The solution set of -x-1-x-x-1-x-12-x- is -

The correct option is D {x | x gt; 0}∪{ 1}We have, |x + 1|/|x| + |x + 1| = |x + 1|^2/|x| Since, |a|+|b| = |a+b|, if ab≥0 ∴(x+1)/x. (x+1)≥ 0 ⇒(x + 1)^2/x≥ 0 ...

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Quantitative Aptitude

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Question

The solution set of $| \frac{x + 1}{x} | + | x + 1 | = \frac{(x + 1)^{2}}{| x |}$ is :

{x | x \geq 0

}

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{x | x > 0} \cup {- 1}

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{- 1, 1}

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{x | x \geq 1 o r x \leq - 1}

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Solution

The correct option is B ${x | x > 0} \cup {- 1}$
We have, $\frac{| x + 1 |}{| x |} + | x + 1 | = \frac{| x + 1 |^{2}}{| x |}$
$S i n c e, | a | + | b | = | a + b |, i f a b \geq 0 ∴ \frac{(x + 1)}{x} . (x + 1) \geq 0 \Rightarrow \frac{(x + 1)^{2}}{x} \geq 0 \Rightarrow x > 0 o r {- 1}$

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The solution set of |x+1x|+|x+1|=(x+1)2|x| is :

The correct option is B {x|x>0}∪{−1}We have, |x+1||x|+|x+1|=|x+1|2|x| Since,|a|+|b|=|a+b|,if ab≥0∴(x+1)x.(x+1)≥0⇒(x+1)2x≥0⇒x>0 or {−1}

The solution set of $| \frac{x + 1}{x} | + | x + 1 | = \frac{(x + 1)^{2}}{| x |}$ is :

The correct option is B ${x | x > 0} \cup {- 1}$
We have, $\frac{| x + 1 |}{| x |} + | x + 1 | = \frac{| x + 1 |^{2}}{| x |}$
$S i n c e, | a | + | b | = | a + b |, i f a b \geq 0 ∴ \frac{(x + 1)}{x} . (x + 1) \geq 0 \Rightarrow \frac{(x + 1)^{2}}{x} \geq 0 \Rightarrow x > 0 o r {- 1}$