# The trigonometric equation $sin^{-1}x=2sin^{-1}a,$ has a solution for Option 1) all real values Option 2) $\left | a \right |< \frac{1}{2}\; \;$ Option 3) $\; \left | a \right |\geqslant \frac{1}{\sqrt{2}}\; \; \;$ Option 4) $\; \frac{1}{2}< \left | a \right |< \frac{1}{\sqrt{2}}$

S Sabhrant Ambastha

As we learnt in

Domains and Ranges of Inverse Trigonometric Functions -

For $\sin ^{-1}x$

Domain $\epsilon \left [ -1, 1 \right ]$

Range $\epsilon \left [ -\frac{\pi }{2}, \frac{\pi }{2} \right ]$

-

$sin^{-1}x=2sin^{-1}a\ \, \, \, .........(1)$

For (1) to have solution,

$-\frac{\pi }{2}\leq 2\sin ^{-1}a\leq \frac{\pi }{2}$

$\Rightarrow -\frac{\pi }{4}\leq \sin ^{-1}a\leq \frac{\pi }{4}$

$\Rightarrow -\sin ^{-1}\frac{\pi }{4}\leq a\leq \sin {\frac{-\pi }{4}}$

$\Rightarrow -\frac{1}{\sqrt{2}}\leq a\leq \frac{1}{\sqrt{2}}$

Thus $|a|\leq \frac{1}{\sqrt{2}}$

Also, for $\sin ^{-1}a\:$ to be defined $|a|\leq 1$.  Thus solution is $|a|\leq \frac{1}{\sqrt{2}}$

None of the solution is correct.

Option 1)

all real values

Option 2)

$\left | a \right |< \frac{1}{2}\; \;$

Option 3)

$\; \left | a \right |\geqslant \frac{1}{\sqrt{2}}\; \; \;$

Option 4)

$\; \frac{1}{2}< \left | a \right |< \frac{1}{\sqrt{2}}$

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