NIMCET 2024 — Mathematics PYQ

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Question

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

Let $Z$ be the set of all integers, and consider the sets

$X = {(x, y) : x^{2} + 2 y^{2} = 3, x, y \in Z}$ and

$Y = \{(x,y) : x > y,\ x, y \in \mathbb{Z} \}$ .

Then the number of elements in $X \cap Y$ is

Choose the correct answer:

Explanation

$X = {(x, y) : x^{2} + 2 y^{2} = 3}$
$x^{2} + 2 y^{2} = 3$
$\Rightarrow \frac{x ^{2}}{3} + \frac{y ^{2}}{3/2} = 1$ (ellipse form)

$Y = \{(x, y) : x > y,\ x, y \in \mathbb{Z} \}$

Now find integer solutions for $x^{2} + 2 y^{2} = 3$ :
Try small integers:
- $(1, 1) : 1^{2} + 2 (1)^{2} = 1 + 2 = 3$
- $(- 1, 1) : 1 + 2 = 3$
- $(1, - 1) : 1 + 2 = 3$
- $(- 1, - 1) : 1 + 2 = 3$

So, $X = {(1, 1), (- 1, 1), (1, - 1), (- 1, - 1)}$

Now apply the condition $x > y$ from set $Y$ :
- $(1, 1) : x = y$
- $(-1,1): x < y$
- $(1,-1): x > y$
- $(- 1, - 1) : x = y$

So, $X \cap Y = {(1, - 1)}$

**Option (a)**

Choose the correct answer:

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