Before You Begin

Question #66916

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(x) = \begin{bmatrix} \int_0^x \cos t \, dt & 1 & 0 \\ 1 + \int_0^x t \cos t \, dt & x & 1 \\ \int_0^x (t^2+2) \cos t \, dt & x^2 & 2x \end{bmatrix}$ . If $f(x) = \det(\text{adj}(A(x)) + A'(x))$ , where $\text{adj}(A(x))$ denotes the adjugate matrix of $A(x)$ and $A'(x)$ is the element-wise derivative of $A(x)$ with respect to $x$ , then the value of $f(\pi)$ is:

Options

A.
$-3\pi^2 - 1$
B.
-2
C.
$3\pi^2 - 1$
D.
0

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 Threshold|Differentiation of Determinants|Leibniz Rule for Differentiation under Integral

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 intersection of Wronskian-like determinant properties, the Leibniz rule, and matrix calculus.

A common trap is noticing that $C_1(x) = \sin x C_2(x) + \cos x C_3(x)$ , meaning $\det(A(x)) = 0$ for all $x$ .

Students incorrectly assume that because $A(x)$ is singular, its adjugate must be the zero matrix, leading them to calculate only $\det(A'(\pi)) = -2$ .

However, $\text{adj}(A) = 0$ only if $\text{rank}(A) \le n-2$ .

Here, $\text{rank}(A) = 2$ , so the adjugate is a non-zero rank-1 matrix.

Another trap is assuming differentiation preserves singularity, falsely concluding $\det(A'(x)) = 0$ .

Bypassing these traps requires rigorous application of the Leibniz rule and direct evaluation of the adjugate and derivative matrices.

Quick Facts About This Question

Question Information

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