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Question #41634

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}(x) = \hat{i} + x\hat{j} + x^2\hat{k}$ , $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ , and $\vec{c}(x) = \hat{i} + f(x)\hat{j} + g(x)\hat{k}$ , where $f(x) = \int_{-x}^{x} \ln\left(t + \sqrt{t^2+1}\right) dt$ $g(x) = \int_{-x}^{x} \sqrt{t^2} dt$ If $V(x)$ represents the volume of the parallelepiped formed by the vectors $\vec{a}(x)$ , $\vec{b}$ , and $\vec{c}(x)$ for $x \in [-3, 3]$ , what is the maximum value of $V(x)$ ?

Options

A.
$72$
B.
$18$
C.
$2$
D.
$24$

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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

Collinearity vs Coplanarity|Geometry Constraint Modeling|Parametric Derivative Endpoint Trap|Even and Odd Function Symmetry

Why This Question Is Tricky

Common Mistakes

This problem models 3D coplanarity and volume constraints into a dynamic optimization problem, deploying two lethal traps.

First, the 'Even Function Symmetry' trap (Trap 283): students routinely assume $\int_{-x}^{x} |t| dt = x^2$ , forgetting that for $x < 0$ , the integral evaluates to $-x^2$ (formally $x|x|$ ).

This error simplifies the volume function to $V(x) = |3x - x^3|$ globally, leading to an incorrect maximum of 18.

Second, the 'Endpoint Trap' (Trap 252): the true volume function for $x < 0$ is $V(x) = |x^3 - 4x^2 + 3x|$ , which has no critical points in $[-3, 0)$ .

The global maximum strictly occurs at the boundary $x = -3$ , yielding 72.

Neglecting boundary conditions or misapplying symmetry guarantees failure.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Collinearity vs Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free