Question

If $x$ is real, then the minimum and maximum values of $\frac{x+2}{2x^2+3x+6}$ are

• A. $-\frac{1}{3}$ & $-\frac{1}{13}$
• B. $\frac{1}{13}$ & $\frac{1}{3}$
• C. $-\frac{1}{3}$ & $\frac{1}{13}$
• D. $-\frac{1}{13}$ & $\frac{1}{3}$

Answer 1

Answer: (D)

$-\frac{1}{13}$ & $\frac{1}{3}$

Let $y=\frac{x+2}{2x^2+3x+6}$.
$\Rightarrow 2y{x}^2+(3y-1)x+6y-2=0$
Since, $x$ is real, so for real roots $D\ge 0$
$\Rightarrow (3y-1{)}^2-16y(3y-1)\ge 0$
$\Rightarrow (13y+1)(3y-1)\le 0$
$\Rightarrow -\frac{1}{13}\le y\le \frac{1}{3}$