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FUNCTIONS, , (II)POLYNOMIAL FUNCTION :, , TOPIC-1-FUNCTIONS, , A relation f : X  Y is said to be function if every, element in the domain X have one and only one image, in the co domain Y and is defined as f x   y, x  X is a pre image of y &  y  Y is a image of x, , A function f : R  R is defined by, , f  x   a0 x0  a1 x1  a2 x 2  ...  an x n , x  R, Is called polynomial function., Domain=R and Range=R, SHORT CUT METHODS, , SHORT CUT METHODS, , Let A and B be the any two sets such that, n A  m & nB  n then the number of functions, from A to B = n, , 1. If x   x     0 , then x   ,  , 2. If x   x     0 , then x   ,  , , m, , 3. If  x    x     0, then x   ,  , , DOMAIN AND RANGE OF A FUNCTION, , 4. If  x    x     0, then x   ,  , , If a function f is defined from a set A to set B then for, f : A  B set A is called the domain of function f and, set B is called the co-domain of function f. The set of, all f-images of the elements of A is called the range of, function f., In other words, we can say, Domain = All possible values of x for which f(x), exists., Range = For all values of x, all possible values of f(x), ., Range, , A, a, b, c, d, , B, p, q, r, s, , f, , Domain, , = {a, b, c, d} = A, ,  f   D f   x : f x  0, , 1., , D, , 2., ,  f , D   D f   D g   x : g  x   0, g, , 1, , f x   a 2  x 2, ,  a, a, , 2, , f x  , ,  a, a , , 3, , f x  , , 4, , f x  , , 5, , f x  x  a b  x if a  b, , a, b, , 6, , f x  , , a, b, , = {p, q, r}, , x y  (a  x) (b  y ) are implicit functions., CHETHAN M G, , f  x, , g  x, , where f(x) and g(x) are polynomial functions of x, defined in a domain, where g(x) ≠ 0., , Range, , function, while x 2  y 2  xy and, 2, , Rational functions are functions of the type, , S.N, , A function is, said to be explicit if it can be expressed directly in, terms of the independent variable. If the function, cannot be expressed directly in terms of the, independent variable or variables, then the function is, said to be implicit. e.g. y  x  log x is explicit, 2, , (III) RATIONAL FUNCTION :, , Co-domain = {p, q, r, s} = B, , (I)EXPLICIT AND IMPLICIT FUNCTIONS :, , 2, , D, , 2,  ,   , if a  0 where D  b  4ac, 4, a, , , , 3., , SOME FUNCTIONS AND THEIR GRAPHS, , 3, ,  D, , &,  4a ,   if a  0, , SHORT CUT METHODS, , Co-domain, , Domain, , 5. Range of f x   ax2  bx  c is, , FUNCTIONS, , DOMAIN, , 1, a  x2, 2, , x2  a2, 1, x2  a2, , 1, if a  b, x  ab  x, , xa, xb, ,  , a   a,  ,  , a    a,  , , , ,  , , , , 7,  /, 7019881906, a  b, , f x   Phone number-8105418762, if a  b, [email protected],, , xa, , , , 1

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5, , f x  x  a b  x if a  b, , a, b, , 6, , f x  , , a, b, , 1, , x  ab  x, , f x  , , 7, , f x  , , 8, , if a  b, , CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, PROPERTIES, , xa, if a  b, xb, ,  , a  b, , , xa, if a  b, bx, , a, b, , 1. x  0 if x  R, 2. x  a  a  x  a, i.e. x   a , a , , , , 3. x  a  a  x  a, i.e. x   a , a, , , , 4. x  a  x  a or x  a i.e. x    , a   a , , , (IV) CONSTANT FUNCTION:, , Let k be a fixed real, number. Then a function f(x) given by f ( x )  k for all, x  R is called a constant function. The domain of the, constant function f ( x )  k is the complete set of real, numbers and the range of f is the singleton set {k}. The, graph of a constant function is a straight line parallel to, x-axis as shown in figure and it is above or below the, x-axis according as k is positive or negative. If k = 0,, then the straight line coincides with x-axis., Y, ,  , , 5. x  a  x  a or x  a i.e. x    , a  a ,  , 6. x  a  b  a  b  x  a  b i.e. x  a  b, a  b, 7. x  a  b  x  a  b or x  a  b i.e. x   , a  b  a  b, , 8., , xy  x y, , 9., , x, x, , y, y, , 10., , x  x x  R, , 11. Triangular inequality, ,  x  y  x  y  x, y  R, f (x) = k, , k, X, ,  x  y  x  y  x, y  R, , X, , O, , Let f ( x)  [ x],, where [x] denotes the greatest integer less than or equal, to x. The domain is R and the range is I. e.g. [1.1] = 1,, [2.2] = 2, [– 0.9] = –1, [– 2.1] = – 3 etc. The, function f defined by f ( x )  [ x ] for all x  R , is, called the greatest integer function., (VII) GREATEST INTEGER FUNCTION:, , YThe function defined by, , (V)IDENTITY FUNCTION :, , f ( x)  x for all x  R , is called the identity function, on R. Clearly, the domain and range of the identity, function is R. The graph of the identity function is a, straight line passing through the origin and inclined at, an angle of 45o with positive direction of x-axis., , 3, 2, 1, , , , f (x) = x, , , , , , -1 1, -2, -3, , , 2, , 3, , PROPERTIES, , X, , X, , 5., ,  x  x, x  I,   x   x, x  I,  x    x  1, x  I,  x      x   ,   I,  x  y    x   y , , 6., ,  x    x , , 1., , O, , 2., 3., , Y, (VI)MODULUS FUNCTION:, , , , -3 -2 -1, , Y, , , , The function defined by, ,  x, when x  0, f ( x) | x | , is called the modulus, -x, when x  0, function. The domain of the modulus function is the set, R of all real numbers and the range is the set of all nonY, negative real numbers., , 4., , 1 , 2, n  1, ,   x    ...   