Question

The range of $\alpha$ for which the points $(\alpha ,\alpha +2)$ and $(\frac{3\alpha }{2},{\alpha }^2)$ lie on opposite sides of the line $2x+3y-6=0$ is

• A. $(-\infty ,-2)\cup (0,1)$
• B. $(-\infty ,-2)\cup (1,\infty )$
• C. $(-\infty ,0)\cup (1,2)$
• D. $(-1,0)\cup (1,2)$

Answer 1

The correct option is A $(-\infty ,-2)\cup (0,1)$

Let, $f\left(x\right)=2x+3y-6$
$L_1=f(\alpha ,\alpha +2)=2\alpha +3\alpha +6-6=5\alpha$
$L_2=f(\frac{3\alpha }{2},{\alpha }^2)=3\alpha +3{\alpha }^2-6$

As the point lies on the opposite sides of the line, so
$L_1\cdot L_2<0$
$\Rightarrow 5\alpha (3\alpha +3{\alpha }^2-6)<0\Rightarrow \alpha ({\alpha }^2+\alpha -2)<0$
$\Rightarrow \alpha (\alpha +2)(\alpha -1)<0$
$\Rightarrow \alpha \in (-\infty ,-2)\cup (0,1)$