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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}}

 

Answers (1)

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}}

Posted by

Sabhrant Ambastha

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