Question

Let $f:R\to R$ be defined by $f\left(x\right)=\frac{3x^2+mx+n}{x^2+1}.$ If the range of $f$ is $[-4,3),$ then the value of $m^2+n^2$ is

Answer 1

Let $y=\frac{3x^2+mx+n}{x^2+1}$
$\Rightarrow x^2(y-3)-mx+(y-n)=0$
Given that $x\in R$
$\therefore D\ge 0$ and $y\ne 3$
$\Rightarrow m^2-4(y-3)(y-n)\ge 0$
$\Rightarrow m^2-4(y^2-ny-3y+3n)\ge 0$
$\Rightarrow 4y^2-4y(n+3)+12n-m^2\le 0\dots \left(1\right)$

Also, given $y\in [-4,3)$
$\Rightarrow (y+4)(y-3)\le 0$
$\Rightarrow y^2+y-12\le 0\dots \left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right),$ we get
$\frac{4}{1}=-\frac{4(n+3)}{1}=\frac{12n-m^2}{-12}$
$\Rightarrow n=-4$ and $m=0$