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Question: If \(\frac{2x}{2x^{2} + 5x + 2} > \frac{1}{x + 1}\), then...

If 2x2x2+5x+2>1x+1\frac{2x}{2x^{2} + 5x + 2} > \frac{1}{x + 1}, then

A

−2>x>−1- 2 > x > - 1

B

−2≥x≥−1- 2 \geq x \geq - 1

C

−2<x<−1- 2 < x < - 1

D

−2<x≤−1- 2 < x \leq - 1

Answer

−2<x<−1- 2 < x < - 1

Explanation

Solution

Given 2x2x2+5x+2−1x+1>0\frac{2x}{2x^{2} + 5x + 2} - \frac{1}{x + 1} > 0

⇒ 2x2+2x−2x2−5x−2(2x+1)(x+2)(x+1)>0\frac{2x^{2} + 2x - 2x^{2} - 5x - 2}{(2x + 1)(x + 2)(x + 1)} > 0 ⇒ −3x−2(2x+1)(x+2)(x+1)>0\frac{- 3x - 2}{(2x + 1)(x + 2)(x + 1)} > 0

⇒ −3(x+2/3)(x+1)(x+2)(2x+1)>0\frac{- 3(x + 2/3)}{(x + 1)(x + 2)(2x + 1)} > 0 ⇒ (x+2/3)(x+1)(x+2)(2x+1)<0\frac{(x + 2/3)}{(x + 1)(x + 2)(2x + 1)} < 0

Equating each factor equal to 0,

We get x=−2,−1,−2/3,−1/2x = - 2, - 1, - 2/3, - 1/2

∴ x∈]−2,−1[∪]−2/3,−1/2[x \in \rbrack - 2, - 1\lbrack \cup \rbrack - 2/3, - 1/2\lbrack⇒ −2/3<x<−1/2- 2/3 < x < - 1/2 or

−2<x<−1- 2 < x < - 1