A) \[\left( 0,\,-\frac{1}{2} \right)\]
B) \[\left( \frac{1}{2},\,0 \right)\]
C) {0}
D) (-1, 0)
Correct Answer: C
Solution :
\[{{\sin }^{-1}}(1-x)-2{{\sin }^{-1}}x=\frac{\pi }{2}\] Þ \[{{\sin }^{-1}}(1-x)=\left( \frac{\pi }{2}-2{{\sin }^{-1}}x \right)\] Þ \[1-x=\sin \left( \frac{\pi }{2}-2{{\sin }^{-1}}x \right)\] Þ \[1-x=\sin \frac{\pi }{2}\cos (2{{\sin }^{-1}}x)-\cos \frac{\pi }{2}\sin (2{{\sin }^{-1}}x)\] Þ \[1-x=\cos {{(2{{\sin }^{-1}}x)}^{2}}\] Þ \[1-x=\cos {{\cos }^{-1}}(1-2x\] Þ \[2{{x}^{2}}-x=0\] \[x=0\], \[x=1/2\]which does not satisfy the equation \ \[x=0\] is only the solution.
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