Question

Consider quadratic equation $a{x}^2+(2-a)x-2=0$, where $a\in R$.

Let $\alpha ,\beta$ be roots of quadratic equation. If there are at least four negative integers between $\alpha$ and $\beta$, then the complete set of values of $a$ is

• A. $(-\frac{7}{2},-3)$
• B. $(0,\frac{1}{2})$
• C. $(-\frac{3}{2},-\frac{1}{2})$
• D. $(3,\frac{7}{2})$

Answer 1

The correct option is B $(0,\frac{1}{2})$

For real root,
$(2-a{)}^2+4(2\left)\right(a)\ge 0$
$(a-2{)}^2+8a\ge 0$
$a^2-4a+4+8a\ge 0$
$a^2+4a+4\ge 0$
$(a+2{)}^2\ge 0$
Hence, discriminant is always greater than 0.
Now,
$\alpha =\frac{a-2\pm (a+2)}{2a}$
$=\frac{a-2+a+2}{2a}=1$
And,
$\beta =\frac{a-2-a-2}{2a}$
$=\frac{-4}{2a}=\frac{-2}{a}$
Hence,
$\alpha =1$ and $\beta =\frac{-2}{a}$.
Now there are atleast $4$ negative integers between the roots.
Hence, one possible interval is $(0,\frac{1}{2}]$
Substituting $a=\frac{1}{2}$, we get $\beta =-4$.
Now there are $4$ negative integers between $1$ and $-4$.
Hence correct answer is Option B.