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Question: The solution set of \(\left| \frac{x + 1}{x} \right|\)+ \| x + 1 \| =\(\frac{(x + 1)^{2}}{|x|}\) is ...

The solution set of ∣x+1x∣\left| \frac{x + 1}{x} \right|+ | x + 1 | =(x+1)2∣x∣\frac{(x + 1)^{2}}{|x|} is -

A

{x | x ³ 0}

B

{x | x > 0}Č {– 1}

C

{ – 1, 1}

D

{x | x ³ 1 or x £ – 1}

Answer

{x | x > 0}Č {– 1}

Explanation

Solution

We have, ∣x+1∣∣x∣\frac{|x + 1|}{|x|}+ | x + 1 | =∣x+1∣2∣x∣\frac{|x + 1|^{2}}{|x|}

Since, | a | + | b | = | a + b |, if ab ³ 0

\(x+1)x\frac{(x + 1)}{x}. (x + 1) ³ 0 Ž (x+1)2x\frac{(x + 1)^{2}}{x} ³ 0

Ž x > 0 or {– 1}