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, ∞)