A) {1, 2}
B) {-1, 2}
C) {-1, 1, 0}
D) {1, 1/2, 0}
Correct Answer: C
Solution :
\[{{\sin }^{-1}}x=2{{\tan }^{-1}}x\] Þ \[{{\sin }^{-1}}x={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}\] \[\Rightarrow \frac{2x}{1+{{x}^{2}}}=x\] Þ \[{{x}^{3}}-x=0\] Þ \[x(x+1)(x-1)=0\] Þ \[x=\left\{ -1,\,\,1,\,\,0 \right\}\].You need to login to perform this action.
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