SOLVED EXAMPLE ON MODULUS

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ques:|x + 1| + \left| {\frac{{x + 1}}{x}} \right| = \frac{{{{(x + 1)}^2}}}{{|x|}}

or, |x + 1| + \frac{{\left| {x + 1} \right|}}{{|x|}} = \frac{{{{(x + 1)}^2}}}{{|x|}}

 since denominator cant be zero therefore |x| \ne 0 \Rightarrow x \ne 0...................(i)
there are two behaviour changing points in respect to mod in this question -1 and 0, first consider,

 1. for x \le - 1  \Rightarrow  |x + 1| = - x - 1 and |x| = - x
\begin{array}{l} x \le - 1\\ - x - 1 + \frac{{ - x - 1}}{{ - x}} = \frac{{{{(x + 1)}^2}}}{{ - x}} \end{array}
\begin{array}{l} {x^2} + x - x - 1 = {x^2} + 1 + 2x\\ - 1 = 1 + 2x \end{array}

 \begin{array}{l} or,2x = - 2\\ or,x = - 1 \end{array}
since , we take x \le - 1 , therefore x=-1,

2. next,  - 1 \le x \le 0  \Rightarrow  |x + 1| = x + 1 and |x| = - x

then,x + 1 + \frac{{x + 1}}{{ - x}} = \frac{{{{(x + 1)}^2}}}{{ - x}}
- {x^2} - x + x + 1 = {x^2} + 1 + 2x
2{x^2} + 2x = 0
x(x + 1) = 0
therefore x=0 or x=-1
but x cant be zero from (i)
therefore x=-1 which belongs to [-1,0] or  - 1 \le x \le 0

3.3rd condition x \ge 0  \Rightarrow  |x + 1| = x + 1 and |x| = x
then, x + 1 + \frac{{x + 1}}{x} = \frac{{{{(x + 1)}^2}}}{x}
{x^2} + x + x + 1 = {x^2} + 1 + 2x
{x^2} + 2x + 1 = {x^2} + 1 + 2x
or 0 = 0
this means equation becomes identity
therefore equation is true for all values of x \ge 0 but x cant be equal to 0(from (i))
therefore , x > 0

 therefore solution of equation combining all three statements is
x \in \{ - 1\} \bigcup {(0,\infty )}

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