Solve - x -3- x - x -2-1A- -5-2 -1- - -3-2 -1- - -5-2 -1- -D- -5-3 -1- -

The correct option is A (−5, −2) ∪ (−1, ∞) nbsp;|x + 3| + xx + 2 gt; 1, x2 ⇒|x + 3| 2x + 2 gt; 0 For x ≥ −3, x+1x+2 gt; 0 ⇒ x ∈ ((−∞, −2) ∪ (−1, ∞)) ...

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Byju's Answer

Standard XII

Mathematics

Inequalities Involving Modulus Function

Solve | x +3|...

Question

Solve $\frac{| x + 3 | + x}{x + 2} > 1$

(- 5, - 2) \cup (- 1, \infty)

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(- 3, - 2) \cup (- 1, \infty)

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(- 5, - 2) \cup (1, \infty)

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(- 5, - 3) \cup (- 1, \infty)

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Solution

The correct option is A $(- 5, - 2) \cup (- 1, \infty)$
$\frac{| x + 3 | + x}{x + 2} > 1, x \neq 2$
$\Rightarrow \frac{| x + 3 | - 2}{x + 2} > 0$

For $x \geq - 3, \frac{x + 1}{x + 2} > 0$
$\Rightarrow x \in ((- \infty, - 2) \cup (- 1, \infty)) \cap [- 3, \infty)$
$\Rightarrow x \in [- 3, - 2) \cup (- 1, \infty)$

For $x < - 3, \frac{x + 5}{x + 2} < 0$
$\Rightarrow x \in (- 5, - 2) \cap (- \infty, - 3)$
$\Rightarrow x \in (- 5, - 3)$

$∴ x \in (- 5, - 2) \cup (- 1, \infty)$

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Solve |x+3|+xx+2>1

The correct option is A (−5,−2)∪(−1,∞) |x+3|+xx+2>1, x≠2 ⇒|x+3|−2x+2>0 For x≥−3, x+1x+2>0 ⇒x∈((−∞,−2)∪(−1,∞))∩[−3,∞) ⇒x∈[−3,−2)∪(−1,∞) For x<−3, x+5x+2<0 ⇒x∈(−5,−2)∩(−∞,−3) ⇒x∈(−5,−3) ∴x∈(−5,−2)∪(−1,∞)

Solve $\frac{| x + 3 | + x}{x + 2} > 1$