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Question #41926

Difficulty: 8/10 | 🟢 Free

The Question

Let the vectors $\vec{v}_1(x)$ , $\vec{v}_2(x)$ , and $\vec{v}_3(x)$ be defined for $x > 0$ as: $\vec{v}_1(x) = \left( 2 \int_{1/e}^x \frac{|\ln t|}{t} dt \right) \hat{i} + \left( 2 \int_1^x \frac{|\ln t|}{t} dt \right) \hat{j} + \hat{k}$ $\vec{v}_2(x) = ((\ln x)^2 - \ln x) \hat{i} + (\ln x + 2) \hat{j} + \hat{k}$ $\vec{v}_3(x) = \hat{i} + 2 \hat{j}$ Let $S$ be the set of all real numbers $x > 0$ for which the vectors $\vec{v}_1(x), \vec{v}_2(x), \vec{v}_3(x)$ are linearly dependent. Find the product of all elements in $S$ .

Options

A.
$e^4$
B.
$e^3$
C.
$e^{\frac{15 - \sqrt{57}}{6}}$
D.
$e^{\frac{27 - \sqrt{57}}{6}}$

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

Linear Independence vs Dependence|Newton-Leibniz Formula with variable limits|Modulus Function Breakdown|Logarithmic Integration Modulus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly evaluate definite integrals involving modulus functions and logarithmic arguments before applying the linear dependence condition.

The lethal trap lies in ignoring the modulus and blindly integrating $\frac{\ln t}{t}$ to $\frac{1}{2}(\ln t)^2$ .

If a student falls for this, they will incorrectly evaluate $2 I_1 = (\ln x)^2 - 1$ and $2 I_2 = (\ln x)^2$ , leading to the erroneous equation $u^2 - 3u = 0$ , which yields $x = 1$ and $x = e^3$ (product $e^3$ ).

The rigorous splitting of the integral at the critical points of the modulus is required to uncover the true piecewise nature of the determinant, revealing a hidden irrational root in the negative domain.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Linear Independence vs Dependence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free