Let $\mathrm{A}=\{(\mathrm{x}, \mathrm{y}) \in \mathbf{R} \times \mathbf{R}:|\mathrm{x}+\mathrm{y}| \geq 3\}$ and $\mathrm{B}=\{(\mathrm{x}, \mathrm{y}) \in \mathbf{R} \times \mathbf{R}:|\mathrm{x}|+|\mathrm{y}| \leq 3\}$. If $\mathrm{C}=\{(\mathrm{x}, \mathrm{y}) \in \mathbf{A} \cap \mathbf{B}: \mathrm{x}=0$ or $\mathrm{y}=0\}$, then $\sum_{(x, y) \in C}|x+y|$ is :

Linear Inequalities — JEE Main · 2025 · 23 Jan 2025 · Shift 2 Question with Solution

Let

A = {(x, y) \in R \times R : ∣ x + y ∣ \geq 3}

and

B = {(x, y) \in R \times R : ∣ x ∣ + ∣ y ∣ \leq 3}

.
If

C = {(x, y) \in A \cap B : x = 0

y = 0}

, then

\sum_{(x, y) \in C} ∣ x + y ∣

is :

Answer: (D)

Set

A

is the region outside the strip between

x + y = - 3

and

x + y = 3

, and set

B

is the square

∣ x ∣ + ∣ y ∣ \leq 3

with vertices

(\pm 3, 0)

(0, \pm 3)

.

On the x-axis (

y = 0

): the conditions

∣ x ∣ \geq 3

and

∣ x ∣ \leq 3

force

∣ x ∣ = 3

, giving the points

(3, 0)

and

(- 3, 0)

.

On the y-axis (

x = 0

): similarly

∣ y ∣ = 3

, giving

(0, 3)

and

(0, - 3)

.

So

C = {(3, 0), (- 3, 0), (0, 3), (0, - 3)}

, and

(x, y) \in C \sum ∣ x + y ∣ = 3 + 3 + 3 + 3 = 12

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option