6 k ) , , , Again, c = −3i − 2 j − 5 k , d = 3i + 4 j + 7 k, \ Equation of line joining C and D is,,  ,  , r = c + µ( d − c ), , ⇒ r = −3i − 2 j − 5 k + µ(6i + 6 j + 12 k ) , , ⇒ x = 7m + 3, y = 5m + 9, z = 4, The coordinates of a general point on CD are, (7m + 3, 5m + 9, 4), If the line AB and CD intersect then they have a common, point. So, for some values of l and m, we must have, 4l = 7m + 3, 6l – 1 = 5m + 9, 2l – 1 = 4, ⇒ 4l – 7m = 3 ...(i), 6l – 5m = 10 ...(ii), 5, and λ =, ...(iii), 2, 5, Substituting λ = in (ii), we get m = 1, 2, 5, Since λ =, and m = 1 satisfy (i), so the given lines AB, 2, and CD intersect., 22. , The given lines are, , r = 3 i + 2 j − 4 k + λ (i + 2 j + 2 k ) and, , r = 5 i − 2 j + µ ( 3 i + 2 j + 6 k ), , ⇒ r = ( 3 + λ ) i + ( 2 + 2 λ ) j + ( 2 λ − 4 ) k ...(i), , and r = ( 5 + 3µ ) i + ( 2 µ − 2 ) j + 6µ k ...(ii), If these lines intersect, they must have a common point., So, we must have, ( 3 + λ ) i + ( 2 + 2 λ ) j + ( 2 λ − 4 ) k = ( 5 + 3µ ) i + ( 2 µ − 2 ) j + 6µ k, 3 + l = 5 + 3m ⇒, 2 + 2l = 2m – 2 ⇒, and 2l – 4 = 6m, ⇒, ⇒ l = –4, m = –2., ⇒, , l – 3m = 2,, l – m = –2,, l – 3m = 2, , ... (ii), , Equations (i) and (ii) will intersect, when, −2i + 3 j + 5k + λ( 9i − 3 j − 6 k ) = −3i − 2 j − 5k, + µ(6i + 6 j + 12 k ), , 21. , The equation of line AB is given by, x−0 y+1 z+1, =, =, = λ (say), 4−0 5+1 1+1, ⇒ x = 4l, y = 6l – 1, z = 2l – 1, The coordinates of a general point on AB are, (4l, 6l –1, 2l –1), The equation of line CD is given by, x−3 y−9 z−4, =, =, = µ (say), 3+4 9−4 4−4, , ... (i), , –2 + 9l =– 3 + 6m ; 3 – 3 l = – 2+ 6m ;, 5 – 6l = – 5 + 12 m, ⇒ 9l – 6m = – 1 ; 3l + 6m = 5 ; 6l + 12m = 10, 1, 2, Solving first two equations, we get λ = , µ =, 3, 3, which also satisfy third equation., ⇒, , 1, in (i), we get the point of intersection of lines, 3, is i + 2 j + 3k ≡ (1, 2 , 3) ., Put λ =, , 24. The given lines are, , r = (8 i − 19 j + 10 k ) + λ ( 3 i − 16 j + 7 k ), , and r = (15 i + 29 j + 5k ) + µ ( 3 i + 8 j − 5k ), Equation of any line through (1, 2, –4) with d.r’s, , l, m, n is r = ( i + 2 j − 4 k ) + t (l i + m j + nk ) ...(i), Since, the required line is perpendicular to both the, given lines., \ 3l – 16m + 7n = 0 and 3l + 8m – 5n = 0, l, m, n, l m n, ⇒, =, =, ⇒ =, =, 80 − 56 21 + 15 24 + 48, 2, 3 6, \ From (i), the required line is, , r = (i + 2 j − 4 k ) + t ( 2i + 3 j + 6 k ), Here, the position vector of passing point is, , , a = i + 2 j − 4 k and parallel vector is b = 2 i + 3 j + 6 k, \ Cartesian equation is given by, x−1 y−2 z+4, =, =, 2, 3, 6, , 25. Here, the lines are, , r = ( i + 2 j + 3k ) + λ (i − 3j + 2 k ) and, , r = (4 i + 5j + 6k ) + µ (2i + 3j + k ), , , Here, a1 = i + 2 j + 3k , b1 = i − 3j + 2 k and, , , a = 4 i + 5j + 6k , b = 2i + 3j + k, 2, , 2

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95, , Three Dimensional Geometry, The shortest distance between the lines is given by,  ,  , (a − a ) ⋅ (b × b ), d = 2 1  1 2, |b1 × b2 |,  , a2 − a1 = 3i + 3j + 3k, , Hence, d =, , | 5 ( −8) − 7( 4 )|, 68 17 5, =, =, units, 5, 4 5, 4 5, x+1 y+1 z+1, 28. , Let l1 :, =, =, −6, 7, 1, x − ( −1) y − ( −1) z − ( −1), ⇒, =, =, −6, 7, 1, x−3 y −5 z−7, and l2 :, =, =, −2, 1, 1, \ Vector equation of lines are, , r = − i − j − k + λ (7i − 6 j + k ), , , i j k, , , b1 × b2 = 1 −3 2 = − 9 i + 3j + 9k, 2 3 1,  , ⇒ |b1 × b2 |= (−9)2 + 32 + 9 2 = 3 19,    , Also, (a2 − a1 ).(b1 × b2 ), = (3i + 3j + 3k ).(−9i + 3j + 9k ), = 3 × (–9) + 3 × 3 + 3 × 9 = 9, 9, 3, unit., ∴ d=, =, 3 19, 19, 26. , The given line is, , ⇒, , x + 2 2y − 7 5 − z, =, =, 2, 6, 6, , 7, x + 2 y − 2 z − 5 , =, =, 3, 2, −6, , ...(i), , Its d.r’s are 2, 3, –6, 2 2 + 32 + ( − 6 )2 = 7, , ∵, , 2 3, 6, , ,−, 7 7, 7, Eq. of a line through (–1, 2, 3) and parallel to (i) is, x+1 y−2 z−3, =, =, = λ (say), −6, 2, 3, \ Vector equation of a line passing through (–1, 2, 3), and parallel to (i) is given by, , r = ( −i + 2 j + 3k ) + λ( 2i + 3 j − 6 k ), \, , Its d.c’s are, , 27. , The given lines are, , r = 2 i − 5 j + k + λ ( 3 i + 2 j + 6 k ) and, , , r = 7 i − 6 k + µ ( i + 2 j + 2 k ), ,  , S.D. between the lines r = a1 + λb1 and, ,  , r = a2 + µ b2 is given by,  , , , ( a2 − a1 ) ⋅ (b1 × b2 ), , , d=, | b1 × b2 |, On comparing, we get, , , a1 = 2 i − 5 j + k , b1 = 3 i + 2 j + 6 k, , , a2 = 7 i − 6 k , b2 = i + 2 j + 2 k, , , ∴ a − a = 5 i + 5 j − 7 k, 2, , , , b1 × b2, , ∴ | b1, , 1, , i j k, = 3 2 6 = −8 i + 4 k, 1 2 2, , × b2 | = ( −8)2 + 4 2 = 4 5, , ( 5 i + 5 j − 7 k ) ⋅ ( −8 i + 4 k ), 4 5, , =, , , and r = 3 i + 5 j + 7 k + µ (i − 2 j + k ), , , We get a1 = −i − j − k , b1 = 7i − 6 j + k, , , and a2 = 3i + 5 j + 7 k , b2 = i − 2 j + k, , , So, a2 − a1 = ( 3 i + 5 j + 7 k ) − ( − i − j − k ) = 4 i + 6 j + 8 k, i j k,  , And, b × b = 7 −6 1 = − 4 i − 6 j − 8k, 1, , 2, , 1 −2 1, Shortest distance between two skew lines is,,  , , , (b1 × b2 ) . ( a2 − a1 ),  , d=, | b1 × b2 |, ⇒, , d=, , ⇒, , d=, , ( − 4 i − 6 j − 8k ).( 4 i + 6 j + 8 k ), ( −4 )2 + ( −6 )2 + ( −8)2, −16 − 36 − 64, ⇒ d = 2 29 units, 116, , 29. , H ere the position vectors of A, B and C are, respectively 2i − j + k , i + j + 2 k and 2i + 3 k ., , \ AB = (i + j + 2 k ) − ( 2i − j + k ) = − i + 2 j + k, , AC = ( 2i + 3k ) − ( 2i − j + k ) = 0 . i + j + 2 k, A vector normal, A, B and C is, i,   , n = AB × AC = −1, 0, , to the plane containing points, j k, 2 1 = 3i + 2 j − k, 1 2, , The required unit vector, , 3i + 2 j − k, n, 1, =  =, =, ( 3i + 2 j − k ), 2, 2, 2, |n|, 14, 3 + 2 + ( −1), , \, , 30. , Let the eq. of the plane through (2, 1, –1) be, a(x – 2) + b(y – 1) + c(z + 1) = 0, Also, point (–1, 3, 4) lies on it., \ –3a + 2b + 5c = 0, Also (i) is ^ to the plane x – 2y + 4z = 10, \ a · 1 + b(–2) + c · 4 = 0, , ...(i), ...(ii)

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CBSE Board Term-II Mathematics Class-12, , 96, ⇒ a – 2b + 4c = 0 , Solving (ii) and (iii), we get, a, b, c, a, b, c, =, =, ⇒, =, =, 8 + 10 5 + 12 6 − 2, 18 17 4, , ...(iii), , \ From (i), required eqn. of the plane is, 18(x – 2) + 17(y – 1) + 4(z + 1) = 0, ⇒ 18x + 17y + 4z = 49, In vector form, eq. of this plane is, , r ⋅ (18 i + 17 j + 4 k ) = 49., 31. , The given lines are, 5 − x y − 7 z + 3 and x − 8 2 y − 8 z − 5, =, =, =, =, −4, −5, 4, 7, 2, 3, y −7 z+3, y−4 z−5, x, −, 5, x, −, 8, ⇒, =, =, and, =, =, 4, 4, −5, 7, 1, 3, The condition of coplanarity of these lines is, x2 − x1 y 2 − y1 z2 − z1, a1, b1, c1, =0, a2, b2, c2, Here, L.H.S. =, , x 2 − x1, a1, a2, , y 2 − y1, b1, b2, , z2 − z1, c1, c2, , 8 − 5 4 − 7 5 − ( −3), 3 −3 8, = 4, 4, −5, = 4 4 −5, 7, 1, 3, 7 1, 3, = 3(12 + 5) + 3(12 + 35) + 8(4 – 28), = 51 + 141 – 192 = 0 = R.H.S., Hence the given lines are coplanar., 32. , Let P be the point with position vector, i + 3 j + 4 k ≡ (1, 3, 4) and the plane is r . ( 2 i − j + k ) + 3 = 0, ⇒ 2x – y + z + 3 = 0, ...(i), Any line ^ to this plane passes through P(1, 3, 4) is, x−1 y−3 z−4, =, =, = λ (say), −1, 2, 1, Any point M on the line is (2l + 1, –l + 3, l + 4), This lies on the plane (i)., \ 2(2l + 1) – (–l + 3) + (l + 4) + 3 = 0, , ⇒ 6l + 6 = 0 ⇒ l = –1., \ Coordinates of point M are (–1, 4, 3)., Let the image of P in the plane (i) be Q(a, b, g), then M, will be the midpoint of PQ., α+1, β+3, γ+4, ∴, = −1,, = 4,, =3, 2, 2, 2, ⇒ a = –3, b = 5, g = 2, Hence, Q(–3, 5, 2) is the image of P(1, 3, 4) in plane (i)., , 33. , Eq. of any plane through (–1, 2, 0) is, a(x + 1) + b(y – 2) + cz = 0 , Also it passes through (2, 2, –1), \ 3a + 0 · b – c = 0 , Further plane (i) is parallel to the line, 1, x−1 y+ 2 z+1, =, =, 1, 1, −1, \ a · 1 + b · 1 + c · (–1) = 0, ⇒ a + b – c = 0, Eliminating a, b and c from (ii) and (iii), we get, a b c, = = = λ(say ), 1 2 3, ⇒ a = l, b = 2l and c = 3l, Put these values in (i), we get, x + 2y + 3z = 3, This is the eq. of the required plane., , ...(i), ...(ii), , ...(iii), , 34. , E quation of plane passing through (3, 2, 0) is, a(x – 3) + b(y – 2) + c(z – 0) = 0 , ...(i), x−3 y−6 z−4, Given line is, =, =, 1, 5, 4, Since plane contains the line, so, a(3 –3) + b(6 – 2) + c(4 – 0) = 0, ⇒ 0·a + 4b + 4c = 0 , ...(ii), and a(1) + b(5) + c(4) = 0, ⇒ a + 5b + 4c = 0 , ...(iii), Solving (ii) & (iii), we get, a, b, c, =, =, = λ (say), 16 − 20 4 − 0 0 − 4, ⇒ a = –4l, b = 4l, c = –4l, Putting values of a, b, c in (i), we get, –4l(x – 3) + 4l(y – 2) – 4l(z – 0) = 0, ⇒ –4x + 12 + 4y – 8 – 4z = 0, ⇒ x – y + z – 1 = 0 is the required equation of plane., 35. The eq. of the line through A(3, 4, 1) and, x−3 y−4 z−1, =, =, B(5, 1, 6) is, 5−3 1− 4 6−1, x−3 y−4 z−1, =, =, , −3, 2, 5, Now (i) meets the XY-plane (whose eq. is z = 0), x−3 y−4 0−1, 1, ∴, =, =−, =, 2, −3, 5, 5, x−3, 1 y−4, 1, ⇒, =− ,, =− , z=0, 2, 5 −3, 5, 2, 3, ⇒ x = 3− ,y = 4+ ,z=0, 5, 5, 13, 23, ⇒ x= ,y=, ,z=0, 5, 5, ⇒, , ...(i), ,  13 23 , ⇒  , , 0  is the point where line (i) meets the,  5 5 , , XY-plane.

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97, , Three Dimensional Geometry, 36. , The given line is, , r = − i + 3 j + k + λ ( 2 i + 3 j − k ), Its cartesian eq. is, x+1 y−3 z−1, =, =, = λ (say) ...(i), 2, 3, −1, Any point Q on (i) is (2l – 1, 3l + 3, – l + 1), Also, the given point is P(5, 4, 2)., Now d.r’s of the line PQ are, (2l – 1 – 5, 3l + 3 – 4, –l + 1 – 2) = (2l – 6, 3l – 1, –l –1)., For PQ to be ^ to (i), we must have, (2l – 6). 2 + (3l – 1) . 3 + (– l – 1). (–1) = 0, ⇒ 14l – 14 = 0 ⇒ l = 1, \ Q is (1, 6, 0), which is the foot of ^ from P on line (i)., Now, PQ = ( 5 − 1)2 + ( 4 − 6 )2 + ( 2 − 0 )2, = 24 = 2 6 units., , Further if R(a, b, g) is the image of P in line (i), then, α+5, β+4, γ+2, = 1,, = 6,, =0, 2, 2, 2, ⇒ a = –3, b = 8, g = –2, \ Image of P in line (i) is R(–3, 8, –2)., 37. The given line is, , r = (2 i − 4j + 2 k ) + λ (3 i + 4j + 2 k ), x−2 y+4 z−2, =, =, = λ (say) ...(i), 3, 4, 2, Any point on it is (3l + 2, 4l – 4, 2l + 2), , This lies on the plane r .( i − 2 j + k ) = 0, ⇒, , x – 2y + z = 0, 3l + 2 – 2 (4l – 4) + 2l + 2 = 0, –3l + 12 = 0 ⇒ l = 4, The point of intersection of (i) and (ii) is, (3 × 4 + 2, 4 × 4 – 4, 2 × 4 + 2) = (14, 12, 10), Its distance from the point (2, 12, 5), ⇒, \, ⇒, \, , ...(ii), , 39. , Any point on the line, x−2 y−2 z−3, =, =, = r (say ) ...(i), 1, 3, 1, is (r + 2, 3r + 2, r + 3), Any point on the line, x−2 y−3 z−4, =, =, = k(say ) ...(ii), 1, 4, 2, is (k + 2, 4k + 3, 2k + 4), For lines (i) and (ii) to intersect, we must have, r + 2 = k + 2, 3r + 2 = 4k + 3, r + 3 = 2k + 4, On solving these, we get, r = k = –1, \ Lines (i) and (ii) intersect and their point of, intersection is (1, –1, 2), Now, required equation of plane is, x−2 y−2 z−3, 1, 3, 1 =0, 1, 4, 2, ⇒ (x – 2)(6 – 4) – (y – 2)(2 – 1) + (z – 3)(4 – 3) = 0, ⇒ 2x – 4 – y + 2 + z – 3 = 0 ⇒ 2x – y + z – 5 = 0, 40. The given lines are, , r = 2 i + j − 3 k + λ (i + 2 j + 5k ) ...(i), , and r = 3 i + 3 j + 2 k + µ ( 3i − 2 j + 5 k ) ...(ii), , , Here, a1 = 2 i + j − 3 k , b1 = i + 2 j + 5k, , , a2 = 3 i + 3 j + 2 k , b2 = 3 i − 2 j + 5k, The plane containing lines (i) and (ii) will pass through, , a1 = 2i + j − 3k ., , , Also the plane is parallel to two vectors b1 and b2 ., \ The plane is normal to the vector, i j k, ,  , n = b × b = 1 2 5 = 20i + 10 j − 8k, 1, , = (14 − 2)2 + (12 − 12)2 + (10 − 5)2, = 144 + 0 + 25 = 169 = 13 units., 38. , V ector equation of a line passing through, (2, 3, 2) and parallel to the line, , r = ( −2 ^i + 3 ^j ) + λ( 2 ^i − 3 ^j + 6 k^) is given by, , r = ( 2 ^i + 3 ^j + 2 k^) + µ( 2 ^i − 3 ^j + 6 k^), , , Now, a1 = −2 ^i + 3 ^j , b = 2 ^i − 3 ^j + 6 k^, , a2 = 2 ^i + 3 ^j + 2 k^, Distance between given parallel lines,  , , ^, ^, ^, ^, ^, ^, (2 i − 3 j + 6 k) ⋅ (4 i + 0 j + 2 k ), b × ( a2 − a1 ), , =, =, |b |, | 4 + 9 + 36 |, 8 + 0 + 12, 20, =, =, units, 7, 49, , 2, , 3 −2 5, \ The vector eqn. of the required plane is,   ,    , (r − a1 ) ⋅ n = 0 ⇒ r ⋅ n = a1 ⋅ n, , ⇒ r ⋅ ( 20 i + 10 j − 8k ) = ( 2 i + j − 3k ) . ( 20 i + 10 j − 8k ), , = 40 + 10 + 24 = 74, , ...(iii), ⇒ r ⋅(10 i + 5 j − 4 k ) = 37 , Its cartesian eqn. is 10x + 5y – 4z = 37, The given line is,, , r = ( 2 i + 5 j + 2 k ) + p( 3 i − 2 j + 5k ) ...(iv), The line (iv) will lie in the plane (iii) if the plane passes, , through the point a = 2 i + 5 j + 2 k on line (iv) and is, parallel to the line (iv)., , Now, a ⋅ (10 i + 5 j − 4 k ), = ( 2 i + 5 j + 2 k ) . (10 i + 5 j − 4 k ) = 37, , \ a lies in the plane (iii), Also, (10i + 5 j − 4 k ) ⋅ ( 3i − 2 j + 5k ), = 10 × 3 + 5 × (–2) – 4 × 5 = 0, \ Line (iv) is parallel to the plane (iii)., \ Line (iv) lies in the plane (iii)., , 

