21. Let when $0 \leqslant x \leqslant 1$, there exists a positive number $q$ such that $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\mathrm{t}}}{q}$ holds, find the smallest positive number $t$ that makes the above inequality true. For this smallest $t$ value, what is the minimum value of $q$ that makes the above inequality true when $0 \leqslant x \leqslant 1$?
21. Since $y=\sqrt{x}$ is strictly concave, for $0<x<1$, we have
$$
q \geqslant 2-(\sqrt{1+x}+\sqrt{1-x}).
$$

After two constant transformations, we get
$$
\begin{array}{l}
q \geqslant x^{1-2}\left(1+\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \\
\cdot\left(1+\sqrt{1-x^{2}}\right).
\end{array}
$$

If $t<2$, then as $x$ approaches 0, the above expression tends to infinity. Therefore, for any fixed $q$, there exists an $x$ value sufficiently close to 0 such that the required inequality does not hold. For $t=2$, when $x=0$, the right-hand side of the above expression equals 4. Thus, $q \geqslant 4$. For $0<x \leqslant 1$, since $\sqrt{1+x}+\sqrt{1-x}<1$ and $\sqrt{1-x^{2}}<1$, when $t=2$, the minimum value of $q$ is 4.