Iffrac{{2x}}{{2{x^2} + 5x + 2}} frac{1}{ | vedclass

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Question

(c) Given $\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}$

==> $\frac{{2x}}{{(2x + 1)(x + 2)}} > \frac{1}{{(x + 1)}}$

==> $\frac

Vedclass · Accepted Answer

$ - 2 < x < - 1$

Vedclass · Answer

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

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

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