Question 1

Let f:(1, ) R be a function defined as f(x) = x-1 x+1 . Let f^ i+1 (x) = f(f^i(x)), i=1, 2, , 25, where f^1(x)=f(x). If g(x) + f^ 26 (x) = 0, x (1, ), then the area of the region bounded by the curves

Accepted Answer

Given f(x) = x-1 x+1 . Let us find the first few compositions of f(x): f^2(x) = f(f(x)) = x-1 x+1 -1 x-1 x+1 +1 = x-1-x-1 x-1+x+1 = -2 2x = - 1 x f^3(x) = f(f^2(x)) = f (- 1 x ) = - 1 x -1 - 1 x +1 = -1-x -1+x = x+1 1-x f^4(x) = f(f^3(x)) = f ( x+1 1-x ) = x+1 1-x -1 x+1 1-x +1 = x+1-1+x x+1+1-x = 2x 2 = x Since f^4(x) = x, the sequence of functions is periodic with a period of 4. Therefore, f^ 26 (x) = f^ 4 6 + 2 (x) = f^2(x) = - 1 x . We are given g(x) + f^ 26 (x) = 0, which implies: g(x) - 1

Question 2

Let f be a polynomial function such that _2(f(x)) = ( _2 (2+ 2 3 + 2 9 + ) ) _3 (1+ f(x) f(1/x) ), x>0 and f(6)=37. Then _ n=1 ^ 10 f(n) is equal to ________.

Accepted Answer

The sum of the infinite geometric progression is given by: S = 2 + 2 3 + 2 9 + = 2 1 - 1 3 = 3 Substituting this into the given equation: _2(f(x)) = _2(3) _3 (1 + f(x) f(1/x) ) Using the base change property _a(b) _b(c) = _a(c): _2(f(x)) = _2 (1 + f(x) f(1/x) ) Equating the arguments: f(x) = 1 + f(x) f(1/x) f(x)f(1/x) - f(1/x) = f(x) f(x)f(1/x) = f(x) + f(1/x) The only polynomial functions satisfying this relation are of the form f(x) = 1 x^n. Given f(6) = 37: 1 6^n = 37 Taking the positive sign

Question 3

Let f: R R be defined as f(x) = 2x^2 - 3x + 2 3x^2 + x + 3 . Then f is :

Accepted Answer

Given f(x) = 2x^2 - 3x + 2 3x^2 + x + 3 To check for one-one, let f(x) = f(y) 2x^2 - 3x + 2 3x^2 + x + 3 = 2y^2 - 3y + 2 3y^2 + y + 3 Cross-multiplying and simplifying, we get: 11x^2y - 11xy^2 - 11x + 11y = 0 11xy(x - y) - 11(x - y) = 0 11(x - y)(xy - 1) = 0 This implies x = y or xy = 1. Since f(x) = f(1/x) for x 0 (for example, f(2) = f(1/2) = 4 17 ), the function is many-one (not one-one). To check for onto, let y = 2x^2 - 3x + 2 3x^2 + x + 3 (3y - 2)x^2 + (y + 3)x + (3y - 2) = 0 For x R , the

Question 4

Let e be the base of natural logarithm and let f: \ 1, 2, 3, 4\ \ 1, e, e^2, e^3\ and g: \ 1, e, e^2, e^3\ \ 1, 1 2 , 1 3 , 1 4 \ be two bijective functions such that f is strictly decreasing and g is

Accepted Answer

Given f: \ 1, 2, 3, 4\ \ 1, e, e^2, e^3\ is a strictly decreasing bijective function. Arranging the domain and codomain in increasing order, we get f(1) = e^3, f(2) = e^2, f(3) = e, and f(4) = 1. Given g: \ 1, e, e^2, e^3\ \ 1, 1 2 , 1 3 , 1 4 \ is a strictly increasing bijective function. Arranging the domain and codomain in increasing order, we get g(1) = 1 4 , g(e) = 1 3 , g(e^2) = 1 2 , and g(e^3) = 1. Evaluating (x) = [f^ -1 \ g^ -1 ( 1 2 ) \ ]^x: From the mapping of g, g^ -1 ( 1 2 ) = e^2.

