If $$0 < x < \dfrac{1}{\sqrt{2}}$$ and $$\dfrac{\sin^{-1}x}{\alpha} = \dfrac{\cos^{-1}x}{\beta}$$, then a value of $$\sin\dfrac{2\pi\alpha}{\alpha + \beta}$$ is

Given $$0 < x < \frac{1}{\sqrt{2}}$$ and $$\frac{\sin^{-1}x}{\alpha} = \frac{\cos^{-1}x}{\beta}$$, we need to find the value of $$\sin\frac{2\pi\alpha}{\alpha + \beta}$$.

Let $$\frac{\sin^{-1}x}{\alpha} = \frac{\cos^{-1}x}{\beta} = k$$ (say)

Then $$\alpha = \frac{\sin^{-1}x}{k}$$ and $$\beta = \frac{\cos^{-1}x}{k}$$.

$$\alpha + \beta = \frac{\sin^{-1}x + \cos^{-1}x}{k} = \frac{\pi/2}{k}$$

(since $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$)

$$\frac{\alpha}{\alpha + \beta} = \frac{\sin^{-1}x / k}{\pi/(2k)} = \frac{2\sin^{-1}x}{\pi}$$

$$\frac{2\pi\alpha}{\alpha + \beta} = 2\pi \cdot \frac{\alpha}{\alpha + \beta} = 2\pi \cdot \frac{2\sin^{-1}x}{\pi} = 4\sin^{-1}x$$

Let $$\theta = \sin^{-1}x$$, so $$\sin\theta = x$$ and $$\cos\theta = \sqrt{1-x^2}$$.

$$\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$$

First: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2x\sqrt{1-x^2}$$

Next: $$\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2x^2$$

Therefore:

$$\sin(4\theta) = 2 \cdot 2x\sqrt{1-x^2} \cdot (1-2x^2) = 4x\sqrt{1-x^2}(1-2x^2)$$

This matches Option B.

Therefore, the correct answer is Option B: $$4x\sqrt{1-x^2}(1-2x^2)$$.