JEE Main 2025 — Mathematics Algebra

$$\mathrm{A}: \mathrm{x}^2+\mathrm{y}^2=25...(i)$$A:x2+y2=25...(i)

B : $\frac{x^2}{144}+\frac{y^2}{16}=1...(ii)$144x2​+16y2​=1...(ii)

C: $x^2+y^2 \leq 4...(iii)$x2+y2≤4...(iii)

Solve (1) & (2)

$\begin{aligned}
& x^2+9\left(25-x^2\right)=144 \
& -8 x^2=144-225=-81 \
& x= \pm \frac{9}{2 \sqrt{2}}
\end{aligned}$

$$\operatorname{By}(1) \Rightarrow y= \pm \sqrt{25-x^2}$$By(1)⇒y=±25−x2​

$$= \pm \sqrt{25-\frac{81}{8}}= \pm \frac{\sqrt{119}}{2 \sqrt{2}}$$=±25−881​​=±22​119​​

$$\therefore \mathrm{D}=\mathrm{A} \cap \mathrm{~B}=$$∴D=A∩ B=

$$\left\{\left(\frac{9}{2 \sqrt{2}}, \frac{\sqrt{119}}{2 \sqrt{2}}\right),\left(\frac{9}{2 \sqrt{2}},-\frac{\sqrt{119}}{2 \sqrt{2}}\right),\left(\frac{-9}{2 \sqrt{2}}, \frac{\sqrt{119}}{2 \sqrt{2}}\right),\left(\frac{-9}{2 \sqrt{2}}, \frac{-\sqrt{119}}{2 \sqrt{2}}\right)\right\}$${(22​9​,22​119​​),(22​9​,−22​119​​),(22​−9​,22​119​​),(22​−9​,22​−119​​)}

No. of elements in set $D=4$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 $\mathrm{C}=13$C=13

Total no. of one-one function from

Set $D$D to $\sec C \Rightarrow 13 \times 12 \times 11 \times 10=17160$secC⇒13×12×11×10=17160