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Question #41268

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \begin{pmatrix} x & y & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} x & y & 1 \\ -1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ . Let $M = A B^T$ . A point $P(x,y)$ moves in the $xy$ -plane such that it satisfies the following conditions:

(i) $\text{sgn}(\det(M)) \cdot |\det(M) - 8| \ge 1$ (ii) $|x + y - 1| + |x - y + 1| \le 4$

Find the area of the region consisting of all such points $P(x,y)$ .

Options

A.
4
B.
8
C.
16
D.
32

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

Determinant Product Rule|Area of Triangle via Determinants|Modulus Inequality Logic|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

The question tests the Determinant Product Rule and coordinate transformations, but the core trap lies in the signum-constrained modulus inequality.

Students often ignore the $\text{sgn}(u)$ term or assume it only flips the sign without restricting the domain.

If $u < 0$ , the inequality becomes $u \ge 9$ , which is a contradiction, meaning $u = \det(M)$ must be strictly positive.

This restricts the locus to only two of the four quadrants in the transformed $XY$ -plane.

Failing to recognize this discontinuity trap leads to calculating the area of the entire square (yielding 16 instead of 8).

Additionally, forgetting the Jacobian of the transformation $(x,y) \to (X,Y)$ provides another path to the trap answer 16.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Determinant Product Rule
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free