The complete set of values of $a$ for which the Inequality $ax^2-(3+2a)x+6>0,a\neq 0$ holds for exactly three negative integer values of $x,$ is
What I try: $\displaystyle ax^2-3x-2ax+6>0$
$x(ax-3)-2(ax-3)=0$
$\displaystyle (x-2)(ax-3)=0\Longrightarrow x=2\quad\text{or}\quad x=\frac{3}{a}$
From above , For exactly one integer $x,$ We have $a=3$
But I did not understand How can I find range of $a$ for exactly $3$ negative integer values of $x$
Please have a look , Thanks
If $a$ is negative, then the function will be positive in the interval $(\frac{3}{a}, 2)$.
If $a$ is positive then function is positive in the interval $(-\infty, 2) \cup (\frac{3}{a}, \infty)$ or $(-\infty, \frac{3}{a}) \cup (2, \infty)$ (based on whichever is higher out of 2, $\frac{3}{a}$)
As its positive for exactly three negative integers then it means $a < 0$. Also if $\frac{3}{a}$ lies in $[-4, -3)$ then the function will be positive for exactly three negative integers $x = -1, -2, -3$.
So we have to choose $a$ such that $\frac{3}{a}$ lies in interval $[-4, -3)$ which implies $a$ should lie in $(-1, -\frac{3}{4}]$.
Note : I first solved the problem considering exactly three integers. I have edited the answer which solves for exactly three negative integers
If we are restricting to $f(x) = ax^2−3x−2ax+6 > 0$ for only $3$ integer $x < 0$, then its clear only the integers $x = -1, -2, -3$ meet the condition.
Firstly, the parabola must be concave downward such that $f(x) > 0$ is bounded, where $f(x_{max}) > 0$, where $x_{max} = \dfrac{3}{2a} + 1$. Because $x = 2$ is always a zero, we must have $3/a < 2$. Furthermore, $3/a < 0$, since $f(\exists x < 0) > 0$.
Secondly, note that $f(0) = 6$ and the $x_{max} = \dfrac{3}{2a} + 1$. If $\dfrac{3}{2a} + 1 > 0$, $a < -3/2$. Let $a = -3/2$. Then $x_{max} = 0$ and $3/a = -2$. However, this does not satisfy the condition. For $a < -3/2$, $x_{max} > 0$, which indicates the number of $\{x: x < 0, f(x) > 0\}$ decrease. So it must be that $\dfrac{3}{2a} + 1 < 0$ and the continuity of $f(x)$ means $x = -1, -2, -3$.
Hence, $-4 \le 3/a < -3 \implies -1 < a \le -3/4$
