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Question #41340

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \left(-\frac{\pi}{2}, 0\right) \to \mathbb{R}$ be defined as $f(x) = 2x^2 + 6x + \cos^{-1}\left(\sin\left(|x| - \frac{\pi}{6}\right)\right)$ . If $g(x)$ is the inverse function of $f(x)$ , then the value of $g'\left(\frac{\pi^2}{18} - \frac{\pi}{2}\right)$ is equal to:

Options

A.
$\frac{3}{21 - 2\pi}$
B.
$\frac{3}{15 - 2\pi}$
C.
$\frac{3}{21 + 2\pi}$
D.
$\frac{3}{15 + 2\pi}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Inverse Function Derivative|Domain and Range Nuance|The Quadrant Adjustment Logic|Modulus Function Symmetry

Why This Question Is Tricky

Common Mistakes

This question tests the derivative of an inverse function combined with piecewise domain restrictions.

The primary trap lies in the modulus function inside the inverse trigonometric composition.

Students often blindly apply the chain rule or standard simplifications without accounting for the negative domain $x \in (-\frac{\pi}{2}, 0)$ where $|x| = -x$ .

Failing to substitute $|x| = -x$ before differentiating (or forgetting the $\text{sgn}(x)$ term in the chain rule) leads to an incorrect local derivative of $-1$ instead of $+1$ for the trigonometric term, resulting in $f'(-\frac{\pi}{6}) = \frac{15 - 2\pi}{3}$ and the classic trap option B.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Inverse Function Derivative
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free