x ,   nx , , n , n, n , , , x  I & n  N, , NOTE, , Fractional part : We know that x  [x ]. The, f (x) =– x, X, , 2, , f (x) = x, , O, , difference between the number ‘x’ and it’s integral, X, , value ‘[x]’ is called the fractional part of x and is, symbolically denoted as {x}. Thus, {x}  x  [x ], , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, Y, , MCP-MATHEMATICS

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SOLVED PROBLEMS, , (VIII) SIGNUM FUNCTION:, , The function defined by,  1, x  0, | x |, , x  0 or, , , f ( x)   0, x  0 is called, f ( x)   x, 1, x  0,  0, x  0, , the signum function. The domain is R and the, range is the set {–1, 0, 1}., , LEVEL-I, , 1. Which of the following is a function?, (a){(2,1), (2,2), (2,3), (2,4)}, (b){(1,4), (2,5), (1,6) , (3,9)}, (c){(1,2), (3,3), (2,3), (1,4)}, (d){ (1,2), (2,2), (3,2), (4,2)}, Solution:(d) We know that for a relation to be, function every element of first set should be, associated with one and only one element of second, set but elements of first set can have same f-image, in second set which is given in (D)., 2. If A contains 10 elements then total number of, functions defined from A to A is, (a)10, (b) 2 10, , Y, (0, 1), X, , O, X, , O, , O (0, –1), Y, , (I)EXPONENTIAL FUNCTION :, , Let a  1 be a positive real number. Then, f : R  (0, ) defined by f (x )  a x called, exponential function. Its domain is R and range, is (0, ) ., Y, a<1, , a>1, (0, 1), X, , f(x) = ax, , x, , f(x) = a, X, , O, , X, , Let a  1 be a, positive real number. Then f : (0, )  R defined by, , (II)LOGARITHMIC FUNCTION :, , f ( x)  log a x is called logarithmic function. Its, domain is (0, ) and range is R., Y, f(x) = loga x, X, , O, , X, (1, 0), , Y, Graph of f (x )  log a x, when a > 1, , SHORT CUT METHODS, FUNCTIONS, , , , DOMAIN, , , , 1, , f x   log a 2  x 2, , 2, , f x   log x 2  a 2, , , , f  x   log  x  a   x  b , , 3, if a  b, , CHETHAN M G, ,  a, a , , , , xSolution:(a)f(x), , Graph of f (x )  a , when a<1, , Graph of f (x )  a , when a > 1, x, , S.N, , (c) 10 10, (d) 210  1, Solution:(c) According to formula, total number of, functions  n n .Here, n = 10. So, total, Y number of functions  1010., 3. Function f(x) = x–2 + x–3 is, (a) a rational function, (0, 1), X(b) an irrational function, O, (c) an inverse function, (d) None of these, Y, , Y, ,  , a    a,  , ,  , a  b, , , FUNCTIONS, , =, , 1, x, , 2, , , , 1, x, , 3, , , , x 1, x3, , = ratio of two, , polynomials  f(x) is a rational function., x, , 4. If f (x) =,  , then f (y) equals, x 1 y, (a) x, (b) x – 1, (c) x + 1, (d) 1 – x, y, (, x, , 1) / x, x 1, Solution:(d) f ( y ) , , ,  1 x, x, , 1, y 1, 1 x 1 x, x, 5. The function which map [–1, 1] to [0, 2] are, (a) One linear function, (b) Two linear function, (c) Circular function, (d) None of these, Solution:(b) Width of both interval is same, which can, mapped by these function y  1  x and y  1  x., | x 3|, 6. Domain and range of f ( x) , are, x 3, respectively, (a) R, [ 1, 1], (b) R  {3}, 1,  1, (c) R  , R, , (d)None of these, , Solution:(b)Domain of f ( x)  R  3 and range, {1, –1}., 7. The domain of the function f ( x)  (2  2 x  x2 ) is, [email protected], Phone number-8105418762 / 7019881906, , 3

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CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, , (a) 3  x  3, (b) 1  3  x  1  3, (c) 2  x  2, (d)None of these, Solution:(b)The quantity square root is, positive, when 1  3  x  1  3., 8. If in greatest integer function, the domain is a set of, real numbers, then range will be set of, (a) Real numbers, (b) Rational numbers, (c) Imaginary numbers, (d) Integers, Solution:(d) Obviously, x3, 9. Domain of the function f ( x) , is, ( x  1) x 2  4, (a)(1, 2), (b) (,  2)  (2, ), (c) (,  2)  (1, ), (d) (, )  {1,  2}, Solution:(b) Obviously, here | x |  2 and x  1, i.e., x  ( ,  2) (2, ) ., 10. If the domain of function f ( x)  x 2  6 x  7 is, (, ) , then the range of function is, , 14. Domain of the function log | x 2  9 | is, (a) R, (b) R  [3, 3], (c) R  {3, 3}, (d) None of these, Solution:(c) For x  3, 3, | x 2  9 |  0, Therefore log| x 2  9 | does not exist at x   3, 3., Hence domain of function is R   3, 3., 15. The largest possible set of real numbers which can, be the domain of f ( x)  1 , (a) (0, 1)  (0, ), (c) (,  