Mathematics - Functions Question with Solution | TestHub

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

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