Consider the sets A=(x,y) R R:x^2+y^2=25,B=(x,y) R R:x^2+9y^2=144, C=(x,y) Z Z:x^2+y^2 4 a…

Question

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

Accepted Answer

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

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

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 $C=13$

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

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

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

Solutions · 1