Inverse Trigonometric Functions Question 39
Question: If $ {{\sin }^{-1}}(1-x)-2{{\sin }^{-1}}x=\pi /2 $ , then x equals
[Orissa JEE 2005]
Options:
A) $ ( 0,-\frac{1}{2} ) $
B) $ ( \frac{1}{2},0 ) $
C) {0}
D) (-1, 0)
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\sin }^{-1}}(1-x)-2{{\sin }^{-1}}x=\frac{\pi }{2} $
Therefore $ {{\sin }^{-1}}(1-x)=( \frac{\pi }{2}-2{{\sin }^{-1}}x ) $
Therefore $ 1-x=\sin ( \frac{\pi }{2}-2{{\sin }^{-1}}x ) $
Therefore $ 1-x=\sin \frac{\pi }{2}\cos (2{{\sin }^{-1}}x)-\cos \frac{\pi }{2}\sin (2{{\sin }^{-1}}x) $
Therefore $ 1-x=\cos {{(2{{\sin }^{-1}}x)}^{2}} $
Therefore $ 1-x=\cos {{\cos }^{-1}}(1-2x $
Therefore $ 2x^{2}-x=0 $
$ x=0 $ , $ x=1/2 $ which does not satisfy the equation \ $ x=0 $ is only the solution.
BETA
