Let [R] be the interior region between the lines 3 x-y+1=0 and x+2 y-5=0 containing the or…

Question

Let $$\mathrm{R}$$ be the interior region between the lines $3 x-y+1=0$ and $x+2 y-5=0$ containing the origin. The set of all values of $a$, for which the points $$\left(a^2, a+1\right)$$ lie in $R$, is :

Accepted Answer

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

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