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Question #41286

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a} = (x^2-4)\hat{i} + x^2\hat{j} + e^x\hat{k}$ , $\vec{b} = (x^2-4)[x]\hat{i} + (x^2[x]+[x])\hat{j} + (e^x[x]+[x])\hat{k}$ , and $\vec{c} = (x^2-4)[x]^2\hat{i} + (x^2[x]^2+[x]^2)\hat{j} + (e^x[x]^2+[x]^2+[x])\hat{k}$ , where $[x]$ denotes the greatest integer less than or equal to $x$ . Let $V(x)$ be the volume of the parallelepiped formed by the vectors $\vec{a}, \vec{b},$ and $\vec{c}$ . Find the value of $\lim_{h \to 0^+} \frac{V(2+h) - V(2-h)}{h}$

Options

A.
12
B.
20
C.
0
D.
16

Check Your Answer

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

Triangular/Diagonal Matrix Shortcut|Scalar Triple Product Volume Logic|Greatest Integer Function Traps

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of matrix row operations, the geometric definition of the scalar triple product, and the piecewise nature of the greatest integer function.

The primary trap lies in the definition of volume as the absolute value of the determinant.

Students who successfully reduce the matrix to a triangular form often forget the absolute value, yielding $V(x) = (x^2-4)[x]^2$ .

When evaluating the limit, this omission leads to a negative volume for $x < 2$ , resulting in the incorrect answer 20.

A secondary trap is treating $[x]$ as a constant across the limit, which incorrectly yields 0.

The problem demands rigorous adherence to the definition $V(x) = |x^2-4|[x]^2$ and careful evaluation of the left and right neighborhoods of $x=2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Triangular/Diagonal Matrix Shortcut
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free