Question

The complete set of values of $a$ for which the inequality $a{x}^2-(3+2a)x+6>0,a\ne 0$ holds good for exactly three integral values of $x$ is

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

Answer 1

The correct option is B $(-3,-\frac{3}{2}]$

Let $f\left(x\right)=a{x}^2-(3+2a)x+6$
$f\left(x\right)=a{x}^2-3x-2ax+6\Rightarrow f\left(x\right)=(ax-3)(x-2)$
$x=2$ is a root of $f\left(x\right)=0$
$f\left(0\right)=6$
For an upward opening parabola there are infinitely many integers where $f\left(x\right)>0$, so the parabola should be downward opening,
A rough graph of the parabola is illustrated,


The three integral values are $-1,0,1$ for which $f\left(x\right)>0$, so
$f(-1)>0\Rightarrow 3a+9>0\Rightarrow a>-3\cdots \left(1\right)f(-2)\le 0\Rightarrow 8a+12\le 0\Rightarrow a\le -\frac{3}{2}\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$a\in (-3,-\frac{3}{2}]$