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Question #41672

Difficulty: 8/10 | 🟢 Free

The Question

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

Options

  • A.

    33

  • B.

    238\frac{23}{8}

  • C.

    44

  • D.

    258\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)=0f'(x) = 0 to find maxima.

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

Students who blindly evaluate the function at these critical points and the boundaries (x=0,3x=0, 3) will incorrectly conclude the maximum is 33.

The true absolute maximum occurs at an interior point of non-differentiability (x=2x=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

SubjectMathematics
ChapterLinear Programming - Class 12
ConceptFeasible Region Constancy
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree