Question

$f\left(x\right)=\cos \left(2{\tan }^{-1}\left(\sin \left(co{t}^{-1}\left(\sqrt{\frac{1-x}{x}}\right)\right)\right)\right)$ then

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

Answer 1

The correct option is B

${(1-x)}^{2}f\text{'}\left(x\right)+2(f{\left(x\right))}^2=0$

Explanation for the correct option:

Step 1. Find the value of given function:

$f\left(x\right)=\cos \left(2{\tan }^{-1}\left(\sin \left(co{t}^{-1}\left(\sqrt{\frac{1-x}{x}}\right)\right)\right)\right)$

$\Rightarrow$$f\left(x\right)=\cos \left(2{\tan }^{-1}\left(\sin \left({\sin }^{-1}\left(\sqrt{x}\right)\right)\right)\right)$ $\left[\mathbf{\because }\mathit{c}\mathit{o}{\mathit{t}}^{\mathbf{-}\mathbf{1}}\sqrt{\frac{\mathbf{1}\mathbf{-}\mathbf{\theta }}{\mathbf{\theta }}}\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{\theta }}\right]$

$\Rightarrow$$f\left(x\right)=\cos \left(2{\tan }^{-1}\left(\sqrt{x}\right)\right)$

$\Rightarrow$$f\left(x\right)=\cos \left({\cos }^{-1}\left(\frac{1-x}{1+x}\right)\right)$ $\left[\mathbf{\because }\mathbf{2}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{1}\mathbf{-}{\mathit{\theta }}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathit{\theta }}^{\mathbf{2}}}\mathbf{\right)}\right]$

$\Rightarrow$$f\left(x\right)=\left(\frac{1-x}{1+x}\right)$

Differentiate it with respect to $x$

$f\text{'}\left(x\right)=\frac{-1\times (1+x)-1\times (1-x)}{{(1+x)}^2}$ $\left[\mathbf{\because }\frac{\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}}\right]$

$$\begin{gathered} =\frac{-1-x-1+x}{{\left(1+x\right)}^2} \\ =\frac{-2}{{\left(1+x\right)}^2} \end{gathered}$$

Step 2. Multiply both sides by ${\left(1-x\right)}^2$

${(1-x)}^{2}f\text{'}\left(x\right)=\frac{-2}{{(1+x)}^2}{(1-x)}^2$

$\Rightarrow$ ${(1-x)}^{2}f\text{'}\left(x\right)=-2{\left(\frac{1-x}{1+x}\right)}^2$

$\Rightarrow$ ${(1-x)}^{2}f\text{'}\left(x\right)=-2{\left(f\left(x\right)\right)}^2$

$\therefore {\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\left(}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right)}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence, Option ‘B’ is Correct.