Question 1

Consider the following statements in respect of the function f(x)=x in the interval (-1,1) : I. The function attains maximum value. II. The function attains minimum value. Which of the statements give

Accepted Answer

Given the function f(x) = x defined on the open interval (-1, 1). For any x (-1, 1), we have -1 Since the interval is open, x can be arbitrarily close to 1 but never equals 1. Thus, f(x) does not attain a maximum value in the given interval. Similarly, x can be arbitrarily close to -1 but never equals -1. Thus, f(x) does not attain a minimum value in the given interval. Therefore, neither statement I nor statement II is correct. Answer: Neither I nor II

Question 2

Passage: Consider the curves y=x^2 and y=2|x|. Question: What is the number of points of intersection of the curves ?

Accepted Answer

Equating the two given curves y = x^2 and y = 2|x|: x^2 = 2|x| Since x^2 = |x|^2, the equation becomes: |x|^2 - 2|x| = 0 |x|(|x| - 2) = 0 This gives |x| = 0 or |x| = 2. Solving for x, we get x = 0, x = 2, and x = -2. The corresponding points of intersection are (0, 0), (2, 4), and (-2, 4). The number of points of intersection is 3. Answer: 3

Question 3

If 2f(x)+f(1-x)=x, then what is f(x) equal to ?

Accepted Answer

Given equation is 2f(x) + f(1-x) = x. Replacing x with 1-x, we get: 2f(1-x) + f(x) = 1-x Multiplying the first equation by 2, we get: 4f(x) + 2f(1-x) = 2x Subtracting the second equation from this, we get: 3f(x) = 2x - (1-x) 3f(x) = 3x - 1 f(x) = x - 1 3 Answer: x- ( 1 3 )

Question 4

For the following two (02) items: Consider the function f(x) = 1 - [3] (x-1)^2 . What is the domain of the function?

Accepted Answer

The given function is f(x) = 1 - [3] (x-1)^2 . The expression inside the cube root is (x-1)^2, which is a polynomial and is defined for all real values of x. Since the cube root of any real number is well-defined for all real numbers, the function f(x) is defined for all x R . Therefore, the domain of the function is (- , ). Answer: (- , )

Question 5

Consider the following for the two (02) items that follow: Let the function y = (1 - x)^ -1 , where x 2n and n is an integer. What is the range of the function?

Accepted Answer

Given function is y = (1 - x)^ -1 = 1 1 - x The range of x is [-1, 1]. Since x 2n , x 1. Therefore, -1 x Multiplying by -1, we get -1 Adding 1, we get 0 Taking the reciprocal, we get 1 1 - x 1 2 Thus, y 0.5 The range of the function is [0.5, ).

Question 6

For the following two (02) items: Consider the function f(x) = 10^x - 10^ -x 10^x + 10^ -x . What is f f f f f(0) equal to?

Accepted Answer

Given the function f(x) = 10^x - 10^ -x 10^x + 10^ -x Substituting x = 0 into the function: f(0) = 10^0 - 10^ -0 10^0 + 10^ -0 f(0) = 1 - 1 1 + 1 = 0 Since f(0) = 0, applying the function f repeatedly to 0 will always result in 0. f(f(0)) = f(0) = 0 f(f(f(0))) = f(0) = 0 f(f(f(f(0)))) = f(0) = 0 f(f(f(f(f(0))))) = f(0) = 0 Therefore, f f f f f(0) = 0 Answer: 0

Question 7

Consider the following for the two (02) items that follow: Let the curve f(x) = |x - 3|. What is the domain of the function f(x)?

Accepted Answer

The given function is f(x) = |x - 3|. The expression x - 3 is a linear polynomial, which is defined for all real values of x. The absolute value function is also defined for all real numbers. Thus, f(x) is defined for all x R . The domain of f(x) is (- , ). Answer: (- , )

Question 8

Consider the following for the two (02) items that follow: Let f = \ (1, 1), (2, 4), (3, 7), (4, 10)\ . If f(x) = px + q, then what is the value of (p + q)?

Accepted Answer

Given f(x) = px + q. From the given set of points, we have f(1) = 1. Substituting x = 1 in the function, we get: p(1) + q = 1 p + q = 1 Alternatively, using f(2) = 4, we get 2p + q = 4. Solving p + q = 1 and 2p + q = 4 gives p = 3 and q = -2. Thus, p + q = 3 - 2 = 1. Answer: 1

Question 9

For the following two (02) items: Consider the function f(x) = x 1 - x (x > 0, x 1). What is (1 - x)f( x ) + xf( x + 1) equal to?

Accepted Answer

Given f(x) = x 1 - x Substitute x in f(x): f( x ) = x 1 - x Substitute x + 1 in f(x): f( x + 1) = x + 1 1 - ( x + 1) = x + 1 - x Now, evaluate the given expression: (1 - x)f( x ) + xf( x + 1) = (1 - x) ( x 1 - x ) + x ( x + 1 - x ) Using 1 - x = (1 - x )(1 + x ): = (1 - x )(1 + x ) ( x 1 - x ) - x ( x + 1) = x (1 + x ) - x ( x + 1) = 0 Answer: 0

Question 10

Consider the following for the two (02) items that follow: The function f(x) satisfies f ( x y ) = f(x) f(y) for all positive real values of x and y, and f(2) = 3. What is f(16) equal to?

