If -x-1-x-x-1-x-12-x- then x

The correct option is A (0,∞) ∪{ 1}|x+1x|+|x+1|=(x+1)^2|x| nbsp; ⇒ |x+1|+|x||x+1|=(x+1)^2, here x0 ⇒ |x+1|+|x^2+x|=|x^2+2x+1| ⇒ (x+1)(x^2+x) ≥0 (∵ |a|+|b ...

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Byju's Answer

Standard XI

Mathematics

Intersection

If |x+1/x|+|x...

Question

If $∣ ∣ ∣ \frac{x + 1}{x} ∣ ∣ ∣ + | x + 1 | = \frac{(x + 1)^{2}}{| x |},$ then $x \in$

(0, \infty) \cup {- 1}

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(- \infty, 0)

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(0, \infty)

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(- \infty, - 1] \cup (0, \infty)

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Solution

The correct option is A $(0, \infty) \cup {- 1}$
$∣ ∣ ∣ \frac{x + 1}{x} ∣ ∣ ∣ + | x + 1 | = \frac{(x + 1)^{2}}{| x |}$
$\Rightarrow | x + 1 | + | x | | x + 1 | = (x + 1)^{2},$ here $x \neq 0$
$\Rightarrow | x + 1 | + | x^{2} + x | = | x^{2} + 2 x + 1 | \Rightarrow (x + 1) (x^{2} + x) \geq 0 (∵ | a | + | b | = | a + b |) \Rightarrow x (x + 1)^{2} \geq 0 \Rightarrow x \in (0, \infty) \cup {- 1}$

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If ∣∣∣x+1x∣∣∣+|x+1|=(x+1)2|x|, then x∈

The correct option is A (0,∞)∪{−1}∣∣∣x+1x∣∣∣+|x+1|=(x+1)2|x| ⇒|x+1|+|x||x+1|=(x+1)2, here x≠0 ⇒|x+1|+|x2+x|=|x2+2x+1|⇒(x+1)(x2+x)≥0 (∵|a|+|b|=|a+b|)⇒x(x+1)2≥0⇒x∈(0,∞)∪{−1}

If $∣ ∣ ∣ \frac{x + 1}{x} ∣ ∣ ∣ + | x + 1 | = \frac{(x + 1)^{2}}{| x |},$ then $x \in$