Solve |x−2|−1|x−2|−2≤0,xϵR.
We have either x−2≥0 or x−2<0
Case I
When x−2≥0.
In this case, x≥2 and |x−2|=x−2.
So, the given inequation becomes
x−2−1x−2−2≥0⇒x−3x−4≤0.
∴(x−3≥0 and x−4<0) or (x−3≤0 and x−4>0)
⇒(x≥3 and x−4<0) or (x<3 and x>4)
⇒3≤x<4 [∵ x<3 and x>4 is not possible]
⇒xϵ[3,4) [ this includes x≥2].
Case II
When x-2< 0.
In this case, x<2 and |x−2|=−(x−2)=−x+2.
So, the given inequation becomes
−x+2−1−x+2−2≤0⇒−x+1−x≤0
[multiplying num. and denom. by -1]
∴(x−1≤0 and x>0) or (x−1≥ and x<0)
⇒(x≤1 and x>0) or (x≥1 and x<0)
⇒0<x≤ [∵x≥1 and x<0 is not possible]
⇒xϵ(0,1] [this includes x<2].
Hence, from the above two case, we get
solution set =(0,1]∪[3,4).