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Question #40321

Difficulty: 8/10 | 🟢 Free

The Question

For each $x \in \mathbb{R}$ , define the following sets: $A_x = \{ y \in \mathbb{R} : 0 \le y \le x^2 - 10x + 29 \}$ $B_x = \{ y \in \mathbb{R} : 0 \le y \le 29 - 4x \}$ $C_x = \{ y \in \mathbb{R} : 0 \le y \le 2x - 1 \}$ Let $f(x) = \mu(A_x \cup B_x \cup C_x)$ , where $\mu(S)$ denotes the Lebesgue measure (length) of the set $S$ . Find the global minimum value of $f(x)$ for $x \in \mathbb{R}$ .

Options

A.
4
B.
9
C.
11
D.
$11 - 2\sqrt{6}$

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

Cardinality Inclusion-Exclusion|Global Extrema on Intervals|Critical Point vs Stationary Point

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to optimize a function defined by the inclusion-exclusion of sets (which simplifies to the maximum of their measures since they are nested intervals).

The lethal trap lies in the distinction between critical points and stationary points.

Students often mechanically differentiate the quadratic component $h_1(x)$ to find a stationary point at $x=5$ with a minimum value of $4$ , incorrectly assuming this is the global minimum.

However, the true global minimum of the upper envelope $f(x)$ occurs at a non-differentiable 'kink' (critical point) where $h_2(x)$ and $h_3(x)$ intersect, yielding a minimum value of $9$ .

Distractors include the false stationary minimum ( $4$ ) and partial intersection minimums ( $11$ and $11 - 2\sqrt{6}$ ).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Cardinality Inclusion-Exclusion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free