Before You Begin

Question #41512

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(a) = \int_0^{\pi/2} \frac{1}{\cos x} \sin^{-1}\left( \frac{2a \cos x}{1 + a^2 \cos^2 x} \right) \, dx$ for $a > 0$ . Find the value of $f'(\sqrt{3})$ .

Options

A.
$\frac{\pi}{2} - 2 \tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$
B.
$\frac{\pi}{2}$
C.
$\frac{\pi}{2} - \sqrt{2} \tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$
D.
$\frac{\pi}{2} - 2 \tan^{-1}(\sqrt{2})$

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

Leibniz Rule for Differentiation under Integral|Algebraic-Trigonometric Graphical Mapping|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Leibniz rule for differentiation under the integral sign when the integrand contains an inverse trigonometric function.

Students commonly apply the chain rule blindly, assuming the derivative of $\sin^{-1}\left(\frac{2y}{1+y^2}\right)$ is universally $\frac{2}{1+y^2}$ .

However, this is only valid on the principal value branch ( $|y| \le 1$ ).

When the parameter $a$ causes the argument to exceed 1, the function transitions to $\pi - 2\tan^{-1}y$ , introducing a sign change in the partial derivative.

Failing to split the integral at the boundary $a \cos x = 1$ leads directly to the trap answer $\frac{\pi}{2}$ , effectively testing the student's vigilance regarding domain constraints during parametric integration.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Leibniz Rule for Differentiation under Integral
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free