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Question #40904

Difficulty: 8/10 | 🟢 Free

The Question

A point $P(x,y)$ moves along the locus $C$ defined by $y = \frac{x^4}{\sin x - x \cos x}$ for $x \in \left(0, \frac{\pi}{2}\right]$ , and $P(0,0)$ is also on $C$ . Let $L$ be the tangent line to the locus $C$ at the origin. Find the area of the triangle bounded by the line $L$ , the vertical line $x = \frac{\pi}{2}$ , and the x-axis.

Options

A.
$\frac{3\pi^2}{8}$
B.
$\frac{\pi^2}{8}$
C.
$0$
D.
$\frac{3\pi^2}{4}$

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

Locus via Distance Constraints|Global Extrema on Intervals|The L'Hôpital's Rule Misuse

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly evaluate the derivative at a boundary point where the function is defined via a limit (a removable singularity).

The trap lies in blindly differentiating the constraint function $y(x)$ using the quotient rule and then attempting to evaluate $y'(0)$ by setting the numerator to zero or misapplying L'Hôpital's rule on the massive derivative expression.

Students who do this often incorrectly conclude that $y'(0) = 0$ (leading to Option C) because $x=0$ is a root of the derivative's numerator.

The correct and much simpler approach is to use the first principle of derivatives $y'(0) = \lim_{x \to 0} \frac{y(x)}{x}$ , which elegantly resolves to 3.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Locus via Distance Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free