25~26高三下·广州一模·第8题

B

设$CD=m,AD=n$，
则$A_1D + \dfrac{\sqrt{2}}{2}CD$$= \sqrt{n^2 + 1} + \dfrac{\sqrt{2}}{2}m$$\ge \sqrt{n^2 + 1} + \dfrac{\sqrt{2}}{2}(2-n)$,
当且仅当$D$在线段$AC$上时取等号.
令$f(x) = \sqrt{x^2 + 1} – \dfrac{\sqrt{2}}{2}x + \sqrt{2},\ x \in (0,2)$，
则$f'(x) = \dfrac{1}{2\sqrt{x^2 + 1}} – \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{x^2 + 1} – \sqrt{2}}{2\sqrt{x^2 + 1}}$.
所以$f(x)$在$(0,1)$上递减，在$(1,2)$上递增，
$f(x) \ge f(1) = \sqrt{2} – \dfrac{\sqrt{2}}{2} + \sqrt{2} = \dfrac{3\sqrt{2}}{2}$.
故$A_1D + \dfrac{\sqrt{2}}{2}CD$的最小值为$\dfrac{3\sqrt{2}}{2}$.
故选：B.