1. Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the opposite angles of $a, b, c$ are $\angle A, \angle B, \angle C$ respectively. Then the range of $\sin B+\cos B$ is ( ).
(A) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(B) $\{, \sqrt{2}]$
(C) $\left(1,1+\frac{\sqrt{3}}{2}\right)$
(D) $\left[\frac{1}{2}, \sqrt{2}\right]$
1. Solution: Since $a, b, c$ form a geometric sequence, we have $b^{2} = ac$.
Thus, $b^{2} = ac = a^{2} + c^{2} - 2ac \cos B$
$$
\begin{aligned}
& \geqslant 2ac - 2ac \cos B, \\
\cos B & \geqslant \frac{1}{2}.
\end{aligned}
$$

Therefore, $0 < B \leqslant 60^{\circ}$.
From $\frac{1}{2} \leqslant \cos B < 1$ and $0 < \sin B \leqslant \frac{\sqrt{3}}{2}$, we get
$$
\frac{1}{2} < \cos B + \sin B < \frac{\sqrt{3}}{2} + 1.
$$

On the other hand, $\sin B + \cos B = \sqrt{2} \sin \left(B + 45^{\circ}\right)$.
Since $45^{\circ} < \angle B + 45^{\circ} \leqslant 105^{\circ}$,
we have $1 < \sin B + \cos B \leqslant \sqrt{2}$.
Combining (1) and (2), we get $1 < \sin B + \cos B \leqslant \sqrt{2}$. Choose B.