4-2.Quadratic Equations and Inequations
medium

यदि $\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}$ तो

A

$ - 2 > x > - 1$

B

$ - 2 \ge x \ge - 1$

C

$ - 2 < x < - 1$

D

$ - 2 < x \le - 1$

(IIT-1987)

Solution

दिया गया है $\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}$

 $\Rightarrow$  $\frac{{2x}}{{(2x + 1)(x + 2)}} > \frac{1}{{(x + 1)}}$

$\Rightarrow$    $\frac{{2x}}{{(2x + 1)(x + 2)}} – \frac{1}{{(x + 1)}} > 0$

$\Rightarrow$  $\frac{{2x(x + 1) – (2x + 1)(x + 2)}}{{(x + 1)(2x + 1)(x + 2)}} > 0$

$\Rightarrow$  $\frac{{2{x^2} + 2x – 2{x^2} – 4x – x – 2}}{{(x + 1)(x + 2)(2x + 1)}} > 0$

$\therefore $ प्रत्येक गुणांक को शून्य के बराबर रखने पर $x =  – 2, – 1, – \frac{2}{3}, – \frac{1}{2}$.

स्पष्टत: $ – \frac{2}{3} < x <  – \frac{1}{2}$ या $ – 2 < x <  – 1$.

Standard 11
Mathematics

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