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# Find the set of all solutions of the equation2|y| - |2y – 1 – 1| 2y – 1 + 1

Navjyot Kalra
7 years ago
Sol. The given equation is,
2|y| - |2y – 1 - 1| = 2y – 1 + 1
On the basis of absolute values involved here (|y| and | 2y 1 – 1), there are two critical pts 0 and l. So we shall consider three cases, when y lies in three different intervals namely (- ∞, 0), [0, 1], (1, ∞)
Case I : y∈ (- ∞, 0) then
|y| = -y and |2y-1 -1| = 1 – 2y – 1
∴ The given equation becomes
2-y – 1 + 2y – 1 = 2y – 1 + 1
⇒ 2-y = 2 ⇒ y = -1 ∈ (- ∞, 0)
Case II : y ∈ [0, 1]
If y = 0 we get 1 - |1/2 – 1| = 1/2 + 1
1 – 1/2 = 1/2 + 1 (not satisfied)
∴ y = 0 is not a soln
If y = 1 we get 2 - |20 – 1| = 20 + 1
⇒ 2 = 2 (satisfied)
∴ y = 1 is a soln
If y ∈ (0, 1) then |y| = y and |2y - 1 – 1| = 1 – 2y-1
∴ the eqn becomes
2y – 1 + 2y – 1 + 1 ⇒ 2y = 2
⇒ y = 1 ∉ (0, 1)
∴ y = 1 is the only soln in this case.
Case III : y ∈ (1, ∞)
Then |y| = 1, |2y – 1 – 1| = 2y – 1 – 1
The given eqn becomes, 2y – 2y - 1 + 1 = 2y – 1 + 1
⇒ 2y – 2y = 0
Which is satisfied for all real values of y but y ∈ (1, ∞)
∴ (1, ∞ ) is the soln in this case.
Combining all the cases, we get the soln as y ∈ {1} ∪ [1, ∞]