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

Step 1: Understanding the Concept:
This problem involves simplifying an expression composed of several nested inverse trigonometric functions and then finding the relationship between the function and its derivative.
Step 2: Detailed Explanation:
Let \(\alpha = \cot^{-1} \sqrt{\frac{1-x}{x}}\). Then \(\cot\alpha = \sqrt{\frac{1-x}{x}}\).
Constructing a right-angled triangle: Adjacent = \(\sqrt{1-x}\), Opposite = \(\sqrt{x}\).
Hypotenuse = \(\sqrt{(\sqrt{x})^2 + (\sqrt{1-x})^2} = \sqrt{x + 1 - x} = 1\).
Then \(\sin\alpha = \frac{\text{Opp}}{\text{Hyp}} = \sqrt{x}\).
Now, \(f(x) = \cos(2\tan^{-1}\sqrt{x})\).
Let \(\beta = \tan^{-1}\sqrt{x}\). Then \(\tan\beta = \sqrt{x}\).
Using the identity \(\cos 2\beta = \frac{1-\tan^2\beta}{1+\tan^2\beta}\):
\[ f(x) = \frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2} = \frac{1-x}{1+x} \]
Differentiate \(f(x)\):
\[ f'(x) = \frac{-(1+x) - (1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2} \]
Calculate the expression in (C):
\[ (1-x)^2 f'(x) + 2(f(x))^2 = (1-x)^2 \left(\frac{-2}{(1+x)^2}\right) + 2\left(\frac{1-x}{1+x}\right)^2 \]
\[ = \frac{-2(1-x)^2}{(1+x)^2} + \frac{2(1-x)^2}{(1+x)^2} = 0 \]
Step 3: Final Answer:
The correct relationship is \((1-x)^2 f'(x) + 2(f(x))^2 = 0\).