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CHAPTER, , 6, , Probability, Recap Notes, , INTRODUCTION, , Extension of Multiplication Theorem, , In earlier classes, we have studied the probability, as a measure of uncertainty of events in a random, experiment. We also learnt the axiomatic theory and, classical theory of probability. In this chapter, we shall, discuss some more important concepts of probability., , (i) If A, B, C are three events associated with a random, experiment, then, P(ABC) or P(A ∩ B ∩ C) = P(A) ⋅ P(B | A) ⋅, P(C | (A ∩ B)) = P(A) P(B | A) P(C | AB), (ii) If A 1, A 2, ......, A n are n events associated with a, random experiment, then, P(A1 ∩ A2 ...... ∩ An) = P(A1) P(A2 | A1) P(A3 | (A1 ∩, , CONDITIONAL PROBABILITY, h, , The probability of an event A is called the conditional, probability of A given that B has already happened., P( A ∩ B), P( A|B) =, , P(B) ≠ 0, P(B), , Properties of Conditional Probability, h, , Property 1 : The conditional probability of an event, A given that B has occurred lies between 0 and 1., , h, , Property 2 : If A and B are any two events of sample, space S, and F is an event of S such that P(F) ≠ 0,, then, P((A ∪ B) | F) = P(A | F) + P(B | F) – P((A ∩ B) |, F), In particular, if A and B are disjoint events, then, P((A ∪ B) | F) = P(A | F) + P(B | F), , h, , Property 3 : P(A′ | B) = 1 – P(A | B), , MULTIPLICATION THEOREM ON PROBABILITY, Let A and B be two events associated with a sample, space S. Then A ∩ B denotes the event that both A, and B have occurred. A ∩ B is also written as AB., Sometimes we need to find the probability of the, event AB, which is given by, P(A) ⋅ P(B|A), provided P(A) ≠ 0., P(AB) = P(A ∩ B) =, P(B) ⋅ P(A|B), provided P(B) ≠ 0., h, , This result is known as multiplication rule of, probability., , A2)) ...... P(An | (A1 ∩ A2 ...... ∩ An–1))., , INDEPENDENT EVENTS, h, , h, , h, , h, , Two events are called independent if the occurrence, or non-occurrence of one does not affect the, occurrence of the other., Two events A and B associated with a random, experiment are said to be independent, if, P(A | B) = P(A), provided P(B) ≠ 0,, P(B | A) = P(B), provided P(A) ≠ 0, Two events A and B associated with a random, experiment are independent if and only if, P(A ∩ B) = P(A) . P(B), If three events A, B, C associated with a random, experiment are independent, then, P(ABC) i.e. P(A ∩ B ∩ C) = P(A) . P(B) . P(C), and in general, if n events A1, A2, ......, An associated, with a random experiment are independent, then, P(A1 ∩ A2 ∩ ..... ∩ An) = P(A1) . P(A2) ..... P(An), , Remarks, (i) Two events A and B are said to be dependent if they, are not independent i.e., if P(A ∩ B) ≠ P(A)⋅P(B), (ii) Three events A, B, C are said to be mutually independent, if P(A ∩ B) = P(A) P(B), P(A ∩ C) = P(A) P(C),, P(B ∩ C) = P(B) P(C), and P(A ∩ B ∩ C) = P(A) P(B) P(C), If at least one of the above four conditions is not, true for three given events, we say that the events, are not mutually independent.

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99, , Probability, (iii) Events A and B are mutually exclusive if and only, if A ∩ B = f., , P(Ei|A) =, , Events A and B are independent if and only if, P(A ∩ B) = P(A)⋅P(B), (iv) If events A and B having non-zero probabilities are, independent, then P(A ∩ B) = P(A)⋅P(B) ≠ 0. Thus, A ∩ B ≠ f. Hence, two independent events having, non-zero probabilities cannot be mutually exclusive., , BAYES' THEOREM, , (a) Ei ∩ Ej = f, i ≠ j, i, j = 1, 2, 3, ...., n, , RANDOM VARIABLES AND ITS PROBABILITY DISTRIBUTIONS, h, , h, , h, , (b) E1 ∪ E2 ∪ .... ∪ En = S and, , (c) P(Ei) > 0 for all i = 1, 2, ...., n, In other words, the events E1, E2, ...., En represent a, partition of the sample space S, if they are pairwise, disjoint, exhaustive and have non-zero probabilities., , Theorem of Total Probability, If E1, E2, E3, ......., En is a partition of the sample space S, of a random experiment and A is any event associated, with S, then P(A) = P(E1) P(A | E1) + P(E2) P(A | E2) +, ........ + P(En) P(A | En)., , Bayes' Theorem, If E1, E2, E3, ......., En constitute a partition of the sample, space S of a random experiment and A is any event, associated with the experiment, then, , ∑ P(E j ) P( A|E j ), , , where i = 1, 2 , 3, ......., n., , j=1, , Partition of a Sample Space, A set of events E 1, E 2, ...., E n is said to represent a, partition of sample space S, if, , P(Ei ) P( A|Ei ), , n, , A random variable is often described as a variable, whose values are determined by the outcomes of a, random experiment., A random variable is called discrete if it assumes, only finite or countable number of values. A, random variable which can assume non-countably, infinite number of values is called non-discrete or, continuous random variable., Let X is a discrete random variable which can, assume values x 1 , x 2 , x 3 , ......., x n (arranged in, increasing order of magnitude) with probabilities, p1, p2, p3, ......., pn., X, , x1, , x2, , ..., , xn, , P(X = x), , p1, , p2, , ..., , pn, , Then the probability distribution of X is the system, of numbers, shown in the table., Note:, (i) P(x k ) i.e., P(X = x k ) lies between 0 and 1 for, k = 1, 2, ......., n, (ii), , n, , ∑ pi = 1, , i =1, , (iii) P(X ≤ xi) = p1 + p2 + ...... + pi,, , P(X ≥ xi) = pi + pi+1 + ...... + pn,, , P(X ≤ xi) = P(X < xi) + P(X = xi) etc.

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Practice Time, OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1. Suppose that five good fuses and two, defective ones have been mixed up. To find the, defective fuses, we test them one-by-one, at, random and without replacement. What is the, probability that we are lucky and find both of, the defective fuses in the first two tests?, 2, 1, 1, 1, (a), (b), (c), (d), 21, 18, 42, 21, 2. If six cards are selected at random (without, replacement) from a standard deck of 52 cards,, then what is the probability that there will be, no pairs (two cards of same denomination)?, (a) 0.28, (b) 0.562 (c) 0.345 (d) 0.832, 3. A die, whose faces are marked 1, 2, 3 in red, and 4, 5, 6 in green, is tossed. Let A be the event, “number obtained is even” and B be the event, “number obtained is red”. Find P(A ∩ B), 1, 1, 1, 1, (a), (b), (c), (d), 4, 2, 6, 3, 7, 9, 12, 4. If P ( A ) =, and P ( A ∪ B) =, , P( B) =, ,, 13, 13, 13, then evaluate P(A | B)., 4, 4, 4, 9, (a), (b), (c), (d), 13, 9, 5, 13, 1, 2, 3, and P ( A ∩ B) = ,, , P( B) =, 5, 5, 10, then P(A′ | B′) is equal to, (a) 5/6, (b) 5/7, (c) 25/42 (d) 1, 5., , If P ( A ) =, , 6. A bag contains 5 red and 3 blue balls. If 3, balls are drawn at random without replacement, the probability of getting exactly one red ball is, 135, 15, 45, 15, (a), (b), (c), (d), 392, 56, 196, 29, 7. Let A and B be independent events with, P(A) = 1/4 and P(A ∪ B) = 2P(B) – P(A). Find, P(B)., 2, 1, 2, 3, (a), (b), (c), (d), 3, 4, 5, 5, , 8. A random variable X has the following, distribution., X, , 1, , 2, , 3, , 4, , 5, , 6, , P(X) 0.15 0.23 0.12 0.10 0.20 0.08, , 7, , 8, , 0.07, , 0.05, , For the event E = {X is prime number}, find P(E)., (a) 0.87, (b) 0.62, (c) 0.35, (d) 0.50, 9., , If A and B are two events such that P(A|B), 1, 5, = p, P(A) = p, P ( B) = and P ( A ∪ B) = , then find, 3, 9, the value of p., (a) 2/3, (b) 4/9, (c) 5/9, (d) 1/3, 10. A bag contains 3 white and 6 black balls, while another bag contains 6 white and 3 black, balls. A bag is selected at random and a ball is, drawn. Find the probability that the ball drawn, is of white colour., 3, 5, 1, 1, (a), (b), (c), (d), 4, 4, 4, 2, 11. Two dice are thrown together. What is the, probability that the sum of the number on the, two faces is neither 9 nor 11 ?, 3, 1, 5, 2, (a), (b), (c), (d), 4, 2, 6, 3, 12. If A and B are two events and A ≠ f, B ≠ f,, then, (a) P(A|B) = P(A) . P(B), P( A ∩ B), (b) P ( A | B) =, P( B), (c) P(A | B) . P(B | A) = 1, (d) P(A | B) = P(A) | P(B), 13. If A and B are two independent events such, that P(A ∪ B) = 0.6 and P(A) = 0.2, then find, P(B)., (a) 0.3, (b) 0.4, (c) 0.1, (d) 0.5, 14. If A and B are two independent events, then, the probability of occurrence of at least one of A, and B is given by, (a) 1 – P(A) P(B), (b) 1 – P(A) P(B′), (c) 1 – P(A′) P(B′), (d) 1 – P(A′) P(B)

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101, , Probability, , 15. The probability distribution of a discrete, random variable X is given below :, X, , 2, , 3, , 4, , 5, , P(X), , 5, k, , 7, k, , 9, k, , 11, k, , The value of k is, (a) 8, (b) 16, , (c) 32, , (d) 48, , 16. An urn contains 10 black and 5 white balls., Two balls are drawn from the urn one after the, other without replacement, then the probability, that both drawn balls are black, is, 3, 5, 2, 1, (a), (b), (c), (d), 7, 7, 7, 7, 17. The probability that student entering, a university will graduate is 0.4. Find the, probability that out of 3 students of the, university none will graduate., (a) 0.216 (b) 0.36, (c) 0.6, (d) 0.1296, 18. If two events A and B are such that, P ( A ) = 0.3, P(B) = 0.4 and P ( A ∩ B) = 0.5 then, P ( B |( A ∪ B)) =, 1, 1, (a), (b) 1, (c) 2, (d), 2, 4, 3, 5, 19. Let X denote the number of hours you study, on a Sunday. Also it is known that, , if x = 0, 0.2, kx, , if x = 1 or 2, , P( X = x ) = , k(4 − x ), if x = 3 or 4, , otherwise, 0, where k is a constant., What is the probability that you study atleast, two hours?, (a) 0.55, (b) 0.15, (c) 0.6, (d) 0.3, 3, 2, 3, , P ( B) = and P ( A ∪ B) = , then, 10, 5, 5, P(B | A) + P(A | B) equals, 7, 1, (a) 1, (b), (c) 5, (d), 12, 3, 4, 12, 20. If P ( A ) =, , 21. Aprobleminmathematicsisgivento3students, 1 1 1, whose chances of solving it are , , . What is, 2 3 4, the probability that the problem is solved ?, (a) 1/5, (b) 1/4, (c) 3/4, (d) 2/3, 22. A random variable X has the following, probability distribution :, , 0, 1, 2, X, P(X) a, 6a 6a, Find the value of a., 1, 1, (a), (b), 48, 47, , 3, 4a, (c), , 4, 8a, , 5, 8a, , 1, 33, , 6, 6a, (d), , 7, 9a, 1, 29, , 23. The probability distribution of a discrete, random variable X is given below :, X, , 0, , 4, k, The value of k is, (a) 8, (b) 16, P(X), , 1, , 2, , 3, , 6, k, , 10, k, , 12, k, , (c) 32, , (d) 48, , 24. The probability distribution of X is, X =x, 0 1, 2 3 4, ., P ( X = x ) k 2k 4k 2k k, , Then find P(X ≤ 1)., (a) 0.1, (b) 0.3, , (c) 0.4, , (d) 0.5, , 25. If A and B are events such that P(A) > 0 and, P(B) ≠ 1, then P(A′ | B′) equals, (a) 1 – P(A|B), (b) 1 – P(A′|B), 1 − P( A ∪ B), (c), (d) P(A′) | P(B′), P ( B′ ), 26. A flashlight has 8 batteries out of which 3, are dead. If two batteries are selected without, replacement and tested, the probability that, both are dead is, 9, 3, 33, 1, (a), (b), (c), (d), 64, 28, 56, 14, 27. You are given that A and B are two, 3, 1, events such that P ( B) = , P ( A | B) =, and, 5, 2, 4, P ( A ∪ B) = , then P(A) equals, 5, 1, 1, 3, 3, (a), (b), (c), (d), 2, 5, 10, 5, 28. A and B are events such that P(A) = 0.4, P(B), = 0.3 and P(A ∪ B) = 0.5. Then P(B′ ∩ A) equals, 1, 1, 3, 2, (a), (b), (c), (d), 5, 2, 10, 3, 29. A box contains 3 orange balls, 3 green balls, and 2 blue balls. Three balls are drawn at, random from the box without replacement. The, probability of drawing 2 green balls and one blue, ball is, 2, 3, 167, 1, (a), (b), (c), (d), 21, 28, 168, 28

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CBSE Board Term-II Mathematics Class-12, , 102, 30., (a), (b), (c), (d), , Two events A and B will be independent, if, A and B are mutually exclusive, P(A′ ∩ B′) = [1 – P(A)] [1 – P(B)], P(A) = P(B), P(A) + P(B) = 1, , 31. Given that, the events A and B are such that, 1, 3, P(A) = , P ( A ∪ B) = and P(B) = p. Then find, 2, 5, the value of p, if A and B are mutually exclusive., (a), , 3, 5, , (b), , 1, 5, , (c), , 2, 3, , (d), , 1, 10, , 3, 2, 3, , P( B) =, and P ( A ∪ B) = ,, 10, 5, 5, then find the value of P(B / A)., 32. If P ( A ) =, , (a), , 2, 3, , (b), , 1, 3, , (c), , 2, 5, , (d), , 1, 4, , 33. If A and B are two events such that P(A) =, 0.2 , P(B) = 0.4 and P(A ∪ B) = 0.5 , then value, of P(A/B) is, (a) 0.1, (b) 0.25, (c) 0.5, (d) 0.08, 34. An urn contains 6 balls of which two are, red and four are black. Two balls are drawn, at random. Probability that they are of the, different colours is, 2, 1, 8, 4, (a), (b), (c), (d), 5, 15, 15, 15, 35. If P(A) = 0.4, P(B) = 0.8 and P(B | A) = 0.6, then, P(A ∪ B) is equal to, (a) 0.24, (b) 0.3, (c) 0.48, (d) 0.96, , Case Based MCQs, Case I : Read the following passage and answer, the questions from 36 to 40., A doctor is to visit a patient. From the past, experience, it is known that the probabilities, that he will come by cab, metro, bike or by other, means of transport are respectively 0.3, 0.2, 0.1, and 0.4. The probabilities that he will be late are, 0.25, 0.3, 0.35 and 0.1 if he comes by cab, metro,, bike and other means of transport respectively., , (a), , 4, , 21, , (b), , 1, 7, , (c), , 5, , 14, , (d), , 2, 21, , 38. When the doctor arrives late, what is the, probability that he comes by bike?, 5, 1, 4, 5, (a), (b), (c), (d), 21, 6, 7, 6, 39. When the doctor arrives late, what is the, probability that he comes by other means of, transport?, 2, 6, 5, 4, (a), (b), (c), (d), 7, 7, 14, 21, 40. What is the probability that the doctor is, late by any means?, 1, 1, (a) 1, (b) 0, (c), (d), 4, 2, Case II : Read the following passage and answer, the questions from 41 to 45., , 36. When the doctor arrives late, what is the, probability that he comes by metro?, 2, 5, (a), , (b), 7, 14, 1, 5, (c), , (d) 6, 21, 37. When the doctor arrives late, what is the, probability that he comes by cab?, , One day, a sangeet mahotsav is to be organised, in an open area of Rajasthan. In recent years, it, has rained only 6 days each year. Also, it is given, that when it actually rains, the weatherman, correctly forecasts rain 80% of the time. When, it doesn’t rain, he incorrectly forecasts rain 20%, of the time., If leap year is considered, then answer the, following questions.

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103, , Probability, , Case III : Read the following passage and answer, the questions from 46 to 50., Varun and Isha decided to play with dice to, keep themselves busy at home as their schools, are closed due to coronavirus pandemic. Varun, throw a dice repeatedly until a six is obtained., He denote the number of throws required by X., 41. The probability that it rains on chosen day, is, 1, 1, 1, 1, (a), (b), (c), (d), 366, 73, 61, 60, 42. The probability that it does not rain on, chosen day is, 1, 5, (a), (b), 366, 366, 360, (c), (d) none of these, 366, 43. The probability that the weatherman, predicts correctly is, 1, 7, 5, 4, (a), (b), (c), (d), 5, 8, 6, 5, , 46. The probability that X = 2 equals, 1, 5, 1, 5, (a), (b), (c), (d), 6, 2, 6, 3, 63, 6, 47. The probability that X = 4 equals, , 3, (c) 5, (d) 5, 4, 6, 6, 64, 6, 48. The probability that X ≥ 2 equals, 1, 25, 25, 5, (a), (b), (c), (d), 36, 36, 216, 6, , (a), , 1, , 4, , (b), , 1, , 6, , 44. The probability that it will rain on the, chosen day, if weatherman predict rain for that, day, is, (a) 0.0625 (b) 0.0725 (c) 0.0825 (d) 0.0925, , 49. The value of P(X ≥ 6) is, , 45. The probability that it will not rain on the, chosen day, if weatherman predict rain for that, day, is, (a) 0.94, (b) 0.84, (c) 0.74, (d) 0.64, , 50. The probability that X > 3 equals, 36, 5, 52, 53, (a), (b), (c), (d), 25, 6, 62, 63, , (a), , 55, , 5, , 6, , 3, 3, 3, (b) 1 – 5 (c) 5 × 61 (d) 5, 65, 65, 64, , Assertion & Reasoning Based MCQs, Directions (Q.-51 to 60) : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose the correct, answer out of the following choices :, , (a), (b), (c), (d), , Assertion, Assertion, Assertion, Assertion, , and Reason both are correct statements and Reason is the correct explanation of Assertion., and Reason both are correct statements but Reason is not the correct explanation of Assertion., is correct statement but Reason is wrong statement., is wrong statement but Reason is correct statement., , 51. Let E1 and E2 be any two events associated, with an experiment, then, Assertion : P(E1) + P(E2) ≤ 1., Reason : P(E1) + P(E2) = P(E1 ∪ E2) , + P(E1 ∩ E2)., 52. Consider the system of equations, ax + by = 0, cx + dy = 0, where a, b, c, d ∈{0, 1}., , Assertion : The probability that the system of, 3, equations has a unique solution is ., 8, Reason : The probability that the system of, equations has a solution is 1., 53. Assertion : Consider the experiment of, drawing a card from a deck of 52 playing cards,, in which the elementary events are assumed to, be equally likely.

