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Q.
If point $P\left(\alpha, \alpha^{2}-2\right)$ lies inside the triangle formed by the lines $x+y=1,y=x+1$ and $y=-1,$ then $\alpha $ belongs to
NTA AbhyasNTA Abhyas 2022
Solution:
Given lines are plotted as shown in the following figure.
If point $P\left(\alpha, \alpha^{2}-2\right)$ lies inside $\triangle A B C$,
then $A$ and $P$ must lie on the same side of $B C$ having equation
$y+1=0$
$
\begin{array}{l}
\Rightarrow(1+1)\left(\alpha^{2}-2+1\right)>0 \\
\Rightarrow \alpha^{2}-1>0 \\
\Rightarrow \alpha \in(-\infty,-1) \cup(1, \infty) \ldots
\end{array}
$
$B$ and $P$ must lie on the same side of $A C$ having equation
$
\begin{array}{l}
x+y-1=0 \\
\Rightarrow(-2-1-1)\left(\alpha+\alpha^{2}-2-1\right)>0 \\
\Rightarrow\left(\alpha^{2}+\alpha-3<0\right. \\
\Rightarrow \frac{-1-\sqrt{13}}{2}<\alpha<\frac{-1+\sqrt{13}}{2} \ldots \text { (2) }
\end{array}
$
$C$ and $P$ must lie on the same side of $A B$.
$
\begin{array}{l}
\Rightarrow(2+1+1)\left(\alpha-\alpha^{2}+2+1\right)>0 \\
\Rightarrow \alpha^{2}-\alpha-3<0 \\
\Rightarrow \frac{1-\sqrt{13}}{2}<\alpha<\frac{1+\sqrt{13}}{2} \ldots(3)
\end{array}
$
$\alpha \in \left(\frac{1 - \sqrt{13}}{2} , - 1\right)\cup\left(1 , \frac{- 1 + \sqrt{13}}{2}\right)$
