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Question #66802

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(x)$ be a $3 \times 3$ matrix defined as: $A(x) = \begin{pmatrix} \text{sgn}(\sin x + \cos 2x) & \ln(1+x^2) & e^{\sin x} \\ -\ln(1+x^2) & \text{sgn}(\sin x + \cos 2x) & \sqrt{1+x^4} \\ -e^{\sin x} & -\sqrt{1+x^4} & \text{sgn}(\sin x + \cos 2x) \end{pmatrix}$ where $\text{sgn}(\cdot)$ denotes the signum function. Let $S$ be the set of all $x \in [0, 2\pi]$ such that the linear transformation $T_x: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T_x(v) = A(x)v$ is not an injective mapping. Find the sum of all elements in $S$ .

Options

A.
$\frac{7\pi}{2}$
B.
$3\pi$
C.
0
D.
$5\pi$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Inverse Existence Criterion|Injective Mapping (One-One)|Signum Function Discontinuity

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the Inverse Existence Criterion linked to Injective Mappings via a parameterized matrix transformation.

The core trap lies in the behavior of the signum function at its roots.

Students often analyze the invertibility by checking if the determinant's limit approaches zero.

At $x = \pi/2$ , the function $f(x) = \sin x + \cos 2x$ touches zero but does not change sign, meaning $\lim_{x \to \pi/2} \det(A(x)) \neq 0$ .

A student relying on limits or continuity will incorrectly assume the transformation remains injective at this point.

However, injectivity strictly requires $\det(A(x)) \neq 0$ at the exact point.

Since $\text{sgn}(0) = 0$ , the matrix becomes singular at $x = \pi/2$ , breaking the injective mapping.

Missing this root leads to the lethal distractor $3\pi$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Inverse Existence Criterion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free