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Question #41364

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function. For any $x \in \mathbb{R}$ , the signed volume of the parallelepiped formed by the vectors $\vec{u} = e^{x^2}\hat{i} - 2xe^{x^2}\hat{j}$ , $\vec{v} = f(x)\hat{i} + f'(x)\hat{j}$ , and $\vec{w} = \hat{k}$ is $4x^3$ . If $f(x)$ is strictly decreasing for all $x \in (0, \infty)$ , then the minimum possible value of $f(0)$ is:

Options

A.
$0$
B.
$1$
C.
$2$
D.
$4$

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

Increasing/Decreasing Strictness|General vs Particular Solutions|Determinant as Geometric Area/Volume

Why This Question Is Tricky

Common Mistakes

This question integrates Linear Algebra (scalar triple product) with Differential Equations and Differential Calculus.

The primary trap is the 'General vs Particular Solution' pitfall: students often forget the constant of integration $C$ when solving the exact differential equation, leading them to incorrectly assume $f(x) = x^4 e^{-x^2}$ and failing to satisfy the monotonicity condition.

Furthermore, it tests the strictness of increasing/decreasing functions; students must recognize that $f'(x) \le 0$ with equality holding only at discrete points (here, $x=1$ ) is sufficient for strict decreasing nature, which establishes the tightest lower bound for the parameter $C$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Increasing/Decreasing Strictness
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free