• # question_answer If $\frac{2x}{2{{x}^{2}}+5x+2}$>$\frac{1}{x+1}$, then [IIT 1987] A) $-2>x>-1$ B) $-2\ge x\ge -1$ C) $-2<x<-1$ D) $-2<x\le -1$

Given $\frac{2x}{2{{x}^{2}}+5x+2}>\frac{1}{x+1}$ Þ  $\frac{2x}{(2x+1)(x+2)}>\frac{1}{(x+1)}$ Þ  $\frac{2x}{(2x+1)(x+2)}-\frac{1}{(x+1)}>0$ Þ $\frac{2x(x+1)-(2x+1)(x+2)}{(x+1)(2x+1)(x+2)}>0$ Þ  $\frac{2{{x}^{2}}+2x-2{{x}^{2}}-4x-x-2}{(x+1)(x+2)(2x+1)}>0$ Þ $\frac{-3x-2}{(x+1)(x+2)(2x+1)}>0$ 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$-2<x<-1$.