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Question #40433

Difficulty: 8/10 | 🟢 Free

The Question

Let $I(x) = \int_{0}^{x} \ln \left( \frac{\sin 2t + \sin 4t + \sin 6t}{\cos 2t + \cos 4t + \cos 6t} \right)^2 dt$ . Find the sum of all $x \in [0, \pi]$ such that $I'(x) = \ln 3$ .

Options

A.
$\frac{3\pi}{2}$
B.
$\frac{11\pi}{6}$
C.
$\frac{5\pi}{3}$
D.
$3\pi$

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

Factor-Product Conversion|Logarithmic Integration via Trig Manipulation|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' within the context of logarithmic integration and trigonometric manipulation.

Students must apply factor-product formulas to simplify the integrand's argument into a tangent function.

The trap occurs when students blindly solve the simplified equation $I'(x) = \ln 3 \implies \tan^2 4x = 3$ without checking the domain of the original unsimplified function.

The canceled factor $(1 + 2 \cos 2x)$ equals zero at specific points in the interval, rendering the original expression $0/0$ and thus undefined.

Failing to exclude these extraneous roots leads to an incorrect sum.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Factor-Product Conversion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free