question_answer

If point \[P(\alpha ,\,{{\alpha }^{2}}-2)\] lies inside the triangle formed by the lines \[x+y=1,\] \[y=x+1\]and \[y=-1,\] then \[\alpha \in /\]

A) \[\left( \frac{1-\sqrt{13}}{2},-1 \right)\]

B)        \[\left( 1,\frac{-1+\sqrt{13}}{2} \right)\]

C) \[\left( \frac{-1-\sqrt{13}}{2},1 \right)\]

D)        None of these

Correct Answer: C

Solution :

[c] Given lines are plotted as shown in the following figure: If point \[p(\alpha ,\,{{\alpha }^{2}}-2)\]lies inside \[\Delta ABC,\] then A and P must lie on the same side of BC having equation\[y+1=0\]. \[\Rightarrow \,\,\,\,(1+1)\,({{\alpha }^{2}}-2+1)>0\] \[\Rightarrow \,\,\,\,{{\alpha }^{2}}-1>0\] \[\Rightarrow \,\,\,\,\alpha \in (-\infty ,-1)\cup (1,\infty )\]                  ??(1) B and P must lie on the same side of AC having equation\[x+y-1=0\]. \[\Rightarrow \,\,\,(-2-1-1)\,(\alpha +{{\alpha }^{2}}-2-1)>0\] \[\Rightarrow \,\,\,\,{{\alpha }^{2}}+\alpha -3<0\] \[\Rightarrow \,\,\,\,\frac{-1-\sqrt{13}}{2}<\alpha <\frac{-1+\sqrt{13}}{2}\]                       ? (2) C and P must lie on the same side of AB. \[\Rightarrow \,\,\,\,(2+1+1)\,(\alpha -{{\alpha }^{2}}+2+1)>0\] \[\Rightarrow \,\,\,{{\alpha }^{2}}-\alpha -3<0\] \[\Rightarrow \,\,\,\frac{1-\sqrt{13}}{2}<\alpha <\frac{1+\sqrt{13}}{2}\]               ? (3) From equations (1), (2) and (3), we get \[\alpha \in \left( \frac{1-\sqrt{13}}{2},-1 \right)\cup \left( 1,\frac{-1+\sqrt{13}}{2} \right)\]