Question

lf $x$ is real, then $\frac{x+2}{2x^2+3x+6}$ lies in the interval

• A. $\left(\frac{1}{13},\frac{1}{3}\right)$
• B. $\left[\frac{-1}{13},\frac{1}{3}\right]$
• C. $(-\infty ,\frac{1}{13})\cup \left(\frac{1}{3},\infty \right)$
• D. $(1,2)$

Answer 1

Answer: (B)

$\left[\frac{-1}{13},\frac{1}{3}\right]$

$Let\frac{x+2}{2x^2+3x+6}=y\Rightarrow 2y{x}^2+(3y-1)x+(6y-2)=0\because x\in R\Rightarrow \Delta \ge 0\Delta ={(3y-1)}^2-4(6y-2).2y\ge 0\Rightarrow -(39y^2-10y-1)\ge 0\Rightarrow 39y^2-10y-1\le 0\Rightarrow (3y-1)(13y+1)\le 0\Rightarrow y\le \frac{1}{3},y\ge -\frac{1}{13}AnsOptionB.$