Question

The range of $f\left(x\right)=\frac{x^2+14x+9}{x^2+2x+3},x\in R$ is

• A. $R-\{1,3\}$
• B. $[-5,4]$
• C. $R$
• D. $R-\{-5,4\}$

Answer 1

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

Let $y=\frac{x^2+14x+9}{x^2+2x+3}$
$\Rightarrow (y-1)x^2+2(y-7)x+(3y-9)=0$
The above equation is quadratic in $x$.
As $x\in R$,
$\therefore D\ge 0$
$\Rightarrow 4(y-7{)}^2-4(y-1\left)\right(3y-9)\ge 0$
$\Rightarrow y^2+y-20\le 0$
$\Rightarrow (y-4)(y+5)\le 0$
$\Rightarrow y\in [-5,4]$

Hence, $R\left(f\right)=[-5,4]$