Inverse Trigonometric Functions Question 88
Question: Solving $ 2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-x^{2}}), $ we get
Options:
A) $ x\in [ \frac{\sqrt{2}}{2},1 ] $
B) $ x=3 $
C) $ x\in [3,4] $
D) $ x=0 $
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Answer:
Correct Answer: A
$ x=\cos y; $ where $ 0\le y\le \pi ,| x |\le 1 $
$ 2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-x^{2}}) $
$ \Rightarrow 2{{\cos }^{-1}}(cosy)=si{n^{-1}}(2cosy\cdot \sqrt{1-{{\cos }^{2}}y}) $
$ \Rightarrow 2{{\cos }^{-1}}(cosy)=si{n^{-1}}(2cosy.siny) $
$ \Rightarrow 2{{\cos }^{-1}}(cosy)=si{n^{-1}}(sin2y) $
$ \Rightarrow {{\sin }^{-1}}(sin2y)=2y $ for $ -\pi /4\le y\le \pi /4 $ and $ 2{{\cos }^{-1}}(cosy)=2y $ for $ 0\le y\le \pi $
Thus, Eq. (i) holds only when, $ y\in [0,\pi /4]\Rightarrow x\in [\sqrt{2}/2,1] $
BETA
