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Question #41356

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \min_{t>0} \left( t + \frac{25}{t} \right)$ and $P = 3 \cdot \min_{x,y,z>0} \frac{(x+y)(y+z)(z+x)}{xyz}$ . Let $\alpha, \beta$ (with $\alpha < \beta$ ) be the roots of the equation $x^2 - Sx + P = 0$ . Consider the function $f(x) = \frac{x^2 - Sx + P}{x - c}$ , where $c \in (\alpha, \beta)$ . Let $A$ be the set of all values of $c \in (\alpha, \beta)$ for which the equation $f'(x) = 0$ has at least one real root in the interval $(\alpha, \beta)$ . Which of the following is true?

Options

A.
A = \emptyset
B.
A = (4, 6)
C.
A = {5}
D.
A = (4, 5] \cup [5.5, 6)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Rolle's Theorem Anchors|Arithmetic Mean Inequality|Denominator Restriction

Why This Question Is Tricky

Common Mistakes

This question tests the critical continuity condition of Rolle's Theorem alongside the AM-GM inequality.

Students often blindly apply Rolle's Theorem when they see $f(\alpha) = f(\beta) = 0$ , incorrectly concluding that $f'(x) = 0$ must have a root in $(\alpha, \beta)$ and thus choosing $A = (4, 6)$ .

However, the rational function $f(x)$ has a vertical asymptote at $x = c \in (\alpha, \beta)$ , breaking the continuity requirement of Rolle's Theorem.

The AM-GM integration forces students to first derive the interval $(4, 6)$ , setting up the trap where the discriminant of the derivative's numerator is strictly negative, proving no such root exists.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Rolle's Theorem Anchors
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free