Question

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

• A. \((1-x)^2 f'(x) - 2(f(x))^2 = 0\)
• B. \((1+x)^2 f'(x) + 2(f(x))^2 = 0\)
• C. \((1-x)^2 f'(x) + 2(f(x))^2 = 0\)
• D. \((1+x)^2 f'(x) - 2(f(x))^2 = 0\)

Hint:

Always simplify the nested inverse trigonometric functions first before attempting differentiation. Use triangle properties to convert between \(\cot^{-1}, \sin, \tan^{-1}\).

Answer 1

The Correct Option is C

We begin by analyzing the given function \(f(x)\):

\[ f(x) = \cos\left(2\tan^{-1} \sin \left(\cot^{-1} \sqrt{\frac{1-x}{x}}\right)\right) \]

First, let's simplify \(\sin \left(\cot^{-1} \sqrt{\frac{1-x}{x}}\right)\):

Let \(\theta = \cot^{-1} \sqrt{\frac{1-x}{x}}\). This implies \(\cot \theta = \sqrt{\frac{1-x}{x}}\), and thus \(\tan \theta = \sqrt{\frac{x}{1-x}}\).

Therefore, \(\sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}\):

\[ \sin \theta = \frac{\sqrt{\frac{x}{1-x}}}{\sqrt{1 + \frac{x}{1-x}}} = \frac{\sqrt{\frac{x}{1-x}}}{\sqrt{\frac{1}{1-x}}} = \sqrt{x} \]

Substitute back into the original function:

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

We use the identity for \(\cos(2\tan^{-1} u)\):

\[ \cos(2\tan^{-1} u) = \frac{1-u^2}{1+u^2} \]

For \(u = \sqrt{x}\), we then have:

\[ f(x) = \frac{1-x}{1+x} \]

Next, we find the derivative \(f'(x)\):

Using the quotient rule:

\[ f'(x) = \frac{(1+x)'(1-x) - (1-x)'(1+x)}{(1+x)^2} \]

Simplifying, we find:

\[ f'(x) = \frac{-2}{(1+x)^2} \]

Now consider the given equation \((1-x)^2 f'(x) + 2(f(x))^2 = 0\):

Substitute for \(f(x)\) and \(f'(x)\):

\[ (1-x)^2 \left(\frac{-2}{(1+x)^2}\right) + 2\left(\frac{1-x}{1+x}\right)^2 = 0 \]

\[ \Rightarrow (1-x)^2 \left(\frac{-2}{(1+x)^2}\right) + 2 \frac{(1-x)^2}{(1+x)^2} = 0 \]

\[ \Rightarrow \frac{-2(1-x)^2}{(1+x)^2} + \frac{2(1-x)^2}{(1+x)^2} = 0 \]

This simplifies both terms to zero, confirming that the equation is satisfied.

Therefore, the correct answer is:

\((1-x)^2 f'(x) + 2(f(x))^2 = 0\)