Question 5

Let f: R R be a function such that f(x) + 3f ( 2 - x ) = x, x R . Let the maximum value of f on R be . If the area of the region bounded by the curves g(x) = x^2 and h(x) = x^3, > 0, is ^2, then 30 ^3

Accepted Answer

Given f(x) + 3f ( 2 - x ) = x Replacing x with 2 - x, we get: f ( 2 - x ) + 3f(x) = ( 2 - x ) = x Multiplying this equation by 3 and subtracting the first equation gives: 9f(x) - f(x) = 3 x - x 8f(x) = 3 x - x f(x) = 3 x - x 8 The maximum value of f(x) is = 3^2 + (-1)^2 8 = 10 8 . Thus, ^2 = 10 64 = 5 32 . The points of intersection of the curves g(x) = x^2 and h(x) = x^3 are given by: x^2 = x^3 x^2(1 - x) = 0 x = 0, x = 1 The area of the region bounded by the curves is: _ 0 ^ 1/ (x^2 - x^3) dx

Question 6

The sum of all the integral values of p such that the equation 3 ^2 x + 12 x - 3 = p, x R , has at least one solution, is:

Accepted Answer

Given equation: 3 ^2 x + 12 x - 3 = p Substituting ^2 x = 1 - ^2 x: 3(1 - ^2 x) + 12 x - 3 = p -3 ^2 x + 12 x = p Let x = t. Since x R , t [-1, 1]. The equation becomes p = -3t^2 + 12t. Let f(t) = -3t^2 + 12t. Differentiating with respect to t: f'(t) = -6t + 12 = 6(2 - t) For t [-1, 1], f'(t) > 0, which means f(t) is strictly increasing in the interval [-1, 1]. The minimum value of f(t) occurs at t = -1: f(-1) = -3(-1)^2 + 12(-1) = -15 The maximum value of f(t) occurs at t = 1: f(1) = -3(1)^2 +

Question 7

Let A = \ 1, 2, 3, 4, 5, 6\ . The number of one-one functions f: A A such that f(1) 3, f(3) 4 and f(2) + f(3) = 5, is __________.

Accepted Answer

Given A = \ 1, 2, 3, 4, 5, 6\ . Since f: A A is a one-one function, f(x) takes distinct values from A for each x A. We are given the condition f(2) + f(3) = 5. The possible pairs for (f(2), f(3)) from the set A are: (1, 4), (2, 3), (3, 2), (4, 1) Notice that for all these pairs, the condition f(3) 4 is automatically satisfied. We also have the condition f(1) 3. Since f is one-one, f(1) cannot take the values already assigned to f(2) and f(3). Let us find the number of choices for f(1) in each ca

Question 8

For the function f:[1, ) [1, ) defined by f(x)=(x-1)^4+1, among the two statements: (I) The set S=\ x [1, ): f(x)=f^ -1 (x)\ contains exactly two elements, and (II) The set S=\ x [1, ): f(x)=f^ -1 (x+

Accepted Answer

Given f(x) = (x-1)^4 + 1 for x 1. To find the inverse function f^ -1 (x), let y = (x-1)^4 + 1. (x-1)^4 = y-1 x-1 = (y-1)^ 1/4 x = (y-1)^ 1/4 + 1. Thus, f^ -1 (x) = (x-1)^ 1/4 + 1. Evaluating the first statement, since f'(x) = 4(x-1)^3 0 for x 1, f(x) is strictly increasing. For a strictly increasing function, the solutions to f(x) = f^ -1 (x) are the same as the solutions to f(x) = x. (x-1)^4 + 1 = x (x-1)^4 - (x-1) = 0. Let t = x-1 0. t^4 - t = 0 t(t-1)(t^2+t+1) = 0. Since t 0, the real roots a

Question 9

Let for some R , f: R R be a function satisfying f(x+y)=f(x)+2y^2+y+ xy for all x,y R . If f(0)=-1 and f(1)=2, then the value of _ n=1 ^ 5 ( +f(n)) is:

Accepted Answer

Given f(x+y) = f(x) + 2y^2 + y + xy Substituting x = 0 and replacing y with x, we get: f(x) = f(0) + 2x^2 + x + (0)x Since f(0) = -1, we have: f(x) = 2x^2 + x - 1 To find , substitute f(x) into the original functional equation: 2(x+y)^2 + (x+y) - 1 = (2x^2 + x - 1) + 2y^2 + y + xy 2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + xy Comparing both sides, we get = 4. We need to find the value of _ n=1 ^ 5 ( +f(n)): _ n=1 ^ 5 (4 + 2n^2 + n - 1) = _ n=1 ^ 5 (2n^2 + n + 3) = 2 _ n=1 ^ 5 n^2

