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Question #40929

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ . Using differentials, the approximate value of $f(0.98) + f(1.02)$ is:

Options

A.
$\pi - 0.04$
B.
$2\pi$
C.
$\pi + 0.04$
D.
$0$

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

Algebraic Simplification|Approximation via Differentials|The Quadrant Adjustment Logic

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly identify the principal value branches of inverse trigonometric functions before applying differentials.

A common trap is blindly differentiating the expression using standard chain-rule formulas, yielding $f'(x) = \frac{4}{1+x^2}$ everywhere.

This ignores the branch cut at $x=1$ , where the argument of the inverse functions crosses the boundary of their principal domains.

By substituting $x = \tan \theta$ , students must apply the Quadrant Adjustment Logic to reveal that $f(x) = 0$ for $x > 1$ , making $f(1.02) = 0$ .

Failing to account for this branch transition leads to the incorrect approximation of $2\pi$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	Algebraic Simplification
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free