If \( 0 < x < \dfrac{1}{\sqrt{2}} \text{ and } \dfrac{\sin^{-1} x}{\alpha} = \dfrac{\cos^{-1} x}{\beta}\) then the value of \(\sin\left(\dfrac{2\pi\alpha}{\alpha+\beta}\right)\) is:
\( 4\sqrt{(\,1-x^2\,)}\,(1-2x^2) \)
\( 4x\sqrt{(\,1-x^2\,)}\,(1-2x^2) \)
\( 2x\sqrt{(\,1-x^2\,)}\,(1-4x^2) \)
\( 4\sqrt{(\,1-x^2\,)}\,(1-4x^2) \)