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Question #62137

Difficulty: 8/10 | 🟢 Free

The Question

Let $I = \int_{0}^{\sqrt{2}+1} \left[ \frac{4}{\pi} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Find the value of $\lim_{n \to \infty} \int_{0}^{I} \frac{x^n}{1+x^n} dx$ .

Options

A.
1
B.
$1 + \sqrt{2}$
C.
$2\sqrt{2}$
D.
2

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

Variable Transformation Substitution|Inequalities in Definite Integrals|Greatest Integer Function (GIF)|Extraneous Solution Rejection

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the student's ability to correctly apply inverse trigonometric substitutions while respecting principal value branches, combined with evaluating limits of integrals using bounding inequalities.

The primary trap lies in the substitution $x = \tan\theta$ for $\sin^{-1}(\frac{2x}{1+x^2})$ .

Students commonly assume this simplifies to $2\tan^{-1}x$ for all $x$ .

However, for $x > 1$ , the correct branch is $\pi - 2\tan^{-1}x$ .

Failing to use the correct branch yields an extraneous integral value of $2+\sqrt{2}$ instead of $2$ .

The secondary phase requires using inequalities (Squeeze Theorem) to evaluate the limit of the definite integral, punishing those who carry forward the incorrect upper bound.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Variable Transformation Subsitution
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free