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Question #40381

Difficulty: 8/10 | 🟢 Free

The Question

Let M(x)=(sgn(x)11−1−sgn(x)−100sgn2(x)−1)M(x) = \begin{pmatrix} \text{sgn}(x) & 1 & 1 \\ -1 & -\text{sgn}(x) & -1 \\ 0 & 0 & \text{sgn}^2(x) - 1 \end{pmatrix} and b(x)=(x2−5x+5+sgn(x)−2x2−5x+4)\mathbf{b}(x) = \begin{pmatrix} x^2 - 5x + 5 + \text{sgn}(x) \\ -2 \\ x^2 - 5x + 4 \end{pmatrix}, where sgn(x)\text{sgn}(x) denotes the signum function. Let SS be the set of all real values of xx for which the system of linear equations M(x)v=b(x)M(x) \mathbf{v} = \mathbf{b}(x) has at least one solution. Find the number of elements in SS.

Options

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

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

Signum Function Discontinuity|Nilpotent Matrix Logic|Cramer's Rule Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the interaction between nilpotent matrices, system consistency, and the point discontinuity of the signum function.

The trap lies in the determinant det⁡(M(x))=−(sgn2(x)−1)2\det(M(x)) = -(\text{sgn}^2(x) - 1)^2.

Students often incorrectly assume sgn2(x)=1\text{sgn}^2(x) = 1 for all real xx, concluding that the matrix is always singular (det⁡=0\det = 0) and nilpotent.

Consequently, they only check the consistency conditions for x≠0x \neq 0, finding x=1x=1 and x=4x=4, but completely miss the singularity trap at x=0x=0.

At x=0x=0, sgn(0)=0\text{sgn}(0)=0, the rank fluctuates, and the matrix regains full rank (det⁡=−1\det = -1), guaranteeing a unique solution and adding x=0x=0 to the solution set.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptSignum Function Discontinuity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree