MCQ
જો $\frac{{2x}}{{2{x^2} + 5x + 2}}$>$\frac{1}{{x + 1}}$ ,તો . . . .
- A$ - 2 > x > - 1$
- B$ - 2 \ge x \ge - 1$
- ✓$ - 2 < x < - 1$
- D$ - 2 < x \le - 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$.
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