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Question #41371

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined as $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) - |x|$ Let $M$ be the global maximum value of $f(x)$ . Then $M$ is equal to:

Options

A.
$\pi - 1$
B.
$\frac{4\pi}{3} - \sqrt{3}$
C.
$\frac{\pi}{3} - \sqrt{3}$
D.
$\pi$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

First Derivative Test Trap|Domain Constraint Intersections|Square Root and Modulus Logic

Why This Question Is Tricky

Common Mistakes

This question tests the 'First Derivative Test Trap' combined with domain constraints of inverse trigonometric identities.

Students commonly assume $\sin^{-1}(\frac{2x}{1+x^2}) = 2\tan^{-1}x$ for all $x \in \mathbb{R}$ .

Differentiating the resulting false function $f(x) = 4\tan^{-1}x - x$ yields a critical point at $x = \sqrt{3}$ , leading to a false maximum of $\frac{4\pi}{3} - \sqrt{3}$ .

However, the identity is only valid for $x \in [-1, 1]$ .

For $x > 1$ , the function simplifies to $\pi - x$ , making $x=1$ a sharp peak (non-differentiable) and the true global maximum.

The modulus and piecewise nature force rigorous boundary analysis.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	First Derivative Test Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free