Given, 2∣y∣−∣2y−1−1∣=2y−1+1
Case I When y∈(−∞,0] ∴2−y+(2y−1−1)=2y−1+1 ⇒2−y=2 ⇒y=−1∈(−∞,0]...(i)
Case II When y∈(0,1] ∴2y+(2y−1−1)=2y−1+1 ⇒2y=2 ⇒y=1∈(0,1]...(ii)
Case III When y∈(1,∞) ∴2y−2y−1+1=2y−1+1 ⇒2y−2.2y−1=0 ⇒2y−2y=0 true for all y>1...(iii)
From Eqs. (i), (ii) and (iii), we get y∈{−1}∪[1,∞).