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Q)

The value of $x$ satisfying $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$ are

$(a)\;0,\large\frac{1}{2}$$\qquad(b)\;1,2\qquad(c)\;0,2\qquad(d)\;None\;of\;these$

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A)
We have $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$
$\Rightarrow \large\frac{\pi}{2}$$-\cos^{-1}x+\large\frac{\pi}{2}$$-\cos^{-1}(1-x)=\cos^{-1}x$
$\Rightarrow 2\cos^{-1}x=\pi-\cos^{-1}(1-x)$
$\Rightarrow \cos^{-1}(2x^2-1)=\cos^{-1}(x-1)$
$\cos^{-1}(-x)=\pi-\cos^{-1}x$
$\Rightarrow 2x^2-1=x-1$
$\Rightarrow x(2x-1)=0$
$x=0$
$2x-1=0$
$2x=1$
$x=\large\frac{1}{2}$
$\therefore x=0,\large\frac{1}{2}$
Hence (a) is the correct answer.
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