The nbsp-set nbsp-of nbsp-possible nbsp-values nbsp-of nbsp-x nbsp-for nbsp-which nbsp-2x2x2-5x-2-1x-1 nbsp-is-are-2- nbsp-1-2- nbsp-1-23- nbsp-3-23- nbsp-12-23- nbsp-12-

The correct options are A 2, nbsp; 1 E [ 23, nbsp;12)2x2x2+5x+2≥1x+1⇒2x2x2+5x+2 1x+1≥0⇒2xx+1 2x2+5x+22x2+5x+2x+1≥0⇒2x2+2x 2x2 5x 22x2+4x+x+2x+1≥0⇒ 3x 22xx ...

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Byju's Answer

Standard X

Mathematics

Method of intervals

The nbsp;se...

Question

$The set of possible values of x for which \frac{2 x}{(2 x^{2} + 5 x + 2)} \geq \frac{1}{x + 1} is / are$

(- 2, - 1)

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(- 2, - 1]

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(- \frac{2}{3}, 3)

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(- \frac{2}{3}, - \frac{1}{2}]

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[- \frac{2}{3}, - \frac{1}{2})

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Solution

The correct options are
A $(- 2, - 1)$
E $[- \frac{2}{3}, - \frac{1}{2})$
$\frac{2 x}{(2 x^{2} + 5 x + 2)} \geq \frac{1}{x + 1} \Rightarrow \frac{2 x}{(2 x^{2} + 5 x + 2)} - \frac{1}{x + 1} \geq 0 \Rightarrow \frac{2 x (x + 1) - (2 x^{2} + 5 x + 2)}{(2 x^{2} + 5 x + 2) (x + 1)} \geq 0 \Rightarrow \frac{2 x^{2} + 2 x - 2 x^{2} - 5 x - 2}{(2 x^{2} + 4 x + x + 2) (x + 1)} \geq 0 \Rightarrow \frac{- 3 x - 2}{(2 x (x + 2) + 1 (x + 2)] (x + 1)} \geq 0 \Rightarrow \frac{3 x + 2}{(2 x + 1) (x + 2) (x + 1)} \leq 0$

Therefore, the values of x is $x \in (- 2, - 1) \cup [- \frac{2}{3}, - \frac{1}{2})$

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