Question

All the values of $a$ for which the quadratic expression $a{x}^2+(a-2)x-2$ is negative for exactly two integral values of $x$ may lie in

• A. $[1,3/2)$
• B. $[3/2,2)$
• C. $[1,2)$
• D. $[-1,2)$

Answer 1

The correct option is C $[1,2)$

$a{x}^2+(a-2)x-2$

$\Rightarrow$ The discriminant $=(a-2{)}^2-4(a\left)\right(-2)$

$=a^2-4a+4+8a$

$=a^2+4a+4$

$=(a+2{)}^2$

That's the positive for all $a$, so there are always two roots.

Now if $a>0,$ then the quadratic is negative between those two real roots.

Using quadratic formula for roots:

$\Rightarrow$ $\frac{2-a-(a+2)}{2a}<x<\frac{2-a+a+2}{2a}$

$\Rightarrow$ $-1<x<\frac{2}{a}$

For it exactly $2$ integer values of $x$ are in that interval, the integer must be $0$ and $+1.$

That requires $1<\frac{2}{a}\le 2.$

Taking reciprocals:

$1>\frac{a}{2}\le \frac{1}{2}$ [ In equality direction is reversed by reciprocal ]

$2>a\ge 1$ [ Multiply by $2$ ]

$\therefore$ $a\in [1,2)$