Inverse Trigonometric Functions Question 89
Question: If $ | {{\sin }^{-1}}x |+| {{\cos }^{-1}}x |=\frac{\pi }{2} $ , then x $ \in $
Options:
A) R
B) $ [ -1,1 ] $
C) $ [ 0,1 ] $
D) $ \phi $
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Answer:
Correct Answer: C
$ | {{\sin }^{-1}}x |+| {{\cos }^{-1}}x |=\frac{\pi }{2} $
$ \Rightarrow | {{\sin }^{-1}}x |+{{\cos }^{-1}}x=\frac{\pi }{2} $ (
$ \therefore {{\cos }^{-1}}x $ is always non-negative)
$ \Rightarrow | {{\sin }^{-1}}x |={{\sin }^{-1}}x $
$ \Rightarrow {{\sin }^{1}}x\ge 0\Rightarrow 0\le {{\sin }^{-1}}x\le \frac{\pi }{2} $
$ \Rightarrow 0\le x\le 1 $
BETA
