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Question #41797

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, \dots, 15\}$ . A number $k$ is chosen at random from $S$ . Let the function $f_k : [-3, 3] \to \mathbb{R}$ be defined as $f_k(x) = \min(9-x^2, 2|x|+k)$ Let $M_k$ be the global maximum value of $f_k(x)$ on the interval $[-3, 3]$ . Let $E$ be the event that $M_k$ is an integer, and $F$ be the event that $f_k(x)$ is non-differentiable at the point(s) where it attains its global maximum. Find the conditional probability $P(F \mid E)$ .

Options

A.
$\frac{2}{9}$
B.
$\frac{2}{3}$
C.
$\frac{1}{3}$
D.
$\frac{1}{4}$

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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

Sample Space Partitioning|Global Extrema on Intervals|Non-Differentiable Extrema

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to partition a sample space based on the differentiability of global extrema of piecewise functions.

The primary trap is twofold: First, students often blindly apply the intersection formula $x_k = \sqrt{10-k}-1$ without checking the domain validity for $k \ge 10$ , thereby missing the cases where the function simplifies entirely to the smooth parabola $9-x^2$ .

Second, the presence of $|x|$ in the definition tempts students to assume the function is always non-differentiable at $x=0$ .

However, for $k \ge 9$ , the absolute value term is strictly bounded above by the parabola, making the effective function globally differentiable.

Failing to recognize this leads to an incorrect count of the sample space for Bayes' Theorem.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Sample Space Partitioning
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free