Question 10

Let [ ] denote the greatest integer function. If the domain of the function f(x) = ^ -1 ( 4x+2[x] 3 ) is [ , ], then 12( + ) is equal to:

Accepted Answer

For the domain of f(x) = ^ -1 ( 4x+2[x] 3 ), the argument must satisfy: -1 4x+2[x] 3 1 -3 4x + 2[x] 3 Let x = I + f, where I = [x] Z and f = \ x\ [0, 1). Substituting x = I + f into the inequality: -3 4(I + f) + 2I 3 -3 6I + 4f 3 Since 0 f If I 1, 6I + 4f 6(1) + 0 = 6 > 3, which gives no solution. If I -2, 6I + 4f Thus, the only possible values for I are 0 and -1. Case 1: I = 0 -3 6(0) + 4f 3 -3 4f 3 - 3 4 f 3 4 Since f [0, 1), we get 0 f 3 4 . Therefore, x = I + f [0, 3 4 ]. Case 2: I = -1 -3 6

Question 11

The number of functions f: \ 1, 2, 3, 4\ \ a, b, c\ , which are not onto, is:

Accepted Answer

Total number of functions from a set of 4 elements to a set of 3 elements is 3^4 = 81. The number of onto functions is given by the inclusion-exclusion principle: 3^4 - ^ 3 C_ 1 2^4 + ^ 3 C_ 2 1^4 = 81 - 3 16 + 3 1 = 81 - 48 + 3 = 36 The number of functions which are not onto is the total number of functions minus the number of onto functions: 81 - 36 = 45 Answer: 45

Question 12

If the domain of the function f(x) = _ (0.6) ( | 2x-5 x^2-4 | ) is (- , a] \ b\ [c, d) (e, ), then the value of a + b + c + d + e is _______.

Accepted Answer

For the function f(x) = _ 0.6 ( | 2x-5 x^2-4 | ) to be defined, two conditions must be satisfied: 1. The argument of the logarithm must be strictly positive: | 2x-5 x^2-4 | > 0 2x - 5 0 x 5 2 Also, the denominator cannot be zero: x^2 - 4 0 x 2 2. The expression inside the square root must be non-negative: _ 0.6 ( | 2x-5 x^2-4 | ) 0 Since the base of the logarithm is 0.6 | 2x-5 x^2-4 | (0.6)^0 | 2x-5 x^2-4 | 1 Since both sides are non-negative, we can square both sides: (2x-5)^2 (x^2-4)^2 (x^2-4)

Question 13

Given below are two statements: Statement I: The function f: R R defined by f(x)= x 1+|x| is one-one. Statement II: The function f: R R defined by f(x)= x^ 2 +4 x-30 x^ 2 -8 x+18 is many-one. In the l

Accepted Answer

Statement I: f(x) = x 1+|x| . For x 0, f(x) = x 1+x is strictly increasing. For x 0, also strictly increasing. The function is continuous at x = 0, so f is strictly increasing on R and hence one-one. Statement I is true. Statement II: f(x) = x^2+4x-30 x^2-8x+18 . The denominator (x-4)^2 + 2 > 0 for all x. f'(x) = -12(x^2 - 8x + 14) (x^2 - 8x + 18)^2 f'(x) = 0 x = 4 2 . Since f' changes sign, f is not monotonic and hence many-one. Statement II is true. Both statements are true.

Question 14

The sum of all the elements in the range of f(x)= Sgn ( x)+ Sgn ( x)+ Sgn ( x)+ Sgn ( x), x n 2 , n Z , where Sgn ( t )= \ aligned 1, & if t >0 \ -1, & if t <0 aligned ., is :

Accepted Answer

x (0, /2) y = 1 + 1 + 1 + 1 = 4 x ( /2, ) y = 1 - 1 - 1 - 1 = -2 x ( , 3 /2) y = -1 - 1 + 1 + 1 = 0 x (3 /2, 2 ) y = -1 + 1 - 1 - 1 = -2 Range of y is \ -2, 0, 4\ Required sum = -2 + 0 + 4 = 2

Question 15

If g(x)=3 x^ 2 +2 x-3, f(0)=-3 and 4 g(f(x))=3 x^ 2 -32 x+72, then f(g(2)) is equal to:

Accepted Answer

Let f(x) = ax + b. Then 4g(f(x)) = 12(ax+b)^2 + 8(ax+b) - 12. Comparing with 3x^2 - 32x + 72: x^2: 12a^2 = 3 a = 1 2 Constant: 12b^2 + 8b - 12 = 72 3b^2 + 2b - 21 = 0 b = 7 3 or b = -3. Since f(0) = -3, b = -3. x: 8a(3(-3)+1) = -32 -8a = -4 a = 1 2 . So f(x) = x 2 - 3. g(2) = 3(4) + 4 - 3 = 13. f(g(2)) = f(13) = 13 2 - 3 = 7 2 .

