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Question #40411

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sqrt{2 + \sqrt{2 + 2\cos 4x}}$ and $g(x) = \sqrt{2 - \sqrt{2 + 2\cos 4x}}$ . Let $A$ be the area of the region bounded by the curves $y = f(x)$ and $y = g(x)$ for $x \in [0, 37\pi]$ . If $N$ is the number of points in the interval $(0, 37\pi)$ where $f(x)$ is non-differentiable, find the value of $A + N$ .

Options

A.
$296\sqrt{2} - 222$
B.
$296\sqrt{2} - 259$
C.
$148\sqrt{2}$
D.
$296\sqrt{2} - 148$

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

Allied Angle Periodicity|Algebraic-Trigonometric Graphical Mapping|Square Root and Modulus Logic

Why This Question Is Tricky

Common Mistakes

This question tests the nested modulus trap in trigonometric simplifications.

Students often blindly write $\sqrt{x^2} = x$ , simplifying $\sqrt{2+2\cos 4x}$ to $2\cos 2x$ .

This fatal error changes the area and eliminates all points of non-differentiability.

Even if the first modulus is caught, evaluating the outer square root requires careful piecewise analysis.

A secondary trap occurs if students assume $f(x) = 2|\cos x|$ globally, which introduces 37 false points of non-differentiability at the axes.

The correct mapping reveals $f(x)$ as a maximum of absolute functions, which is surprisingly smooth at the axes and only sharp at odd multiples of $\pi/4$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Allied Angle Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free