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Question #41866

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a} = |x^2 - 4|\hat{i} - 2x\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ . Let $\lambda = \text{sgn}(\vec{a} \cdot \vec{b})$ , where $\text{sgn}(\cdot)$ denotes the signum function. Consider the vectors $\vec{u} = \lambda \hat{i} - \hat{j} - \hat{k}$ , $\vec{v} = -\hat{i} + \lambda \hat{j} - \hat{k}$ , and $\vec{w} = -\hat{i} - \hat{j} + \lambda \hat{k}$ . If $\vec{u}, \vec{v}, \vec{w}$ are linearly dependent, then the set of all possible values of $x$ is:

Options

A.
$(\sqrt{6}-1, 3)$
B.
$(-\infty, \sqrt{6}-1) \cup (3, \infty)$
C.
$(-1, 3)$
D.
$(1/2, 3)$

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

Modulus Sign Dependency|Dot Product Projection Sign|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of vector linear dependence with modulus-based inequalities.

The primary trap lies in the determinant condition: students often assume $\lambda = 1$ yields dependence (as in standard cyclic matrices), but the negative signs in the off-diagonals shift the valid root to $\lambda = -1$ .

This forces the projection sign to be negative, requiring the solution of $|x^2 - 4| < 2x - 1$ .

A secondary trap occurs during the modulus inequality resolution: students may forget the domain restriction $2x - 1 > 0$ or incorrectly solve the piecewise components, leading to erroneous intervals like $(-1, 3)$ or the inverted case $(-\infty, \sqrt{6}-1) \cup (3, \infty)$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	Modulus Sign Dependency
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free