Let `f(x)=sin^(-1)x+|sin^(-1)x|+sin^(-1)|x|` If the equation f(x) = k has two solutions, then true set of values of k is

Options

A. `k in (0,(pi)/(2))`
B. `k in[0,(pi)/(2)]`
C. `k in(0,(pi)/(2)]`
D. `k in [0,(pi)/(2))`

Answer

C

Explanation

`f(x)={{:(3sin^(-1)x,,x in[0,1]),(-sin^(-1)x,,x in[-1,0)):}`
The graph of the function y = f(x) is an shown in the following figure

From the graph f(x) = x has only one solution, x = 0
Range of y = f(x) is `[0, 3pi//2]`
For f(x) = k having two solutions, `k in (0, pi//2]`