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Question #41156

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \int_1^x \arcsin\left(\frac{2xt}{x^2+t^2}\right) dt$ for $x > 0$ . Find the value of $f'\left(\frac{1}{2}\right) + f'(2)$ .

Options

A.
$\pi - 2\ln 2$
B.
$\pi + \ln\left(\frac{25}{16}\right)$
C.
$\pi + \ln\left(\frac{16}{25}\right)$
D.
$-2\ln 2$

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

Principal Value Branch Boundaries|Newton-Leibniz Formula with variable limits|Substitution (Change of Variable)

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Newton-Leibniz formula combined with the hidden domain restrictions of Inverse Trigonometric Functions (ITFs).

When differentiating the integrand $\arcsin\left(\frac{2xt}{x^2+t^2}\right)$ with respect to $x$ , an absolute value term $|x^2-t^2|$ emerges from $\sqrt{1-u^2}$ .

Students commonly ignore this absolute value, assuming $\sqrt{(x^2-t^2)^2} = x^2-t^2$ .

However, the sign depends on the integration limits: for $x=2$ , $t \in [1,2]$ so $t < x$ , but for $x=1/2$ , $t \in [1/2,1]$ so $t > x$ .

This causes a sign flip in the partial derivative, acting as a hidden discontinuity trap that perfectly filters algorithmic solvers from those who check domain constraints.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Principal Value Branch Boundaries
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free