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Question #41543

Difficulty: 8/10 | 🟢 Free

The Question

Let $P_1: 3x + 4y + 5z = 14$ and $P_2: 4x + 3y - 5z = 14$ be two planes. Let $\Pi$ be the plane bisecting the obtuse angle between $P_1$ and $P_2$ . Let $M(h, k, l)$ be the image of the point $N(2, 0, 0)$ in the plane $\Pi$ .

Let $R$ be the region in the $xy$ -plane bounded by the parabola $x = y^2 - y$ and the line $x = ky$ . Find the area of the region $R$ .

Options

A.
$\frac{9}{2}$
B.
$\frac{4}{3}$
C.
$\frac{1}{6}$
D.
$\frac{32}{3}$

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

Horizontal vs Vertical Strips|Angle between two planes|Image of a point in a plane|Angle Bisector Direction

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate 3D geometry with 2D area calculations, specifically targeting the distinction between the image of a point and its orthogonal projection.

The core trap lies in the formula for the reflection of a point across a plane, which requires a factor of $-2$ , whereas the foot of the perpendicular (projection) uses a factor of $-1$ .

If a student incorrectly calculates the projection instead of the image, they will obtain $k=1$ instead of $k=2$ , leading to a mathematically valid but incorrect integration region and an area of $4/3$ (Option B).

Furthermore, the area calculation is designed to heavily favor horizontal strips ( $dy$ integration); using vertical strips requires splitting the integral at the vertex of the parabola, testing the student's strategic choice of integration variables.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Horizontal vs Vertical Strips
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free