Question

If $\alpha +\frac{1}{\alpha }\&2-\beta -\frac{1}{\beta }(\alpha ,\beta >0)$ 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. 5.0
• B. 05
• C. 5.00

Answer 1

Answer: (A), (B), (C)

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

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

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

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