Question
The solution set of $$\left| {\frac{{x + 1}}{x}} \right| + \left| {x + 1} \right| = \frac{{{{\left( {x + 1} \right)}^2}}}{{\left| x \right|}}$$ is
A.
$$\left\{ {x\left| {x \geqslant 0} \right.} \right\}$$
B.
$$\left\{ {x\left| {x > 0} \right.} \right\} \cup \left\{ { - 1} \right\}$$
C.
$$\left\{ { - 1,1} \right\}$$
D.
$$\left\{ {x\left| {x \geqslant 1} \right.\,\,{\text{or, }}x \leqslant - 1} \right\}$$
Answer :
$$\left\{ {x\left| {x > 0} \right.} \right\} \cup \left\{ { - 1} \right\}$$
Solution :
$$\eqalign{
& \frac{{\left| {x + 1} \right|}}{{\left| x \right|}} + \left| {x + 1} \right| = \frac{{\left| {x + 1} \right|^2}}{{\left| x \right|}} \cr
& \Rightarrow \,\,\left| {x + 1} \right|\left\{ {\frac{1}{{\left| x \right|}} + 1 - \frac{{\left| {x + 1} \right|}}{{\left| x \right|}}} \right\} = 0 \cr
& \therefore \,\,\left| {x + 1} \right| = 0\,\,\,{\text{or, }}1 + \left| x \right| - \left| {x + 1} \right| = 0. \cr
& \left| {x + 1} \right| = 0 \cr
& \Rightarrow \,\,x = - 1. \cr
& {\text{If }}x < - 1,1 + \left| x \right| - \left| {x + 1} \right| = 0 \cr
& \Rightarrow \,\,1 - x + x + 1 = 0 \cr
& \Rightarrow \,\,2 = 0\left( {{\text{absurd}}} \right). \cr
& {\text{If }} - 1 \leqslant x < 0,1 + \left| x \right| - \left| {x + 1} \right| = 0 \cr
& \Rightarrow \,\,1 - x - \left( {x + 1} \right) = 0 \cr
& \Rightarrow \,\,x = 0\left( {{\text{not possible}}} \right). \cr
& {\text{If }}x \geqslant 0,1 + x - \left( {x + 1} \right) = 0 \cr
& \Rightarrow \,\,0 = 0 \cr} $$
⇒ $$x$$ can have any value in the interval.
$$\therefore \,\,x = - 1,x > 0\left( {\because \,\,x \ne 0} \right).$$