Question

If $x$ is real then the range of the function $f\left(x\right)=\frac{x^2+14x+9}{x^2+2x+3}$ is

• A. $(4,5)$
• B. $(-4,5)$
• C. $[-5,4]$
• D. None of the above

Answer 1

The correct option is C $[-5,4]$

Let $f\left(x\right)=y$
$\therefore \frac{x^2+14x+9}{x^2+2x+3}=y$
$\Rightarrow x^2+14x+9=x^{2}y+2xy+3y$
$\Rightarrow x^2(y-1)+2x(y-7)+3y-9=0$

For $x$ to be real

$\therefore 4(y-7{)}^2-4(3y-9\left)\right(y-1)\ge 0$
$\Rightarrow 4(y^2+49-14y)-4(3y^2+9-12y)\ge 0$
$\Rightarrow (y+5)(y-4)\le 0$

$\therefore$ range of $f\left(x\right)$ will be $[-5,4]$