Question

If $\alpha +\frac{1}{\alpha }$ and $2-\beta -\frac{1}{\beta }$ $(\alpha ,\beta >0)$ are the roots of the quadratic equation $x^2-2(a+1)x+a-3=0,$ then the sum of integral values of $a$ is

• A. $5$
• B. $6$
• C. $12$
• D. $17$

Answer 1

The correct option is A $5$

We know that $\alpha +\frac{1}{\alpha }\ge 2$ $(\because \alpha >0)$
and $-\beta -\frac{1}{\beta }\le -2$
$\Rightarrow 2-\beta -\frac{1}{\beta }\le 0$
So, one root is less than or equal to $0$ and other root is greater than or equal to $2.$


$f\left(0\right)\le 0$
$\Rightarrow a-3\le 0$
$\Rightarrow a\le 3\dots \left(1\right)$

and $f\left(2\right)\le 0$
$\Rightarrow 4-4(a+1)+a-3\le 0$
$\Rightarrow -3a-3\le 0$
$\Rightarrow a\ge -1\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get $a\in [-1,3]$
$\therefore$ Sum of integral values of $a=-1+0+1+2+3=5$