The solution set of the equation $|\frac{x+1}{x}|+|x+1|=\frac{(x+1)^2}{|x|}$, is
Answer & explanation
Correct answer: option 2
We have,
$|\frac{x+1}{x}|+|x+1|=\frac{(x+1)^2}{|x|}$
$⇒|\frac{x+1}{x}|+|x+1|=\frac{|x+1|^2}{|x|}$
$⇒|x+1|\left\{\begin{matrix}\frac{1}{|x|}+1-\frac{|x+1|}{|x|}\end{matrix}\right\}=0$
$⇒|x+1|=0$ or, $\frac{1}{|x|}+1=\frac{|x+1|}{|x|}$
$⇒|x+1|=0$ or, $1+|x|=|x+1|$
$⇒x=-1$ or, $1+|x|=|x+1|$
In order to solve the equation $1 +|x|=|x+1|$, we consider the following cases:
CASE I When $x <-1$
In this case, we have
$|x|=-x$ and $|x+1|=-(x+1)$
$∴1+|x|=|x+1|$
$⇒1-x=-(x+1)$, which is absurd.
CASE II When $-1≤x<0$
In this case, we have
$|x|=-x$ and $|x+1|=x+1$
$∴1+|x|=|x+1|⇒1-x=x+1⇒x=0$
But, $-1≤x<0$
So, there is no solution in this case.
CASE III When x ≥ 0.
In this case, we have
$|x|=x$ and $|x+1|=x+1$
$∴1+|x|=|x+1|$
$⇒1+x=x+1$, which is true for all x.
Clearly, the given equation is meaningful for x ≠ 0.
Hence, the solution set of the given equation is $\{x:x>0\}∪\{-1\}$.