Question

The set of values of $x$ for which $\frac{2x}{2x^2+5x+2}>\frac{1}{x+1}$ are

• A. $\left(\frac{-2}{3},\frac{-1}{3}\right)$
• B. $\left(\frac{-2}{3},\frac{-1}{2}\right)\cup (-2,-1)$
• C. $\left(\frac{-2}{3},-1\right)$
• D. $(-1,1)$

Answer 1

Answer: (B)

$\left(\frac{-2}{3},\frac{-1}{2}\right)\cup (-2,-1)$

$\frac{2x}{2x^2+5x+2}>\frac{1}{x+1}\frac{2x}{2x^2+5x+2}-\frac{1}{x+1}>0\frac{2x(x+1)-(2x^2+5x+2)}{(2x^2+5x+2)(x+1)}>0\frac{-(3x+2)}{(2x+1)(x+2)(x+1)}>0\frac{(3x+2)}{(2x+1)(x+2)(x+1)}<0$
Using wavy curve method,
$x\in (-2,-1)\cup (-\frac{2}{3},-\frac{1}{2})$