1)  (0, ), , 1, is, x, , (b) (1, 0)  (1, ), (d) (, 0)  (1, ), , Solution:(b) 1  1  0  x  1 . Also, x  0 ., x, ,  Required interval  (, 0)  (1, ) ., , (a) (, ), , (b) [2, ), , 16. Domain of the function f ( x) , , (c) (2, 3), , (d) (,  2), , (a) {x : x  R, x  3}, (b) {x : x  R, x  2}, , Solution:(b) x  6 x  7  ( x  3)  2, Obviously, minimum value is – 2 and maximum  ., Hence range of function is [–2, ]., 2, , 11. Domain of the function, (a) (–3, 1), (c) (–3, 2], Solution:(c) (i) x  2, , 2, , 1, , 2 x , , 9  x2, , is, , (b) [–3, 1], (d) [–3, 1), , (ii) 9  x 2  0  | x |  3 or 3  x  3., Hence domain is (  3, 2], 12. The natural domain of the real valued function, 2, 2, defined by f (x )  x  1  x  1 is, (a) 1  x  , (b)   x  , (c)   x  1, (d) (, )  (1, 1), , Solution:(d) f ( x)  x 2  1  x 2  1,  f ( x)  y1  y2 Domain of y1 , , x2 1, ,  x 2  1  0  x 2  1  x  ( , )  (1, 1) and, Domain of y 2 is real number, Domain of, f ( x)  (, )  (1, 1) ., 13. Domain of the function 1  x  1  x is, x, , (a) (–1, 1), (c) [–1, 1], 4, , Solution:(d) 1  x  0  x  1 ;, 1  x  0  x  1, x  0, Hence domain is [1, 1]  {0} ., , (b) (–1, 1)–{0}, (d) [–1, 1]–{0}, , x 2  3x  2, is, x2  x  6, , (c) {x : x  R}, (d) {x : x  R, x  2, x  3}, Solution:(d) f (x ) , , (x  2) (x  1), (x  2) (x  3), , Hence domain is x : x  R, x  2, x  3., 17. Domain of f ( x)  ( x 2  1)1/2 is, (a) (,  1)  (1, ), (b) (,  1]  (1, ), (c) (,  1]  [1, ), (d) None of these, Solution:(a) Here | x |  1, therefore, x  ( ,  1)  (1, )., 18. The domain of the function f :  defined by, f  x   x 2  7 x  12 is, , (a)  3, 4 , (c)  ,3   4,  , , (b)  ,3   4,  , , (d)  ,3   4,  , , [KCET- 19-1m] [Only one option correct], , Solution:(d), , x 2  7 x  12  0, ,  x  4 x  3  0, x   ,3   4,  , , 12, 4  3, , 1, x , 19. f  x    tan   ,  1  x  1 and, 2,  2 , , g  x , , 3  4 x  4 x  . Find domain of, 2, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , f g, , MCP-MATHEMATICS

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FUNCTIONS, , (a)  1 ,1, 2, , , (c)  1 , 3 ,  2 2, , , , (iv) f 1  b  0 ,hence the point 1,0  is not lies on the, graph of the function., 21. If f (x )  a cos( bx  c)  d , then range of f (x ) is, (a) [d  a, d  2a], (b) [a  d, a  d], (c) [d  a, a  d], (d) [d  a, d  a], Solution:(d) f (x )  a cos( bx  c)  d …(i), For minimum cos( bx  c)  1, from (i), f (x )  a  d  (d  a), For maximum cos( bx  c)  1, from (i), f (x )  a  d  (d  a), Range of f (x)  [d  a, d  a], 22. Range of f (x )  [x ]  x is, (a) [0, 1], (b) (–1, 0], (c) R, (d) (–1, 1), Solution:(b)As shown in graph  Range is (–1, 0]., , (b)  1 ,1, , 2 , , 1, , (d)  1,1, , [KCET-2015-1m][Only one option correct], , Solution:(a) (i) Domain of f is 1  x  1, (ii) Domain of g, , g  x , , 3  4x  4x , 2, , is defined when, , 3  4 x  4 x2  0,  4x2  4x  3  0,  4 x2  6 x  2 x  3  0, ,  2 x  2 x  3   1 2 x  3   0, ,   2 x  1 2 x  3  0, , 3,  1   ,  2 x    2 x    0, 2, 2,   , ,  4 on both sides, , O (1,0) (2,0), , , 3,  1   ,   x    x    0, 2,  2 , , ,  Domain of g is, , y=–1, , 1, 3, x, 2, 2, , 23. The range of f (x )  cos( x / 3) is, (a) (1 / 3, 1 / 3), (b) [1, 1], , (iii) D  f  g   D  f   D  g , , (c) (1 / 3,  1 / 3), (d) (3, 3), Solution:(b) f (x)  cos(x/3), We know that 1  cos(x/3)  1 ., ,  1 3 , D  f  g    1,1   , ,  2 2,  1 , D  f  g    ,1, 2 , , 24. The range of the function f (x ) , (a) {0, 1}, (c) R, , 20. Which one of the following is not correct for the, features of exponential function given by, f  x   b x where b  1?, (a) The domain of the function is R, the set of, real numbers., (b) The range of the function is the set of all, positive real numbers., (c) For very large negative values of x, the, function is very close to 0., (d) The point 1,0  is always on the graph of the, , Solution:(b) f (x ) , , (b) {–1, 1}, (d) R  {2}, x 2, | x  2|, , 1,, f (x )  ,  1,, , x  2, x  2, ,  Range of f (x ) is {1,1} ., 25. The range of the function f  x   9  x 2 is, (a)  0,3, , function., [KCET-2014-1m] [Only one option correct], , Solution:(d) (i) Domain of the exponential function is, real number set is correct, (ii) Range of the exponential function is positive real, number set is correct, (iii) For very large negative values of x, the function is, very close to 0 is also correct, 1 1, i.e f   x   b  x  x   0, b, , CHETHAN M G, , x2, is, | x  2|, , (c)  0,3, , (b)  0,3, , (d)  0,3, , [KCET-2017-1M] [One option correct type], [NCERT-DIRECT QUESTION], , Solution:(b) y  9  x 2  0  1 Squaring on both sides, , y 2  9  x2, x2  9  y 2, , x  9  y2, , [email protected], Phone number-8105418762 / 7019881906, , 5

