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Question #40727

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = (x^2 - 6x + 8)^{e^x}$ be a real-valued differentiable function defined on its maximal domain $D \subset \mathbb{R}$ . The logarithmic derivative $\frac{1}{y}\frac{dy}{dx}$ can be expressed in the form $e^x \ln(u) + e^x \frac{v}{u}$ , where $u$ and $v$ are coprime polynomials in $x$ and the leading coefficient of $u$ is $1$ . If a moving point $P(u, v)$ traces a curve in the $uv$ -plane for all $x \in D$ , which of the following represents the complete locus of $P$ ?

Options

A.
$v^2 = 4(u + 1)$ for $u > 0$
B.
$v^2 = 4(u + 1)$ for $u \ge -1$
C.
$v^2 = 4(u + 1)$ for $u \in \mathbb{R}$
D.
$v^2 = 4(u + 1)$ for $u > -1$

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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

Locus of a Moving Point|Completing the Square in Quadratics|Logarithmic Differentiation Domain

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of algebraic locus finding and calculus-based domain restrictions.

The primary trap lies in completing the square to find the parabolic locus $v^2 = 4(u + 1)$ and blindly accepting the entire curve (or just the vertex constraint $u \ge -1$ ).

Students often forget that the variables $u$ and $v$ originated from a logarithmic differentiation, which strictly requires the argument of the logarithm to be positive ( $u > 0$ ).

Failing to intersect the algebraic locus with the domain constraint of the original function leads to selecting the unconstrained parabola.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Locus of a Moving Point
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free