Question

The set of values of $\alpha$ for which the point $P(\alpha ,{\alpha }^2-2)$ lies inside the triangle formed by the lines $x+y=1,$ $y=x+1$ and $y=-1,$ is

• A. $(-\sqrt{3},\sqrt{3})$
• B. $(\frac{1-\sqrt{13}}{2},-1)\cup (1,\frac{-1+\sqrt{13}}{2})$
• C. $[-1,1]$
• D. $(\frac{1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2})$

Answer 1

The correct option is B $(\frac{1-\sqrt{13}}{2},-1)\cup (1,\frac{-1+\sqrt{13}}{2})$

Let
$AB:y=x+1BC:y+1=0AC:x+y=1$
Finding the vertices of triangle
$A=(0,1),B=(-2,-1),C=(2,-1)$

We know that $P(\alpha ,{\alpha }^2-2)$ lies inside the $\Delta ABC$

$\left(i\right)A\text{and}P$ must lie on the same side of $BC$
$(1+1)({\alpha }^2-2+1)>0\Rightarrow {\alpha }^2-1>0\Rightarrow \alpha \in (-\infty ,-1)\cup (1,\infty )$

$\left(ii\right)B\text{and}P$ must lie on the same side of $AC$.
$(-2-1-1)(\alpha +{\alpha }^2-2-1)>0\Rightarrow {\alpha }^2+\alpha -3<0$
Now ${\alpha }^2+\alpha -3=0$
$\Rightarrow \alpha =\frac{-1\pm \sqrt{13}}{2}\Rightarrow \frac{-1-\sqrt{13}}{2}<\alpha <\frac{-1+\sqrt{13}}{2}$

$\left(iii\right)C\text{and}P$ must lie on the same side of $AB$.
$(2+1+1)(\alpha -{\alpha }^2+2+1)>0\Rightarrow {\alpha }^2-\alpha -3<0\Rightarrow \frac{1-\sqrt{13}}{2}<\alpha <\frac{1+\sqrt{13}}{2}$

From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$\alpha \in (\frac{1-\sqrt{13}}{2},-1)\cup (1,\frac{-1+\sqrt{13}}{2})$