Inverse Trigonometric Functions Question 139
Question: The trigonometric equation $ {{\sin }^{-1}}x=2{{\sin }^{-1}}a $ has a solution for
Options:
A) $ \frac{1}{2}<| a |<\frac{1}{\sqrt{2}} $
B) All real values of a
C) $ | a |<1/2 $
D) $ | a |\ge \frac{1}{\sqrt{2}} $
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Answer:
Correct Answer: C
$ {{\sin }^{-1}}x=2{{\sin }^{-1}}a $
$ \therefore -\frac{\pi }{2}\le {{\sin }^{-1}}x\le \frac{\pi }{2}, $
$ \frac{-\pi }{2}\le 2{{\sin }^{-1}}a\le \frac{\pi }{2} $
$ \Rightarrow \frac{-\pi }{4}\le {{\sin }^{-1}}a\le \frac{\pi }{4} $
$ \therefore \frac{-1}{\sqrt{2}}\le a\le \frac{1}{\sqrt{2}} $
$ \therefore | a |\le \frac{1}{\sqrt{2}}(as\frac{1}{\sqrt{2}}>\frac{1}{2}) $
Out of the given four options no one is absolutely correct, but option (3) could be take into consideration.
If $ | a |<1/2 $ is taken as correct, then its domain is not satisfied for $ a=1/\sqrt{3} $ , but the equation is satisfied.
BETA
