12、锐角$\triangle ABC$中，角$A$，$B$，$C$所对的边分别为$a$，$b$，$c$，已知$a = 1$，$b\cos A - \cos B = 1$，则$\sqrt{3}\sin B + 2\sin^{2}A$的取值范围（ ）
A．$\left( 0,\sqrt{3} + 1 \right)$ B．$\left( 2,\sqrt{3} + 1 \right)$ C．$\left( 1,3 \right]$ D．$\left( 2,3 \right]$
解：$\because a = 1$，$\frac{a}{\sin A} = \frac{b}{\sin B}$
$\therefore b\sin A = \sin B$
$\because b\cos A - \cos B = 1$
$\therefore \frac{\sin A}{\cos A} = \frac{\sin B}{1 + \cos B}$
$\therefore \sin B\cos A - \sin A\cos B = \sin A$
$\therefore \sin(B - A) = \sin A$
$\because 0 < A < \frac{\pi}{2},0 < B < \frac{\pi}{2}$
$\therefore B = 2A$
$\therefore \begin{cases} 0 < 2A < \frac{\pi}{2} \\ 0 < \pi - 3A < \frac{\pi}{2} \end{cases}$
$\therefore \frac{\pi}{6} < A < \frac{\pi}{4}$
$\therefore \frac{\pi}{6} < 2A - \frac{\pi}{6} < \frac{\pi}{3}$
$\therefore \frac{1}{2} < \sin\left( 2A - \frac{\pi}{6} \right) < \frac{\sqrt{3}}{2}$
$\therefore \sqrt{3}\sin B + 2\sin^{2}A = \sqrt{3}\sin 2A + 1 - \cos 2A = 2\sin\left( 2A - \frac{\pi}{6} \right) + 1 \in \left( 2,\sqrt{3} + 1 \right)$ 选 B