Question

If the range of $\frac{x^2+x+d}{x^2+2x+d}$ is $[\frac{5}{6},\frac{3}{2}]$ for all real $x$, then the value of $d$ is

Answer 1

Let, $y=\frac{x^2+x+d}{x^2+2x+d}$
$\Rightarrow x^2(1-y)+x(1-2y)+d(1-y)=0$ has real roots
$\Rightarrow D=(1-2y{)}^2-4d(1-y{)}^2\ge 0$
$\Rightarrow y^2(4d-4)+(4-8d)y+(4d-1)\le 0\cdots \left(1\right)$

Given, $\frac{5}{6}\le y\le \frac{3}{2}$
$\Rightarrow (y-\frac{5}{6})(y-\frac{3}{2})\le 0$
$\Rightarrow y^2-\frac{7y}{3}+\frac{5}{4}\le 0\cdots \left(2\right)$
Compare $\left(1\right)$ and $\left(2\right)$
$\Rightarrow \frac{4d-1}{4d-4}=\frac{5}{4}$
$\Rightarrow d=4$