Question

The value(s) of $x$ satisfying the equation $\left(sin\right)^{- 1} \left(1 - x\right)-2\left(sin\right)^{- 1} ⁡ x=\frac{\pi }{2}$ is/are

• A. $0$
• B. $\frac{1}{2}$
• C. $0,\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

We have, $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}+2 \sin ^{-1} x$
$\Rightarrow 1-x=\sin \left(\frac{\pi}{2}+2 \sin ^{-1} x\right)$
$\Rightarrow 1-x=\cos \left(2 \sin ^{-1} x\right)$
$\Rightarrow 1-x=\cos \left\{\cos ^{-1}\left(1-2 x^{2}\right)\right\}\left[\because 2 \sin ^{-1} x=\cos ^{-1}\left(1-2 x^{2}\right)\right]$
$\Rightarrow 1-x=\left(1-2 x^{2}\right)$
$\Rightarrow x=2 x^{2} \Rightarrow x(2 x-1)=0 \Rightarrow x=0, \frac{1}{2}$
For, $x=\frac{1}{2},$ we have $LHS =\sin ^{-1}(1-x)-2 \sin ^{-1} x$
$=\sin ^{-1} \frac{1}{2}-2 \sin ^{-1} \frac{1}{2}=-\sin ^{-1} \frac{1}{2}=\frac{-\pi}{6} \neq R.H.S.$
So, $x=\frac{1}{2}$ is not a root of the given equation. Clearly, $x=0$ satisfies the equation

Here, $x=0$ is the root of the given equation.