Consider the sets $\mathrm{A}=\left{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=25\right}$, $\mathrm{B}=\left{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:{\mathrm{x}}^{2}+9{\mathrm{y}}^{2}=144\right},$$\mathrm{C}={(\mathrm{x},\mathrm{y})$$\left.\in \mathrm{ℤ}\times \mathrm{ℤ}:{x}^{2}+{y}^{2}\leq 4\right}$, 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, 4 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: c) $17160$. $\begin{matrix}A={(x,y)\in R\times R:{x}^{2}+{y}^{2}=25}, \ \ B={(x,y)\in \mathrm{ℝ}\times \mathrm{ℝ}:{x}^{2}+9{y}^{2}=144}\end{matrix}$ ${x}^{2}+9{y}^{2}−({x}^{2}+{y}^{2})=144−25$ Plug in ${y}^{2}=\frac{119}{8}$ into either equation to find x. $\begin{matrix}{x}^{2}=25−\frac{119}{8} \ {x}^{2}=\frac{200−119}{8} \ {x}^{2}=\frac{81}{8} \ x=\pm \sqrt{\frac{81}{8}},y=\pm \sqrt{\frac{119}{8}}\end{matrix}$ Now, $C={(x,y)\in Z\times Z:{x}^{2}+{y}^{2}\leq 4}$ Valid points are $(−2,0),(−1,−1),(−1,0),(−1,1),(0,−2),$ $(0,−1),(0,0),(0,1),(0,2),(1,−1),(1,0),(1,1)$ $∴$ Total valid points in $C=13$ $\Rightarrow$ There are 4 distinct real points in set D $∴$ The number of one-one functions from D to C $\begin{matrix}\Rightarrow 13{P}_{4}\Rightarrow \frac{13!}{(13−4)!}=\frac{13!}{9!}=17160\end{matrix}$

Consider the sets \(\mathrm{A}=\left{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:{\mathrm{x}}^{2}+{\mathrm{y}…

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