The value of `a` for which `a x^2+sin^(-1)(x^2-2x+2)+cos^(-1)(x^2-2x+2)=1` has a real solution is `pi/2` (b) `-pi/2` (c) `2/pi` (d) `-2/pi`

Options

A. `(pi)/(2)`
B. `-(pi)/(2)`
C. `(2)/(pi)`
D. `-(2)/(pi)`

Answer

B

Explanation

The given equation is `ax^(2) + sin^(-1) ((x -1)^(2)+1) + cos^(-1) ((x -1)^(2) + 1) = 0`
Now, `-1 le (x -1)^(2) + 1 le 1 rArr x = 1`
So, we have `a + (pi)/(2) = 0 " or " a = -(pi)/(2)`