Question

If the point $(\alpha ,{\alpha }^2)$ lies between $x+y-2=0$ and $4x+4y=3,$ then the range of $\alpha$ is

• A. $(-2,-\frac{3}{2})\cup (\frac{1}{2},1)$
• B. $(-2,-\frac{3}{2})\cup (\frac{1}{2},2)$
• C. $(2,-\frac{3}{2})\cup (\frac{1}{2},1)$
• D. $(-2,-\frac{1}{2})\cup (\frac{1}{2},1)$

Answer 1

The correct option is A $(-2,-\frac{3}{2})\cup (\frac{1}{2},1)$

Let two given lines be
$L_1:4x+4y=3$ and $L_2:x+y=2$


As the point lies between both the lines, we get
$(\alpha +{\alpha }^2-2)(4\alpha +4{\alpha }^2-3)<0\Rightarrow (\alpha +2)(\alpha -1)(2\alpha +3)(2\alpha -1)<0\therefore \alpha \in (-2,-\frac{3}{2})\cup (\frac{1}{2},1)$