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Question #63423

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined as $f(x) = x^5 + 2x^3 + 3x$ . Let $g(x) = \frac{12}{\pi} \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ . If $H(x) = f^{-1}(g(x))$ , then the value of $\lim_{x \to 1^+} H'(x)$ is

Options

A.
$-\frac{6}{7\pi}$
B.
$\frac{6}{7\pi}$
C.
$-\frac{12}{7\pi}$
D.
$\frac{12}{7\pi}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Derivative of Implicit Inverse Forms|Inverse Function Derivative|Composition Non-Commutativity|Algebraic-Trigonometric Graphical Mapping

Why This Question Is Tricky

Examiner Levers

Concept Bridge

Cross-chapter synthesis (e.g., using Logarithms in Electrochemistry).

Difficulty Boost : +5

Common Mistakes

This question tests the student's vigilance regarding the domain and principal branches of inverse trigonometric functions during composition.

The primary trap lies in the algebraic-trigonometric mapping $g(x)$ .

Students commonly memorize $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$ without the constraint $|x| \le 1$ .

For the right-hand limit $x \to 1^+$ , the input $x$ exceeds 1, requiring the branch adjustment $\pi - 2\tan^{-1}x$ .

Failing to apply this non-commutative property of inverse composition results in a sign error, leading directly to the trap option.

Furthermore, it rigorously tests the implicit derivative of an inverse function evaluated at a mapped point.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Derivative of Implicit Inverse Forms
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free