JEE Main 2025 — Sets And Relations Question with Solution

$$\begin{array}{rl}& A=\left\{(x,y)\in R\times R:x^2+y^2=25\right\},B=\{(x,y)\in \mathbb{R}\times \\ & \mathbb{R}:x^2+9y^2=144\}\\ & x^2+9y^2-\left(x^2+y^2\right)=144-25\\ & \text{ Plug in }{y}^2=\frac{119}{8}\text{ into either equation to find }x.\\ & x^2=25-\frac{119}{8}\\ & x^2=\frac{200-119}{8}\\ & x^2=\frac{81}{8}\\ & x=\pm \sqrt{\frac{81}{8}},y=\pm \sqrt{\frac{119}{8}}\end{array}$$

Now, $C=\left\{(x,y)\in \mathbb{Z}\times \mathbb{Z}:x^2+y^2\le 4\right\}$

Valid points are $(-2,0),(-1,-1),(-1,0),(-1,1)$, $(0,-2),(0,-1),(0,0),(0,1),(0,2),(1,-1),(1,0)$, $(1,1)$

$\therefore$ Total valid points in $C=13$

$\Rightarrow$ There are 4 distinct real points in set $D$

$\therefore$ The number of one-one functions from $D$ to $C$

$$\Rightarrow 13P_4\Rightarrow \frac{13!}{(13-4)!}=\frac{13!}{9!}=17160$$