Question

The set of values of $a$ for which $a{x}^2-(4-2a)x-8<0$ for exactly three integral values of $x$ is -

• A. $2\le a<4$
• B. $1\le a<2$
• C. $1\le a<2$
• D. $0\le a<1$

Answer 1

The correct option is A $2\le a<4$

$a{x}^2-4x+2ax-8=(x+2)(ax-4)<0$
Clearly, one root is -2.
Now, here $a>0$ (for $a<0$, there are infinite integers possible)
Now, the other root is $\frac{4}{a}$. for exactly three integers, $1<\frac{4}{a}\le 2$ (-1, 0, 1 come between the roots)
or $-6\le \frac{4}{a}<-5$ (-5, -4, -3 come between the roots)

The latter case is not possible as a > 0
$1<\frac{4}{a}\le 2$
$\Rightarrow \frac{1}{2}\le \frac{a}{4}<1$
$\Rightarrow 2\le a<4$

So for this inequality $1<\frac{4}{a}\le 2\Rightarrow 2\le a<4$