A) \[\frac{1}{\sqrt{x(1-x)}}\]
B) \[\frac{-1}{\sqrt{x(1-x)}}\]
C) \[\frac{1}{\sqrt{x(1+x)}}\]
D) None of these
Correct Answer: B
Solution :
\[{{\sin }^{-1}}\sqrt{1-x}={{\sin }^{-1}}\sqrt{1-{{(\sqrt{x})}^{2}}}={{\cos }^{-1}}\sqrt{x}\] \[\therefore y=2{{\cos }^{-1}}\sqrt{x}\]or\[\frac{dy}{dx}=2.\frac{-1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}\]etc. Aliter: \[y={{\sin }^{-1}}\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}\] Þ \[\frac{dy}{dx}=\frac{1}{\sqrt{1-1+x}}.\frac{-1}{2\sqrt{1-x}}-\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}\] \[=\frac{-2}{2\sqrt{x}\sqrt{1-x}}=\frac{-1}{\sqrt{x}\sqrt{1-x}}\].
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