l 4x-2 l

|4x-2| < |x-2|.

When x
< 1/2 ,  |4x-2| < |x-2|  implies 2-4x <
2-x.

0 <  3x

x >
0 when  x < 1/2. So 0 < x < 1/2 is a
solution.

When x = 1/2,  0 < |x-2|
==> 0 < 2-x .  So x < 2. Therefore , x = 1/2
is  a solution. (Please note that the symbol ==> stands for
implies, as the symbol , => looks like equal or greater
than.)

When 1/2 < x < 2, |4x-2| <
|x-2| ==>  4x-2 < 2-x , or  5x < 4 . So x  < 4/5 is a
solution. Therefore 1/2 < x < 4/5 is a
solution.

When x = 2. |4x-2| < |x-2|
==> 4x-2 < 0, or x < 2/4. So  x = 1/2. But x = 1/2 is a solution
already covered. But x = 1/2 when x = 2 is
contradiction.

When x > 2, |4x-2| < |x-2|
==> 4x-2 > x-2 , Or 4x-x = 0, or x= 0 does not go together with  when x
> 2.

So the solution is    0 <
x 1/2  or  1/2 < x <
4/5.

Combining  we get:  0
< x < 4/5 is a
solution.

Therefore  solution  set is   ] 0 
, 4/5 [ which is both left open and right open interval
.

Tally :

We take a value x
< 0, say  x = -2,  |4x-2| = |4*-2-2| = |-10| = 10. The RHS = |-4-2| = 6. So 10
> 6 the inequality does not hold.

At x = 0, we get
LHS |4*0-2|  = 2 and RHS |0-2 | = 2. So it becomes an equality.