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CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, , 9 y  0, 2, ,  y 9  0, 2, ,  y 2  32  0, ,   y  3 y  3  0, ,   y   3   y  3  0, ,  3  y  3   2 , From equations (1&2), 0  y  3 or 0,3, [ y is the range of a function], 26. Let f :  2,    R be the function defined by, , f  x   x 2  4 x  5, then the range of f, (a) 5,  , , (b)  ,  , , (c) 1,  , , (d) 1, , , LEVEL-II, , 1. The domain of f(x) =, , 1, is, x x, 3, , (a) R – {–1, 0, 1}, (b) R, (c) R – {0, 1}, (d) None of these, Solution:(a)Domain = {x; x R; x3 – x ≠ 0}, = R – {–1, 0, 1}, 1, 2. The domain of the function f  x  , is, x x, , [KCET-2020-1M] [One option correct type], , Solution:(c) f  x   x 2  4 x  5, , f  x   x2  4 x  4  1   x  2  1  1, x  2,  , 2, , (a) (1, 1), , (b) [1, 1), , a)  , , b) 0,  , c)  ,0, d)  ,   0, Solution:(c) For domain, | x |  x  0  | x |  x . This, is possible, only when x  R  ., ,  , (c)  , , , (d) [ 2 , 2 ], , 3. The domain of the function f ( x) , , Hence range  f  x   1,  , , 27. The range of f (x )  cos x  sin x is, ,  2 2, , Solution:(d) f (x )  cos x  sin x, , (a) R, , 1,  1, , f ( x)  2  cos x  sin x , 2, 2, , , ,  , , , , f ( x)  2 cos cos x  sin sin x   2 cos   x , 4,  4, , 4 , , , 1  cos   x   1, 4, , , ,  2  2 cos   x   2, 4, , , 28. Range of the function, , 4. The domain of definition of f x  , , (a) [1, 3], (c) (1, 3), , (d)  , 1 , 1, 3, , | x |  x ., , log2 x  3, is, x 2  3x  2, , d)  3,    1,2, log ( x  3), log 2 ( x  3), Solution:(d)Here f ( x)  2 2, , x  3x  2 ( x  1)( x  2), exists if, Numerator x  3  0  x  3 … (i), and denominator ( x  1)( x  2)  0  x  1,  2, Thus, from (i) and (ii); we have domain of f ( x ) is, (3, )  {1,  2} ., 5. The domain of the definition of the function y(x), given by the equation 2 x  2 y  2 is, a) 0  x  1, b) 0  x  1, c)    x  0, d)    x  1, , , , 1, , sin 3 x  [1, 1], Solution:(b) f (x ) , 2  sin 3 x, , 1, Hence f (x ) lies in  ,, 3, , b)  2,  , , a) R   1,2, c) R   1,2,3, , 1, is, | x | x, (d) R, , (c) R0, , , This is possible, only when x  R ., , 1, is, 2  sin 3 x, 1 , (b)  , 1, 3 , , (b) R, , Solution:(b) For domain, | x |  x  0, ,  2  f  x  2, , , 1 ., , , Solution:(d) 2 y  2  2 x y is real if 2  2 x  0,  22,  1  x  x  (, 1), x, , 29. Range of the function f (x )  sin (x )  cos (x ) is, 2, , 6, , (a) (, ), (b){1}, (c)(–1, 1), (d)(0, 1), 2, 4, Solution:(b) f (x )  sin (x )  cos 2 (x 4 )  1 .Hence, range  1., 30. Range of the function f (x )  9  7 sin x is, (a)(2, 16), (b)[2, 16], (c)[–1, 1], (d)(2, 16], Solution:(b) y  f (x )  9  7 sin x . Range  [2, 16 ]., , 4, , 2, , 4, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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FUNCTIONS, , 6. The domain of function f(x) = 2  3 is, (a) (– , 0], (b) R, (c)[0, ), (d) No value of x, Solution:(a) Domain = {x ; 2x – 3x  0} = {x ; (2/3)x 1}, = x (– ,0], x, , 7. Domain of the function, (a) (,  1)  (1, ), , 1, x 1, 2, , x, , is, , (b) (,  1]  (1, ), , (c) (,  1)  [1, ), , (d)None of these, , Solution:(a) For domain, x 2  1  0,  ( x  1)( x  1)  0  x  1 or x  1,  x  (, 1)  (1, ), 8. The domain of the function, , ,  5x  x 2, Solution:(b) We have f (x )  log 10 , 4, , , , , , , , , 1/2, , ..(i), , From (i), clearly f ( x ) is defined for those values of x for, 5x  x 2 , 0,  4, , , which log 10 , , 2, 2,   5 x  x   100   5 x  x   1, , , , 4, , , , , , 4, , , ,  x  5 x  4  0  (x  1)(x  4 )  0, Hence domain of the function is [1, 4]., 13. Domain of the function log (5 x  x 2 ) / 6 is, 2, , f ( x)  x  x 2  4  x  4  x is, (a) [4, ), (b) [–4, 4], (c) [0, 4], (d) [0, 1], , Solution:(d) f (x )  x  x  4  x  4  x, Clearly f (x ) is defined, if 4  x  0,  x  4  4  x  0  x  4, x(1  x)  0  x  0 and x  1  Domain of, f  (, 4]  [4, )  [0,1]  [0,1] ., 2, , 9. Domain of f ( x)  log | log x | is, (a) (0,  ), , 1/2, , ,  5x  x2 , 12. Domain of the function f ( x)  log10 ,   is,  4 , , (a)   x  , (b) 1  x  4, (c) 4  x  16, (d) 1  x  1, , (b) (1,  ), , (c) (0, 1)  (1, ), (d) ( , 1), Solution:(c) f ( x)  log | log x | , f ( x ) is defined if, | log x | 0 and x  0 i.e., if x  0 and x  1, ( | log x |  0 if x  1)  x  (0,1)  (1, )., , (a) (2, 3), (c) [1, 2], , (b) [2, 3], (d) [1, 3], 2, 2, , , Solution:(b) log  5 x  x   0  5 x  x  1, 6,  6 , or x 2  5 x  6  0 or (x  2)(x  3)  0 .Hence, 2  x  3., , 14. The domain of the function f ( x)  log3 x ( x2 1) is, (a) (3,  1)  (1, ), (b) [3,  1)  [1, ), (c) (3,  2)  (2,  1)  (1, ), (d) [3,  2)  (2,  1)  [1, ), Solution:(c), , f ( x ) is to be defined when, , x  1  0  x 2  1,  x  1 or x  1 and, 3  x  0  x  3 and x  2,  Df  (3,  2)  (2, 1)  (1, ) ., 2, , 1, 10. The domain of the function f ( x)  log, is, | sin x |, (a) R  {2n , n  I }, (b) R  {n , n  I }, (c) R  { ,  }, (d) (, ), Solution:(b) f ( x) , , log, , 1, | sin x |, ,  x  n  (1)n 0  x  n ., Domain of f  x   R  {n , n  I } ., 11. The domain of the function, f ( x)  log( x  4  6  x ) is, (a) [4, ), (b) (, 6], (c) [4, 6], (d) None of these, Solution:(c) f ( x)  log, , , , x4  6 x, , , ,  x  4  0 and 6  x  0  x  4 and x  6,  Domain of f (x )  [4, 6] ., CHETHAN M G, , 15. The domain of the function log( x2  6 x  6) is, (a) (, ), (b) (, 3  3)  (3  3, ), (c) (, 1]  [5, ), (d) [0, ), Solution:(c) The function f (x )  log( x 2  6 x  6 ) is, defined when log( x 2  6 x  6)  0  x 2  6 x  6  1,  ( x  5)( x  1)  0 This inequality holds if x  1 or, x  5 . Hence, the domain of the function will be, (,1]  [5, ) ., 16. Domain of definition of the function, 3, f ( x) ,  log10 ( x 3  x) , is, 4  x2, [email protected], Phone number-8105418762 / 7019881906, , 7

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CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, , (b) (1, 0)  (1, 2), (d) (1, 0)  (1, 2)  (2, ), , (a) (1, 2), (c) (1, 2)  (2, ), , 3, Solution:(d) f ( x) ,  log10 ( x 3  x) ., 2, 4 x, , +, –1, , –, 0, , , ,  5 x  3  2 x 2  0 or, , 3, , ( x  1)  x    0  D [1, 3 / 2] ., 2, , 18. The domain of the function, , , , , , 1,  log10 x3  x ,is, 2, 4 x, (a)  2, 1   1,0    2,  , (b) 1, 2    2,  , (c)  1,0  1,2   2,  , (d)  1,0   1,2   3,  , , Solution:(c) Given f  x   g  x   h  x , , g is defined when 4  x  0, 4  x 2  0  x2  4  x   4  2, Hence D  g   R  2, 2, , , , (ii) h  x   log10 x3  x, , (b)  2 , 1    3 , 4 , ,  5 2   5 5, (c)  2 ,1  1, 4 , ,  5   5, (d)  0, 1    2 , 4 , 3, 5 5, , , , , , , , , [JEE-MAIN-8/1/2020-SHIFT-II], [One option correct type-4M], ,  x, if 1  x  2,  2, Solution:(b) f  x    x  1,  2 x if 2  x  3,  x 2  1,  2 1 ,  5 , 2  if 1  x  2, , ,  f  x  ,  3 , 4  if 2  x  3,  5 5 ,  2 1   3 4,  Range of f   ,    , ,  5 2   5 5, 1 if 1  x  2 , ,  if 2  x  3, ,  x   2, , , , , ,  x  x  1 x  1  0, , +ve, , -ve, , 1, 0, -1, , Hence D  h    1,0   1,  , (iii) W.K.T. If f  x   g  x   h  x , , x2, is, 1  x2, , x2, y, f(x) = y = , 1  x2, y, x, ....(1), (1  y ), , h is defined when x 3  x  0,  x x2  1  0, , 8, , , , (a) R – {1}, (b) R+  {0}, (c) [0, 1], (d) None of these, Solution:(d)Range is containing those real numbers y, for which f(x) = y where x is real number. Now, , 2, , -ve, , , , 20. The range of function f(x) =, , 1, 4  x2, , , , , , , , , , [JEE-MAIN-9/4/2019-SHIFT-II], [One option correct type-4M], , (i) g  x  , , : 1,3  R then range of f  x  is, , (a)  0, 1    3 , 7 , 2, 5 5, , 17. The domain of the function f ( x)  exp( 5 x  3  2 x 2 ) is, 3 , (a) 1,  3 , (b)  ,  , , , 2, 2 , , (c) [ , 1], (d) 1, 3 ,  2, , , , f  x , , x2  1, , [Where . denoted greatest integer function], , +, 1, , 5 x  3 2 x 2, , x  x, , 19. Let f  x  , ,  D  (1,0)  (1, )  { 4} i.e.,, D  (1, 0)  (1, 2)  (2, ) ., , Solution:(d) f ( x)  e, ,  D  f  x    R  2, 2   1,0  1,  ,  D  f  x    1,0  1, 2    2,  , , So, 4  x 2  0  x   4 and, x3  x  0  x( x 2  1)  0  x  0, x  1, –, ,  D  f  x   D  g  x   D h  x , , by (1) clearly y  1, and for x to be real, , +ve, , , y,  0  y  0 and y < 1., 1 y, 21. Range of the function f (x ) , , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , x2  x  2, x2  x 1, , ; x  R is, , MCP-MATHEMATICS

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(a) (1, ), (c) (1, 7 / 3], , SOLVED PROBLEMS, , (b) (1, 11 / 7], (d) (1, 7 / 5], , Solution:(c) f (x )  1 , , 1, , LEVEL-I, , 1. If f : R+  R+, f(x) = x2 + 2 and g : R+  R+,, g(x) = x  1 then (f + g) (x) equals, , 2, , 1, 3, , x   , 2, 4, , ,  Range  (1, 7 / 3] ., 22. Range of f (x ) , , x 2  34 x  71, x 2  2x  7, , Solution:(b) Let, , x 2  34 x  71, x  2x  7, 2, , is, , y, ,  x 2 (1  y )  2 (17  y ) x  (7 y  71 )  0, 2, For real value of x , B  4 AC  0, ,  y  14 y  45  0  y  9, y  5 ., 2, , 23. If x is real, then value of the expression, x  14 x  9, 2, , x 2  2x  3, , lies between, (b) 5 and –4, (d) None of these, , (a)5 and 4, (c)– 5 and 4, Solution:(c), , x 2  14 x  9, x 2  2x  3, , y, ,  x 2  14 x  9  x 2 y  2 xy  3 y,  x 2 (y  1)  2 x (y  7)  (3 y  9)  0, 2, For real value of x , B  4 AC  0,  4( y  7)2  4(3 y  9)( y  1)  0, ,  4( y 2  49  14 y )  4(3 y 2  9  12 y )  0,  4 y 2  196  56 y  12 y 2  36  48 y  0,  8 y 2  8 y  160  0  y 2  y  20  0,  (y  5)(y  4 )  0 ;  y lies between – 5 and 4., TOPIC-2-ALGEBRA ON FUNCTIONS, , Let f  x  & g  x  be any two function, then, a., b., c., ,  f  g  x   f  x   g  x ,  f  g  x   f  x   g  x ,  f  g  x   f  x   g  x , , d.  f   x   f  x  , g  x   0, g  x, g, FORMULA FOR DOMAIN OF A FUNCTION, , a. D f  g   D f   Dg , b. D f  g   D f   Dg , , (a) x 2  3, , (b) x + 3, , (c) x  2  (x  1), , (d) x2 + 2 + (x  1), , 2, , (b) (, 5]  [9, ), (d) None of these, , (a) [5, 9], (c) (5, 9), , Solution:(d) (f + g) (x) = f(x) + g(x) = x2 + 2 +, 2., , x 1, , x| x |, , then f (1) , If f (x ) , | x|, , (a)1, , (b)– 2, (c)0, (d)2, 1 | 1| 1  1, Solution:(b) f (1) , ,  2 ., | 1|, 1, , 1,  y, , 3. If f (y )  log y, then f ( y )  f   is equal to, (a)2 (b)1, (c)0, (d)– 1, Solution:(c) Given f ( y )  log y,  f (1/ y )  log (1/ y) , then, , 1, f ( y )  f    log y  log(1/ y )  log1  0.,  y, x, f (a), 4. If f ( x) , , then, , x 1, f (a  1), 1, (a) f (  a ), (b) f  , a, (c) f (a 2 ), (d) f   a ,  a 1 , f (a ), a / (a  1), a2, ,  2  f (a 2 ) ., f (a  1) (a  1) / a a  1, , Solution:(c), , 5. The equivalent function of log x 2 is, (a) 2 log x, 2, , (c) | log x |, , (b) 2 log | x |, (d) (log x) 2, , Solution:(b) As log x is defined for only positive, values of x. But log x 2 defined for all real values of, x, also log | x | is also defined  real x. Hence, log x 2 and 2 log | x | are identical functions., , 6. If  ( x)  a x , then { ( p)}3 is equal to, (a)  (3 p ), (b) 3 ( p ), (c) 6 ( p ), (d) 2 ( p ), x, Solution:(a)  ( x)  a   ( p)  a p,  [ ( p)]3  [a p ]3  a3 p   (3 p), , 7. If y  f ( x) , (a) 1/ f ( x), , CHETHAN M G, , FUNCTIONS, , ax  b, , then x is equal to, cx  a, (b) 1 / f ( y ), , [email protected], Phone number-8105418762 / 7019881906, , 9

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CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, , (c) yf ( x), , (d) f ( y ), , ax  b,  x(cy  a)  b  ay, cx  a, , Solution:(d) y , ,  x  ay  b  f ( y ) ., cy  a, , 3, 8. Numerical value of the expression 3x  1, , 2 x2  2, , for x  3 is, (a)4, (b)2, (c)3, (d)0, 3, 3x  1, 3(27)  1 80, Solution:(a), , ,  4., 2, 2 x  2 x 3, 2(9)  2, 20, x2, ,then, x 1, , 9. If y  f (x ) , , x, (b) 2 f (y ), , (a) f (y ), (c), , 1, f (y ), , (d)None of these, , Solution:(a) y , , x2, 3, y2,  x , 1 ,  f (y ) ., x 1, y 1, y 1, , 2 x ; x  3 , , , 10. Let f : R  R be defined by f  X    x 2 ; 1  x  3, 3x ; x  1 , , , Then f  1  f  2  f  4  is, (a)9, , (b)14, , (c)5, , (d)10, , [KCET-2018-1M] [One option correct type], , Solution:(a) f  1  f  2  f  4, ,  3  1  22  2  4 ,  3  4  8  12  3  9, f  3.8  f  4 , 11. If f  x   2x 2 , find, 3.8  4, (a) 1.56, (b) 156, (c) 15.6, (d) 0.156, [KCET-2015-1M] [One option correct type], , Solution:(c), , f  3.8  f  4 , 3.8  4, , 2  3.8   2  4 , 2, , , , 2, , 3.8  4, 28.88  32 3.12, , ,  15.6, 0.2, 0.2, LEVEL-II, , 1, 1. If f  x   2 f    3x, x  0 ; and, x, S  {x  R : f  x   f   x }; then S, (a) A null set, (b) A singleton set, (c) contains exactly two elements, (d) contains more than two elements, , Solution:(c) f  x   2 f  1   3 x, x  0  1, x, 1, 3, 1, put x   f    2 f  x     2 , x, x, x, 6, eq(1)-2eq(2)  3 f  x   3 x , x, 2,  f  x   x, x, Given f  x   f   x , , 2, 2, 4,  x  x   2x    x 2  2, x, x, x, x 2,  S  { 2,  2}, 2. If f(x) = 2|x – 2| – 3|x – 3|, then the value of f(x), when 2 < x < 3 is, (a) 5 – x, (b) x – 5, (c) 5x – 13, (d) None of these, Solution: (c), 2 < x < 3, |x – 2| = x – 2 & |x – 3| = 3 – x, f(x) = 2 (x – 2) – 3 (3 – x) = 5x – 13., 1, 1, 3. If f ( x) , for, , x  2 2x  4, x  2 2x  4, x  2, then f (11) , 5, (a) 7, (b) 5, (c) 6, (d), 7, 6, 6, 7, 1, 1, Solution:(c) f ( x) , , x  2 2x  4, x  2 2x  4, , , f (11) , , , 1, 11  2 18, , , , 1, 11  2 18, , 1, 1, 3 2 3 2 6 ., , , , , 7, 7, 7, 3 2 3 2, , 4. The value of b and c for which the identity, f ( x  1)  f ( x)  8 x  3 is satisfied, where, , f ( x)  bx 2  cx  d , are, (a) b  2, c  1, (b) b  4, c  1, (c) b  1, c  4, (d) b  1, c  1, Solution:(b) f ( x  1)  f ( x)  8 x  3,  [b ( x  1)2  c ( x  1)  d ]  (bx 2  cx  d )  8x  3,  (2b) x  (b  c)  8 x  3, ,  2b  8, b  c  3  b  4, c  1., , 5. If f(x) =, , 2x  2x, ,then f (x + y) .f (x – y) is equal, 2, , to, 1, 2, , (a) [f (x+ y) + f(x – y)], , [JEE-main Offline-16-4m], [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , 10, , MCP-MATHEMATICS

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FUNCTIONS, , (c) , , , 1, (b) [f (2x) + f (2y)], 2, 1, (c) [f(x+ y). f(x – y)], 2, , (d) None of these, , 2 x  y  2 x  y 2 x  y  2 x  y, ., 2, 2, , Solution:(b) f(x + y).f(x – y) =, , 22 x  22 y  22 x  22 y, 4, 2x, , 1 2  22 x 22 y  22 y ,  , , , 2, 2, 2, , , , , =, , 1, [ f (2x) + f(2y) ], 2, , 8. If e x  y  1  y 2 , then y =, e x  e x, 2, , (b), , (c) e x  e  x, , e x  e x, 2, , x, 2, 2, Squaring both the sides, (e  y)  (1  y ), , e 2 x  y 2  2ye x  1  y 2  e 2 x  1  2ye x, , e 2x  1, 2, y, ,  2y  e x  e  x, , ex, e x  e x, ., 2, , 1,  is, x, , 9. If f ( x)  4 x3  3x 2  3x  4 , then x 3 f , , 1, f ( x), , f (u, v) , , x2  y 2, u 2  v2,  f ( x, y ) , ., 4, 4, , f ( x), , 11. If e, , 10  x, , x  (10, 10) and, 10  x, ,  200 x , , then k , f ( x )  kf , 2 ,  100  x , (a) 0.5, (b) 0.6, (c) 0.7, (d) 0.8, 10, , x, Solution:(a) e f ( x ) , , x  (10,10), 10  x,  f ( x)  log  10  x ,  10  x , 10 , , 200 x , , 2, , 100  x 2 ,  10  x ,  2 log ,   2 f ( x),  10  x ,  f ( x)  1 f  200 x 2   k  1  0.5., 2  100  x , 2, , ,  e x  y  1  y2, , (b), , 2, 2, x2  y 2, (d) x 2 y, 4, a, Solution:(c) Given f ( x  ay, x  ay )  axy ...