Question

Find range of $f\left(x\right)=\frac{x^2+2x+3}{x}$, the range is $R-(a,b)$, find ab?(If the answer is m. Find -m)

Answer 1

The domain of the given function is $R-\left\{0\right\}$
Now $y=\frac{x^2+2x+3}{x}$
$xy=x^2+2x+3$
$x^2+x(2-y)+3=0$
$x=\frac{y-2\pm \sqrt{y^2-4y+4-12}}{2}$
However, $xϵR-\left\{0\right\}$
Hence the above equation has no real root.
This is possible if
$y^2-4y+4-12<0$
$y^2-4y-8<0$ ...(i)
Let the roots be $\alpha ,\beta$
Then $(y-\alpha )(y-\beta )<0$
Hence $yϵR-\{\alpha ,\beta \}$
From equation (i)
$\alpha .\beta =-8$ and $\alpha +\beta =4$
Thus $ab=\alpha .\beta =-8$.