If 2x-2x2 - 5x - 2 - 1 -x - 1- then

Explanation for the correct option:Given, 2x/(2x^2 + 5x + 2) > 1 /(x + 1)⇒2x/(2x^2 + 4x + x + 2) > 1 /(x + 1)⇒2x/(2x + 1)(x + 2) 1 /(x + 1) > 0⇒2x(x + 1) 2x ...

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Standard XII

Mathematics

Logarithmic Inequalities

If 2x/2x2 + 5...

Question

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

$- 2 > x > - 1$

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$- 2 \geq x \geq - 1$

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$- 2 < x < - 1$

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$- 2 < x \leq - 1$

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Solution

The correct option is C
$- 2 < x < - 1$

Explanation for the correct option:
Given, $\frac{2 x}{(2 x^{2} + 5 x + 2)} > \frac{1}{(x + 1)}$
$\Rightarrow$ $\frac{2 x}{(2 x^{2} + 4 x + x + 2)} > \frac{1}{(x + 1)}$
$\Rightarrow$ $\frac{2 x}{(2 x + 1) (x + 2)} - \frac{1}{(x + 1)} > 0$
$\Rightarrow$ $\frac{2 x (x + 1) - 2 x^{2} - 5 x - 2}{(2 x + 1) (x + 2) (x + 1)} > 0$
$\Rightarrow$ $\frac{- (3 x + 2)}{(2 x + 1) (x + 2) (x + 1)} > 0$
$\Rightarrow$ $x \in (- 2, - 1) \cup (- \frac{2}{3}, - \frac{1}{2})$
$∴ - \frac{2}{3} < x < \frac{1}{2} o r - 2 < x < - 1$
Hence, Option ‘C’ is Correct.

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If 2x2x2+5x+2>1(x+1), then

The correct option is C –2<x<–1Explanation for the correct option:Given, 2x2x2+5x+2>1(x+1)⇒2x2x2+4x+x+2>1(x+1)⇒2x2x+1x+2-1(x+1)>0⇒2x(x+1)-2x2-5x-22x+1x+2(x+1)>0⇒-(3x+2)2x+1x+2(x+1)>0⇒x∈(−2,−1)∪(−23,−12)∴-23<x<12or-2<x<-1Hence, Option ‘C’ is Correct.

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