Let $$f(x) = \log_e \sin x$$, $$0 < x < \pi$$ and $$g(x) = \sin^{-1}(e^{-x})$$, $$(x \geq 0)$$. If $$\alpha$$ is a positive real number such that$$a=f\circ g'(\alpha)$$ and $$b = f \circ g(\alpha)$$, then

We are given $$f(x) = \log_e(\sin x)$$ for $$0 < x < \pi$$ and $$g(x) = \sin^{-1}(e^{-x})$$ for $$x \geq 0$$.

We are told that $$\alpha$$ is a positive real number such that $$a = (f \circ g)'(\alpha)$$ and $$b = (f \circ g)(\alpha)$$.

Step 1: Find $$(f \circ g)(x)$$

$$(f \circ g)(x) = f(g(x)) = \log_e(\sin(\sin^{-1}(e^{-x})))$$

Since $$e^{-x} \in (0, 1]$$ for $$x \geq 0$$, we have $$\sin^{-1}(e^{-x}) \in (0, \pi/2]$$, so $$\sin(\sin^{-1}(e^{-x})) = e^{-x}$$.

Therefore:

$$(f \circ g)(x) = \log_e(e^{-x}) = -x$$

Step 2: Find $$(f \circ g)'(x)$$

$$(f \circ g)'(x) = \frac{d}{dx}(-x) = -1$$

Step 3: Compute $$a$$ and $$b$$

$$a = (f \circ g)'(\alpha) = -1$$

$$b = (f \circ g)(\alpha) = -\alpha$$

Step 4: Check each option

Substituting $$a = -1$$ and $$b = -\alpha$$ into Option D: $$a\alpha^2 - b\alpha - a$$

$$= (-1)(\alpha^2) - (-\alpha)(\alpha) - (-1)$$

$$= -\alpha^2 + \alpha^2 + 1$$

$$= 1$$

This matches Option D: $$a\alpha^2 - b\alpha - a = 1$$.

The correct answer is Option D.