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Question #40805

Difficulty: 8/10 | 🟢 Free

The Question

Let $H_e$ be a family of hyperbolas with a fixed transverse axis of length $2$ and eccentricity $e \in (1, \infty)$ . Let $L(e)$ denote the length of the conjugate axis of $H_e$ . A function $f: (0, \infty) \to \mathbb{R}$ is defined as $f(x) = x^2 - 4x + 3$ . Consider the composite function $h(e) = f(L(e))$ . If $S$ is the set of all real values of $k$ for which the equation $h(e) = k$ has exactly two distinct solutions for $e \in (1, \infty)$ , then the set $S$ is:

Options

A.
$(-1, 3)$
B.
$[-1, 3)$
C.
$(-1, 3]$
D.
$(-1, \infty)$

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

Hyperbola Conjugate Axes|Function Definition (The 'One-to-One' Constraint)|Composite Injectivity/Surjectivity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Symmetry Fallacy' and composite function injectivity.

Students often assume that because the hyperbola's eccentricity uniquely determines its shape, any function of its parameters will be monotonic.

By mapping the conjugate axis length $L(e)$ into a non-monotonic quadratic $f(x)$ , the composite function $h(e)$ fails the 'One-to-One' constraint.

The trap lies in failing to restrict the domain of the inner function $L(e) \in (0, \infty)$ when analyzing the outer function's intersections, leading students to incorrectly include $k \ge 3$ or boundary points like $k = -1$ and $k = 3$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Conic Sections - Class 11
Concept	Hyperbola Conjugate Axes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free