Question

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

• A. $m=0$
• B. $n=4$
• C. $m=2$
• D. $n=-4$

Answer 1

Answer: (A), (D)

$n=-4$

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

Given range is $[-4,3]$
So, $y^2+y-12\le 0\dots \left(2\right)$

Compare $\left(1\right)$ and $\left(2\right),$ we get
$\frac{4}{1}=-\frac{4(n+3)}{1}=\frac{12n-m^2}{-12}$
$\Rightarrow m=0$ and $n=-4$