11. Given that when $|x| \leqslant 1$, we always have
$$
2 p x^{2}+q x-p+1 \geqslant 0 \text {. }
$$

Find the maximum value of $p+q$.
II. 11. Solution 1 Let
$$
f(x)=2 p x^{2}+q x-p+1(x \in[-1,1]) \text {. }
$$
(1) First consider the case when $p>0$.
(i) If $-1 \leqslant-\frac{q}{4 p} \leqslant 1$, i.e., $-4 p \leqslant q \leqslant 4 p$, then by the problem, $f\left(-\frac{q}{4 p}\right) \geqslant 0$, i.e.,
$$
\begin{array}{l}
2 p\left(-\frac{q}{4 p}\right)^{2}+q\left(-\frac{q}{4 p}\right)-p+1 \geqslant 0 \\
\Rightarrow q^{2}+8\left(p-\frac{1}{2}\right)^{2} \leqslant 2 .
\end{array}
$$

Let $q=r \cos \theta, p-\frac{1}{2}=\frac{r \sin \theta}{2 \sqrt{2}}$, where $0 \leqslant r \leqslant \sqrt{2}, \theta \in[0,2 \pi]$.
Then $p+q=r\left(\frac{1}{2 \sqrt{2}} \sin \theta+\cos \theta\right)+\frac{1}{2}$.
Let $\varphi \in\left(0, \frac{\pi}{2}\right)$, and $\tan \varphi=2 \sqrt{2}$. Then
$$
\begin{array}{l}
p+q=\frac{3}{2 \sqrt{2}} r \sin (\theta+\varphi)+\frac{1}{2} \\
\leqslant \frac{3}{2 \sqrt{2}} \times \sqrt{2} \times 1+\frac{1}{2}=2,
\end{array}
$$

When $r=\sqrt{2}, \sin \theta=\frac{1}{3}, \cos \theta=\frac{2 \sqrt{2}}{3}$, i.e., $p=\frac{2}{3}, q=\frac{4}{3}$, the equality holds.
(ii) If $-\frac{q}{4 p}4 p$, then by
$$
\begin{array}{l}
f(-1)=p-q+1 \geqslant 0 \Rightarrow q \leqslant p+1 \\
\Rightarrow 4 p1$, i.e., $q<-4 p$, then
$$
p+q \leqslant-3 p<0<2 \text {. }
$$
(2) When $p \leqslant 0$, by
$$
f(-1)=p-q+1 \geqslant 0 \Rightarrow q \leqslant p+1 \text {. }
$$

Thus, $p+q \leqslant 2 p+1<2$.
In summary, the maximum value of $p+q$ is 2.
Solution 2 In the inequality (1), take the special value $x=-\frac{1}{2}$, we get $p+q \leqslant 2$.
When and only when $p=\frac{2}{3}, q=\frac{4}{3}$,
$$
\begin{array}{l}
2 p x^{2}+q x-p+1=\frac{4}{3} x^{2}+\frac{4}{3} x+\frac{1}{3} \\
=\frac{4}{3}\left(x+\frac{1}{2}\right)^{2} \geqslant 0 .
\end{array}
$$

Thus, the maximum value of $p+q$ is 2.