Question

If exactly two integers lie between the roots of the equation $x^2+ax-1=0$, then possible integral value(s) of $a$ is (are)

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (A), (C)

$1$

Let the roots be $\alpha ,\beta .$
$\because f\left(0\right)=-1\Rightarrow 0$ lies in between the roots of $f\left(x\right)=0$, as for an upward parabola $f\left(x\right)<0$ for all values in between the roots.
Since, exactly two integers lie between the roots, other integer apart from $0,$ should be $1$ or $-1$

Case $1:$ When $-1$ lies between the roots


Required conditions
$f(-1)<0\Rightarrow a>0f(-2)>0\Rightarrow a<\frac{3}{2}f\left(1\right)>0\Rightarrow a>0\therefore a\in (0,\frac{3}{2})\cdots \left(1\right)$

Case $2:$ When $1$ lies between the roots


Required conditions
$f(-1)>0\Rightarrow a<0f\left(2\right)>0\Rightarrow a>-\frac{3}{2}f\left(1\right)<0\Rightarrow a<0\therefore a\in (-\frac{3}{2},0)\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$a\in (-\frac{3}{2},\frac{3}{2})-\left\{0\right\}$
$\therefore$ Possible integral values of $a$ are $-1,1$

$\underset{–}{\text{Alternate}:}$
Let the roots be $\alpha ,\beta .$
$\alpha \beta =-1,$ so roots are opposite in nature.
So, $0$ will lie in between the roots and other integer will be either $-1$ or $1.$
So let $\alpha >0$ and $\beta <0$
For exactly two integers, $-2$ and $2$ will not lie inside the roots.
So when $a<0$, $\alpha +\beta >0$
$\Rightarrow 1<\alpha <2$ and $-1<\beta <0$
$\therefore f\left(2\right)>0\Rightarrow a>-\frac{3}{2}\therefore a\in (-\frac{3}{2},0)\cdots \left(1\right)$
So when $a>0$, $\alpha +\beta <0$
$\Rightarrow 0<\alpha <1$ and $-2<\beta <-1$
$\therefore f(-2)>0\Rightarrow a<\frac{3}{2}\therefore a\in (0,\frac{3}{2})\cdots \left(2\right)$
When $a=0,$ then $x=\pm 1$ only one integer in between the roots, so $a=0$ is not possible.
From $\left(1\right)$ and $\left(2\right),$
Possible integral values of $a$ are $-1,1.$