Question

find the interval of
$\frac{2x}{2 {x}^{2} + 5x + 2} > \frac{1}{x + 1}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{2x}{2 {x}^{2} + 5x + 2} > \dfrac{1}{x + 1}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{2x}{2 {x}^{2} + 5x + 2} - \dfrac{1}{x + 1} > 0$

$\rm :\longmapsto\:\dfrac{2x(x + 1) - ( {2x}^{2} + 5x + 2)}{(2 {x}^{2} + 5x + 2)(x + 1)} > 0$

$\rm :\longmapsto\:\dfrac{2 {x}^{2} + 2x - {2x}^{2} - 5x - 2}{(2 {x}^{2} + 4x + x+ 2)(x + 1)} > 0$

$\rm :\longmapsto\:\dfrac{- 3x - 2}{(2x(x + 2)+1(x+ 2))(x + 1)} > 0$

$\rm :\longmapsto\:\dfrac{- (3x + 2)}{(2x + 1)(x+ 2)(x + 1)} > 0$

$\rm :\longmapsto\:\dfrac{ (3x + 2)}{(2x + 1)(x+ 2)(x + 1)} < 0$

For domain to be defined,

$\rm :\longmapsto\:x \: \ne \: - 1, \: - 2, \: - \: \dfrac{1}{2}$

Now, let we find the critical points and check the sign in intervals.

$\rm :\longmapsto\:x = - 2, \: - 1, \: - \dfrac{2}{3}, \: - \: \dfrac{1}{2}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Intervals & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x < - 2 & \sf + ve \\ \\ \sf - 2 < x < - 1 & \sf - ve\\ \\ \sf - 1 < x \leqslant - \dfrac{2}{3} & \sf + ve\\ \\ \sf - \dfrac{2}{3} \leqslant x < - \dfrac{1}{2} & \sf - ve\\ \\ \sf x > - \dfrac{1}{2} & \sf + ve \end{array}} \\ \end{gathered}$

$\bf :\implies\: \:x \: \in \: ( - 2, \: - 1) \: \cup \: \bigg[ - \: \dfrac{2}{3}, - \: \dfrac{1}{2}\bigg)$

Additional Information :-

If a and b are two positive real numbers such that a < b, then

$\boxed{ \sf{ \:(x - a)(x - b) < 0 \: \rm \implies\:\:a < x < b}}$

$\boxed{ \sf{ \:(x - a)(x - b) \leqslant 0 \: \rm \implies\:\:a \leqslant x \leqslant b}}$

$\boxed{ \sf{ \:(x - a)(x - b) > 0 \: \rm \implies\:\:x < a \: \: or \: \: x > b}}$

$\boxed{ \sf{ \:(x - a)(x - b) \geqslant 0 \: \rm \implies\:\:x \leqslant a \: \: or \: \: x \geqslant b}}$