Question

A cubic function f (x) = x³ + ax² + bx + c . If f (x) is an odd function and f (x) = 0 at x = −1. Now the domain of function is reduced, so as to make f (x) invertible such that f ⁻¹ (x) remains in 2nd and 4th quadrant. Then

1.   |f^-1(x)| + f^-1(|x|) = 0 has

(A)  no solution

(B)  exactly one solution

(C)  infinite solutions

(D)  exactly three solutions

2.  Range of f x ⁻¹ |x| is

(A)  (−∞, 0) 

(B)  (−∞, −1)

(C)  (0 ,∞)

(D)  (1, ∞) 

Answer 1

Correct option 1. (c) ,2. (B)

1, See Fig. f⁻¹ remains in the 2nd and 4th quadrants. So, f (x) is defined as

2 . See Fig. f x ⁻¹ (x) is defined as