Question

If \(f(x) = \cot^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)\) and \(g(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)\), then \(\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}\), \(0<a<1/2\), is

• A. \(\frac{3}{2(1 + a^2)}\)
• B. \(\frac{3}{2(1 + x^2)}\)
• C. \(\frac{3}{2}\)
• D. \(-\frac{3}{2}\)

Hint:

\(\cot^{-1}(\tan\theta) = \pi/2 - \theta\); \(\cos^{-1}(\cos 2\theta) = 2\theta\) for appropriate range.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Simplify \(f(x)\) and \(g(x)\) using trigonometric identities.
Step 2: Detailed Explanation:
Put \(x = \tan\theta\): \(f(x) = \cot^{-1}(\tan 3\theta) = \pi/2 - 3\theta = \pi/2 - 3\tan^{-1}x\)
\(g(x) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x\)
Limit = \(\lim_{x \to a} \frac{(\pi/2 - 3\tan^{-1}x) - (\pi/2 - 3\tan^{-1}a)}{2\tan^{-1}x - 2\tan^{-1}a}\)
= \(\frac{-3(\tan^{-1}x - \tan^{-1}a)}{2(\tan^{-1}x - \tan^{-1}a)} = -\frac{3}{2}\)
Step 3: Final Answer:
Limit = \(-3/2\).