IIT JEE , AIEEE Forum - Mathematics , Algebra

sriram.a

in this questions we have 3 cases

x>1 (x/x-1)+x=x²/x-1 x not equal to 1

we get 0=0 i.e... true for all x>1

0<x<1 (x/1-x)+x=x²/1-x

we get x=0

x<0 (x/x-1)-x=x²/1-x

we get x=0

so solution is {0}U{(1,infinity)}. cheeeeeerrrrrrssss

hsbhatt

The approach can be simplified a bit.

We are given that

$\frac{|x|}{|x-1|} + |x| = \frac{x^2}{|x-1|}$

Noting that x = 0 is a solution, let us now divide both sides by |x|. We get

$\frac{1}{|x-1|} + 1 = \frac{|x|}{|x-1|}$

Now, we use the property that $|a|+|b| \ge |a+b|$ with equality occuring only when ab>0

Here we have

$\left |\frac{1}{x-1} + 1 \right| = \left | \frac{x}{x-1}\right |$

This means (x-1).1>0 or x>1

Thus x = 0 or x>1 are the solutions.