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Question #41329

Difficulty: 8/10 | 🟢 Free

The Question

Let k=22k = 2\sqrt{2} and consider the function F(x)=∫−kx4−t4/32t1/3dtF(x) = \int_{-k}^{x} \frac{4 - t^{4/3}}{2 t^{1/3}} dt defined for x∈[−k,k]x \in [-k, k]. Let mm be the number of points c∈(−k,k)c \in (-k, k) such that F′(c)=0F'(c) = 0, and let nn be the number of points in (−k,k)(-k, k) where F(x)F(x) is not differentiable. Find the value of 2m+3n2m + 3n.

Options

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    5

Check Your Answer

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 Root Existence|Newton-Leibniz Formula with variable limits|Even and Odd Function Symmetry

Why This Question Is Tricky

Common Mistakes

This question tests the blind application of Rolle's Theorem and symmetry properties.

Students observe that the integrand is an odd function, leading to F(−k)=F(k)=0F(-k) = F(k) = 0.

Assuming continuity and differentiability, they incorrectly invoke Rolle's Theorem to conclude there must be at least one root c∈(−k,k)c \in (-k, k) where F′(c)=0F'(c) = 0.

Furthermore, since F′(x)F'(x) is an odd function, they might falsely assume F′(0)=0F'(0) = 0, yielding m=1m=1 and n=0n=0.

The 'lethal collision' occurs because they fail to check the continuity of the integrand at t=0t=0, which translates to a point of non-differentiability for F(x)F(x) at x=0x=0.

Thus, Rolle's Theorem is invalid, F′(0)F'(0) is undefined, and the actual roots lie exactly on the boundaries, resulting in m=0m=0 and n=1n=1.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterContinuity and Differentiability - Class 12
ConceptRolle's Theorem Root Existence
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree