Consider the sets \(\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}: \mathrm{x}^{2}+\mathrm{y}^{2}=25\right\}\), \(\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}: \mathrm{x}^{2}+9 \mathrm{y}^{2}=144\right\}\), \(\mathrm{C}=\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 4\}\), and \(\mathrm{D}=\mathrm{A} \cap \mathrm{B}\). The total number of one-one functions from the set D to the set C is:

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Sets Relations and Functions

Consider the sets

A = {(x, y) \in R \times R : x^{2} + y^{2} = 25}

B = {(x, y) \in R \times R : x^{2} + 9 y^{2} = 144}

C = {(x, y) \in Z \times Z : x^{2} + y^{2} \leq 4}

, and

D = A \cap B

. The total number of one-one functions from the set D to the set C is:

Answer: (C)

The circle

x^{2} + y^{2} = 25

and the ellipse

\frac{x ^{2}}{144} + \frac{y ^{2}}{16} = 1

meet where both equations hold together. Replacing

x^{2}

25 - y^{2}

x^{2} + 9 y^{2} = 144

gives

25 + 8 y^{2} = 144

, so

y^{2} = \frac{119}{8}

and

x^{2} = \frac{81}{8}

. Each of

x

and

y

then takes two signs, so the intersection

D

has four points. The set

C

of integer points with

x^{2} + y^{2} \leq 4

contains

5

points on

x = 0

3

each on

x = \pm 1

and

1

each on

x = \pm 2

, which is

13

points in all. A one-one map from the

4

points of

D

into the

13

points of

C

can be made in

13 \times 12 \times 11 \times 10

ways, which equals

17160

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option