Question 16

Let f be a function such that 3 f(x)+2 f ( m 19 x )=5 x, x 0, where m= _ i=1 ^ 9 (i)^ 2 . Then f(5)-f(2) is equal to

Accepted Answer

m = _ i=1 ^ 9 i^2 = 9 10 19 6 = 285, so m 19 = 15. The equation becomes 3f(x) + 2f ( 15 x ) = 5x ...(i) Replace x by 15 x : 3f ( 15 x ) + 2f(x) = 75 x ...(ii) From 3 (i) - 2 (ii): 5f(x) = 15x - 150 x , so f(x) = 3x - 30 x . f(5) = 15 - 6 = 9, f(2) = 6 - 15 = -9. f(5) - f(2) = 9 - (-9) = 18.

Question 17

If the domain of the function f(x)= _ (10 x^ 2 -17 x+7 ) (18 x^ 2 -11 x+1 ) is (- , a) (b, c) (d, )-\ e\ , then 90(a+b+c+d+e) equals:

Accepted Answer

For f(x) = _ (10x^2-17x+7) (18x^2-11x+1), we need: Base > 0: 10x^2-17x+7 > 0 10(x-1)(x-7/10) > 0 x 1 Base 1: 10x^2-17x+6 = 0 x = 1/2 or x = 6/5 Argument > 0: 18x^2-11x+1 > 0 18(x-1/2)(x-1/9) > 0 x 1/2 Intersecting conditions 1 and 3: (- , 1/9) (1/2, 7/10) (1, ) Removing base = 1 points: x = 1/2 is already excluded; x = 6/5 (1, ) must be removed. Domain: (- , 1/9) (1/2, 7/10) (1, ) - \ 6/5\ So a = 1/9,\; b = 1/2,\; c = 7/10,\; d = 1,\; e = 6/5. 90(a+b+c+d+e) = 90 ( 1 9 + 1 2 + 7 10 +1+ 6 5 ) = 10

Question 18

Consider two sets A =\ x Z :|(|x-3|-3)| 1\ and B = \ x R -\ 1,2\ : (x-2)(x-4) x-1 _ e (|x-2|)=0 \ . Then the number of onto functions f: A B is equal to

Accepted Answer

For set A: |(|x-3|-3)| 1 2 |x-3| 4. If x 3: 5 x 7 x \ 5,6,7\ . If x So A = \ -1,0,1,5,6,7\ , |A|=6. For set B: (x-2)(x-4) x-1 _e(|x-2|) = 0 with x R -\ 1,2\ . Either (x-2)(x-4) x-1 =0 x=4 (since x 1,2), or _e|x-2|=0 |x-2|=1 x=3 (since x 1). So B = \ 3,4\ , |B|=2. Number of onto functions from A to B = 2^6 - 2 = 62.

Question 19

Let the domain of the function f(x)= _ 3 _ 5 (7- _ 2 (x^ 2 -10 x+85 ) )+ ^ -1 ( | 3 x-7 17-x | ) be ( , ]. Then + is equal to :

Accepted Answer

For _3 _5( ) we need _5( )>0 So 7- _2(x^2-10x+85)>1, giving x^2-10x+85 (x-3)(x-7) 3 For ^ -1 ( | 3x-7 17-x | ) we need |3x-7| |17-x| i.e, (x-6)(x+5) 0 -5 x 6. Intersection: (3,6], so =3, =6. + =9

Question 20

Let f and g be functions satisfying f(x+y)=f(x) f(y), f(1)=7 and g(x+y)=g(x y), g(1)=1, for all x, y N . If _ x=1 ^ n ( f(x) g (x) )=19607, then n is equal to:

Accepted Answer

f(x+y)=f(x)f(y) with f(1)=7 gives f(x)=7^x. For g: put y=1 in g(x+y)=g(xy) to get g(x+1)=g(x). So g(x)=g(1)=1 for all x N . _ x=1 ^ n f(x) g(x) = _ x=1 ^ n 7^x= 7(7^n-1) 6 =19607 7^ n+1 -7=117642 7^ n+1 =117649=7^6 n+1=6 n=5

Functions — Previous Year Questions & Solutions