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 $\frac{1}{2}$

Let $y=\frac{x^2+2x+4}{2x^2+4x+9}$

$⟹2y=\frac{2x^2+4x+8}{2x^2+4x+9}$

$⟹2y-1=\frac{-1}{2x^2+4x+9}$

Also, $7y=\frac{7x^2+14x+28}{2x^2+4x+9}$

$⟹7y-3=\frac{x^2+2x+1}{2x^2+4x+9}$

$⟹7y-3=\frac{(x+1{)}^2}{2x^2+4x+9}$

Now, $(2y-1)(7y-3)=\frac{-(x+1{)}^2}{(2x^2+4x+9{)}^2}\le 0$

$\therefore (2y-1)(7y-3)\le 0$

$⟹\frac{3}{7}\le y\le \frac{1}{2}$

Thus, the greatest value of $y$ is $\frac{1}{2}$.