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CBSE Board Term-II Mathematics Class-12, , 104, If E and F denote the events the card drawn is a, spade and the card drawn is an ace respectively,, 1, 1, then P (E | F ) = and P ( F |E ) = ., 13, 4, Reason : E and F are two events such that the, probability of occurrence of one of them is not, affected by occurrence of the other. Such events, are called independent events., 54. Assertion : Let E and F be events associated, with the sample space S of an experiment. Then,, we have P(S|F) = P(F|F) = 1., Reason : If A and B are any two events, associated with the sample space S and F is an, event associated with S such that P(F) ≠ 0, then, P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F), 55. Let A and B be two events associated with, an experiment such that P(A ∩ B) = P(A)P(B)., Assertion : P(A|B) = P(A) and P(B|A) = P(B), Reason : P(A ∪ B) = P(A) + P(B)., 56. Let H1, H2, ..., Hn be mutually exclusive and, exhaustive events with P(Hi) > 0, i = 1, 2, ..., n., Let E be any other event with 0 < P(E) < 1, Assertion : P(Hi | E) > P(E | Hi)P(Hi) for, i = 1, 2, ..., n, Reason :, , n, , ∑, , i =1, , P (Hi ) = 1., , 57. Assertion : Bag I contains 3 red and 4 black, balls while another bag II contains 5 red and 6, black balls. One ball is drawn at random from, one of the bags and it is found to be red. Then,, the probability that it was drawn from bag II is, 35, ., 68, Reason : Given, three identical boxes I, II and, III, each containing two coins. In box I, both, coins are gold coins, in box II, both are silver, , coins and in the box III, there is one gold and, one silver coin. A person chooses a box at random, and takes out a coin. If the coin is of gold, then, the probability that the other coin in the box is, 1, also of gold is ., 2, 58. Assertion : An urn contains 5 red and 5, black balls. A ball is drawn at random, its colour, is noted and is returned to the urn. Moreover,, 2 additional balls of the colour drawn are put, in the urn and then a ball is drawn at random., Then, the probability that the second ball is red, 1, is ., 2, Reason : A bag contains 4 red and 4 black balls,, another bag contains 2 red and 6 black balls., One of the two bags is selected at random and a, ball is drawn from the bag which is found to be, red. Then, the probability that the ball is drawn, 2, form the first bag is ., 3, 59. Assertion : The probability that candidates, 1, 2, A and B can solve the problem is and , then, 5, 5, probability that problem will be solved is given, 12, by, ., 25, Reason : If events A & B are independent, then, P (A ∩ B) = P (A) × P (B)., 60. A man P speaks truth with probability, p and an other man Q speaks truth with, probability 2p., Assertion : If P and Q contradict each other, with probability 1/2, then there are two values, of p., Reason : A quadratic equation with real, coefficients has two real roots., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. A bag contains 10 white and 6 black balls., 4 balls are successively drawn out without, replacement. What is the probability that they, are alternately of different colours?, 2. A die is thrown repeatedly until a six comes, up. Write the sample space for this experiment., , 3. A coin is tossed. If it shows a tail, we draw, a ball from a box which contains 2 yellow and 3, red balls. If it shows head, we throw a die. Write, the sample space for this experiment., 4. Find the total number of elementary events, associated to the random experiment of throwing, three die together.

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105, , Probability, , 5. A bag contains 4 identical red balls and 3, identical black balls. The experiment consists of, drawing one ball, then putting it into the bag, and again drawing a ball. Write the possible, outcomes of this experiment., 6. Two coins (a ` 2 coin and a ` 5 coin) are, tossed once. Find the total number of elements, in sample space., 7. Let the sample space associated with an, experiment is S = {1, 2, 3, 4, 5, 6} and an event, is E = {1, 3, 5}, then find E′ or E., 8. Suppose 3 bulbs are selected at random, from a lot. Each bulb is tested and classified, , as defective (D) or non-defective (N). Write the, sample space of this experiment., 9. The numbers 1, 2, 3 and 4 are written, separately on four slips of paper. The slips are, put in a box and mixed thoroughly. A person, draws two slips from the box, one after the other,, without replacement. Describe the sample space, for the experiment., 10. If A and B are two events associated with the, 3, same random experiment such that P(A ∪ B) = ,, 1, 2, 4, P(A ∩ B) =, and P ( A) = , then find P(B)., 4, 3, , Short Answer Type Questions (SA-I), 11. Three events A, B and C have probabilities, 1, 1, 2 1, , and 2 respectively. Given that P(A ∩ C) = 5, 5 3, 1, and P(B ∩ C) = , find the value of P(C/B) and, 4, , 15. A pair of dice is rolled. If the outcome is a, doublet, a coin is tossed. Find the total number, of possible outcomes for this experiment., , P ( A ∩ C) ., , 17. A card is drawn from a well shuffled deck of 52, cards, then find the probability of red king card., , 12. Two thirds of the students in a class are boys, and the rest girls. It is known that the probability, of a girl getting a first class is 0.25 and that of a, boy getting a first class is 0.28. Find the probability, that a student chosen at random will get first class, marks in the subject., 13. Two dice are rolled. Let A, B, C be the events, of getting a sum of 2, a sum of 3 and a sum of 4, respectively. Then, show that, (i) A is a simple event, (ii) B and C are compound events, (iii) A and B are mutually exclusive events., 14. Find the probability of getting the sum as a, prime number when two dice are thrown together., , 16. What is the probability that all L’s come, together in the word PARALLEL?, , 18. In a single throw of a die, find the probability, of getting an even prime number., 1, 19. If A and B are events such that P(A) = ,, 3, 1, 1, , then find P(not A, P(B) =, and P(A ∩ B) =, 12, 4, and not B)., 20. A university has to select an examiner from, a list of 50 persons, 20 of them are women and, 30 men, 10 of them knowing Hindi and 40 not,, 15 of them being teachers and the remaining, 35 not. What is the probability of the university, selecting a Hindi knowing woman teacher?, , Short Answer Type Questions (SA-II), 21. Six new employees, two of whom are married, to each other, are to be assigned six desks that, are lined up in a row. If the assignment of, employees to desks is made randomly, what is, the probability that the married couple will have, non-adjacent desks?, , 22. A coin is tossed three times, consider the, following events :, A : ‘No head appears’, B : ‘Exactly one head appears’, C : ‘At least two heads appear’, Do they form a set of mutually exclusive and, exhaustive events?

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106, 23. A bag contains 9 balls of which 4 are red, 3, are blue and 2 are yellow. The balls are similar, in shape and size. A ball is drawn at random, from the bag. Calculate the probability that it, will be:, (i) red , (ii) not blue, (iii) either red or blue, 24. What is the probability that:, (i) a non-leap year has 53 Tuesdays?, (ii) a leap year has 53 Wednesdays?, (iii) a leap year has 53 Fridays and 53 Saturdays?, 25. A shopkeeper sells three types of seeds A1,, A2 and A3. They are sold as a mixture where, the proportions are 4 : 4 : 2 respectively. The, germination rates of three types of seeds are, 45%, 60% and 35%., Calculate the probability, (i) that it will not germinate given that the seed, is of type A3., (ii) of a randomly chosen seed to germinate., (iii) that it is of type A2 given that a randomly, chosen seed does not germinate., 26. A bag A contains 4 black and 6 red balls, and bag B contains 7 black and 3 red balls. A, die is thrown. If 1 or 2 appears on it, then bag, A is chosen, otherwise bag B. If two balls are, drawn at random (without replacement) from, the selected bag, find the probability of one of, them being red and another black., 27. Assume that each born child is equally likely, to be a boy or a girl. If a family has two children,, what is the conditional probability that both are, girls? Given that, (i) the youngest is a girl., (ii) atleast one is a girl., 28. A couple has 2 children. Find the probability, that both are boys, if it is known that, (i) atleast one of them is a boy,, (ii) the older child is a boy., 29. A bag contains 3 red and 7 black balls. Two, balls are selected at random one-by-one without, replacement. If the second selected ball happens, to be red, what is the probability that the first, selected ball is also red?, , CBSE Board Term-II Mathematics Class-12, 30. P speaks truth in 70% of the cases and Q in, 80% of the cases. In what percent of cases are, they likely to agree in stating the same fact?, 31. The probabilities of two students A and, 3, 5, B coming to the school in time are, and, 7, 7, respectively. Assuming that the events, ‘A coming, in time’ and ‘B coming in time’ are independent,, find the probability of only one of them coming, to the school in time., 32. In a shop X, 30 tins of ghee of type A and, 40 tins of ghee of type B which look alike, are, kept for sale. While in shop Y, similar 50 tins of, ghee of type A and 60 tins of ghee of type B are, there. One tin of ghee is purchased from one of, the randomly selected shop and is found to be of, type B. Find the probability that it is purchased, from shop Y., 33. Three persons A, B and C apply for a job of, Manager in a Private Company. Chances of their, selection (A, B and C) are in the ratio 1 : 2 : 4., The probabilities that A, B and C can introduce, changes to improve profits of the company are, 0.8, 0.5 and 0.3 respectively. If the change does, not take place, find the probability that it is due, to the appointment of C., 34. Let X denote the number of colleges where, you will apply after your results and P(X = x), denotes your probability of getting admission in, x number of colleges. It is given that, kx, 2kx, P( X = x) = , k(5 − x ), , 0, , , , if, , if, , if, , if, , x = 0 or 1, x =2, x = 3 or 4, x>4, , where k is a positive constant. Find the value, of k. Also find the probability that you will get, admission in, (i) exactly one college, (ii) atmost 2 colleges, (iii) atleast 2 colleges., 35. A bag contains 4 balls. Two balls are drawn, at random (without replacement) and are found, to be white. What is the probability that all balls, in the bag are white ?

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107, , Probability, , Long Answer Type Questions (LA), 36. In a hockey match, both teams A and B, scored same number of goals up to the end of, the game, so as to decide the winner, the referee, asked both the captains to throw a die alternately, and decided that the team, whose captain gets, a six first, will be declared the winner. If the, captain of team A was asked to start, find their, respective probabilities of winning the match., 37. Assume that the chances of a patient, having a heart attack is 40%. Assuming that a, meditation and yoga course reduces the risk of, heart attack by 30% and prescription of certain, drug reduces its chance by 25%. At a time a, patient can choose any one of the two options, with equal probabilities. It is given that after, going through one of the two options, the patient, selected at random suffers a heart attack. Find, the probability that the patient followed a course, of meditation and yoga., 38. Of the students in a college, it is known that, 60% reside in hostel and 40% are day scholars, (not residing in hostel). Previous year results, , OBJECTIVE TYPE QUESTIONS, 1. (d) : Let D1, D2 be the events that we find a defective, fuse in the first and second test respectively., \, , Required probability = P(D1 ∩ D2), , 2 1 1, ⋅ =, 7 6 21, 2. (c) : Let Ei be the event that the first i cards have, no pair among them. Then we want to compute P(E6),, which is actually the same as P(E1 ∩ E2 ∩ ... ∩ E6), since, E6 ⊂ E5 ⊂ ... ⊂ E1, implying that E1 ∩ E2 ∩ ... ∩ E6 = E6., = P(D1 )P(D2 |D1 ) =, , \, , , P(E1 ∩ E2 ∩ ... ∩ E6) = P(E1) P(E2|E1), , 52 48 44 40 36 32, =, = 0.345, 52 51 50 49 48 47, 3. (c) : We have, S = {1, 2, 3, 4, 5, 6}, , P(E3|E1 ∩ E2)...., , Let A be the event that number is even = {2, 4, 6}, and B be the event that number is red = {1, 2, 3}, Now, A ∩ B = {2}, , report that 30% of all students who reside in, hostel attain ‘A’ grade and 20% of day scholars, attain ‘A’ grade in their annual examination., At the end of the year, one student is chosen, at random from the college and he has an ‘A’, grade, what is the probability that the student, is a hosteler?, 39. Consider the experiment of tossing a coin., If the coin shows head, toss it again, but if it, shows tail, then throw a die. Find the conditional, probability of the event that ‘the die shows a, number greater than 4’ given that ‘there is at, least one tail’., 40. In a group of 400 people, 160 are smokers, and non-vegetarian, 100 are smokers and, vegetarian and the remaining are non-smokers, and vegetarian. The probabilities of getting a, special chest disease are 35%, 20% and 10%, respectively. A person is chosen from the group, at random and is found to be suffering from the, disease. What is the probability that the selected, person is a smoker and non-vegetarian?, , ∴ P( A ∩ B) =, , 1, 6, 7, 9, 12, , P(B) =, and P( A ∪ B) =, 13, 13, 13, , 4., , (b) : Given, P( A) =, , ∴, , P(A ∩ B) = P(A) + P(B) – P(A ∪ B), 7, 9 12, 4, +, −, ⇒ P( A ∩ B) =, 13 13 13, 13, P( A ∩ B) 4 / 13 4, =, =, P ( A|B) =, P(B), 9 / 13 9, =, , ∴, , (b) : Given, P( A) = 2 , P(B) = 3 , P( A ∩ B) = 1, 5, 10, 5, Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B), 5., , , \, , =, , 2 3 1 1, +, − =, 5 10 5 2, , P( A′ ∩ B′) = P(( A ∪ B)′) = 1 − P( A ∪ B) = 1 −, , Also, P(B′) = 1 − P(B) = 1 −, , 3, 7, =, 10 10, , 1 1, =, 2 2

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CBSE Board Term-II Mathematics Class-12, , 108, ∴, , P( A′|B′) =, , P( A′ ∩ B′), 1/2 5, =, =, P(B′), 7 / 10 7, , 6. (c) : Required probability = P{(RBB), (BRB), (BBR)}, = P(RBB) + P(BRB) + P(BBR), 5 3 2 3 5 2 3 2 5, 5 15, = × × + × × + × × = 3×, =, 8 7 6 8 7 6 8 7 6, 56 56, 7. (d) : We have, P(A) = 1/4, Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B), = P(A) + P(B) – P(A) P(B)( A, B are independent), = 1/4 + P(B) – (1/4) P(B) = 2P(B) – 1/4, (Given), ⇒ P(B) = 2/5, 8. (b) : P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7), = 0.23 + 0.12 + 0.20 + 0.07 = 0.62, 5, 1, and P( A ∪ B) =, 9, 3, P( A ∩ B), p, Now, P( A|B) =, = p ⇒ P( A ∩ B) =, P(B), 3, Since, P(A ∪ B) = P(A) + P(B) – P(A ∩ B), 5, 1 p, 5 − 3 2p, 2 3 1, ⇒, =p+ −, ⇒, =, ⇒ p= × =, 9, 3 3, 9, 3, 9 2 3, 9., , (d) : We have, P( A) = p, P(B) =, , 10. (d) : Let E1 = event that bag I is selected, E2 = event that bag II is selected, E = event that the ball drawn is of white colour, By rule of total probability,, P(E) = P(E1)·P(E|E1) + P(E2)·P(E|E2), 1 3 1 6 9 1, = ⋅ + ⋅ =, =, 2 9 2 9 18 2, 11. (c) : If two dice are thrown, then total number of, cases = 36, Number of cases of total of 9 or 11, = {(3, 6), (4, 5), (6, 3), (5, 4), (6, 5), (5, 6)}, 6, 1, P(total 9 or 11) =, =, 36 6, P(neither total of 9 nor 11) = 1 – P(total 9 or 11), 1 5, =, 6 6, 12. (b) : By multiplication theorem,, P(A ∩ B) = P(A|B) × P(B) = P(B|A) × P(A), P( A ∩ B), ⇒ P( A|B) =, P(B), 13. (d) : If A and B are two independent events, then, P( A ∩ B) = P( A) × P(B), It is given that P( A ∪ B) = 0.6, P(A) = 0.2, \ P(A ∪ B) = P(A) + P(B) – P( A ∩ B), ⇒ P(A ∪ B) = P(A) + P(B) – P(A) × P(B), ⇒ 0.6 = 0.2 + P(B)(1 – 0.2), ⇒ 0.4 = P(B) (0.8), 0.4, 1, ⇒ P(B) =, ⇒ P(B) = = 0.5, 0.8, 2, =1−, , 14. (c) : P(atleast one of A and B), = P(A ∪ B) = P(A) + P(B) – P(A ∩ B), = P(A) + P(B) – P(A) P(B), [ A, B are independent], = P(A) + P(B) [1 – P(A)] = [1 – P(A′)] + P(B) P(A′), = 1 – P(A′) + P(B) P(A′) = 1 – P(A′) [1 – P(B)], = 1 – P(A′) P(B′), 15. (c) : We have, SP(X) = 1, ⇒, , 5 7 9 11, + + +, =1 ⇒, k k k k, , 32, = 1 ⇒ k = 32, k, , 16. (d) : Let E and F denote respectively the events that, first and second ball drawn are black. We have to find, P(E ∩ F)., 9, 10, Now, P(E) =, , P(F|E) =, 14, 15, By multiplication rule of probability, we have, 10 9 3, P(E ∩ F) = P(E).P(F|E) =, ×, =, 15 14 7, 17. (a) : Let X denote the number of students who, graduated., Now, the probability that a student graduates = 0.4, \ Probability that a student not graduates =,  1 – 0.4, = 0.6, \ P (none will graduate) = (0.6)3 = 0.216, 18. (d) : P( A) = 0.3 ⇒ P( A) = 0.7 , P(B) = 0.4 ⇒ P(B) = 0.6, P( A ∩ B) = 0.5 ⇒ P( A ∩ B) = P( A) − 0.5 = 0.2, P(B|( A ∪ B)) =, =, , P[B ∩ ( A ∪ B)], P( A ∪ B), , P(B ∩ A), 0.2 1, 0.2, =, =, =, P( A) + P(B) − P( A ∩ B) 0.7 + 0.6 − 0.5 0.8 4, , 19. (c) : From the given information, we find that the, probability distribution of X is, X, , 0, , 1, , 2, , 3, , 4, , P(X), , 0.2, , k, , 2k, , k, , 0, , We know that Spi = 1, ⇒ 0.2 + k + 2k + k + 0 = 1 ⇒ 4k = 1 – 0.2 = 0.8, ⇒ k = 0.2, Now, P(you study atleast two hours) = P(X ≥ 2), = P(2) + P(3) + P(4) = 2k + k + 0 = 3k = 3 × 0.2 = 0.6, 3, 2, 3, , P(B) = , P( A ∪ B) =, 10, 5, 5, Now, P(A ∩ B) = P(A) + P(B) – P(A ∪ B), 3 2 3 3+4−6 1, =, + − =, =, 10 5 5, 10, 10, P( A ∩ B) P( A ∩ B), +, Now, P(B| A) + P( A|B) =, P( A), P(B), 20. (d) : Given, P( A) =, , =, , 1 / 10 1 / 10 1 1 7, +, = + =, 3 / 10 2 / 5 3 4 12