Accepted Answer

Given f ( x y ) = f(x) f(y) Substituting x = 1 and y = 1, we get f(1) = 1 Substituting x = 1, we get f ( 1 y ) = f(1) f(y) = 1 f(y) Replacing y with 1 y in the given equation, we get: f(xy) = f(x) f ( 1 y ) = f(x)f(y) Using this property, f(16) = f(2^4) = (f(2))^4 Given f(2) = 3, we have f(16) = 3^4 = 81 Answer: 81

Question 11

Consider the following for the two (02) items that follow: The function f(x) satisfies f ( x y ) = f(x) f(y) for all positive real values of x and y, and f(2) = 3. What is f(1)f(4) equal to?

Accepted Answer

Given f ( x y ) = f(x) f(y) for all x, y > 0. Substituting x = 1 and y = 1, we get: f(1) = f(1) f(1) f(1) = 1 Substituting x = 4 and y = 2, we get: f ( 4 2 ) = f(4) f(2) f(2) = f(4) f(2) f(4) = (f(2))^2 Since f(2) = 3, we have f(4) = 3^2 = 9. Therefore, f(1)f(4) = 1 9 = 9. Answer: 9

Question 12

Consider the following for the two (02) items that follow: A function f is such that f(xy) = f(x+y) for all real values of x and y, and f(5) = 10. What is f(20) + f(-20) equal to?

Accepted Answer

Given f(xy) = f(x+y) for all real values of x and y. Substituting y = 0, we get f(x 0) = f(x + 0), which simplifies to f(0) = f(x) for all real x. This implies that f(x) is a constant function for all real values of x. Since f(5) = 10, the constant value of the function is 10. Thus, f(x) = 10 for all x R . We need to find the value of f(20) + f(-20). f(20) = 10 f(-20) = 10 f(20) + f(-20) = 10 + 10 = 20 Answer: 20

Question 13

For the following two (02) items: Consider the function f(x) = x 1 - x (x > 0, x 1). What is f(x) f(x+1) equal to?

Accepted Answer

Given f(x) = x 1 - x Substitute x+1 for x to find f(x+1): f(x+1) = x+1 1 - (x+1) = x+1 -x Now, calculate the ratio f(x) f(x+1) : f(x) f(x+1) = x 1 - x x+1 -x f(x) f(x+1) = x 1 - x -x x+1 = -x^2 1 - x^2 Since f(x^2) = x^2 1 - x^2 , we get: f(x) f(x+1) = -f(x^2) Answer: -f(x^2)

Question 14

For the following two (02) items: Consider the function f(x) = 10^x - 10^ -x 10^x + 10^ -x . What is the inverse of the function?

Accepted Answer

Let y = f(x) = 10^x - 10^ -x 10^x + 10^ -x Multiplying the numerator and the denominator by 10^x, we get: y = 10^ 2x - 1 10^ 2x + 1 Applying componendo and dividendo: 1 + y 1 - y = 10^ 2x + 1 + 10^ 2x - 1 10^ 2x + 1 - (10^ 2x - 1) 1 + y 1 - y = 2 10^ 2x 2 10^ 2x = 1 + y 1 - y Taking logarithm to the base 10 on both sides: 2x = _ 10 ( 1 + y 1 - y ) x = 1 2 _ 10 ( 1 + y 1 - y ) Replacing y with x to find the inverse function: f^ -1 (x) = 1 2 _ 10 ( 1 + x 1 - x ) Answer: 1 2 _ 10 ( 1 + x 1 - x )

Question 15

Consider the following for the two (02) items that follow: Let f = \ (1, 1), (2, 4), (3, 7), (4, 10)\ . Consider the following statements: I. f is one-one function. II. f is onto function if the codom

Accepted Answer

The given function is f = \ (1, 1), (2, 4), (3, 7), (4, 10)\ . The domain of f is \ 1, 2, 3, 4\ and the range of f is \ 1, 4, 7, 10\ . Since each element in the domain has a distinct image in the range, f is a one-one function. Statement I is correct. If the codomain is the set of natural numbers, the range \ 1, 4, 7, 10\ is a proper subset of the codomain. Since the range is not equal to the codomain, f is not an onto function. Statement II is incorrect. Therefore, only Statement I is correct.

Question 16

Consider the following for the two (02) items that follow: A function f is such that f(xy) = f(x+y) for all real values of x and y, and f(5) = 10. What is f(0) equal to?

Accepted Answer

Given the functional equation f(xy) = f(x+y) for all real values of x and y. Substituting y = 0 into the given equation: f(x 0) = f(x + 0) f(0) = f(x) This shows that f(x) is a constant function, meaning it takes the same value for all real x. We are given that f(5) = 10. Since f(x) = f(0) for all x, substituting x = 5 gives: f(5) = f(0) Therefore, f(0) = 10. Answer: 10

Functions — Previous Year Questions & Solutions