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Question #41672

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: [0, 3] \to \mathbb{R}$ be defined as $f(x) = \int_0^x (4t + 1 - 6\lfloor t \rfloor) dt$ where $\lfloor t \rfloor$ denotes the greatest integer less than or equal to $t$ . Find the absolute maximum value of $f(x)$ on the interval $[0, 3]$ .

Options

A.
$3$
B.
$\frac{23}{8}$
C.
$4$
D.
$\frac{25}{8}$

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

Feasible Region Constancy|Increasing/Decreasing Strictness|Step Function Discontinuities

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to optimize a function whose derivative contains step discontinuities.

The primary trap is the standard calculus heuristic of setting $f'(x) = 0$ to find maxima.

Here, $f'(x) = 0$ only yields local minima ( $x = 5/4$ and $x = 11/4$ ).

Students who blindly evaluate the function at these critical points and the boundaries ( $x=0, 3$ ) will incorrectly conclude the maximum is $3$ .

The true absolute maximum occurs at an interior point of non-differentiability ( $x=2$ ), forcing the student to transition from smooth gradient-based optimization to interval-based analysis and LHL/RHL evaluation at the step boundaries.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Feasible Region Constancy
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free