Question

For real $x$, the greatest value of $\frac{x^2+2x+4}{2x^2+4x+9}$ is

• A. $1$
• B. $-1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{2}$

$y=\frac{x^2+2x+4}{2x^2+4x+9}$
$\Rightarrow 2x^{2}y+4xy+9y=x^2+2x+4$
$\Rightarrow (2y-1)x^2+(4y-2)x+(9y-4)=0$

For real $x$, $D\ge 0$
$\Rightarrow (4y-2{)}^2-4(2y-1\left)\right(9y-4)\ge 0$
$\Rightarrow 4(2y-1{)}^2-4(2y-1\left)\right(9y-4)\ge 0$
$\Rightarrow 4(2y-1)[2y-1-9y+4]\ge 0$
$\Rightarrow (2y-1)(-7y+3)\ge 0$
$\Rightarrow (2y-1)(7y-3)\le 0$
$\Rightarrow y\in [\frac{3}{7},\frac{1}{2}]$
$\Rightarrow y_{max}=\frac{1}{2}$