QJEEM24Maed04: JEE Main 2024 Mathematics - Class 11 JEEM24Maed04 Solution: Let be the interior region between the lines and containing the …

$\begin{aligned}& \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\& \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0\end{aligned}$

Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_1$

$\begin{aligned}& \Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\& \therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0\end{aligned}$

$\begin{aligned}& \Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\& \mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \text{..... (1)}\end{aligned}$

Let $L_2: x+2 y-5=0$

Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_2$

$\begin{aligned}& \Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\& \Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\& \Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\& \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0\end{aligned}$

$\therefore \mathrm{a} \in(-3,1)\quad \text{..... (2)}$

Intersection of (1) and (2)

$\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$