(i), Let x  ay  u and x  ay  v, uv, uv, Then x , and y , 2, 2a, Substituting the value of x and y in (i), we obtain, , , , (d) e x  e  x, , (a) f (  x ), , 3 3, 1, 4, , Solution:(d) x 3 f    x 3  3  2   4 , x, x,  x, x, , 2, 3,  4  3 x  3 x  4 x  f ( x) ., 10. If f ( x  ay, x  ay )  axy , then f ( x, y ) is, equal to, (a) xy, (b) x 2  a 2 y 2, ,  f  200 x   log , 100  x 2   log 10(10  x) , , , 10(10  x) , 2 , 200 x ,  100  x , , , 10 , , , x, 2, Solution:(b)  e  y  1  y, , Hence, y , , (d) f ( x ), , (c), , 6. If f : R  R. f (x) = 2x + |x| , then, f (3x) – f (–x) – 4x equals, (a) f(x), (b) – f(x), (c) f (–x), (d) 2f(x), Solution:(d) f (3x) – f (–x) – 4x, = 6x + |3x| – {–2x + | –x|} – 4x, = 6x + 3 |x| + 2x – |x| – 4x, = 4x + 2 |x| = 2 f(x)., 7. Let f : R  R be defined by f ( x)  2 x  | x | ,, then f (2 x)  f ( x)  f ( x) , a) 2x, (b) 2 | x | (c) 2x, (d) 2 | x |, Solution:(b) f (2 x)  2(2 x) | 2 x |  4 x  2 | x | ,, f (x)  2 x | x |  2 x | x | ,, f (2 x)  f ( x)  f ( x)  4 x  2 | x |  | x | 2 x  2 x  | x |  2 | x |, , (a), ,  1 , f  ,  x , , 2, , 12. The graph of the function y  f ( x) is, symmetrical about the line x  2 , then, (a) f ( x)   f ( x), (b) f (2  x)  f (2  x), (c) f ( x)  f ( x), (d) f ( x  2)  f ( x  2), Solution:(b) f ( x)  f ( x)  f (0  x)  f (0  x) is, Symmetrical about x  0 .  f (2  x)  f (2  x) is, symmetrical about x  2 ., 13. The function f satisfies the functional equation,  x  59 , 3 f (x )  2 f ,   10 x  30 for all real x  1 .,  x 1 , , The value of f (7) is, CHETHAN M G, , [email protected], Phone number-8105418762 / 7019881906, , 11

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(a)8, , (b)4, , (c)–8, , CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, TOPIC-3-ODD AND EVEN FUNCTIONS, , (d)11, , Solution:(b) 3 f ( x)  2 f  x  59   10 x  30, ,  x 1 , 3, f, (, 7, )  2 f (11)  70  30  100, x, , 7, For, ,, For x  11 , 3 f (11)  2 f (7)  140, f (7), f (11), 1, , ,  f (7)  4,  20  220, 94, ,  1 x , 14. If f  x   log e ,  , x  1 ,then,  1 x , is equal to:, (a) 2 f ( x 2 ), (c)  f  x  , ,  2x , f, 2 ,  1 x , , (d) 2 f ( x), ,  1 x , ,  1 x ,  1 x  2x , , , , , 1  x2 , , log, , e, , 2,  1 x  2x , , , , ,  1  x2 , ,  1  x  ,  1  x  ,  2 f  x,  log e ,  1  x    2log e  1  x  , , , , , 15. Let f  x  be a quadratic polynomial such that, 2, , f  1  f  2   0 . If one of the roots of f  x   0, is 3,then its other root lies in., (a)  3, 1, (b) 1,3, (d)  0,1, , [JEE-MAIN-02/09/2020-SHIFT-II], [One option correct type-4M], , Solution:(c) Let f  x   a  x  3 x     0, Let  be the other root of f  x   0 ., Consider f  1  f  2   0, , a  1  3 1     a  2  3 2     0, 4a 1     a  1 2     0, 4 1      1 2     0, , 5  2   , , 12, , 2, b. G x   1  f x   f  x  is an odd function, 2, , 1,  f x   f  x   1  f x   f  x , 2, 2, , LEVEL-I, , 2, ,  1  x 2 ,  1  x2  2x , ,  log e ,   log e , 2, 2, , 1, , x, , ,  1 x  2x , , , , 4  4  2    0, , a. F  x   1  f  x   f  x  is an even function, , SOLVED PROBLEMS, , Solution:(d) f  x   log e , , (c)  1,0 , , 2. Given a function f x  , the function is defined by, , f x  , , [JEE-MAIN-8/4/2019-SHIFT-I], [One option correct type-4M], , 2x, , 1, 2, , 2, x, , , f,  log e  1  x, 2 ,  1 x ,  1 2x 2,  1 x, , e. f x  is even function if f  x   f x , f. f x  is odd function if f  x   f x, , c. Every function f x  can be expressed a sum of an, even & odd function, , (b) 2 f ( x), 2, , 1. Let f x  be any function ,then, , 2,  0.4   1, 0 , 5, , 1. Which of the following function is even function, (a) f ( x)  sin x, (b) f ( x)  tan x, (c) f ( x)  cos ecx, (d) f ( x)  cos x, Solution:(d) f ( x)  cos   x   cos x  f  x , 2. Which of the following function is odd function, (a) f ( x)  sin x, (b) f ( x)   tan x, (c) f ( x)  cos ec 2 x, , (d) f ( x)  cos x, , Solution:(a) f ( x)  sin   x    sin  x    f ( x), 3. Which of the following function is even function, (a) f ( x)  sin 2 x  cos x, (b) f ( x)  sin x  tan x, (c) f ( x)  x3  tan x  x, (d) f ( x)  x 5, Solution:(a) f ( x)  sin 2 x  cos x, , f ( x)  sin 2   x   cos   x , , f ( x)  sin 2  x   cos  x   f ( x), 4. The function sin 3 x cos10 x , is, (a) An even function, (b) An odd function, (c) Neither an even nor odd function, (d) None of these, Solution:(b) f  x   sin 3 x  cos10 x, ,  f   x   sin 3   x   cos10   x ,  f x   f  x, 5. The function sin 7 x , is, (a) An even function, (b) An odd function, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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FUNCTIONS, , (c) Neither an even nor odd function, (d) None of these, Solution:(b) Let f  x   sin 7 x, , 1 x , 1 x ,   log, , 1 x , 1 x , , and f ( x )  log, , 1 x ,   log,    f (x ), 1 x , , f   x   sin   x    sin  x    f  x , 7, , 7, ,  f (x ) is an odd function., , 6. The function sin 5 x cos 4 x , is, (a) An even function, (b) An odd function, (c) Neither an even nor odd function, (d) None of these, Solution:(b) f  x   sin 5 x cos4 x, , 3. The function f (x )  sin log(x  x 2  1 ) is, , , Solution:(b) f (x )  sin  log (x  1  x 2 ), , , f   x    sin 5  x  cos4  x    f  x , , , ,  f ( x )  sin[log ( x  1  x 2 )], , , 2, ,  f ( x )  sin log ( 1  x  x ), , 7. The function x 2  cos x , is, (a) An even function, (b) An odd function, (c) Neither an even nor odd function, (d) None of these, Solution:(a) f  x   x 2  cos x, , , , , , ( 1  x 2  x ) , , ( 1  x 2  x) , , , ,  (x  1  x 2 ) , 1, ,  f ( x )  sin log , , 2, , 2 1 , ,  f ( x )  sinlog(x  1  x ) , , f   x   x 2  cos  x   f  x , , 2 , ,  f ( x )  sin log(x  1  x ), , , , f   x     x   cos   x , , (a) f (x ) , , , ,  a 1, , (b) f (x )  x  x, , , a 1, , x, , a 1, , a 1, , (c) f (x ) , , x, , a x  a x, , (d) f (x )  sin x, , a x  a x, , Solution:(b) In (a),, a x  1 1  a x, ax  1, f ( x )   x, , , ,   f (x ), a 1 1  ax, ax 1, So, it is an odd function., In (b), f ( x )  ( x ), , a, a, , x, , 1, , x, , 1, ,  x, , 1a, , x, , 1a, , x, , x, , a 1, x, , ax  1, , a, , x, , a, , x, ,   f (x ) So,, ,  f (x ), , it is an odd function., , 1 x, , then f (x ) is, 1x, , (a) Even function, f (x 1 ),  f (x 1  x 2 ), f (x 2 ), , (b) f (x 1 ) f (x 2 )  f (x 1  x 2 ), (d) Odd function, , 1  x , , Solution:(d) Here, f (x )  log, 1  x , CHETHAN M G, ,  f ( x )   f (x ),  f (x ) is odd function., 4. The function f (x )  log(x  x 2  1 ) , is, (a) An even function, (b) An odd function, (c) A Periodic function, (d) Neither an even nor odd function, f ( x )   log(x  x 2  1 )   f (x ), , In (d), f ( x )  sin(  x )   sin x   f (x ) So, it is an odd, function., 2. If f (x )  log, , , , Solution:(b) f (x )  log(x  x 2  1 ) and, , So, it is an even function., a x  a x, , , ,  f ( x )   sinlog(x  1  x 2 ), , 1. Which of the following function is even function, x, , , , , , LEVEL-II, , (c), , , , (a) Even function, (b) Odd function, (c) Neither even nor odd (d) Periodic function, , f   x   sin 5   x  cos4   x , , In (c), f ( x ) , , 1, ,  f ( x) is odd function., 5. Let f  x   a x  a  0  be written a, , f  x   f1  x   f 2  x  where f1  x  is an even, function and f 2  x  is an odd function. Then, , f1  x  y   f1  x  y  equals, (a) 2 f1 ( x) f1 ( y ), , (b) 2 f1 ( x  y) f 2 ( x  y), , (c) 2 f1 ( x) f 2 ( y), , (d) 2 f1 ( x  y ) f1 ( x  y ), , [JEE-MAIN-8/4/2019-SHIFT-II], [One option correct type-4M], , Solution:(a) (i) W.K.T Every function can be expressed, as sum of an even and odd function, i.e. f  x  , , 1, 1, f  x   f   x    f  x   f   x  , , 2, 2, Even, Odd, , [email protected], Phone number-8105418762 / 7019881906, , 13

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CONCEPTS OF MATHEMATICS FOR KCET-JEE MAIN AND ADVANCED, , Compare with f  x   f1  x   f 2  x , , , , 1, 1, f  x   f   x   a x  a x, , 2, 2, 1, 1 x x, & f2  x    f  x   f   x   a  a, 2, 2, (ii) f1  x  y   f1  x  y ,  f1  x  , , , , , , , , , , , , , , , 1 x y, 1,  a  a  x  y    a x  y  a  x  y , 2, 2, 1 x y, 1,  a  a  x  y    a x  y  a  x  y , 2, 2, 1 y x, 1, a  a  a  x   a  y  a x  a  x , 2 , 2, 1 x,  a  a  x   a y  a  y   f1  y   1  a y  a  y  , , , 2, 2, , , f1 ( x)2 f1 ( y ), ,  2 f1 ( x) f1 ( y ), TOPIC-4-PERIODIC FUNCTIONS, , If a function f is periodic with period P, then for all x, in the domain of f and all positive integers n ,, f  x  nP   f  x , SOLVED PROBLEMS, LEVEL-I, , 1. Function f ( x)  sin x . This function is, (a) A periodic function whose period is 2, (b) A periodic function whose period is, , 1, 2, , (c) A periodic function whose period is 1, (d) Not a periodic function, Solution:(a), 2. Function f ( x)  tan x . This function is, (a) A periodic function whose period is , (b) A periodic function whose period is, , 1, 2, , (c) A periodic function whose period is 1, (d) Not a periodic function, Solution:(a), 3. Function f ( x)  x  [ x], where [] shows a, greatest integer. This function is, (a) A periodic function, (b) A periodic function whose period is, , 1, 2, , (c) A periodic function whose period is 1, (d) Not a periodic function, Solution:(c) It is well known fact that fractional, function always a periodic function whose period is 1., , 14, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS