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Question #41186

Difficulty: 8/10 | 🟢 Free

The Question

Let $I = \int_{1}^{\sqrt{3}} \frac{1}{x} \sin^{-1}\left( \frac{2x}{1+x^2} \right) dx$ . Then $I$ lies in the interval:

Options

A.
$\left( \frac{\pi}{6} \ln 3, \frac{\pi}{4} \ln 3 \right)$
B.
$\left( \frac{\pi}{4} \ln 3, \frac{\pi}{3} \ln 3 \right)$
C.
$\left( \frac{\pi}{8} \ln 3, \frac{\pi}{6} \ln 3 \right)$
D.
$\left( \frac{\pi}{3} \ln 3, \frac{\pi}{2} \ln 3 \right)$

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

Algebraic-Trigonometric Graphical Mapping|Inverse Function Integration|Inequalities in Definite Integrals

Why This Question Is Tricky

Common Mistakes

This question tests the 'Lethal Collision' between piecewise inverse trigonometric identities and definite integral inequalities.

The primary trap is the rote assumption that $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$ universally.

If a student makes this error, they will integrate $\frac{2\tan^{-1}x}{x}$ , leading directly to the bounds in Option B.

The rigorous path forces the student to correctly map the inverse function to its algebraic domain ( $\pi - 2\tan^{-1}x$ for $x \ge 1$ ) before applying monotonicity to bound the non-integrable function $\frac{\tan^{-1}x}{x}$ .

This evaluates both domain-specific piecewise definitions and the application of bounding techniques in calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Algebraic-Trigonometric Graphical Mapping
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free