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109, , Probability, 21. (c) : Let A, B, C be the respective events of solving, 1, 1, 1, the problem. Then, P( A) = , P(B) = and P(C ) = ., 4, 2, 3, Clearly A, B, C are independent events and the problem, is solved if at least one student solves it., \ Required probability = P(A ∪ B ∪ C), , 30. (b) : Two events A and B are independent, if, P(A ∩ B) = P(A) P(B), \ P(A′ ∩ B′) = P(A ∪ B)′ = 1 – P(A ∪ B), = 1 – [P(A) + P(B) – P(A ∩ B)], = 1 – P(A) – P(B) + P(A) P(B), = [1 – P(A)][1 – P(B)], , = 1 − P( A)P(B)P(C ), , 31. (d) : When A and B are mutually exclusive, then, A ∩ B = f ⇒ P(A ∩ B)= 0, \ P(A ∪ B) = P(A) + P(B) – P(A ∩ B), 3 1, 3 1 6−5 1, ⇒, = +p−0 ⇒ p= − =, =, 5 2, 5 2, 10, 10, , 1, 1, 1, 1 3, , = 1 − 1 −  1 −  1 −  = 1 − =, , , , , , , 2, 3, 4, 4 4, 22. (b) : We know that, SPi = 1, ⇒ a + 6a + 6a + 4a + 8a + 8a + 6a + 9a = 1, 1, ⇒ 48a = 1 ⇒ a =, 48, , 3, 2, 3, , P(B) = , P( A ∪ B) =, 10, 5, 5, Now, P(A ∩ B) = P(A) + P(B) – P(A ∪ B), 3 2 3 3+4−6 1, =, + − =, =, 10 5 5, 10, 10, 32. (b) : Given, P( A) =, , 23. (c) : We have SP(X) = 1, 4 6 10 12, ⇒, + +, +, =1, k k k, k, 32, ⇒, = 1 ⇒ k = 32, k, 24. (b) : ∵, , 4, , ∑, , Now, P(B / A) =, , P(X = x) = 1, , x=0, , ⇒ P(X = 0) + P(X = 1) + P(X = 2), + P(X = 3) + P(X = 4) = 1, ⇒ k + 2k + 4k + 2k + k = 1, ⇒ 10k = 1 ⇒ k = 0.1, \ P(X ≤ 1) = P(X = 0) + P(X = 1), ⇒ P(X ≤ 1) = k + 2k = 3k = 0.3, 25. (c) : By definition, P( A′ | B′) =, =, , P( A′ ∩ B′), P(B′), , P(( A ∪ B)′) 1 − P( A ∪ B), =, P(B′), P(B′), , 26. (d) : Required probability = P(DD) =, 27. (c) : We have, P(B) =, , 3 2 3, × =, 8 7 28, , 3, 1, 4, , P( A|B) = , P( A ∪ B) =, 5, 2, 5, , 1 3 3, ⋅ =, 2 5 10, Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B), 4 3 1, 4, 3 3, ⇒ = P( A) + −, ⇒ P( A) = −, =, 5 10 2, 5, 5 10, P( A ∩ B) = P( A|B)P(B) =, , 28. (d) : P(A) = 0.4, P(B) = 0.3, P(A ∪ B) = 0.5, P(A ∩ B) = P(A) + P(B) – P(A ∪ B), = 0.4 + 0.3 – 0.5 = 0.2, 1, 5, 29. (a) : Required probability = P{GGB, GBG, BGG}, = P(GGB) + P(GBG) + P(BGG), Now, P(B′ ∩ A) = P(A) – P(A ∩ B) = 0.4 – 0.2 = 0.2 =, , =, , 3 2 2 3 2 2 2 3 2, 3, × × + × × + × × =, 8 7 6 8 7 6 8 7 6 28, , P( A ∩ B) 1 / 10 1, =, =, P( A), 3 / 10 3, , 33. (b) : We have, P(A) = 0.2, P(B) = 0.4 and , , P(A ∪ B) = 0.5, P(A ∩ B) = P(A) + P(B) – P( A ∪ B), , = 0.2 + 0.4 – 0.5 = 0.1, (, ), P A ∩ B 0.1 1, Now, P(A/B) =, =, = = 0.25, P(B), 0.4 4, 34. (c) : Total number of possible outcomes = 6C2 = 15, Possible outcomes = 2C1 4C1 = 2 × 4 = 8, 8, \ Required Probability =, 15, 35. (d) : Given, P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6., Clearly, P(A ∩ B) = P(B|A)P(A) = 0.6 × 0.4 = 0.24, Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B), = 0.4 + 0.8 – 0.24 = 0.96, (36- 40) :, Let E be the event that the doctor visit the patient late, and let A 1, A 2, A 3, A 4 be the events that the doctor, comes by cab, metro, bike and other means of transport, respectively., P(A1) = 0.3, P(A2) = 0.2, P(A3) = 0.1, P(A4) = 0.4, P(E|A1) = Probability that the doctor arriving late when, he comes by cab = 0.25, Similarly, P(E | A2) = 0.3, P(E | A3) = 0.35, and P(E | A4) = 0.1, 36. (b) : P(A2|E) = Probability that the doctor arriving, late and he comes by metro, P ( A2 )P(E| A2 ), =, ΣP ( Ai )P(E| Ai ), =, , (0.2)(0.3), (0.3)(0.25) + (0.2)(0.3) + (0.1)(0.35) + (0.4)(0.1), , =, , 0.06 2, =, 0.21 7

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CBSE Board Term-II Mathematics Class-12, , 110, 37. (c) : P(A1|E) = Probability that the doctor arriving, late and he comes by cab, P( A1 )P(E| A1 ), =, ΣP( Ai )P(E| Ai ), (0.3)(0.25), (0.3)(0.25) + (0.2)(0.3) + (0.1)(0.35) + (0.4)(0.1), 0.075 5, =, =, 0.21 14, 38. (d) : P(A3|E) = Probability that the doctor arriving, late and he comes by bike, P( A3 )P(E| A3 ), =, ΣP( Ai )P(E| Ai ), (0.1)(0.35), =, (0.3)(0.25) + (0.2)(0.3) + (0.1)(0.35) + (0.4)(0.1), =, , 0.035 1, =, 0.21 6, 39. (c) : P(A4|E) = Probability that the doctor arriving, late and he comes by other means of transport, P( A4 )P(E| A4 ), =, ΣP( Ai )P(E| Ai ), (0.4)(0.1), =, (0.3)(0.25) + (0.2)(0.3) + (0.1)(0.35) + (0.4)(0.1), , 45. (a) : Required probability = 1 – P(A1 | E), = 1 – 0.0625, = 0.9375 ≈ 0.94, 46. (b) : P(X = 2) = (Probability of not getting six at first, throw) × (Probability of getting six at second throw), 5 1 5, = × = 2, 6 6 6, 5 5 5 1 53, 47. (c) : P(X = 4) = × × × =, 6 6 6 6 64, 48. (c) : P(X ≥ 2) = 1 – P(X < 2), 1 5, = 1 – P(X = 1) = 1 − =, 6 6, 5, , 2,  55  1   5  5, 55 , 5 5, = 6 1 + +   + ...∞  = 6 , = , 6 6,  6 1 − 5   6 , 6 , , 6, , =, , 0.04 4, =, 0.21 21, 40. (a) : Probability that the doctor is late by any means, 2 5 1 4, = +, + +, =1, 7 14 6 21, 41. (d) : Since, it rained only 6 days each year, therefore,, 1, 6, probability that it rains on chosen day is, =, 366 61, 42. (c) : The probability that it does not rain on chosen, =, , 1 60 360, =, =, 61 61 366, 43. (c) : It is given that, when it actually rains, the, weatherman correctly forecasts rain 80% of the time., 6, 363 8 4, 80, ×, +, ×, =, \ Required probability =, 100 366 366 10 5, 44. (a) : Let A1 be the event that it rains on chosen day,, A2 be the event that it does not rain on chosen day and, E be the event the weatherman predict rain., day = 1 −, , Then we have, P(A1) =, P(E | A1) =, , 6, , 366, , , P(A2) =, , 8 and P(E | A )= 2, 2, , 10, 10, Required probability = P(A1 | E), P( A1 ) ⋅ P(E| A1 ), =, P( A1 ) ⋅ P(E| A1 ) + P( A2 ) ⋅ P(E| A2 ), 6, 8, ×, 48 = 0.0625, 366 10, =, =, 8 360 2 768, 6, ×, +, ×, 366 10 366 10, , 360, 366, , ,, , 6, , 1 5, 1, 5, 49. (a) : P(X ≥ 6) =   × +   × + ...∞, 6, , , 6, 6, 6, , 3, , 4, , 50. (d) : P(X ≥ 4) =  5  . 1 +  5  . 1 + ......, 6 6 6 6, =, , 2, , 53   5   5 , 1, +, +, + ......, , , , , , 4, , , , , 6, 6, , 6, , 53, 53  1   53 , =  4  .6 = 3, , 4, 6, 6 1 −  5    6 , 6, , 51. (d) : Reason is a standard result. It is addition, theorem of probabilities. However, the first result is, untrue as we can have, P(E1) + P(E2) > 1, For example, when a dice is rolled once and, E1 : ‘a number < 5’ shows up,, E2 : ‘a number > 1’ shows up, 4 2, 5, then P(E1 ) = = and also P(E2 ) = ., 6 3, 6, Here, P(E1) + P(E2) > 1., =, , 52. (b) : The given system of equations can be written as, a b , x, AX = O, where A = ,  , X = y ., c, d, , ,  , As a, b, c, d ∈ {0, 1}, therefore, matrix A can be chosen in, 24 = 16 ways., The given system has a unique solution namely the zero, solution (i.e. x = 0, y = 0) only if | A | ≠ 0., i.e. if A is any one of the following matrices,  1 0   0 1   1 0   1 1   1 1   0 1,  0 1  ,  1 0  ,  1 1  ,  0 1 ,  1 0  ,  1 1  ., ,  ,  ,  ,  ,  , , \, , P(system has a unique solution) =, , 6 3, = ., 16 8, , Thus, the Assertion is correct., Further, we know that a homogeneous system is always, consistent and admits of atleast one solution namely, the zero solution.

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111, , Probability, So, the probability that the system has a solution is 1., Thus Reason is also correct., However, Reason is not a correct explanation of Assertion., 13 1, 4, 1, 53. (a) : We have, P(E) =, = and P(F ) =, =, 52 4, 52 13, Also, E ∩ F denote the event the card drawn is the ace, of spades., 1, ∴ P(E ∩ F ) =, 52, 1, 1, P, E, F, (, ), ∩, 52, Hence, P(E|F ) =, =, =, 1, 4, P(F ), 13, 1, Since, P(E) = = P(E|F ), we can say that the occurrence, 4, of event F has not affected the probability of occurrence, of the event E. We also have,, 1, 1, P(E ∩ F ) 52, P(F|E) =, =, =, = P(F ), 1, 13, P(E), 4, 1, Again, P(F ) =, = P(F|E) shows that occurrence of, 13, event E has not affected the probability of occurrence, of the event F. Thus, E and F are two events such that the, probability of occurrence of one of them is not affected, by occurrence of the other., Such events are called independent events., 54. (b) : Assertion : We know that,, P(S ∩ F ) P(F ), =, =1, P (S|F ) =, P(F ), P(F ), P(F ∩ F ) P(F ), =, =1, P(F ), P(F ), Thus, P(S | F) = P(F | F) = 1, Reason : We have,, P[( A ∪ B) ∩ F ] P[( A ∩ F ) ∪ (B ∩ F )], P (( A ∪ B)|F ) =, =, P(F ), P(F ), (by distributive law of intersection over union), P( A ∩ F ) + P(B ∩ F ) − P( A ∩ B ∩ F ), =, P(F ), Also, P (F|F ) =, , P( A ∩ F ) P(B ∩ F ) P(( A ∩ B) ∩ F ), =, +, −, P(F ), P(F ), P(F ), = P(A | F) + P(B | F) – P((A ∩ B) | F), 55. (c) : Since, P(A ∩ B) = P(A)P(B), therefore, A and B, are independent events., P( A ∩ B) P( A)P(B), ∴ P( A|B) =, =, = P( A), P(B), P(B), Similarly, P(B|A) = P(B)., Thus, Assertion is correct., However, Reason is not correct for independent events., For example, when a dice is rolled once, then the events, A : ‘an even number’ shows up, , and B : ‘a multiple of 3’ show up, are independent as, 3 2 1, P( A)P(B) = × = = P( A ∩ B), 6 6 6, (... A = {2, 4, 6} and B = {3, 6}), But P(A ∪ B) = P({2, 3, 4, 6}), 4, 3 2 5 4, , = ≠ P( A) + P(B), ∵ P( A) + P(B) = + = ≠ , 6, 6 6 6 6, , 56. (d) : P(Hi | E) > P(E | Hi) × P(Hi), P(H i ∩ E), > P(E ∩ H i ) ⇒ P(H ∩ E)(1 – P(E)) > 0, ⇒, i, P(E), ⇒ P(Hi ∩ E) > 0, This leads to a contradiction 0 > 0 if Hi ∩ E = f for any i., 57. (c) : Assertion : Let E1 be the event of choosing the, bag I, E2 be the event of choosing the bag II and A be, the event of drawing a red ball., 1, Then, P(E1 ) = P(E2 ) =, 2, 3, Also, P(A|E1) = P(drawing a red ball from bag I) =, 7, 5, and P(A|E2) = P(drawing a red ball from bag II) =, 11, Now, the probability of drawing a ball from bag II,, being given that it is red, is P(E2|A), By using Bayes’ theorem, we have, P(E2 )P( A|E2 ), P(E2 | A) =, P(E1 )P( A|E1 ) + P(E2 )P( A|E2 ), 1 5, ×, 35, 2, 11, =, =, 1 3 1 5 68, × + ×, 2 7 2 11, Reason : Let E1, E2 and E3 be the events that boxes I, II, and III are chosen, respectively., 1, Then, P(E1 ) = P(E2 ) = P(E3 ) =, 3, Also, let A be the event that ‘the coin drawn is of gold’, 2, Then, P(A|E1) = P(a gold coin from bag I) = = 1, 2, P(A|E2) = P(a gold coin from bag II) = 0, 1, P(A|E3) = P(a gold coin from bag III) =, 2, Now, the probability that the other coin in the box is, of gold, = the probability that gold coin is drawn from the box I., = P(E1|A), By Bayes’ theorem, we know that, P(E1 )P( A|E1 ), P(E1| A) =, P(E1 )P( A|E1 ) + P(E2 )P( A|E2 ) + P(E3 )P( A|E3 ), 1, ×1, 2, 3, =, =, 1, 1 1 3, 1, ×1+ ×0+ ×, 3, 3, 3 2

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CBSE Board Term-II Mathematics Class-12, , 112, 58. (b) : Assertion : The urn contains 5 red and 5 black, balls., Let E1 : a red ball is drawn in the first attempt and, E2 : a black ball is drawn in the first attempt., 5 1, =, Then, P(E1) = P(E2) =, 10 2, Now, let E : red ball is drawn in second attempt., 7, 5, Then, P(E|E1 ) =, and P(E|E2 ) =, 12, 12, Now, probability of drawing second ball as red is, P(E) = P(E1) ⋅ P(E | E1) + P(E2)⋅ P(E|E2), 1, 1 7 1 5 1 7, 5 1, ×, + ×, =  +  = ×1=, 2, 2 12 2 12 2  12 12  2, Reason : Let E1 : first bag is selected, E2 : second bag, is selected., 1, Then, P(E1 ) = P(E2 ) =, 2, Now, let E : ball drawn is red., Then, P(E|E1) = P (drawing a red ball from first bag), =, , 4 1, =, 8 2, P(E|E2) = P (drawing a red ball from the second bag), =, , ⇒, , 2 1, =, 8 4, \ Required probability, P (E|E1 ) P(E1 ), P (E1|E) =, P (E|E1 ) P(E1 ) + P (E|E2 ) P(E2 ), =, , 1, 1 1, 1, 1, ×, 1 8 2, 2 2, =, = 4 = 4 = 4= × =, 3 4 3 3, 1 1 1 1 1 1 2+1, × + ×, +, 8, 8, 2 2 2 4 4 8, 59. (d) : Probability of solving the problem by A & B is, , = 1 – P (None of them can solve the problem), = 1 − P ( A ∩ B ) = 1 − P ( A) ⋅ P ( B ), , = 1 – [1 – P(A)] [1 – P(B)], 4 3, 13, = 1− × =, ., 5 5, 25, 60. (c) : Reason is false., P(P and Q contradict each other), = p(1 – 2p) + 2p(1 – p) = 1/2, ⇒ 8p2 – 6p + 1 = 0., ⇒ (2p – 1)(4p – 1) = 0, ⇒ p = 1/2, 1/4., , SUBJECTIVE TYPE QUESTIONS, Required probability = P(BWBW) + P(WBWB), 6 10 5 9 10 6 9 5, 90, 45, =, ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ =, =, ., 16 15 14 13 16 15 14 13 728 364, 2. The sample space is, S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6),, (1, 3, 6), ....}, 1., , It can be observed that infinite number of possibilities, occur., 3. Let H and T represent a head and a tail of a coin,, respectively., Also, let red balls are represented by R1, R2 and R3 and, yellow balls are represented by Y1 and Y2. Then, the, sample space,, S = {TR1, TR2, TR3, TY1, TY2, H1, H2, H3, H4, H5, H6}, 4. When three dice are tossed together, then the total, number of possible outcomes = 63 = 6 × 6 × 6 = 216, 5. The sample space for this experiment is, S ={RR, RB, BR, BB}, where R denotes the red ball and, B denotes the black ball., 6. The sample space is S = {HH, HT, TH, TT}, Thus, n(S) = 4, 7. Given that, S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, \ E = S – E = {2, 4, 6}., 8. The sample space S for selecting three bulbs at, random from a lot is given by, S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN }, where D indicates a defective bulb and N a nondefective bulb., 9. The sample space S for the given experiment is, S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2),, (3, 4), (4, 1), (4, 2), (4, 3)}., 3, 1, 2, , P(A ∩ B) =, and P( A) =, 4, 4, 3, 2, 2, 1, Now, P( A) =, ⇒ 1 – P(A) =, ⇒ P(A) = ., 3, 3, 3, Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B), 3 1, 1, 3 1 1 2, ⇒, = + P(B) −, ⇒ P(B) = + − = ., 4 3, 4, 4 4 3 3, 10. Given, P(A ∪ B) =, , 2, 1, 1, , P(B) = , P(C) = ,, 5, 3, 2, 1, 1, Also, P(A ∩ C) =, and P(B ∩ C) =, 5, 4, P(C ∩ B) 1 / 4 3, =, =, \ P(C/B) =, P(B), 1/3 4, 11. We have, P(A) =, , and, P( A ∩ C ) = P( A ∪ C ), 2 1 1 3, = 1 – {P(A) + P(C) – P(A ∩ C)} = 1 –  + −  =,  5 2 5  10, 12. Let E1, E2 and A be the events defined as follows:, E1 = a boy is chosen from the class, E2 = a girl is chosen, from the class and, A = the student gets first class marks., Then, P(E 1) = 2/3, P(E 2) = 1/3, P(A/E 1) = 0.28 and, P(A/E2) = 0.25., Thus, P(A) = P(E1) P(A/E1) + P(E2) P(A/E2) = 0.27, 13. We have, A = {(1, 1)}, B = {(1, 2), (2, 1)}, and C = {(1, 3), (3, 1), (2, 2)}, (i) Since A consists of a single sample point, it is a, simple event.

