Question

Range of the rational expression $y=\frac{x+3}{2x^2+3x+9},x\in R$ is

• A. $[-\frac{1}{21},\frac{1}{21}]$
• B. $[-\frac{1}{3},\frac{1}{3}]$
• C. $[\frac{1}{21},\frac{1}{3}]$
• D. $[-\frac{1}{21},\frac{1}{3}]$

Answer 1

The correct option is D $[-\frac{1}{21},\frac{1}{3}]$

$y=\frac{x+3}{2x^2+3x+9}$
$\Rightarrow 2y{x}^2+(3y-1)x+(9y-3)=0$

$\because x\in R\Rightarrow \Delta \ge 0\Rightarrow (3y-1{)}^2-4(2y\left)\right(9y-3)\ge 0$
$\Rightarrow (3y-1)\left[\right(3y-1)-24y]\ge 0$
$\Rightarrow (3y-1)(-21y-1)\ge 0$
$\Rightarrow (21y+1)(3y-1)\le 0\Rightarrow y\in [-\frac{1}{21},\frac{1}{3}]$