10. If the function $f(x)=-\frac{1}{2} x^{2}+\frac{13}{2}$ has a minimum value of $2a$ and a maximum value of $2b$ on the interval $[a, b]$, find $[a, b]$.
10. (1) If $0 \leqslant a < b$, then $f(x)$ attains its minimum value $2a$ at $x=a$, i.e., $2a = -\frac{1}{2}a^2 + \frac{13}{2}$, yielding $a = -2 - \sqrt{17}$;
(3) When $a < b \leqslant 0$, $f(x)$ is monotonically increasing on $[a, b]$, so $f(a) = 2a, f(b) = 2b$, i.e., $2a = -\frac{1}{2}a^2 + \frac{13}{2}, 2b = -\frac{1}{2}b^2 + \frac{13}{2}$. However, the roots of the equation $\frac{1}{2}x^2 + 2x - \frac{13}{2} = 0$ have opposite signs, so there is no interval satisfying $a < b < 0$.
In summary, $[a, b] = [1, 3]$ or $\left[-2 - \sqrt{17}, \frac{13}{4}\right]$.