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Question #41233

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = |x| + x^2$ for $x \in [-1, 1]$ . Consider the matrix $M(x) = \begin{pmatrix} f'(x) & 2 & 2 \\ 1 & -1 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ defined for all $x \in (-1, 1) \setminus \{0\}$ . Let $C$ be the set of all $c \in (-1, 1) \setminus \{0\}$ such that $M(c)$ is a singular matrix. If $C$ is non-empty, let $S = \sum_{c \in C} c$ . If $C$ is empty, let $S = 10$ . Find the value of $S$ .

Options

A.
0
B.
10
C.
$\frac{1}{2}$
D.
$-\frac{1}{2}$

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

Singular vs Non-Singular Matrices|Rolle's Theorem Root Existence|Modulus Function Symmetry

Why This Question Is Tricky

Common Mistakes

This question tests the misapplication of Rolle's Theorem and domain constraints.

Advanced students often integrate the first row of $M(x)$ to form a new determinant $h(x) = 4 - 2f(x)$ .

Since $h(-1) = h(1) = 0$ , they falsely assume by Rolle's Theorem that a root $c$ exists where $h'(c) = \det(M(c)) = 0$ .

Assuming symmetry, they guess the sum of roots is 0.

Alternatively, if they expand the determinant directly, they find false roots $x = \pm 1/2$ and fail to check the domain constraints of the signum function, again arriving at a sum of 0.

The lethal trap is that $f(x)$ is non-differentiable at $x=0$ (a cusp), which completely invalidates Rolle's Theorem and leaves the set of singular points empty.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Singular vs Non-Singular Matrices
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free