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113, , Probability, (ii) Since both B and C contain more than one sample, point, therefore each one of them is a compound event., (iii) Since A ∩ B = φ., \ A and B are mutually exclusive events., 14. Total number of possible outcomes = 36, The sum should be prime number i.e., 2, 3, 5, 7, 11., Sum 2 ≡ (1, 1), Sum 3 ≡ (1, 2), (2, 1), Sum 5 ≡ (1, 4), (4, 1), (2, 3), (3, 2), Sum 7 ≡ (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), Sum 11 ≡ (5, 6), (6, 5), \ Number of favourable outcomes = 15, 15 5, Hence, required probability =, =, 36 12, 15. Corresponding to each doublet we get two, outcomes., For example, corresponding to (1, 1) we get (1, 1, H),, (1, 1, T)., Thus, total number of possible outcomes = 36 + 6 = 42., 16. Total number of possible outcomes =, , 8!, 3! 2 !, , Total number of favourable outcomes when (3L’s, 6!, consider as one letter) =, 2!, 3, 6!, =, Hence, required probability =, 8 !/ 3 ! 28, 17. The outcomes in the sample space S are 52., \ n(S) = 52, Let E be the event of getting a red king card, Out of 4 kings, 2 are red and 2 are black, \ n(E) = 2, n(E), 2, 1, Hence, required probability =, =, =, n(S) 52 26, 18. Let S = {1, 2, 3, 4, 5, 6} be the sample space and E, = {2} be event of getting an even prime number., ⇒ n(S) = 6 & n(E) = 1, n(E), 1, Hence, required probability =, =, n(S), 6, 1 1 1, = P( A ∩ B), 19. Here P(A)⋅P(B) = × =, 3 4 12, ⇒ Events A and B are independent., ⇒ Events A and B are also independent., Now P( A ∩ B) = P( A) P(B), ( ∵ A and B are independent events), = (1 – P(A))(1 – P(B)), 1, 1 2 3 1, , = 1 −  1 −  = × =, , , , 3, 4 3 4 2, 20. Let E 1 : “A woman is selected”, E 2 : “A Hindi, knowing person is selected” and E 3 : “A teacher is, , selected”, then P (E1 ) =, , 20 2, 10 1, = , P (E2 ) =, = , and, 50 5, 50 5, , 15 3, =, 50 10, \ Required probability = P(E1 ∩ E2 ∩ E3 ), 2 1 3, 3, =, = P(E1) P (E2) P (E3 ) = × ×, 5 5 10 125, 21. Six employees can be seated in row in six desks in, 6! ways. Married couple can occupy adjacent seats in, the following 5 ways., 1 – 2, 2 – 3, 3 – 4, 4 – 5, 5 – 6, Also, they can interchange their seats and the remaining, 4 seats can be occupied by remaining 4 employees in, 4! ways., \ Number of ways in which married couple will have, adjacent seats = 5 × 2! × 4!, So, number of ways in which married couple will have, non-adjacent seats = 6! – 5 × 2! × 4! = 480, 480 2, =, Hence, required probability =, 720 3, 22. If S be the sample space of tossing a coin three times, then S = {HHH, HHT, HTH, THH, HTT, THT, TTH,, TTT }, Now, A (Event of getting no head) = {TTT }, B (Event of getting exactly one head), = {HTT, THT, TTH }, C (Event of getting atleast two heads), = {HHT, HHH, HTH, THH }, Clearly, A ∩ B = f, B ∩ C = f and A ∩ C = f, So, A, B and C are mutually exclusive events., Also, A ∪ B ∪ C = S, So, A, B and C are exhaustive events., 23. (i) Total number of possible outcomes = 9, Number of red balls = 4, \ Number of favourable outcomes = 4, Hence, required probability = 4/9, (ii) Number of balls which are not blue = 4 + 2 = 6, \ Number of favourable outcomes = 6, Hence, required probability = 6/9 = 2/3, (iii) Number of balls which are either red or blue, =4+3=7, \ Number of favourable outcomes = 7, Hence, required probability = 7/9, 24. (i) Non leap year contains 365 days i.e.,, 52 weeks + 1 day, 52 weeks contain 52 Tuesdays., 53 Tuesdays means the remaining day is a Tuesday., Total number of possibilities for remaining day = 7, Number of favourable outcomes = 1, 1, \ Probability for 53 Tuesdays =, 7, (ii) Leap year contains 366 days = 52 weeks + 2 days, 52 weeks contain 52 Wednesdays., P (E3 ) =

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CBSE Board Term-II Mathematics Class-12, , 114, 53 Wednesdays means that out of 2 remaining days one, will be Wednesday., Total number of possibilities for remaining two days, (SM, MT, TW, WTh, ThF, FSat, SatS) = 7, Number of favourable outcomes = 2, Hence, probability of 53 Wednesdays in a leap, 2, year =, 7, (iii) Total possibilities for remaining two days of leap, year (SM, MT, TW, WTh, ThF, FSat, SatS) = 7, Number of favourable outcomes = 1, 1, Hence, required probability = ., 7, 25. Consider the following events:, Ei = Seed chosen is of type Ai, i = 1, 2, 3, A = Seed chosen germinates., 4, 4, 2, We have, P(E1) =, , P(E2) =, and P(E3) =, 10, 10, 10, 45, 60, 35, P(A/E1) =, , P(A/E2) =, , P(A/E3) =, 100, 100, 100, (i) Required probability = P(A/E3) = 1 – P(A/E3), 35, 65, =1–, =, = 0.65, 100 100, (ii) Required probability = P(A) = P(E1) P(A/E1) + , , P(E2) P(A/E2) + P(E3) P(A/E3), 4, 45, 4, 60, 2, 35, 490, =, ×, +, ×, +, ×, =, = 0.49, 10 100 10 100 10 100 1000, (iii) Required probability = P(E2/A), P(E2 ∩ A) P(E2 )P( A / E2 ) P(E2 )(1 − P( A / E2 )), =, =, =, P( A), P( A), 1 − P( A), 60 , 4 , 4, 40, × 1−, ,  10 × 100 16, 100, 10, =, =, =, 49, 51, 51, 1−, 100, 100, , 2 1, 26. Probability of choosing bag A = P(A)= =, 6 3, 4 2, Probability of choosing bag B = P(B) = =, 6 3, Let E1 and E2 be the events of drawing a red and a black, ball from bag A and B respectively., 6×4, , \, , P(E1) =, , \, , Required probability = P(A) × P(E1) + P(B) × P(E2), 1 6 × 4 2 7 × 3 8 14 22, =, +, =, = × 10, + ×, 3, C 2 3 10C 2 45 45 45, , 10, , C2, , and P(E2) =, , 7×3, 10, , C2, , 27. Let Gi (i = 1, 2) and Bi (i = 1, 2) denote the ith child, is a girl or a boy respectively., Then sample space is,, S = {G1G2, G1B2, B1G2, B1B2}, Let A be the event that both children are girls, B be the, event that the youngest child is a girl and C be the event, , that at least one of the children is a girl., Then A = {G1G2}, B = {G1G2, B1G2}, and C = {B1G2, G1G2, G1B2}, ⇒ A∩B = {G1G2} and A∩C = {G1G2}, (i) Required probability = P(A/B), P( A ∩ B) 1 / 4 1, =, =, =, P(B), 2 /4 2, (ii) Required probability = P(A/C), P( A ∩ C ) 1 / 4 1, =, =, =, P(C ), 3/4 3, 28. Let Bi(i = 1, 2) and Gi(i = 1, 2) denote the ith child, is a boy or a girl respectively., Then sample space is,, S = {B1B2, B1G2, G1B2, G1G2}, Let A be the event that both are boys, B be the event that, at least one of them is a boy and C be the event that the, older child is a boy., A = {B1B2}, B = {G1B2, B1G2, B1B2}, C = {B1B2, B1G2} ⇒ A∩B = {B1B2} and A∩C = {B1B2}, (i) Required probability = P(A/B), P( A ∩ B) 1 / 4 1, =, =, =, P(B), 3/4 3, (ii) Required probability = P(A/C), P( A ∩ C ) 1 / 4 1, =, =, =, P(C ), 2 /4 2, 29. Let A be the event of drawing a red ball in first, draw and B be the event of drawing a red ball in second, draw., ∴ P( A) =, , 3, , C1, , 10, , C1, , =, , 3, 10, , ⇒, , P ( A) =, , 7, 10, , Now, P(B/A) = Probability of drawing a red ball in the, second draw, when a red ball already has been drawn, in the first draw =, , 2, , C1, , 9, , C1, , =, , 2, B 3, , P  =,  A 9, 9, , Required probability = P(A/B), P(B / A) ⋅ P( A), =, P(B / A) ⋅ P( A) + P(B / A) ⋅ P( A), 2 3, ×, 6 2, 9 10, =, =, =, 2 3 3 7, 27 9, ×, + ×, 9 10 9 10, 30. Let E be the event that P speaks truth and F be the, event that Q speaks truth. Then, E and F are independent, 70, 7, 80 4, events such that P(E) =, =, and P(F ) =, =, 100 10, 100 5, P and Q will agree to each other in stating the same fact, in the following mutually exclusive ways:, (I) P speaks truth and Q speaks truth i.e. E ∩ F, (II) P tells a lie and Q tells a lie i.e. E ∩ F., \ P(P and Q agree to each other)

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115, , Probability, = P(E) P(F ) + P(E) P(F ) =, , 7 4 , 7 , × + 1 −  1 −, 10 5 , 10  , , 4, , 5, , 28 3 31 62, =, +, =, =, ., 50 50 50 100, Hence, in 62% of the cases P and Q are likely to agree, in stating the same fact., 31. Let E be the event that A is coming in time;, 3, P(E) =, 7, and F be the event that B is coming in time,, 5, P(F) =, 7, Also E and F are given to be independent events., \ Probability of only one of them coming to the, school in time = P(E) ⋅ P(F ) + P(E) ⋅ P(F ), 3 , 5 , 3 5, = ⋅ 1 −  + 1 −  ⋅, , , , 7, 7, 7 7, 3 2 4 5 26, = ⋅ + ⋅ =, 7 7 7 7 49, 32. Let E1 be the event of getting ghee from shop X, E2, be the event of getting ghee from shop Y and A be the, event of getting ghee of type B., 1, 1, 40 4, \ P(E1) = , P(E2) = , P(A|E1) =, = ,, 2, 2, 70 7, 60, 6, P(A|E2) =, =, 110 11, Using Bayes’ Theorem, we have, P(E2 )P( A|E2 ), P(E2|A) =, P(E1 )P( A|E1 ) + P(E2 )P( A|E2 ), ⇒, , 1 6, ×, 2 11, P(E2|A) =, 1 4 1 6, × + ×, 2 7 2 11, 6, 42, 42 21, = 11 =, =, =, 4 6, 44, +, 42, 86, 43, +, 7 11, , 33. Let I be the event that changes take place to improve, profits., 1, Probability of selection of A, P(A) =, 7, 2, Probability of selection of B, P(B) =, 7, 4, 7, Probability that A does not introduce changes,, P( I /A) = 1 − 0.8 = 0.2, Probability that B does not introduce changes,, P( I /B) = 1 − 0.5 = 0.5, Probability that C does not introduce changes,, P( I / C ) = 1 − 0.3 = 0.7, Probability of selection of C, P(C) =, , So, required probability = P(C / I ), =, , P(C )P( I /C ), P( A)P( I /A) + P(B)P( I /B) + P(C )P( I /C ), , 4, × 0.7, 7, =, = 0.7, 1, 2, 4, × 0.2 + × 0.5 + × 0.7, 7, 7, 7, 34. The probability distribution of X is, X, , 0, , 1, , 2, , 3, , 4, , k, k, P(X), 0, 4k, 2k, The given distribution is a probability distribution., \, , 4, , ∑ pi = 1, , i =0, , 0 + k + 4k + 2k + k = 1 ⇒ 8k = 1, 1, ⇒ k=, = 0.125, 8, (i) P (getting admission in exactly one college), = P(X = 1) = k = 0.125, (ii) P (getting admission in atmost 2 colleges), = P(X ≤ 2) = 0 + k + 4k = 5k = 0.625, (iii) P (getting admission in atleast 2 colleges), = P(X ≥ 2) = 4k + 2k + k = 7k = 0.875, ⇒, , 35. Consider the following events., E : Two balls drawn are white, A : There are 2 white balls in the bag, B : There are 3 white balls in the bag, C : There are 4 white balls in the bag, 1, P(A) = P(B) = P(C) =, 3, P (E/A) =, P (E/C ) =, , 2, , C2, , =, , 4, , C2, , 4, , C2, , 4, , C2, , ∴ P (C /E) =, , 3, 1, 3 1, C, , P (E/B) = 4 2 = =, 6, C2 6 2, , =1, , P(C ). P(E/C ), ., P( A) P(E/A) + P(B). P(E/B) +, P(C ). P(E/C ), , 1, ×1, 3, 3, =, =, 1 1 1 1 1, × + × + ×1 5, 3, 6 3 2 3, , 36. Probability of getting a six by the captains of both, the teams A and B is, 1, P( A) = = P(B), 6, 1 5, \ P( A) = P(B) = 1 − =, 6 6

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CBSE Board Term-II Mathematics Class-12, , 116, Since A starts the game, he can throw a six in the, following mutually exclusive ways :, ( A), ( ABA), ( ABABA), ..., Probability that A wins, = P( A) + P( ABA) + P( ABABA) + ..., , \, , = P( A) + P( A)P(B)P( A) + P( A)P(B)P( A)P(B)P( A) + ..., , 1 5 5 1 5 5 5 5 1, = + ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + ..., 6 6 6 6 6 6 6 6 6, 2, 4, 1 5, 1 1 5, = + ⋅   + ⋅   + ..., 6 6 6, 6 6, , 60, 30, ×, 6×3, 18, 9, 100, 100, =, =, =, =, 60, 30, 40, 20, 6 × 3 + 4 × 2 18 + 8 13, ×, +, ×, 100 100 100 100, , 2, , 1, 5, This is an infinite G.P., with a = and r =  , 6, 6, Hence the probability of the team A winning the match, 1, 6, 6, =, =, 2, 11, 5, 1−, 6, Since the total probability is unity, the probability of, 6, 5, team B winning the match = 1 −, =, ., 11 11, 37. Let A, E1 and E2 respectively be the events that a, person has a heart attack, the selected person followed, the course of yoga and meditation and the person, adopted the drug prescription., 1, 40, P( A) =, = 0.40, P(E1 ) = P(E2 ) =, 2, 100, P(A/E1) = 0.40 × 0.70 = 0.28,, P(A/E2) = 0.40 × 0.75 = 0.30, Probability that the patient suffering from heart attack, followed the course of meditation and yoga is, P(E1 )P( A/E1 ), P(E1 /A) =, P(E1 )P( A/E1 ) + P(E2 )P( A/E2 ), 1, × 0.28, 0.14, 14, 2, =, =, =, 1, 1, +, 0, 14, 0, 15, 29, ., ., × 0.28 + × 0.30, 2, 2, Now, P(E2 /A) =, , P(E2 )P( A/E2 ), P(E1 )P( A/E1 ) + P(E2 )P( A/E2 ), , 1, × 0.30, 0.15, 15, 2, =, =, =, 1, 1, +, 0, 14, 0, 15, 29, ., ., × 0.28 + × 0.30, 2, 2, , 30, 20, ; P(S/E2 ) =, 100, 100, Required probability, P(E1 ) ⋅ P (S/E1 ), = P (E1 /S ) =, P(E1 ) ⋅ P (S/E1 ) + P(E2 ) ⋅ P (S/E2 ), , P(S/E1 ) =, , 39. The sample space S of the given random experiment, is, S = {( H , H ), ( H , T ), ( T , 1), ( T , 2), ( T , 3), ( T , 4),, (T, 5), (T, 6)}, Let A be the event that the die shows a number greater, than 4 and B be the event that there is at least one tail., \ A = {(T, 5), (T, 6)}, and B = {(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6),(H,T)}, A∩B ={(T, 5), (T, 6)}, \ P(B) = P{(T, 1)} + P{(T, 2)} +P{(T, 3)}, + P{(T, 4)} + P{(T, 5)} + P{(T, 6)} + P{(H, T)}, 1, 1, 1, 1, 1, 1 1 3, =, +, +, +, +, +, + =, 12 12 12 12 12 12 4 4, 1 1 1, P( A ∩ B) = P{(T , 5)} + P{(T , 6)} = + =, 12 12 6, , \, , Required probability = P( A/B) =, 1 /6 2, = 3 / 4 = 9, , P( A ∩ B), P(B), , 40. Let A, B, C and E are respectively the events that, a person is smoker and non-vegetarian, smoker and, vegetarian, non-smoker and vegetarian, and the selected, person is suffering from the disease., Here, n(A) = 160, n(B) = 100,, n(C) = 400 – (160 + 100) = 140., 160, 100, 140, Also, P( A) =, , P(B) =, , P(C ) =, 400, 400, 400, and P(E / A) =, , \, , 35, 20, 10, , P(E / B) =, , P(E / C ) =, 100, 100, 100, , Required probability, , = P( A/E) =, , P( A) ⋅ P(E/A), P( A) ⋅ P(E/A) + P(B) ⋅ P(E/B) + P(C ) ⋅ P(E/C ), , 160 35, ×, 400 100, =, 160 35 100 20 140 10, ×, +, ×, +, ×, 400 100 400 100 400 100, , 38. Let E1, E2 and S be the following events :, E1 : The student resides in hostel, E2 : The student is a day-scholar, S : The student attains A grade., 60, 40, P(E1 ) =, ; P(E2 ) =, 100, 100, , =, , , , 5600, 5600 28, =, =, 5600 + 2000 + 1400 9000 45

