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Question #41551

Difficulty: 8/10 | 🟢 Free

The Question

Let $M$ be the global maximum value of the function $g(t) = \int_0^t 3(1 - s^2) ds$ on the interval $[-2, 2]$ . Let $A$ be the area of the region bounded by the curve $y = |x^3 - 3x|$ and the line $y = M$ . Find the value of $A$ .

Options

A.
3
B.
4
C.
5
D.
8

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Symmetry Utilization|Global Extrema on Intervals|Critical Point vs Stationary Point

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between stationary points and non-differentiable critical points when setting up definite integrals for area.

Students must find the area bounded by $y = |x^3 - 3x|$ and $y = 2$ .

Utilizing symmetry simplifies the integral to $2 \int_0^2 (2 - |x^3 - 3x|) dx$ .

The primary trap lies in evaluating the absolute value: the integrand has a non-differentiable critical point (a cusp) at $x = \sqrt{3}$ where it changes sign.

A common mistake is to rely solely on rote differentiation, finding only the stationary point ( $f'(x)=0$ at $x=1$ ), and failing to partition the integral at the cusp.

Bypassing this domain split leads to an incorrect area calculation of 4 instead of the true area 3.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Symmetry Utilization
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free