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Question #40440

Difficulty: 8/10 | 🟢 Free

The Question

Let $P_1$ and $P_2$ be two planes defined by the equations: $P_1: x \sqrt{1-\sin 2\theta} + y \sqrt{1+\sin 2\theta} + z = \sqrt{2}$ $P_2: x \sqrt{1+\sin 2\theta} + y \sqrt{1-\sin 2\theta} - z = \sqrt{2}$ where $\theta \in \left[0, \frac{\pi}{2}\right]$ . Let $L(\theta)$ be the line of intersection of the planes $P_1$ and $P_2$ , and let $\phi(\theta)$ be the acute angle between the line $L(\theta)$ and the $xy$ -plane. The value of the integral $\int_0^{\pi/2} \tan^2 \phi(\theta) \, d\theta$ is:

Options

A.
$\frac{\pi}{2} - 1$
B.
$\frac{\pi}{2}$
C.
$\frac{\pi}{2} + 1$
D.
$\pi - 2$

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

Square Root and Modulus Logic|Angle between two planes|Collinearity vs Coplanarity|Locus via Distance Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of square root definitions ( $\sqrt{x^2} = |x|$ ) within 3D coordinate geometry.

The primary trap lies in simplifying $\sqrt{1-\sin 2\theta}$ as simply $\cos\theta - \sin\theta$ .

Students who ignore the modulus will incorrectly assume $f(\theta) = 2\cos\theta$ for the entire interval $[0, \pi/2]$ , leading to an integral of $1-\cos 2\theta$ over the full domain, which yields the trap answer $\pi/2$ .

The problem forces students to split the integral at the boundary condition $\theta = \pi/4$ where the dominant trigonometric term switches, correctly linking algebraic constraints to geometric properties (angle between a line and a plane).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Square Root and Modulus Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free