25~26高三下·山东滨州一模·第14题

$[-\sqrt{2},1)$

由$|\overrightarrow{PA}| = |\overrightarrow{PB}| = |\overrightarrow{PC}|$，所以点$P$为$\triangle ABC$的外心.

因为$\angle A = \dfrac{\pi}{4}$，所以$\angle BPC = 2 \times \dfrac{\pi}{4} = \dfrac{\pi}{2}$.

设$|\overrightarrow{PA}| = |\overrightarrow{PB}| = |\overrightarrow{PC}| = r$，再以点$P$为原点，分别以$PB$、$PC$所在直线为$x$、$y$轴建立平面直角坐标系.

则$B(r,0)$，$C(0,r)$，$\overrightarrow{PB} = (r,0)$，$\overrightarrow{PC} = (0,r)$，

所以$\overrightarrow{PA} = \lambda\overrightarrow{PB} + \mu\overrightarrow{PC} = (\lambda r,0) + (0,\mu r) = (\lambda r,\mu r)$.

又因为$|\overrightarrow{PA}| = r$，

所以$(\lambda r)^2 + (\mu r)^2 = r^2$，即$\lambda^2 + \mu^2 = 1$。

又因为$\angle A = \dfrac{\pi}{4}$，

所以点$A$在优弧$BC$上，所以$\overrightarrow{PA}$落在角$\theta \in \left(\dfrac{\pi}{2},2\pi\right)$的终边上.

由三角函数的定义有$\lambda r = r\cos\theta$，$\mu r = r\sin\theta$，即$\lambda = \cos\theta$，$\mu = \sin\theta$。

所以$\lambda + \mu = \cos\theta + \sin\theta = \sqrt{2}\sin\left(\theta + \dfrac{\pi}{4}\right)$.

因为$\theta \in \left(\dfrac{\pi}{2},2\pi\right)$，所以$\theta + \dfrac{\pi}{4} \in \left(\dfrac{3\pi}{4},\dfrac{9\pi}{4}\right)$，
$\sin\left(\theta + \dfrac{\pi}{4}\right) \in \left[-1,\dfrac{\sqrt{2}}{2}\right)$，$\sqrt{2}\sin\left(\theta + \dfrac{\pi}{4}\right) \in [-\sqrt{2},1)$，
所以$\lambda + \mu \in [-\sqrt{2},1)$.