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PRACTICE PAPER 1, Time allowed : 2 hours, , Maximum marks : 40, , General Instructions :, 1. This question paper contains two parts A and B. Each part is compulsory. Part-A carries 8 marks and Part-B, carries 32 marks., 2., , Part-A has Objective Type Questions and Part-B has Descriptive Type Questions., , 3., , Both Part-A and Part-B have internal choices., , Part - A :, 1., , It consists of two Sections-I and II., , 2., , Section-I comprises of 4 MCQs., , 3., , Section-II contains 1 case study-based questions., , Part - B :, 1., , It consists of four Sections-III, IV, V and VI., , 2., , Section-III comprises of 5 questions of 1 mark each., , 3., , Section-IV comprises of 4 questions of 2 marks each., , 4., , Section-V comprises of 3 questions of 3 marks each., , 5., , Section-VI comprises of 2 questions of 5 marks each., , 6. Internal choice is provided in 1 question of Section-III, 1 question of Section-IV, 1 question of Section-V and 2, questions of section-VI. You have to attempt only one of the alternatives in all such questions., , PART - A, Section - I, 1., , The value of, (a), , 2., , 4., , x, x2 − x, , 2 x +1 + C, , dx is, (b), , 2 x −1 + C, , Solution of the differential equation, , (c), , x +1 + C, , (d) none of these, , dy, = 1 − x + y − xy is, dx, , x3, x3, + C , +C, (b) log | 1 − y | = x +, 7, 3, x2, (c) log | 1 + y | = x −, + C , (d) none of these, 2, , , ^ ^ ^, Find the projection of a = 2 i − j + k on b = i^ − 2 ^j + k^ ., 3, 4, 1, (a), (b), (c), (d), 6, 6, 6, (a), , 3., , ∫, , log | 1 + y | = x +, , 2, , 2, ,  dy  d y, What is the degree of the differential equation 5x   −, − 6 y = log x ?,  dx  dx 2, (a) 1, (b) 2, (c) 3, (d) 4, *The paper is for practice purpose. CBSE has yet not released the official sample paper., So, the pattern is suggestive only. For latest information visit www.cbse.gov.in., , 5, 6

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CBSE Board Term-II Mathematics Class-12, , 118, , Section - II, Case study-based question is compulsory. Attempt any 4 sub parts. Each sub-part carries 1 mark., 5., , A rumour on whatsapp spreads in a population of 5000 people at a rate proportional to the product, of the number of people who have heard it and the number of people who have not. Also, it is, given that 100 people initiate the rumour and a total of 500 people know the rumour after 2 days., Based on the above information, answer the following questions., , (i) If y(t) denote the number of people who know the rumour at an instant t, then maximum value, of y(t) is, (a) 500, (b) 100, (c) 5000, (d) none of these, (ii), , dy, is proportional to, dt, (a) (y – 5000), (b) y(y – 500), , (c), , y(500 – y), , (d) y(5000 – y), , (iii) The value of y(0) is, (a) 100, , (b) 500, , (c), , 600, , (d) 200, , (iv) The value of y(2) is, (a) 100, , (b) 500, , (c), , 600, , (d) 200, , , , (b), , y=, , , , (d), , y=, , (v) The value of y at any time t is given by, (a), , y=, , (c), , y=, , 5000, , e, , −5000kt, , +1, , 5000, , 49e, , −5000kt, , +1, , 5000, , 1 + e 5000kt, 5000, , 49(1 + e −5000kt ), , PART - B, Section - III, , π/ 4, , ∫, , (sec2 x + cosec2 x ) dx, , 6., , Evaluate :, , 7., ,  dy ,  dy , Find the order of the differential equation x +   = 1 +   .,  dx ,  dx , , π /6, 2, , OR, Find the solution of the differential equation, , dy, = x 3 e −2 y ., dx, , 8., , Find the equation of a plane with intercepts 2, 3 and 4 on the X, Y and Z-axes respectively., , 9., , If P(A) =, , 5, 3, 3, , P(B) = and P(A ∪ B) = , then find P(A | B) ., 8, 4, 8

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119, , Practice Paper - 1, , 10. Find the area enclosed by the ellipse, , x2, a2, , +, , y2, b2, , = 1., , Section - IV, ,  , ,  , 11. If a = 2i + 3j − k and b = i + 2 j + 3k , find a × b and | a × b | ., 12. Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line, 5x – 25 = 14 – 7y = 35z., OR, ^, , ^, , ^, , Find the cartesian equation of the plane passing through a point having position vector 2 i + 3 j + 4 k, ^, , ^, , ^, , and perpendicular to the vector 2 i + j − 2 k ., , 1, 1, 1, 13. If A and B are events such that P(A) = , P(B) =, and P(A ∩ B) =, , then find P(not A and, 4, 12, 3, not B)., 14. A box contains N coins, of which m are fair and the rest are biased. The probability of getting, 2, 1, head when a fair coin is tossed is , while it is, when a biased coin is tossed. A coin is drawn, 3, 2, from the box at random and is tossed twice. The first time it shows head and the second time it, shows tail. Find the probability that the coin drawn is fair., , Section - V, 15. Evaluate : ∫, , x2 + 9, x 4 + 81, , dx, OR, , Evaluate :, , ∫x, , 2, , sin 2 x dx, , , ,  , ,    , , 16. Vectors a , b and c are such that a + b + c = 0 and a = 3, b = 5 and c = 7 . Find the angle between, , , a and b . , 17. Three machines E1, E2 and E3 in a certain factory producing electric bulbs, produce 50%, 25% and 25%, respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by, each of machines E1 and E2 are defective and that 5% of those produced by machine E3 are defective. If, one bulb is picked up at random from a day’s production, calculate the probability that it is defective., , Section - VI, 18. If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 pass through a straight line, then, find the value of a2 + b2 + c2 + 2abc., OR, Find the equation of the plane passing through the points (2, 2, –1) and (3, 4, 2) and parallel to, the line whose direction ratios are 7, 0, 6., 19. Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line, y = x and the circle x2 + y2 = 32., OR, Using integration, find the area of the region bounded by the line x – y + 2 = 0, the curve x =, and y-axis., , y

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CBSE Board Term-II Mathematics Class-12, , 120, , 1., , =∫, 2., , (b) : Let I = ∫, , x, x(x − 1), , x, 2, , x −x, , dx = ∫, , (c) : We have, , dx, x −1, , 7., , = 2 x −1 + C, , 3., , dy, , ^, , ^, , ^, , dy, = x 3 e −2 y ⇒ e2y dy = x3dx, dx, e2 y x 4, On integrating, we get, =, + C′, 2, 4, We have,, , ∫ 1 + y = ∫ (1 − x )dx ⇒ log|1 + y|= x −, , , , , x2, +C, 2, ^, , ^, , ⇒ 2 e2y = x4 + C, where C = 4 C′, ^, , (d) : We have, a = 2 i − j + k and b = i − 2 j + k, , ,  a ⋅ b, , Projection of a on b is , |b |,  , ^ ^ ^ ^, ^ ^, a ⋅ b = (2 i − j + k ) ⋅ (i − 2 j + k ) = 2 + 2 + 1 = 5, , | b | = 12 + (−2)2 + 12 = 1 + 4 + 1 = 6, , 5, , ., \ Projection of a on b is, 6, , 4. (a) : Since greatest power of highest order derivative, is 1, therefore degree of the given differential equation, is 1., 5. (i) (c) : Since, size of population is 5000., \ Maximum value of y(t) is 5000., (ii) (d) : Clearly, according to given information,, , dy, = ky(5000 − y ) , where k is th e const ant of, dt, , proportionality., , (iii) (a) : Since, rumour is initiated with 100 people., \ When t = 0, then y = 100, Thus y(0) = 100, (iv) (b) : Since, rumour is spread in 500 people, after, 2 days., \ When t = 2, then y = 500., Thus, y(2) = 500, (v) (c) : We know that, when t = 0, then y = 100, This condition is satisfied by option (c) only., 6., , π/ 4, , ∫, , π /6, , (sec2 x + cosec2 x )dx = [tan x − cot x]ππ//64, ,  1,  2, = (1 − 1) − , − 3 =, .,  3, , 3, , We have, x + , , OR, , dy, = (1 − x )(1 + y ), dx, , dy, ⇒, = (1 − x )dx, 1+ y, ⇒, , 2, ,  dy ,  dy , = 1+  , ,  dx ,  dx , dy, Highest order derivative is, . So, its order is 1., dx, , dx, , 8. As the plane has intercepts 2, 3 and 4 on X, Y and, Z axes respectively., \ The required equation of the plane is, , 9., , x y z, + + = 1 ⇒ 6x + 4y + 3z = 12, 2 3 4, , P(A ∪ B) = P(A) + P(B) – P(A ∩ B), , 3 5 3, 3 1, + − = 1− =, 8 8 4, 4 4, P ( A ∩ B) 1/4 2, ∴ P ( A | B) =, =, = ., P ( B), 5/8 5, , \ P(A ∩ B) =, , 10. We know that, the area enclosed by the ellipse, , x2, , a2, , +, , y2, , b2, , = 1 is pab sq. units., , , , , 11. Given, a = 2i + 3j − k and b = i + 2 j + 3k, , i j k, , , Now, a × b = 2 3 −1, 1 2 3, = (9 + 2) i − (6 + 1)j + (4 − 3)k = 11i − 7 j + k .,  , ∴ | a × b | = (11)2 + (−7)2 + (1)2 = 171, 12. Vector equation of the line passing through, (1, 2, –1) and parallel to the line, 5x – 25 = 14 – 7y = 35z, , x −5 y −2, z, x −5 y −2 z, =, =, or, =, =, −5, 1 / 5 −1 / 7 1 / 35, 7, 1, , is r = (i + 2 j − k ) + λ(7i − 5j + k ), i.e.,, , OR, Here, (x1, y1, z1) = (2, 3, 4), a = 2, b = 1, c = –2, Cartesian equation is, a(x – x1) + b(y – y1) + c(z – z1) = 0, ⇒ 2(x – 2) + 1(y – 3) –2(z – 4) = 0, ⇒ 2x – 4 + y – 3 – 2z + 8 = 0, ⇒ 2x + y – 2z = –1

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121, , Practice Paper - 1, , 1 1 1, 13. Here, P(A) . P(B) = × =, = P ( A ∩ B), 3, , 4, , 12, , ⇒ Events A and B are independent., ⇒ Events A and B are also independent., ,  − cos 2 x ,  − cos 2 x , − ∫ 2x ⋅ , = x2 , dx, ,  2 , , , 2, =, , −1 2, x cos 2 x + ∫ x cos 2 x dx, 2, I II, , =, ,   sin 2 x , −1 2, sin 2 x , x cos 2 x +  x , −∫, dx , , 2, 2,   2 , , , Now, P ( A ∩ B ) = P ( A) P ( B ), (Q A and B are independent events), = (1 – P(A))(1 – P(B)), ,  1 1 2 3 1, = 1 −  1 −  = × =,  3 4 3 4 2, 14. Let E be the event that the coin tossed twice shows, first head and then tail and F be the event that the coin, drawn is fair., , P (F ) . P (E/F ), , P (F / E ) =, , P (F ) . P (E/F ) + P (F ) P (E/F ), m.1.1, m/4, N 2 2, =, =, m . 1 . 1 N − m . 2 . 1 m/4 + 2(N − m)/9, +, N 2 2, N 3 3, 9m, =, ., m + 8N, x2 + 9, 1 + 9 / x2, dx ⇒ I = ∫, dx, 15. Let I = ∫, 81, x 4 + 81, x2 +, x2, 2, 1+ 9 / x, ⇒ I= ∫, dx, 2, 9, , , x 2 +   − 18 + 18, x, =∫, , 1 + 9 / x2, 2, , 9, ,  x −  + 18, x, , Let x −, , dt, 2, , t + 18, , ⇒ I=∫, , dt, ,  t , tan −1 , +c,  3 2 , 3 2, , ⇒ I=, ,  x2 − 9 , tan −1 , +c,  3 2x , 3 2, 1, , OR, , ∫I, , II, , 2, , t 2 + (3 2 ), , 1, , Let I = x sin 2 x dx, , ⇒ cos θ =, , 15 1, π, = ⇒ θ=, = 60°, 30 2, 3, , 17. Let A be the event that the bulb is defective., , 50, 25, 25, , P(E2) =, , P(E3) =, 100, 100, 100, 4, 4, 5, P(A/E1) =, , P(A/E2) =, , P(A/E3) =, 100, 100, 100, \ P(E1) =, , P(A) = P(E1)P(A/E1) + P(E2)P(A/E2) + P(E3)P(A/E3), , ⇒ I=, , 2, , ⇒ 2 × 3 × 5 × cosq = 49 – 34 = 15, , \ Required probability,, , dx, , 9, 9 , , = t ⇒ 1 +,  dx = dt, x, , x2 , , ∴ I=∫, , −1 2, x sin 2 x 1, x cos 2 x +, + cos 2 x + c, 2, 2, 4, 2, cos 2 x, −x, x, ∴ I=, cos 2 x + sin 2 x +, +c, 2, 2, 4, , , ,   , 16. Given a + b + c = 0 and a = 3, b = 5, c = 7,   , We have a + b + c = 0, ,  2, 2,  , ⇒ a + b = −c ⇒ a + b = −c,  , 2, 2 2, ⇒ a + b + 2(a ⋅ b ) = c,  , ⇒ 9 + 25 + 2 a b cos θ = 49, =, , =, , 50, 4, 25, 4, 25, 5, ×, +, ×, +, ×, 100 100 100 100 100 100, , =, , 200 + 100 + 125, 425, 17, =, =, 10000, 10000 400, , 18. Given planes are, x – cy – bz = 0, ...(i), cx – y + az = 0, ...(ii), bx + ay – z = 0, ...(iii), ,, The d.r. s of normal to plane (i), (ii) and (iii) are, (1, –c, –b), (c, –1, a) and (b, a, –1) respectively., All planes pass through same line, then the line is, perpendicular to each of the three normals., The d.r’s. of line from planes (i) and (ii) are, , −c −b, 1, ,−, −1 a, c, , −b 1, ,, a, c, , −c, −1, , i.e., – ac – b, – a – bc, –1 + c2...(iv)

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CBSE Board Term-II Mathematics Class-12, , 122, ,, The d.r. s of line from planes (ii) and (iii) are, , −1 a, c, ,−, a −1, b, , \ Required area = Area BACB + Area OABO, , −1, a, , a, c, ,, −1 b, , =, , i.e., 1 – a2, c + ab, ac + b...(v), ,, The d.r. s in (iv) and (v) are in proportion, then, , −a − bc −1 + c, =, =, 2, c + ab, ac + b, 1− a, −ac − b −a − bc, ⇒, =, c + ab, 1 − a2, , OR, The equation of a plane passing through (2, 2, –1) is, a(x – 2) + b(y – 2) + c(z + 1) = 0 , ...(i), This plane also passes through (3, 4, 2)., \ a(3 – 2) + b(4 – 2) + c(2 + 1) = 0, ⇒ a + 2b + 3c = 0 , ...(ii), Now, plane (i) is parallel to the line whose direction, ratios are 7, 0, 6, Therefore, 7a + 0(b) + 6c = 0 , ...(iii), Solving (ii) and (iii) by cross-multiplication method,, we get, , =, , −ac − b, , ∫, , 4, , =, , 2, , 4 2, , 4, , y1dx + ∫ y2dx =, 0, , 4, , 4, , 32 − x 2 dx + ∫ x dx, 0, , 4 2, , 2, 4, 2, ∫ (4 2 ) − x dx + ∫0 x dx, 4 2, ,  x 32 − x 2 32, ,  x , =, + sin −1 ,  4 2   4, 2, 2, , 4 2 ×0, 2, , π , π, π π, = 16 ⋅ −  8 + 16 ⋅  + 8 = 16  −  = 4 π sq.units, , , 2 4, 2, 4, OR, We have curves x – y + 2 = 0 and x =, , a, b, c, ⇒, =, =, = λ (say), 12 15 −14, , x – x2 + 2 = 0, , 19. The given equation of the circle is x2 + y2 = 32 and, the line is y = x, These intersect at A(4, 4) in the first quadrant. The, required area is shown shaded in the figure. Points, B(4, 0) and C 4 2 , 0, , y., , x = y ⇒ y = x 2, which is a parabola with vertex, at origin., , ⇒ a = 12l, b = 15l, c = –14l, Substituting the values of a, b, c in (i), we get, 12l(x – 2) + 15l(y – 2) – 14l(z + 1) = 0, ⇒ 12x – 24 + 15y – 30 – 14z – 14 = 0, [Q l ≠ 0], ⇒ 12x + 15y – 14z = 68, which is the required equation, of plane., , 4, ,  x2 , + ,  2 0, ,  4×4, 1  1 2, + 16 sin −1 1 − , + 16 sin −1,  + (4 − 0),  2, 2 2, , a, b, c, =, =, (2) (6) − (0) (3) (7) (3) − (6) (1) (0)(1) − (2) (7), , ), , ∫, , 4, , ⇒ ac2 + bc + a2bc + ab2 = a + bc – a3 – a2bc, ⇒ c2 + abc + b2 = 1 – a2 – abc, ⇒ a2 + b2 + c2 + 2abc = 1, , (, , 4 2, , From the given equations, we get, ⇒ (x – 2) (x + 1) = 0, ⇒ x = 2 or x = –1, ⇒ x=2, [Q x ≠ – 1, x is positive], When x = 2, y = 4, So, the point of intersection is (2, 4)., 2, , 2, , 0, , 0, , 2, \ Required area = ∫ ( x + 2)dx − ∫ x dx, 2, , = ∫ (x + 2 − x 2 )dx, 0, , 2, , 3,  2, 8 10, =  x + 2 x − x  = 2 + 4 − = sq. units, 3 3, 3 0, 2, , 

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PRACTICE PAPER 2, Time allowed : 2 hours, , Maximum marks : 40, , General Instructions :, 1. This question paper contains two parts A and B. Each part is compulsory. Part-A carries 8 marks and Part-B, carries 32 marks., 2., , Part-A has Objective Type Questions and Part-B has Descriptive Type Questions., , 3., , Both Part-A and Part-B have internal choices., , Part - A :, 1., , It consists of two Sections-I and II., , 2., , Section-I comprises of 4 MCQs., , 3., , Section-II contains 1 case study-based questions., , Part - B :, 1., , It consists of four Sections-III, IV, V and VI., , 2., , Section-III comprises of 5 questions of 1 mark each., , 3., , Section-IV comprises of 4 questions of 2 marks each., , 4., , Section-V comprises of 3 questions of 3 marks each., , 5., , Section-VI comprises of 2 questions of 5 marks each., , 6. Internal choice is provided in 1 question of Section-III, 1 question of Section-IV, 1 question of Section-V and 2, questions of section-VI. You have to attempt only one of the alternatives in all such questions., , PART - A, Section - I, 1., ,  d3 y , Find the degree of the differential equation , ,  dx 3 , (a) 1, , 2., , 3., , Evaluate :, , (b) 2, , ∫, , 2 /3, , +5−2, , (c), , 3, , d2 y, dx 2, , =0., (d), , 2, 3, , dx, x 2 − 3x + 2, , (a), , 3, , log  x −  + x 2 − 3x + 2 + C, , 2, , (b), , 3, log x + + x 2 − 3x + 2 + C, 2, , (c), , 3, log x − − x 2 − 3x + 2 + C, 2, , (d), , 3, log x + − x 2 − 3x + 2 + C, 2, , , ,    , If a − b = a = b = 1, then find the angle between a and b ., (a), , π, 2, , (b), , π, 6, , (c), , π, 3, , (d), , *The paper is for practice purpose. CBSE has yet not released the official sample paper., So, the pattern is suggestive only. For latest information visit www.cbse.gov.in., , π, 4

