【答案】B。【解析】由参数方程 $x = \cos t$，$y = t + \sin 2t$，得 $\dfrac{dx}{dt} = -\sin t$，$\dfrac{dy}{dt} = 1 + 2\cos 2t$，故 $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{1 + 2\cos 2t}{-\sin t}$。代入 $t = \dfrac{\pi}{2}$：$\cos 2t = \cos \pi = -1$，$\sin t = \sin \dfrac{\pi}{2} = 1$，所以 $\left.\dfrac{dy}{dx}\right|_{t=\frac{\pi}{2}} = \dfrac{1 + 2(-1)}{-1} = \dfrac{-1}{-1} = 1$，对应选项 B。选项 A、C、D 均因代入计算错误或导数公式误用导致结果偏差。