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Question #41357

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(x) = \begin{pmatrix} x^2+x & x & 1 \\ 2x & x^2 & 1 \\ 1 & 1 & 0 \end{pmatrix}$ and $f(x) = \det(A(x))$ . By Lagrange's Mean Value Theorem applied to $f(x)$ on the interval $[0, 2]$ , there exists a point $c \in (0, 2)$ such that the tangent to the curve $y = f(x)$ at $x=c$ is parallel to the chord joining the endpoints of the interval. If $M = \text{adj}(A(c))$ , which of the following matrix equations is satisfied by $M$ ?

Options

A.
$M^2 + 2M = O$
B.
$M^2 - 2M = O$
C.
$M^2 + M = O$
D.
$M^2 = O$

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

LMVT Tangent Parallelism|Differentiation of Determinants|Singular vs. Non-Singular Threshold

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of differential calculus (LMVT) and advanced linear algebra (properties of the adjugate of a singular matrix).

The 'trap' lies in the threshold of invertibility: students often compute the derivative of a determinant using Jacobi's formula, but here they must explicitly find the critical point $c=1$ and recognize that $A(1)$ becomes singular.

The lethal collision occurs when students fail to realize that for a singular matrix of rank $n-1$ , its adjugate has rank 1, unlocking the characteristic equation $M^2 = \text{tr}(M)M$ .

Blindly calculating the full $3 \times 3$ adjugate matrix and squaring it wastes precious time, whereas leveraging the singular threshold and rank properties provides an elegant, immediate solution.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	LMVT Tangent Parallelism
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free