MCQ
If $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ then $x$ is
- A$-\frac{1}{2}$
- B1
- C
- D$\frac{1}{2}$
$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$
$\therefore \sin ^{-1}(1-x)=\frac{\pi}{2}+2 \sin ^{-1} x$
$\therefore 1-x=\sin \left(\frac{\pi}{2}+2 \sin ^{-1} x \right)$
$\therefore 1- x =\cos \left(2 \sin ^{-1} x \right) \ldots\left[\cdot\left[\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta\right]\right.$
$\therefore 1-x=1-2\left[\sin \left(\sin ^{-1} x\right)\right]^2 \quad \ldots .\left[\because \cos 2 \theta=1-2 \sin ^2 \theta\right]$
$\therefore 1-x=1-2 x^2$
$\therefore 2 x^2-x=0$
$\therefore x(2 x-1)=0$
$\therefore x=0$ or $x=\frac{1}{2}$
When $x=\frac{1}{2}$
LHS $=\sin ^{-1}\left(1-\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)$
$=\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)$
$=-\sin ^{-1}\left(\frac{1}{2}\right)$
$=-\sin ^{-1}\left(\sin \frac{\pi}{6}\right)$
$=-\frac{\pi}{6} \neq \frac{\pi}{2}$
$\therefore x \neq \frac{1}{2}$
Hence, x = 0.
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