Inverse Trigonometric Functions Question 44
Question: The solution of $ {{\sin }^{-1}}x-{{\sin }^{-1}}2x=\pm \frac{\pi }{3} $ is
[Karnataka CET 2005]
Options:
A) $ \pm \frac{1}{3} $
B) $ \pm \frac{1}{4} $
C) $ \pm \frac{\sqrt{3}}{2} $
D) $ \pm \frac{1}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\sin }^{-1}}2x={{\sin }^{-1}}x-{{\sin }^{-1}}\frac{\sqrt{3}}{2} $
$ {{\sin }^{-1}}2x={{\sin }^{-1}}( x\sqrt{( 1-\frac{3}{4} )}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}} ) $
$ 2x=( \frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-x^{2}} ) $
$ \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}=\frac{x}{2}-2x=\frac{-3x}{2} $
$ \frac{3(1-x^{2})}{4}=\frac{9x^{2}}{4} $
$ \Rightarrow 3-3x^{2}=9x^{2} $
Therefore $ x^{2}=\frac{1}{4}\Rightarrow x=\pm \frac{1}{2} $ .
BETA