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CBSE Board Term-II Mathematics Class-12, , 124, 4., ,  e −2 x, y  dx, Find the integrating factor of the differential equation , −,  = 1.,  x, x  dy, (a), , e x, , (b), , e2 x, , (c), , ex, , (d) ex/2, , Section - II, Case study-based question is compulsory. Attempt any 4 sub parts. Each sub-part carries 1 mark., 5., , In a family there are four children. All of them have to work in their family business to earn their, livelihood at the age of 18., , Based on the above information, answer the following questions., (i) Probability that all children are girls, if it is given that elder child is a boy, is, (a) 3/8, (b) 1/8, (c) 5/8, (d) none of these, (ii) Probability that all children are boys, if two elder children are boys, is, (a) 1/4, (b) 3/4, (c) 1/2, (d) none of these, (iii) Find the probability that two middle children are boys, if it is given that eldest child is a girl., (a) 0, (b) 3/4, (c) 1/4, (d) none of these, (iv) Find the probability that all children are boys, if it is given that at most one of the children is a, girl., (a) 0, (b) 1/5, (c) 2/5, (d) 4/5, (v) F,  ind the probability that all children are boys, if it is given that at least three of the children are, boys., (a) 1/5, (b) 2/5, (c) 3/5, (d) 4/5, , PART - B, Section - III, 6., , Evaluate :, , 1, , ∫, , 0, , {, , e x + sin, , }, , πx, dx, 4, , 7., , Find the area enclosed between the curve x2 + y2 = 16 and the coordinate axes in the first quadrant., , 8., , Find the direction cosines of the line that makes equal angles with the three axes in space., OR, Find the vector equation of the symmetrical form of equation of straight line, , 9., ,  dy , Solve the differential equation sin   = a .,  dx , , x −5 y +4 z −6, ., =, =, 3, 7, 2

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125, , Practice Paper - 2, , 10. The probability distribution of a random variable X is given below :, X, , 2, , 3, , 4, , 5, , P(X), , 5, k, , 7, k, , 9, k, , 11, k, , Find the value of k., , Section - IV, , dy, = 2 xy., dx,    ,      , 12. If a + b + c = 0, then prove that a × b = b × c = c × a ., 11. Solve the differential equation ( x − 1), , OR, π, If the angle between i + k and i + j + ak is , then find the values of a., 3, 13. Find the solution of the differential equation, , 2, 2, dy x + y + 1, satisfying y(1) = 1., =, 2 xy, dx, , 14. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing, through P and perpendicular to OP., , Section - V, 15. Probability of solving specific problem independently by A and B are, try to solve the problem independently, find the probability that, (i) The problem is solved, (ii) exactly one of them solved the problem., , 1, 1, and respectively. If both, 2, 3, , OR, Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random., What is the probability of this person being male ? Assume that there are equal number of males, and females., 16. Find the area lying in the first quadrant bounded by the circle x2 + y2 = 4 and the lines x = 0 and, x = 2., , , , 17. If a = 2i − 3j + k , b = −i + k , c = 2 j − k are three vectors, find the area of the parallelogram having,  ,  , diagonals (a + b ) and (b + c )., , Section - VI, , , 18. Find the equation of the plane containing the lines r = i + j + λ (i + 2 j − k ) and r = i + j + µ (−i + j − 2 k )., Find the distance of this plane from origin and also from the point (1, 1, 1)., OR, , Find the length of the perpendicular drawn from the point (2, 4, –1) to the line r = i + λ(2i + j + 2k ) ., 2, dx, 19. Evaluate : ∫1, x(1 + x 2 ), , OR, Evaluate :, , ∫, , 3x + 1, 5 − 2x − x 2, , dx

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CBSE Board Term-II Mathematics Class-12, , 126, , 1., ,  d3 y , (b) : We have, , ,  dx 3 , 2, , ⇒, , 2 /3, , =2, , d2 y, dx 2, , ∴ P (E|F ) =, , −5, , 3, ,  d3 y ,  d2 y, ,  3  = 2 2 − 5 ,  dx ,  dx, , , (On cubing both sides), , Clearly, degree is 2., [∵ Power of highest order derivative is 2], 2., , =∫, , 3., , (a) : We have,, , ∫, , dx, 2, , 3 1, ,  x −  −  , 2, 2, , 2, , dx, x 2 − 3x + 2, , =∫, , dx, 9 1,  2,  x − 3x +  −, 4, 4, , 3, , = log  x −  + x 2 − 3x + 2 + C, , 2, ,  , , , , , , (c) : Given, a − b = a = b = 1, ,  2, , 2, , 2, ,  , ,  , ⇒ 1 = 1 + 1 − 2 a b cosθ, , , , , , (where q is angle between a and b ), , 4., , π, 1, ⇒ θ=, 2, 3, , (b) : We have,, , dy 1, e −2 x, +, y=, dx, x, x, , This is a linear differential equation of the form, , dy, 1, e −2 x, + Py = Q, where P =, ,Q =, dx, x, x, 1, , \, , (ii) (a) : Let E = All are boys., \ E = {BBBB} i.e., n(E) = 1, F = Two elder children are boys, \ F = {BBBB, BBBG, BBGB, BBGG} i.e., n(F) = 4, Now, n(E ∩ F) = 1, , ∴ P (E|F ) =, , n(E ∩ F ) 1, =, n(F ), 4, , (iii) (c) : Let E = Two middle children are boys., \ E = {BBBB, BBBG, GBBB, GBBG} i.e., n(E) = 4, F = Eldest child is a girl, \ F = {GBBB, GBBG, GBGB, GBGG, GGBB, GGBG,, GGGB, GGGG} i.e., n(F) = 8, Now, n(E ∩ F) = 2, , ∴ P (E | F ) =, , Now, a − b = a + b − 2a ⋅ b, , ⇒ cosθ =, , n(E ∩ F ), =0, n(F ), , ∫ dx, Pdx, = e2 x, I.F. = e ∫, ⇒ I.F. = e x, , 5. Let B and G denote the boy and girl respectively., If a family has 4 children then each of four children can, either boy or girl., Sample space is given by, S = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB,, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG,, GGGB, GGGG}, (i) (d) : Let E = All children are girls., \ E = {GGGG} i.e., n(E) = 1, F = Elder child is a boy, \ F = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG,, BGGB, BGGG} i.e., n(F) = 8, Now, n(E ∩ F) = 0, , 2 1, =, 8 4, , (iv) (b) : Let E = All are boys., \ E = {BBBB} i.e., n(E) = 1, F = At most one child is girl., \ F = {BBBB, BBBG, BBGB, BGBB, GBBB}, i.e., n(F) = 5, Now, n(E ∩ F) = 1, , ∴ P (E | F ) =, , 1, 5, , (v) (a) : Let E = All are boys., \ E = {BBBB} i.e., n(E) = 1, F = At least three of the children are boys., \ F = {BBBB, BBBG, BBGB, BGBB, GBBB} i.e., n(F) = 5, Now, n(E ∩ F) = 1, , ∴ P (E | F ) =, , 1, 5, , We have,, , 6., , 1, , ∫, , 0, , 1, , = e x 0 +, , {, , e x + sin, 1, , 4, πx , 4, 4, − cos  = e − 1 −, +, , π, 4 0, 2π π, , 7. Given curve is a circle, with centre (0, 0) and, radius 4., \ Required area, 4, , = ∫ 16 − x 2 dx, 0, , }, , πx, dx, 4

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127, , Practice Paper - 2, 4, , 16, x, x, =  16 − x 2 + sin −1  = 4 π sq. units, 2, 4 0, 2, 8., , Since l = m = n and l2 + m2 + n2 = 1, , 1, , ⇒ l =m=n=±, , OR, , x − x1 y − y1 z − z1, is, =, =, a, b, c, , The vector form of, , , r = (x1i + y1 j + z1 k ) + λ(ai + b j + ck ), \, , Required equation in vector form is, , , r = (5i − 4 j + 6k ) + µ(3i + 7 j + 2k ), 9., ,  dy , We have, sin   = a,  dx , , ⇒, , dy, = sin −1 a ⇒ dy = sin −1 a dx, dx, , ⇒, , −1, , a dx, , π (i + k ) ⋅ (i + j + ak ), We have, cos =, , 3, 2, 2 1+1+ a, , ⇒, , 32, 5 7 9 11, =1, + + + =1 ⇒, k, k k k k, , ⇒, , k = 32, , dy, = 2xy, dx, dy 2 xy, ⇒ (x – 1)dy = 2xydx ⇒, =, dx x − 1, 11. We have, (x – 1), , x −1+1, dx, x −1, , ⇒, , ∫ y dy = 2∫ x − 1 dx = 2∫, , ⇒, , log | y | + C = 2 [x + log |x – 1|], OR, , 1 (1 + a)2, 1, 1+ a, ⇒, =, =, 4 2(2 + a2 ), 2, 2 2 + a2, , ⇒, , 2 + a2 = 2(1 + a2 + 2a) ⇒ a2 + 4a = 0 ⇒ a = 0, –4, , ⇒, , a = 0 as a = –4 doesn't satisfy the given condition., 2, 2, dy x + y + 1, =, 2 xy, dx, , ⇒, ⇒, ⇒, , 2xydy = (x2 + y2 + 1)dx ⇒ 2xydy – y2dx = (x2 + 1)dx, xd(y2) – y2dx = (x2 + 1)dx, , xd( y 2 ) − y 2dx, x2, ,  y2 , 1 , 1, , , = 1 +  dx ⇒ d   = d  x − , 2, , , , x, x,  x , , y2, 1, = x − + C ⇒ y2 = x2 – 1 + Cx, x, x, Now, given that y(1) = 1, \ 1=1–1+C⇒C=1, Thus, curve becomes y2 = x2 – 1 + x, 14. The direction ratios of OP are < 1 – 0, 2 – 0, – 3 – 0 > i.e., <1, 2, –3 >, ∴ The equation of the plane passing through P and, perpendicular to OP is, (1) (x – 1) + (2) (y – 2) + (– 3) (z + 3) = 0, ⇒ x – 1 + 2y – 4 – 3z – 9 = 0, ⇒ x + 2y – 3z – 14 = 0, 15. Let X and Y denote the respective events of solving, the given specific problem by A and B,, , 1, 1, and P (Y ) =, 2, 3, , dy, 5dy, We have, 5, = e xy 4 ⇒, = exdx, dx, y4, , then P(X) =, , On integrating both sides, we get, , = P ( X ∪ Y ) = 1 − P ( X ) P (Y ), , ∫, , 5. y, , −4, , x, , dy = ∫ e dx, , y −3, −5, = ex + C ⇒, = ex + C, 3, (−3), 3y,      ,    , 12. Given, a + b + c = 0 ⇒ a × (a + b + c ) = a × 0,       ,    , ⇒ a ×a +a ×b +a ×c = 0 ⇒ a ×b = c ×a, ⇒, , 5., ,  , , a ⋅b , ∵ cos θ =  .  , |a | |b |, , , ⇒, , 10. Clearly, SP(X) = 1, , x, ,  , , Integrating both sides, we get, , y = x⋅sin–1a + c, , 1, ,   , ,       ,    , ⇒ b ×a +b ×b +b ×c = 0 ⇒ a ×b =b ×c,      , Thus, we get a × b = b × c = c × a, , 13. Given,, , On integrating both sides, we get, , ∫ dy = ∫ sin, , , , , , OR, , ., , 3, ,   , , Again, a + b + c = 0 ⇒ b × (a + b + c ) = b × 0, , (i), , P (problem is solved), , 1 2 2,  1 1, = 1 − 1 −  1 −  = 1 − × = .,  2  3, 2 3 3, , (ii) P (Exactly one of A and B solves the problem), , P ( X ) ⋅ P (Y ) + P ( X ) ⋅ P (Y ), 1  1  1 1 1 2 1 1 1 2 1 1, = 1 −  + 1 −  ⋅ = ⋅ + ⋅ =  +  =, 2  3  2 3 2 3 2 3 2 3 3 2

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CBSE Board Term-II Mathematics Class-12, , 128, OR, , i j k, , , b1 × b2 = 1 2 −1 = (− 4 + 1)i − (−2 − 1)j + (1 + 2)k, −1 1 −2, , , ⇒ b1 × b2 = − 3i + 3j + 3k, , Let E1 be the event of choosing a male,, E2 be the event of choosing a female and, A be the event that a person has grey hair., Then, P(E1) = P (E2 ) =, , 1, 2, , Therefore, the equation of plane is,, , , (r − (i + j )) ⋅ (−3i + 3j + 3k ) = 0, , , r ⋅ (−3i + 3j + 3k ) + 3 − 3 = 0 ⇒ r ⋅ (−i + j + k ) = 0, , 5, 0.25, and P ( A/E1 ) =, , P ( A/E2 ) =, 100, 100, , ⇒, This is the equation of the required plane., We know that the distance of a point P with position, , Required probability, , \, , = P (E1 /A) =, , P (E1 ) P ( A/E1 ), , , , P (E1 ) P ( A/E1 ) + P (E2 ) P ( A/E2 ), , 1, 5, ×, 5, 5, 500 20, 2 100, =, =, =, =, =, 1, 5 1 0.25 5 + 0.25 5.25 525 21, ×, + ×, 2 100 2 100, 16. Required area, , Y, , 2, , = ∫ 4 − x 2 dx, , Now, distance from origin is 0 and distance from the, point (1, 1, 1) i.e., i + j + k is, , |(i + j + k ) ⋅ (−i + j + k )|, , (0, 2), , (0, 0), , 0, , 12 + 12 + 12, (2, 0) X, ,  , a + b = 2i − 3j + k + ( −i + k ) = i − 3j + 2k ,,  , b + c = (−i + k ) + 2 j − k = −i + 2 j, , (, , ), , (, , ), , i, j,  ,  , ∴ (a + b ) × (b + c ) = 1 −3, −1 2, , k, 2 = −4i − 2 j − k, 0, ,  , , \ Area of a parallelogram whose diagonals are a + b, ,  , and b + c, =, =, ,   1, 1  , a + b × b + c = −4i − 2 j − k, 2, 2, , (, , ) (, , ), , 1, (−4 )2 + (−2)2 + (−1)2 = 21 sq. units., 2, 2, , 18. We are given the equation of lines as, , , r = i + j + λ (i + 2 j − k ) and, , r = i + j + µ (−i + j − 2k ), We know that the equation of plane containing, , =, , | −1 + 1 + 1 |, 3, , =, , 1, 3, , units, , OR, Let M be the foot of the perpendicular drawn from, , , P (2i + 4 j − k ) on the line r = i + λ(2i + j + 2k ), Let the position vector of M be i + λ(2i + j + 2k ), , 2, , x, 4,  x , =, 4 − x 2 + sin −1   ,  2  0, 2, 2, π, = 0 + 2 sin −1 (1) − 0 = 2 × = π sq. uniits., 2,     , , 17. Here, a = 2i − 3 j + k ; b = −i + k ; c = 2 j − k, \, ,  , | a ⋅n − d |, , |n |, ,  , , vector a from the plane r ⋅ n = d is given by, , , , ,  ,   ,  , r = a + λb1 and r = a + λb2 is (r − a ) ⋅ (b1 × b2 ) = 0, , , , Here, a = i + j, b1 = i + 2 j − k , b2 = − i + j − 2k, , = (1 + 2 λ)i + (λ)j + (2 λ)k ( M lies on the line), , Then PM = [(1 + 2 λ)i + (λ)j + (2 λ)k ] − (2i + 4 j − k ), , ∴ PM = (−1 + 2 λ)i + (−4 + λ)j + (1 + 2 λ)k, , Since PM is perpendicular to the given line which is, , parallel to b = (2i + j + 2k ),  , ∴ PM ⋅ b = 0, , , ,   , i.e., [(−1 + 2 λ)i + (−4 + λ) j + (1 + 2 λ)k] ⋅ (2i + j + 2k ) = 0, ⇒, , 2(–1 + 2l) + 1 (–4 + l) + 2(1 + 2l) = 0, , ⇒, , 9l – 4 = 0 ⇒ λ =, , 4, 9, , Putting the value of l, we obtain the position vector of, ,  17  4  8  , i + j + k, 9, 9, 9 , , M as , , , , , Now, PM =, , ∴, , −1  32  17 , i− j+ k, 9, 9, 9, , , Required length PM =, , , , =, 2, , 19. Let I =, , dx, , ∫ x (1 + x 2 ), 1, , 1 1029 289, +, +, 81 81, 81, , 1319, units, 9

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129, , Practice Paper - 2, Consider,, , 1, , x (1 + x 2 ), , =, , where px + q is given by, , A Bx + C, +, x 1 + x2, , ⇒ 1 = A (1+ x 2 ) + (Bx + C ) ⋅ x, ⇒ 1 = x2(A + B) + Cx + A, On equating the coefficient of x2, x and the constant term, from both sides, we get A = 1, B = –1 and C = 0, 2, , \, , 2, , −x, 1, dx, I = ∫ dx + ∫, x, 1 + x2, 1, , 2, , = [ log x ]1 − ∫, , 2, , x, , 11+ x, , 2, , dx = log 2 − ∫, , x dx, , 1 1+ x, , 2, , I = log 2 −, , 5, , 1 1, 1, 5, dt = log 2 −  log t , ∫, 2, 2 t, 2, , 5 − 2x − x 2, , ∫, , 3x + 1 = A( – 2 –2x) + B, , ⇒, , 3x + 1 = – 2Ax + (–2A + B), , 3, 2, , ⇒, , B=–2, , \, , Given integral can be rewritten as, , 1, , − +1 , 3  (5 − 2 x − x 2 ) 2 , 2, =− , dx, −∫, 1, 2, , ( 6 )2 − (x + 1)2, − +1, , , 2, , dx, , This integral is of the form, , ⇒, , 3, − (−2 − 2 x ), −2, I=∫ 2, dx + ∫, dx, 5 − 2x − x 2, 5 − 2x − x 2, , OR, 3x + 1, , d, (5 − 2 x − x 2 ) + B, dx, ,  3, 2, , 1, = log 2 − [log 5 − log 2], 2, 1, 5 1, 8, = log 2 − log   = log  , , , 5, 2, 2, 2, , ∫, , 3x + 1 = A, , and 1 = −2 A + B ⇒ 1 = −2  −  + B,  , , 2, , Let I =, , Now, we write given integrand as, , 3 = −2 A ⇒ A = −, , Put 1 + x2 = t ⇒ 2x dx = dt, When x = 1, t = 2 and when x = 2, t = 5, \, , d, (ax 2 + bx + c) + B, dx, , On equating the coefficients of x and constant term both, sides, we get, , 1, , 2, , px + q = A, , px + q, ax 2 + bx + c, , dx ,, , =−, , , ,  x +1, 3 ( 5 − 2x − x 2 ), − 2 sin −1 , +C,  6 , 2, 1/ 2

