The solution set of 2 x-x2-9-1-x-2 isA- -2 -3- -B- -3 -2-3C- -3-0 -3- -D- -3-2 -3- -

The correct option is B ( ∞, 3) ∪ ( 2, 3)2xx^2 9 ≤1x+2 ⇒2xx^2 9 1x+2≤0 ⇒x^2 +4x+9(x^2 9)(x+2)≤ 0 ⇒(x+2)^2+5(x+3)(x 3)(x+2)≤ 0 ⇒1(x+3)(x 3)(x+2)≤ 0 ...

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Standard XII

Mathematics

Method of Intervals

The solution ...

Question

The solution set of $\frac{2 x}{x^{2} - 9} \leq \frac{1}{x + 2}$ is

(- \infty, - 2) \cup (3, \infty)

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(- \infty, - 3) \cup (- 2, 3)

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(- 3, 0) \cup (3, \infty)

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(- 3, - 2) \cup (3, \infty)

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Solution

The correct option is B $(- \infty, - 3) \cup (- 2, 3)$
$\frac{2 x}{x^{2} - 9} \leq \frac{1}{x + 2} \Rightarrow \frac{2 x}{x^{2} - 9} - \frac{1}{x + 2} \leq 0 \Rightarrow \frac{x^{2} + 4 x + 9}{(x^{2} - 9) (x + 2)} \leq 0 \Rightarrow \frac{(x + 2)^{2} + 5}{(x + 3) (x - 3) (x + 2)} \leq 0 \Rightarrow \frac{1}{(x + 3) (x - 3) (x + 2)} \leq 0 (∵ (x + 2)^{2} + 5 > 0, \forall x \in R)$

$∴ x \in (- \infty, - 3) \cup (- 2, 3)$

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Method of Intervals

Standard XII Mathematics

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The solution set of 2xx2−9≤1x+2 is

The correct option is B (−∞,−3)∪(−2,3)2xx2−9≤1x+2⇒2xx2−9−1x+2≤0⇒x2+4x+9(x2−9)(x+2)≤0⇒(x+2)2+5(x+3)(x−3)(x+2)≤0⇒1(x+3)(x−3)(x+2)≤0 (∵(x+2)2+5>0, ∀ x∈R) ∴x∈(−∞,−3)∪(−2,3)

The solution set of $\frac{2 x}{x^{2} - 9} \leq \frac{1}{x + 2}$ is