JEE Mathematics 2025 Question with Solution

Given:

• $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + y| \geq 3\}$
• $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| + |y| \leq 3\}$
• $C = \{(x, y) \in A \cap B : x = 0 \text{ or } y = 0\}$

Find:

$$\sum_{(x, y) \in C} |x| + |y|$$

For points in $C$, either $x = 0$ or $y = 0$, and they must satisfy both set conditions.

If $x = 0$, then from set $A$:

$$|y| \geq 3$$

From set $B$:

$$|0| + |y| \leq 3 \Rightarrow |y| \leq 3$$

Hence,

$$|y| = 3$$

So the possible points are $(0, 3)$ and $(0, -3)$.

If $y = 0$, then from set $A$:

$$|x| \geq 3$$

From set $B$:

$$|x| + |0| \leq 3 \Rightarrow |x| \leq 3$$

Hence,

$$|x| = 3$$

So the possible points are $(3, 0)$ and $(-3, 0)$.

Therefore,

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

Now compute the required sum:

$$|0| + |3| = 3$$ $$|0| + |-3| = 3$$ $$|3| + |0| = 3$$ $$|-3| + |0| = 3$$

Thus,

$$3 + 3 + 3 + 3 = 12$$

Therefore, the correct option is D, and the required sum is $12$.

Given: the points must satisfy both $|x + y| \geq 3$ and $|x| + |y| \leq 3$, with $x = 0$ or $y = 0$.

Find: the total of $|x| + |y|$ over all such points.

On the coordinate axes, the condition $x = 0$ or $y = 0$ reduces the diamond

$$|x| + |y| \leq 3$$

to the axis endpoints only when combined with

$$|x + y| \geq 3$$

Because on the axes, $|x + y|$ becomes either $|x|$ or $|y|$. So both conditions together force equality:

$$|x| = 3 \quad \text{or} \quad |y| = 3$$

Hence the only points are the four axis intercepts:

$$(\pm 3, 0), (0, \pm 3)$$

Each contributes

$$|x| + |y| = 3$$

So the total is

$$4 \times 3 = 12$$

Therefore, the correct option is D.