Let f(x) = cos left(2 tan ^{-1} sin left(cot | vedclass

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(C) Given $f(x) = \cos \left(2 	an ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}ight)ight)$.Let $	heta = \cot ^{-1} \sqrt{\frac{1-x}{x}}$. Th

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$(1-x)^{2} f^{\prime}(x)+2(f(x))^{2}=0$

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(C) Given $f(x) = \cos \left(2 	an ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}ight)ight)$.Let $	heta = \cot ^{-1} \sqrt{\frac{1-x}{x}}$. Then $\cot 	heta = \sqrt{\frac{1-x}{x}}$,which implies $	an 	heta = \sqrt{\frac{x}{1-x}}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1+	an^2 	heta}} = \frac{\sqrt{x/(1-x)}}{\sqrt{1+x/(1-x)}} = \sqrt{x}$.So,$\cot ^{-1} \sqrt{\frac{1-x}{x}} = \sin ^{-1} \sqrt{x}$.Substituting this into $f(x)$:$f(x) = \cos \left(2 	an ^{-1} \sqrt{x}ight)$.Using the identity $2 	an ^{-1} u = \cos ^{-1} \left(\frac{1-u^2}{1+u^2}ight)$:$f(x) = \cos \left(\cos ^{-1} \left(\frac{1-x}{1+x}ight)ight) = \frac{1-x}{1+x}$.Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:$f^{\prime}(x) = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.We need to check the options. Consider $(1-x)^2 f^{\prime}(x) + 2(f(x))^2$:$(1-x)^2 \left(\frac{-2}{(1+x)^2}ight) + 2 \left(\frac{1-x}{1+x}ight)^2 = \frac{-2(1-x)^2}{(1+x)^2} + \frac{2(1-x)^2}{(1+x)^2} = 0$.Thus,$(1-x)^{2} f^{\prime}(x)+2(f(x))^{2}=0$ is correct.

Let $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$ for $0 < x < 1$. Then :

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