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Question #67584

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(t) = \begin{pmatrix} \sin t + |t| & \sin t + t & 1 \\ 2 & 2 & 3 \\ 3 & 3 & 5 \end{pmatrix}$ . Let $S$ be the set of all real numbers $t$ for which the matrix $A(t)$ is invertible. For any $t \in S$ , let $P(t)$ be a $3 \times 3$ real matrix satisfying $P(t)^T A(t) P(t) = (A(t))^2$ and $\det(P(t)) > 0$ . If $x \in S$ satisfies $\int_{x}^{0} \det(P(t)) dt = \frac{8}{3}$ , then the value of $x$ is:

Options

A.
-2
B.
-4
C.
2
D.
-8

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

Inversion and Determinant Product|Singular vs Non-Singular Matrices|Modulus Sign Dependency

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the student's ability to handle modulus sign dependency within matrix determinants and its implications on invertibility.

The primary trap lies in the piecewise nature of the determinant function $\det(A(t)) = |t| - t$ .

Students often mechanically compute determinants without analyzing the domain of validity for the inverse matrix.

If they fail to establish that $A(t)$ is only non-singular for $t < 0$ , they cannot correctly resolve the modulus to $-t$ , leading to an impasse or imaginary roots when solving $\det(P(t)) = \sqrt{-2t}$ .

The recursive logic forces them to first identify the non-singular domain $S$ to unlock the correct integrand, effectively bridging linear algebra with calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Inversion and Determinant Product
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free