Consider the sets $\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} imes \mathbb{R}: \mathrm{x}^2+\mathrm{y}^2=25 ight\}$, $\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} imes \mathbb{R}: \mathrm{x}^2+9 \mathrm{y}^2=144 ight\}, \mathrm{C}=\{(\mathrm{x}, \mathrm{y})$…

Question

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

Aicharya · Accepted Answer

$\mathrm{A}: \mathrm{x}^2+\mathrm{y}^2=25...(i)$
B : $\frac{x^2}{144}+\frac{y^2}{16}=1...(ii)$
C: $x^2+y^2 \leq 4...(iii)$
Solve (1) \& (2)
$\begin{aligned}
& x^2+9\left(25-x^2\right)=144 \\ & -8 x^2=144-225=-81 \\ & x= \pm \frac{9}{2 \sqrt{2}}
\end{aligned}$
$\operatorname{By}(1) \Rightarrow y= \pm \sqrt{25-x^2}$
$= \pm \sqrt{25-\frac{81}{8}}= \pm \frac{\sqrt{119}}{2 \sqrt{2}}$
$\therefore \mathrm{D}=\mathrm{A} \cap \mathrm{~B}=$
$\left\{\left(\frac{9}{2 \sqrt{2}}, \frac{\sqrt{119}}{2 \sqrt{2}}\right),\left(\frac{9}{2 \sqrt{2}},-\frac{\sqrt{119}}{2 \sqrt{2}}\right),\left(\frac{-9}{2 \sqrt{2}}, \frac{\sqrt{119}}{2 \sqrt{2}}\right),\left(\frac{-9}{2 \sqrt{2}}, \frac{-\sqrt{119}}{2 \sqrt{2}}\right)\right\}$
No. of elements in set $D=4$

$\begin{aligned}
& \because C=\left\{(x, y) \in \mathrm{Z} \times \mathrm{Z}: \mathrm{x}^2+\mathrm{y}^2 \leq 4\right\} \\ & =\{(0,2),(2,0),(0,-2),(-2,0),(1,1),(-1,-1), \\ & (1,-1),(-1,1),(1,0),(0,1),(-1,0),(0,-1), \\ & (0,0)\}
\end{aligned}$
No. of elements in set $\mathrm{C}=13$
Total no. of one-one function from
Set $D$ to $\sec C \Rightarrow 13 \times 12 \times 11 \times 10=17160$

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

Solution