Question

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

• A. $(1-x{)}^2{f}^{\prime }(x)-2(f\left(x\right){)}^2=0$
• B. $(1-x{)}^2{f}^{\prime }(x)+2(f\left(x\right){)}^2=0$
• C. $(1+x{)}^2{f}^{\prime }(x)-2(f\left(x\right){)}^2=0$
• D. $(1+x{)}^2{f}^{\prime }(x)+2(f\left(x\right){)}^2=0$

Answer 1

The correct option is B $(1-x{)}^2{f}^{\prime }(x)+2(f\left(x\right){)}^2=0$

$f\left(x\right)=\cos \left(2{\tan }^{-1}\sin \left({\cot }^{-1}\sqrt{\frac{1-x}{x}}\right)\right)$
Let $\frac{\sqrt{1-x}}{\sqrt{x}}=\cot \theta$
$\Rightarrow f\left(x\right)=\cos \left(2{\tan }^{-1}\sin \theta \right)=\cos \left(2{\tan }^{-1}\right(\sqrt{x}\left)\right)$

Let $\sqrt{x}=\tan \theta$
$\Rightarrow f\left(x\right)=\cos 2\theta =2{\cos }^2\theta -1=\frac{2}{1+x}-1=\frac{1-x}{1+x}$
$\Rightarrow f^{\prime }\left(x\right)=-\frac{2}{(1+x{)}^2}$
$\Rightarrow (1-x{)}^2{f}^{\prime }(x)=-2{\left(\frac{1-x}{1+x}\right)}^2$
$\Rightarrow (1-x{)}^2{f}^{\prime }(x)+2f^2(x)=0$