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Question #41779

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all vectors $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ in $\mathbb{R}^3$ such that $x, y, z \ge 0$ and $|\vec{v}| = 1$ . Let $\vec{v}_1, \vec{v}_2, \vec{v}_3 \in S$ be three distinct vectors. If for each $i \in \{1, 2, 3\}$ , the components of $\vec{v}_i$ satisfy the inequality $x_i + y_i + z_i \le 1 + \int_{0}^{x_i y_i z_i} e^{t^2} dt$ and $M$ is the $3 \times 3$ matrix whose rows are $\vec{v}_1, \vec{v}_2, \vec{v}_3$ , then the value of $|\det(M)| + \text{Tr}(M M^T)$ is:

Options

A.
2
B.
3
C.
4
D.
5

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

Non-Negativity Restriction|Direction Cosines constraint|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

This problem tests the interaction between the spherical manifold constraint ( $x^2+y^2+z^2=1$ ), non-negativity restrictions, and a transcendental inequality.

The core trap is the illusion of a continuous feasible region.

Students often attempt to use Lagrange multipliers or assume the inequality defines a 2D surface patch on the unit sphere, failing to realize that the non-negativity constraint forces the lower bound of $x+y+z$ to grow much faster than the integral term.

By rigorously bounding both sides, the region collapses to exactly three discrete points (the standard basis vectors).

This perfectly targets the JEE Advanced methodology where complex analytical constraints disguise trivial geometric intersections, testing whether candidates verify linear independence boundaries within restricted octants.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Non-Negativity Restriction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free