question_answer

The range of values of a for which the points \[(\alpha ,\,\,2+\alpha )\] and\[\left( \frac{3\alpha }{2},\,\,{{\alpha }^{2}} \right)\]lie on opposite sides of the line\[2x+3y=6\], is

A) \[(-2,\,\,1)\]                      

B)  \[(-\infty ,\,\,-2)\cup (0,\,\,1)\]

C)  \[(-2,\,\,0)\cup (1,\,\,\infty )\] 

D)  \[(-1,\,\,0)\cup (2,\,\,\infty )\]

Correct Answer: B

Solution :

If the points \[(\alpha ,\,\,2+\alpha )\] and \[\left( \frac{3\alpha }{2},\,\,{{\alpha }^{2}} \right)\]are on the opposite sides of\[2x+3y-6=0\],then                 \[(2\alpha +6+3\alpha -6)(3\alpha +3{{\alpha }^{2}}-6)<0\] \[\Rightarrow \]               \[15({{\alpha }^{2}}+\alpha -2)<0\] \[\Rightarrow \]               \[\alpha (\alpha +2)(\alpha -1)<0\] \[\Rightarrow \]               \[\alpha \in (-\infty ,\,\,-2)\cup (0,\,\,1)\]