Before You Begin

Question #40799

Difficulty: 8/10 | 🟢 Free

The Question

Let $L$ be the latus rectum of the parabola $y^2 = 4x$ . Consider the integral evaluated along $L$ : $I = \int_{-2}^{2} \frac{y}{y^2+4} \ln\left( \frac{y^4 - 24y^2 + 16}{(y^2+4)^2} \right) dy$ Which of the following represents the correct evaluation of $I$ ?

Options

A.
0
B.
$\frac{\pi}{4} \ln 2$
C.
$-\frac{\pi}{4} \ln 2$
D.
$\text{Undefined in } \mathbb{R}$

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

Latus Rectum Geometry|Tangent Half-Angle Substitution|Even and Odd Function Symmetry|Logarithmic Differentiation Domain

Why This Question Is Tricky

Common Mistakes

This question tests the student's vigilance regarding domain restrictions within definite integrals, a common trap in JEE Advanced.

The integrand is explicitly designed to look like an odd function $f(-y) = -f(y)$ , baiting students into blindly applying the symmetry property $\int_{-a}^{a} f(x) dx = 0$ to quickly answer '0'.

However, by using the tangent half-angle substitution (Weierstrass substitution), the complex algebraic argument of the logarithm elegantly collapses into $\cos 2\theta$ .

This reveals that over the latus rectum's interval, the argument becomes negative, creating a logarithmic singularity and rendering the integral undefined in the real domain.

Failing to check the domain of the integrand before applying symmetry properties is the specific mathematical trap.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Conic Sections - Class 11
Concept	Latus Rectum Geometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free