Question

Let $f:R\to R$ be a function 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

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

But it is given that range is $[-4,3)$
$\Rightarrow y\in [-4,3)$
$\Rightarrow (y+4)(y-3)\le 0,y\ne 3$
$\Rightarrow y^2+y-12\le 0,y\ne 3$
$\Rightarrow 4y^2+4y-48\le 0,y\ne 3\cdots \left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$, we get
$n+3=-1\Rightarrow n=-4$
and $12n-4m^2=-48$
$\Rightarrow -48-4m^2=-48$
$\Rightarrow m^2=0$

Hence, $m^2+n^2=0+16=16$