Before You Begin

Question #73615

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors in $\mathbb{R}^3$ . A scalar field $f(\vec{r})$ with $f(\vec{0}) = 0$ has its gradient given by $\nabla f(\vec{r}) = g(\vec{r} \cdot (\vec{a} \times \vec{b})) (\vec{a} \times \vec{b})$ where $g(x) = \int_0^x (1 - 2\{t\}) dt$ and $\{t\}$ denotes the fractional part of $t$ . Find the maximum value of $f(\vec{r})$ in the spherical region $|\vec{r}| \le \frac{n}{|\vec{a} \times \vec{b}|}$ , where $n$ is a positive integer.

Options

A.
$\frac{n}{6}$
B.
$\frac{n^2}{2} - \frac{n^3}{3}$
C.
$\frac{1}{6}$
D.
$\frac{n}{6} |\vec{a} \times \vec{b}|^2$

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

Geometric Interpretation of Cross Product|Objective Function Gradient|Integral Part and Fractional Part

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 integration of periodic functions involving the fractional part within a vector calculus optimization context.

The primary trap is ignoring the fractional part {t}, which leads to the incorrect gradient magnitude $g(x) = x - x^2$ and a non-periodic objective function, yielding option B.

Additionally, students must recognize that because $g(x) \ge 0$ , the objective function lacks interior local maxima and is strictly maximized at the boundary of the feasible region, preventing the erroneous assumption of a constant maximum (Option C).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Geometric Interpretation of Cross Product
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free