Question

The range of values of $\alpha$ 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. $(-1,0)\cup (2,\infty )$

Answer 1

Answer: (B)

$(-\infty ,-2)\cup (0,1)$

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
$(2\alpha +6+3\alpha -6)(3\alpha +3{\alpha }^2-6)<0$
$\Rightarrow 15\alpha ({\alpha }^2+\alpha -2)<0$
$\Rightarrow \alpha (\alpha +2)(\alpha -1)<0$
$\Rightarrow \alpha ϵ(-\infty ,-2)\cup (0,1)$