Question

$If\frac{2x}{2x^2+5x+2}$ > $\frac{1}{x+1},then$

• A. -2 > x > -1
• B.
• C. -2 < x < -1
• D. -2 < x -1

Answer 1

The correct option is C

-2 < x < -1

Given $\frac{2x}{2x^2+5x+2}$ > $\frac{1}{x+1}$

$\Rightarrow \frac{2x}{(2x+1)(x+2)}$ > $\frac{1}{(x+1)}$

$\Rightarrow \frac{2x}{(2x+1)(x+2)}-\frac{1}{(x+1)}$ > 0

$\Rightarrow \frac{2x(x+1)-(2x+1)(x+2)}{(x+1)(2x+1)(x+2)}$ > 0

$\Rightarrow \frac{2x^2+2x-2x^2-4x-x-2}{(x+1)(x+2)(2x+1)}$ > 0

$\Rightarrow \frac{-3x-2}{(x+1)(x+2)(2x+1)}$ > 0

Equating each factor equal to 0, we have x = -2, -1, $-\frac{2}{3},-\frac{1}{2}$

It is clear that $-\frac{2}{3}$ < x < $-\frac{1}{2}$ or -2 < x< -1.