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

Sol. The given equation is,

2^|y| - |2^{y – 1} - 1| = 2^{y – 1} + 1

On the basis of absolute values involved here (|y| and | 2^y – 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 |2^y-1 -1| = 1 – 2^{y – 1}

∴ The given equation becomes

2^-y – 1 + 2^{y – 1} = 2^{y – 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 |2^{y - 1} – 1| = 1 – 2^y-1

∴ the eqⁿ becomes

2^y – 1 + 2^{y – 1} + 1 ⇒ 2^y = 2

⇒ y = 1 ∉ (0, 1)

∴ y = 1 is the only solⁿ in this case.

Case III : y ∈ (1, ∞)

Then |y| = 1, |2^{y – 1} – 1| = 2^{y – 1} – 1

The given eqⁿ becomes, 2^y – 2^{y - 1} + 1 = 2^{y – 1} + 1

⇒ 2^y – 2^y = 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, ∞]