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Question #41905

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(x) = \begin{pmatrix} |x-1| & x^2 \\ 2 & |x| \end{pmatrix}$ and $B(x) = \begin{pmatrix} x & -|x-1| \\ |x| & 2x \end{pmatrix}$ . Define the matrix $M(x) = A(x)B(x)$ . Let $D_1 = \det(M'(1^+))$ and $D_2 = \det(M'(1^-))$ , where $M'(1^+)$ and $M'(1^-)$ denote the right-hand and left-hand derivatives of the matrix $M(x)$ with respect to $x$ at $x=1$ , respectively. Find the value of $D_1 + D_2$ .

Options

A.
-28
B.
-30
C.
-32
D.
-24

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Modulus Function Breakdown|Differentiation of Determinants|Matrix Multiplication Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests two critical concepts: the non-commutativity of the matrix product rule and the piecewise differentiation of modulus functions.

A common trap is assuming $\det(M'(x)) = \frac{d}{dx}\det(M(x))$ , which incorrectly yields -32.

Another trap is applying the product rule commutatively as $A'B + B'A$ , which yields -30.

Finally, ignoring the modulus bifurcation at $x=1$ leads to -24.

The student must rigorously apply $M' = A'B + AB'$ while carefully evaluating the left and right derivatives of the modulus components.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Modulus Function Breakdown
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free