Let f(x) = logₑ(sin x), (0 < x < π) and g(x) = sin⁻¹(e⁻ˣ), (x ≥ 0). If α is a positive real number such that a = (f∘g)'(α) and b = (f∘g)'(α), then which of the following equations is true?

Given f(x) = logₑ(sin x), (0 < x < π) and g(x) = sin⁻¹(e⁻ˣ), (x ≥ 0).
We have (f∘g)(x) = f(g(x)) = logₑ(sin(sin⁻¹(e⁻ˣ))) = logₑ(e⁻ˣ) = -x.
Then (f∘g)'(x) = -1.
Since a = (f∘g)'(α) and b = (f∘g)'(α), we have a = -1 and b = -1.
Substituting a = -1 and b = -1 into the given options:
Option A: α² - bα - a = -2; α²
(-1)α² - (-1)α - (-1) = -2
-α² + α + 1 = -2
-α² + α + 3 = 0
This is not necessarily true for all α.
Option B: aα² + bα + a = 0
(-1)α² + (-1)α + (-1) = 0
-α² - α - 1 = 0
α² + α + 1 = 0
The discriminant is 1 - 4 = -3 < 0, so there are no real solutions for α.
Option C: aα² - bα - a = 0
(-1)α² - (-1)α - (-1) = 0
-α² + α + 1 = 0
α² - α - 1 = 0
This equation has real solutions for α.
Option D: aα² - bα - a = 1
(-1)α² - (-1)α - (-1) = 1
-α² + α + 1 = 1
-α² + α = 0
α(-α + 1) = 0
α = 0 or α = 1
Since α is a positive real number, α = 1 is a valid solution.
However, option C is also true for some α values. Let's check the range of g(x):
Since 0 ≤ e⁻ˣ ≤ 1 for x ≥ 0, the range of g(x) is [0, π/2]. Therefore, 0 < g(x) < π/2.
The domain of f(x) is (0, π), so we must have 0 < g(x) < π. Since the range of g(x) is [0, π/2], this condition is met.
Therefore, the correct option is C: aα² - bα - a = 0