Solution: `sin^(-1) (1-x) -2 sin^(-1) x = pi/2`
Putting `pi/2 = sin^(-1) (1-x)^2 + cos^(-1) (1-x)`
or `sin^(-1) (1-x) -2 sin^(-1) x = sin^(-1) (1-x) + cos^(-1)`
`(1-x) => -2 sin^(-1) x = cos^(-1) (1-x)`
Let `sin^(-1) x = alpha :. sin alpha = x`
`:. -2 sin^(-1) x = -2 alpha = cos^(-1) (1-x)`
or `cos 2 alpha =1 -x :. cos (- theta) = cos theta`
`:. 1-2 sin^2 alpha = 1-x`
Putting `sin alpha = x => 1-2x^2 =1-x`
or `2x^2 -x= 0 ` `x (2x -1) =0 :. x=0, 1/2`
But `x= 1/2` does not satisfy the equation
`:. x= 0`
`:.` Option (c) is correct.
Correct Answer is `=>` (C) `0`