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Question #67175

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u}, \vec{v}, \vec{w}$ be three non-coplanar vectors such that the volume of the parallelepiped formed by them is $V_0$ . A new variable parallelepiped is formed by the vectors: $\vec{p} = t\vec{u} + c\vec{v} + c\vec{w}$ $\vec{q} = c\vec{u} + t\vec{v} + c\vec{w}$ $\vec{r} = c\vec{u} + c\vec{v} + t\vec{w}$ where $c > 0$ is a constant and $t$ is a real parameter. If $t$ is constrained by the inequality $t^2 + ct \le 12c^2$ , what is the maximum possible volume of the parallelepiped formed by $\vec{p}, \vec{q}, \vec{r}$ ?

Options

A.
$50c^3 V_0$
B.
$20c^3 V_0$
C.
$4c^3 V_0$
D.
$54c^3 V_0$

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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

Determinant Scaling Property|Geometry Constraint Modeling|First Derivative Test Trap

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the 'First Derivative Test Trap' combined with the determinant scaling property of volumes.

Students must model the volume as $V(t) = |\det(M)| V_0$ .

The trap occurs when optimizing this function: students often find the local maximum at $t = -c$ (yielding $4c^3 V_0$ ) and incorrectly assume it is the global maximum.

Furthermore, even if they check the boundaries of the constraint $t \in [-4c, 3c]$ , they may evaluate the determinant at $t = -4c$ as $-50c^3 V_0$ and discard it because it is negative, forgetting that volume is the absolute value of the determinant.

This forces a rigorous application of extreme value theorem on absolute functions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Determinant Scaling Property
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free