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Question #40347

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sqrt{1 - \sin 2x} + \sqrt{1 + \sin 2x}$ for $x \in \left[0, \frac{\pi}{2}\right]$ . Let $S$ be the set of all points $c \in \left(0, \frac{\pi}{2}\right)$ such that $f'(c) = \frac{f(\pi/2) - f(0)}{\pi/2 - 0}$ . Find the number of elements in $S$ .

Options

A.
0
B.
1
C.
2
D.
3

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

Constant Function Logic|Rolle's Theorem Anchors|Non-Differentiable Extrema

Why This Question Is Tricky

Common Mistakes

This question tests the applicability of Rolle's Theorem and the Mean Value Theorem on functions with hidden non-differentiable points.

Students often blindly apply the chain rule to $f(x)$ , yielding $f'(x) = \cos 2x \left( \frac{1}{\sqrt{1+\sin 2x}} - \frac{1}{\sqrt{1-\sin 2x}} \right)$ , and incorrectly conclude $f'(\pi/4) = 0$ by setting the numerator to zero.

However, the denominator of the derivative vanishes at $x = \pi/4$ , creating a sharp corner (non-differentiable extremum).

The trap penalizes rote differentiation without checking domain validity or piecewise definitions, a hallmark of JEE Advanced calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Constant Function Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free