Question

If $\sin^{-1}x+\cos^{-1}(1-x)=\sin^{-1}(-x)$ , then $x$ satisfies the equation

• A. $2x^2-x+2=0$
• B. $2x^2-3x=0$
• C. $2x^2+x-1=0$
• D. none of these

Answer 1

Answer: (B)

$sin^{-1}x+cos^{-1}\left(1-x\right) = sin^{-1}\left(-x\right) $
$ \Rightarrow cos^{-1}\left(1-x\right) = - sin^{-1}\left(x\right)+sin^{-1}\left(-x\right)$
$\Rightarrow cos^{-1} \left(1-x\right) = -2 sin^{-1}x $
$ \Rightarrow 1-x = cos \left(-2 sin^{-1}x\right)$
$ = cos \left(2 sin^{-1}x\right) = 1-2sin^{2}\left(sin^{-1}x\right) = 1-2x^{2} $
$\Rightarrow $ Either $x= 0$ or $x= 1/2 $
But $x= 1/2$ does not satisfy the given equation
$ \therefore x=0 $