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

Answer: (C)

$\frac{1}{2}$

Answer 2

Let $y=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$
$\Rightarrow 2 x^{2} y+4 x y+9 y=x^{2}+2 x+4$
$\Rightarrow(2 y-1) x^{2}+(4 y-2) x+9 y-4=0$
$\Rightarrow x=\frac{-(4 y-2) \pm \sqrt{-4 y-2)^{2}{-4(2 y-1)(9 y-4)}}}{2(2 y-1)}$
Since, $x$ is real number.
$\therefore (4 y-2)^{2}-4(2 y-1)(9 y-4) \geq 0$
$\Rightarrow 4(2 y-1)^{2}-4(2 y-1)(9 y-4) \geq 0$
$\Rightarrow 4(2 y-1)(2 y-1-9 y+4) \geq 0$
$\Rightarrow 4(2 y-1)(3-7 y) \geq 0$
$\Rightarrow (2 y-1)(7 y-3) \leq 0$