Question

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

• A. $\left(\frac{-1-\sqrt{13}}{2},\frac{1-\sqrt{13}}{2}\right)$
• B. $\left(\frac{-1+\sqrt{13}}{2},\frac{1+\sqrt{13}}{2}\right)$
• C. $\left(\frac{1-\sqrt{13}}{2},-1\right)$
• D. None of these

Answer 1

Answer: (C)

Given lines are plotted as shown in the following figure.

If point $P\left(\alpha ,{\alpha }^2-2\right)$ lies inside $△ABC$,
then $A$ and $P$ must lie on the same side of $BC$ having equation
$y+1=0$
$\begin{array}{l}\Rightarrow (1+1)\left({\alpha }^2-2+1\right)>0\\ \Rightarrow {\alpha }^2-1>0\\ \Rightarrow \alpha \in (-\infty ,-1)\cup (1,\infty )\dots \end{array}$
$B$ and $P$ must lie on the same side of $AC$ having equation
$\begin{array}{l}x+y-1=0\\ \Rightarrow (-2-1-1)\left(\alpha +{\alpha }^2-2-1\right)>0\\ \Rightarrow ({\alpha }^2+\alpha -3<0\\ \Rightarrow \frac{-1-\sqrt{13}}{2}<\alpha <\frac{-1+\sqrt{13}}{2}\dots \text{ (2) }\end{array}$
$C$ and $P$ must lie on the same side of $AB$.
$\begin{array}{l}\Rightarrow (2+1+1)\left(\alpha -{\alpha }^2+2+1\right)>0\\ \Rightarrow {\alpha }^2-\alpha -3<0\\ \Rightarrow \frac{1-\sqrt{13}}{2}<\alpha <\frac{1+\sqrt{13}}{2}\dots (3)\end{array}$
$\alpha \in \left(\frac{1-\sqrt{13}}{2},-1\right)\cup \left(1,\frac{-1+\sqrt{13}}{2}\right)$