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PRACTICE PAPER 3, Time allowed : 2 hours, , Maximum marks : 40, , General Instructions :, 1., ,  is question paper contains two parts A and B. Each part is compulsory. Part-A carries 8 marks and Part-B carries, Th, 32 marks., , 2., , Part-A has Objective Type Questions and Part-B has Descriptive Type Questions., , 3., , Both Part-A and Part-B have internal choices., , Part - A :, 1., , It consists of two Sections-I and II., , 2., , Section-I comprises of 4 MCQs., , 3., , Section-II contains 1 case study-based questions., , Part - B :, 1., , It consists of four Sections-III, IV, V and VI., , 2., , Section-III comprises of 5 questions of 1 mark each., , 3., , Section-IV comprises of 4 questions of 2 marks each., , 4., , Section-V comprises of 3 questions of 3 marks each., , 5., , Section-VI comprises of 2 questions of 5 marks each., , 6., , Internal choice is provided in 1 question of Section-III, 1 question of Section-IV, 1 question of Section-V and 2, questions of section-VI. You have to attempt only one of the alternatives in all such questions., , PART - A, Section - I, 1., , Evaluate :, , tan x dx, , (b) 2, , −1, , x, , (b) e cos, , −1, , x, , (d), , dy, y, +, = x., dx, 1 − x2, 1 − x2, , (c), , 1, log 2, 2, , 1, , (d), , 1 − x2, , ^, , →, ^, ^, Find the unit vector in the direction of vector a = 2 i + 3 j + 4 k ., , (a) (2i + 3j + 9k ) / 29, 4., , (c) log 2, , Find the integrating factor of the differential equation, (a) e sin, , 3., , ∫, 0, , (a) 1, 2., , π/ 4, , (b) 2i + 3j + 9k, , (c) −(2i + 3j + 9k ) / 29 (d) (2i + 3j + 9k ) / 25, , Two dice are thrown together. What is the probability that the sum of the numbers on the two, faces is neither 9 nor 11 ?, (a), , 1, 6, , (b), , 5, 6, , (c), , 2, 3, , (d), , *The paper is for practice purpose. CBSE has yet not released the official sample paper., So, the pattern is suggestive only. For latest information visit www.cbse.gov.in., , 1, 3

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131, , Practice Paper - 3, , Section - II, Case study-based question is compulsory. Attempt any 4 sub parts. Each sub-part carries 1 mark., 5., , Ishaan left from his village on weekend. First, he travelled up to temple. After this, he left for the, zoo. After this he left for shopping in a mall. The positions of Ishaan at different places is given, in the following graph., y, 10, 9, 8, , D, , 7, 6, , Shopping, mall, , Zoo, C, , 5, 4, 3, , B, , Village, A, , 2, , Temple, , 1, x', , O, , y', , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9 10, , x, , Based on the above information, answer the following questions., (i), , Position vector of B is, (a) 3i + 5j, , (b) 5i + 3j, , (c) −5i − 3j, , (d) −5i + 3j, , (b) 3i + 5j, , (c) 8i + 9 j, , (d) 9i + 8 j, , (b) i + 2 j, , (c) 2i + j, , (d) 2i − j, , (b), , (c) 90 units, , (d) 100 units, , (ii) Position vector of D is, , (a) 5i + 3j, , , , (iii) Find the vector BC in terms of i, j ., , (a), , i − 2 j, , , , (iv) Length of vector AD is, , (a), (v), , 67 units, , 85 units, , , If M = 4 j + 3k , then its unit vector is, (a), , 4 3, j+ k, 5, 5, , (b), , 4 3, j− k, 5, 5, , (c) −, , 4 3, j+ k, 5, 5, , (d) −, , 4 3, j− k, 5, 5, , PART - B, Section - III, 2, , cos 2 x + 2 sin x dx, cos2 x, , 6., , Evaluate :, , 7., , Prove that the area bounded by the parabola y2 = 4ax and the line x = a and x = 4a is, units., , ∫, , 56a 2, sq., 3

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CBSE Board Term-II Mathematics Class-12, , 132, 8., , Find the distance from the origin to the plane x + 3y – 2z + 1 = 0., , 9., ,  d 3 y   d 2 y  dy, Find order and degree of the equation ,  +,  + + 4 y = sin x .,  dx 3   dx 2  dx, , 4, , 3, , OR, Solve the differential equation, ^ ^, , ^, , ^ ^, , dy, y−x, =2, ., dx, ^, , ^ ^, , ^, , 10. Find the value of i ⋅ ( j × k ) + j⋅ (i × k ) + k⋅ (i × j)., , Section - IV, 11. A bag contains 12 white pearls and 18 black pearls. Two pearls are drawn in succession without, replacement. Find the probability that the first pearl is white and the second is black., 12. Find the direction cosines of the line passing through the two points (–2, 4, –5) and (1, 2, 3)., 13. Evaluate :, , π /2, , sin x, , ∫ 1 + cos2 x dx, , OR, , 0, , Evaluate :, , ∫, , 16 + (log x )2, , dx, x, 14. A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective, (8%). Produced parts get passed through an automatic inspection machine, which is able to detect, any part that is obviously defective and discard it. What is the probability of the parts that make, it through the inspection machine and get shipped?, Section - V, 2, 1, 15. If A and B are two independent events such that P ( A ∩ B ) =, and P ( A ∩ B ) = , then find, 15, 6, P (B) – P(A)., OR, , 3, 1, 1, If P ( A) = , P ( B) = and P ( A ∩ B) = , then find P ( A | B ) and P ( B | A)., 8, 2, 4, 3, 2, , 16. Evaluate : ∫ | x cos πx | dx, 0, , 17. Solve the initial value problem 2 xy + y 2 − 2 x 2, , dy, = 0, y (1) = 2 ., dx, , Section - VI, 18. Find the equation of the plane passing through the point A(1, 2, 1) and perpendicular to the, line joining the points P(1, 4, 2) and Q(2, 3, 5). Also, find the distance of this plane from the line, x +3 y −5 z −7, =, =, ., 2, −1, −1

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133, , Practice Paper - 3, , OR, Find the coordinates of the points on the line, 2 units from the point (–2, –1, 3)., , x + 2 y +1 z − 3, , which are at a distance of, =, =, 3, 2, 6, x2 y2, x y, +, = 1 and the line + = 1 ., 3 2, 9, 4, , 19. Find the area of the smaller region bounded by the ellipse, OR, , Draw the graph of y = |x + 1|and using integration, find the area below y = |x + 1|, above x - axis, and between x = – 4 to x = 2., , π/ 4, , 1. (d) : We have, , ∫, 0, , = log sec, , tan x dx = [log | sec x |]0π/ 4, , 1, π, − log | sec 0 | = log | 2 | − log | 1 | = log 2, 4, 2, , dy, y, 2. (a) : The given equation is, +, = x,, dx, 1 − x2, 1, where P =, and Q = x, 1 − x2, Pdx, ∴ I.F. = e ∫, =e, , ∫, , dx, 1− x2, , = esin, , −1, , x, , , 3. (a) : The unit vector in the direction of a vector a is, , a, , given by a^ =  . Now, | a | = (2)2 + (3)2 + (4)2 = 29, |a |, Therefore, a^ =, , ^, , 2 ^i + 3 ^j + 4 k, 29, , =, , 2 ^, 3 ^ 4 ^, i+, j+, k, 29, 29, 29, , 4. (b) : If two dice are thrown, then total number of, , cases = 36, Cases for total of 9 or 11 are {(3, 6), (4, 5), (6, 3), (5, 4),, (6, 5), (5, 6)}, i.e., 6 in number., \, , 6 1, P(total 9 or 11) =, =, 36 6, , P(sum is neither 9 nor 11) = 1 – P(sum is 9 or 11), , 1 5, =1− =, 6 6, 5. (i) (b) : Here (5, 3) are the coordinates of B., \, P.V. of B = 5i + 3j, (ii) (d) : Here (9, 8) are the coordinates of D., \, , P.V. of D = 9i + 8 j, , (iii) (b) : P.V. of B = 5i + 3j and P.V. of C = 6i + 5j, , , BC = (6 − 5)i + (5 − 3)j = i + 2 j, , \, , (iv) (b) : Since P.V. of A = 2i + 2 j, P.V. of D = 9i + 8 j, \, , ⇒, (v), \, \, , , AD = (9 − 2)i + (8 − 2)j = 7i + 6 j, , | AD |2 = 72 + 62 = 49 + 36 = 85, , | AD | = 85 units, , (a) : We have, M = 4 j + 3k ,, , | M | = 42 + 32 = 16 + 9 = 25 = 5, , ^, 4 j + 3k 4  3 , M, M=  =, = j+ k, |M|, 5, 5, 5, , 6. Let I =, , =∫, =∫, , ∫, , cos 2 x + 2 sin2 x, cos2 x, , 2 cos2 x − 1 + 2 sin2 x, cos2 x, 2(cos2 x + sin2 x ) − 1, 2, , cos x, , dx, , dx, dx = ∫ sec2 x dx = tan x + c, , 7. Required area = 2 × area of region PSRQP, 4a, , =2∫, a, , 8 ( 3/ 2 3/ 2 ), a 8a − a, 3, 56a 2, =, sq. units, 3, =, , 4a, ,  x 3/ 2 , 4ax dx = 4 a , ,  3 / 2 a

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CBSE Board Term-II Mathematics Class-12, , 134, 8. Required distance,, d=, , 0 + 0 − 0 +1, (1)2 + (3)2 + (−2)2, , =, , 1, 14, ,  d3 y , 9. Highest order derivative is ,  . So, its order is,  dx 3 , 3 and degree is 4., OR, , dy dx, dy, y−x, We have,, ⇒ y = x, =2, dx, 2, 2, Integrating both sides, we get, , −2 − y −2 − x, =, +C, log 2 log 2, –2–y + 2–x = C log2 = k(say) ⇒ 2–x – 2–y = k, , ⇒, , ( ), , 10. i ⋅ i + j − j + k ⋅ k = 1 − 1 + 1 = 1, , 11. Let A and B be the events of getting a white, pearl in the first draw and a black pearl in the, second draw respectively., Now, P(A) = P(getting a white pearl in the first draw), , =, , 12 2, =, 30 5, , P ( B | A) =, , 18, 29, , By multiplication rule of probability, we have, , 2 18 36, P ( A ∩ B) = P ( A). P ( B | A) = × =, 5 29 145, 12. We know that the direction cosines of the line, passing through two points P(x1, y1, z1) and Q(x2, y2, z2), are given by, x 2 − x1 y 2 − y1 z 2 − z1, ,, ,, ., PQ, PQ, PQ, where, PQ =, , ( x 2 − x1 )2 + ( y 2 − y1 )2 + (z 2 − z1 )2, , Here, P is (–2, 4, –5) and Q is (1, 2, 3)., So, PQ =, , (1 − (−2))2 + (2 − 4)2 + (3 − ( −5))2 = 77, , Thus, the direction cosines of the line joining two points, are, , π /2, , sin x, , ∫, , 2, 0 1 + cos x, , dx, , Put cos x = t ⇒ – sin x dx = dt, When x = 0, t = 1 and when x =, \I= −, , 0, , dt, , ∫ 1 + t 2 = − tan, , −1, , 1, , π, ,t = 0, 2, , 0, , t 1 = −[tan −1 0 − tan −1 1] =, , π, 4, , OR, 16 + (log x )2, dx, ∫, x, 1, Put logx = t ⇒ dx = dt, x, Let I =, , ∴, , I = ∫ 16 + t 2 dt, =, , ∴, , 16, t, 16 + t 2 + log | t + 16 + t 2 | + c, 2, 2, , 1, I = log x 16 + (log x )2, 2, +8 log | log x + 16 + (log x )2 + c, , When second pearl is drawn without replacement,, the probability that the second pearl is black is the, conditional probability of the event B occurring when, A has already occurred., \, , 13. Let I =, , 3, −2, 8, ,, ,, ., 77 77 77, , 14. Let G, SD, OD be the events that a randomly chosen, part is good, slightly defective, obviously defective, respectively., Then, P(G) = 0.90, P(SD) = 0.02, and P(OD) = 0.08, Required probability = P(G | ODc), =, , P (G ∩ OD c ), c, , P (OD ), , =, , 0.90, 90, P (G), =, =, = 0.978, −, 1, 0, ., 08, 92, 1 − P (OD), , 15. Since A and B are independent events, therefore,, , A c and B are independent and also A and B c are, independent., \, P(Ac ∩ B) = P (Ac) P(B) = (1 – P(A)) P(B), and P(A ∩ Bc) = P(A) P (Bc), = P (A) (1 – P(B)) ( A = Ac and B = Bc), , 2, ...(i), 15, 1, and P (A) (1 – P (B)) = P ( A ∩ B ) = ...(ii), 6, ⇒, , (1 – P (A)) P (B) = P ( A ∩ B) =, , Subtracting (ii) from (i), we obtain, P (B) – P (A) =, ⇒, , 2 1, −, 15 6, , P (B) – P (A) = −, , ⇒, , 1, 30, , P (B) – P (A) =, , 4−5, 30

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135, , Practice Paper - 3, OR, , \, , We have, P ( A ∩ B) = P ( A ∪ B), ⇒, , P ( A ∩ B) = 1 − P ( A ∪ B), , ⇒, , P ( A ∩ B) = 1 − {P ( A) + P (B) − P ( A ∩ B)}, , ⇒, , 3 1 1  3, P ( A ∩ B) = 1 −  + −  =, 8 2 4  8, 5, 1, Also, P ( A) = 1 − P ( A) =, and P ( B) = 1 − P ( B) =, 8, 2, ⇒, , 3, P ( A ∩ B) 8 3, Now, P A | B =, = =, 1 4, P ( B), 2, 3, P ( A ∩ B) 8 3, and P B | A =, = =, 5 5, P ( A), 8, , (, , (, , ), , 2x, = log x − 1, y, 2x, ⇒ y=, 1 − log x, Clearly, y is defined if x ≠ 0, and 1 – log|x| ≠ 0., Now, 1 – log |x| = 0 ⇒ log|x| = 1 ⇒ |x| = e, ⇒ x = ± e., 2x, , where x ≠ 0, ± e gives the solution, Hence y =, 1 − log x, ,  x cos π x ; 0 < x < 1, , 2, ∵ | x cos πx | = , 1, 3, − x cos π x ;, <x<, , 2, 2, , of the given differential equation., , 18. The line joining the given points, , ∴ I = ∫ (x cos πx )dx − ∫ x cos π x dx, 1, , 3, ,  x sin πx cos πx  2  x sin πx cos πx  2, +, =, +, ,  −, π2  0  π, π2  1,  π, , 2, ,  1, 1     −3,   1,  , , =  + 0 − 0 + 2  −  + 0 −  + 0,   2π,  , , π     2π,   2π, =, , 1, 1, 4, 5, 1 5π − 2, − 2+, =, − 2=, 2π π, 2π 2π π, 2 π2, , 17. We have 2 xy + y 2 − 2 x 2, , ⇒, , dy, =0, dx, , dy 2 xy + y 2, =, dx, 2x 2, , Putting y = vx ⇒, , dy, dv, =v+x, dx, dx, , (i) becomes –1 = 0 + c ⇒ c = –1, , −, , 0, , 1, 2, , dv 2v + v 2, =, −v, dx, 2, 1, 1, 2 ∫ 2 dv = ∫ dx, x, v, 2, − = log x + c, v, 2x, −, = log x + c ...(i), y, x, , Putting c = –1 in (i), we get, , 16. Let I = ∫ | x cos πx | dx, , 0, , ⇒, \, , 3, 2, , 3, 2, , ⇒, , dv 2v + v 2, =, dx, 2, , It is given that y(1) = 2 i.e., y = 2 when x = 1, , ), , 1, 2, , ⇒, , v+x, , P(1, 4, 2) and Q(2, 3, 5) has direction ratios, <1 – 2, 4 – 3, 2 – 5> i.e., <– 1, 1, –3>, The plane through (1, 2, 1) and perpendicular to the, line PQ is –1(x – 1) + 1(y – 2) – 3(z – 1) = 0, ⇒ x – y + 3z – 2 = 0, , x +3, , y −5, , z −7, , Now, direction ratios of line, are, =, =, −, 2, 1, −, 1, 2, –1, –1., Since 2(1) + (–1) (–1) + (3) (–1) = 2 + 1 – 3 = 0, \, Line is parallel to the plane., Since, (–3, 5, 7) lies on the given line., \, Distance of the point (–3, 5, 7) from plane is, , d=, , −3 − 5 + 3(7) − 2, , 1+1+ 9, 11, ⇒ d=, = 11 units., 11, OR, x + 2 y +1 z − 3, =, =, is the given line, 3, 2, 6, , Let P(–2, –1, 3) lies on the line., The direction ratios of line (i) are 3, 2, 6, , ...(i)

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CBSE Board Term-II Mathematics Class-12, , 136, \, , The direction cosines of line are, , 3 2 6, , ,, 7 7 7, , Equation (i) may be written as, , x + 2 y +1 z − 3, =, =, 3, 2, 6, 7, 7, 7, , ..(ii), , Coordinates of any point on the line (ii) may be taken as, , 2, 6, 3, ,  7 r − 2, 7 r − 1, 7 r + 3 , Let, , 3, , 3, , 2, 2 x, 9,  x  2 , =  9 − x 2 + sin −1    −  (3 − x ) ,  3   0 3  −2 , 3 2, 2, 0, , 2  9,   9,  1, =   0 + sin −1(1)  −  0 + sin −1(0)   + 02 − 9,  3,   2, 3  2, =, , 3π, − 3 sq. units., 2, OR, ,  x + 1, if, − x − 1, if, , Here y = |x + 1| = , , 2, 6, 3, , Q ≡  r − 2, r − 1, r + 3 , 7, , 7, 7, , x ≥ −1, x < −1, , Thus we get two lines –x + y = 1 ...(i), x + y = –1...(ii), Their graphs are as shown and the area to be calculated, is shaded., , Given |r| = 2, \ r = ± 2, Putting the value r, we have, ,  −8 −3 33 ,  −20 −11 9 , Q ≡  , ,  or Q ≡ , ,, ,,  7 7 7,  7, 7 7 , x2 y2, x y, +, = 1 ...(i) and + = 1 ...(ii), 9, 4, 3 2, x2 y2, Curve (i) is an ellipse of the form, +, =1 ., a 2 b2, That means its major axis is along x – axis. Also this, ellipse is symmetrical about the x – axis., , 19. We have, , Hence, the required area, , =, , 2, , ∫ (x + 1)dx +, , −1, , 2, , −1, , ∫ (− x − 1)dx, , −4, , −1, , , ,  2, , =  x + x −  x + x,  −4,  −1  2, 2, 2, , Required area =, , 3, , 3, , 0, , 0, , 2, 2, (3)2 − x 2 dx − ∫ (3 − x )dx, 3∫, 3, ,  9 9,  1   1 , = (2 + 2) −  − 1 −   − 1 − (8 − 4) = +,  2   2 ,  2 2, = 9 sq. units., , , , Visit https://telegram.me/booksforcbse for more books.