(2) If the function $f(x)=a^{x}\left(a^{x}-3 a^{2}-1\right)(a>0$ and $a \neq 1)$ is increasing on the interval $[0,+\infty)$, then the range of the real number $a$ is ( ).
(A) $\left(0, \frac{2}{3}\right]$
(B) $\left[\frac{\sqrt{3}}{3}, 1\right)$
(C) $(0, \sqrt{3}]$
(D) $\left[\frac{3}{2},+\infty\right)$
(2) B Hint: Use the monotonicity of composite functions to solve. Let $a^{x}=t$, then
$$
y=g(t)=t^{2}-\left(3 a^{2}+1\right) t .
$$
(1) When $a>1$, $t$ is an increasing function of $x$. To make $f(x)$ an increasing function on $x \in[0, +\infty)$, $g(t)$ needs to be an increasing function on $t \in[1,+\infty)$. Hence, $\frac{3 a^{2}+1}{2} \leqslant 1$, which means $a^{2} \leqslant \frac{1}{3}$, a contradiction.
(2) When $0<a<1$, $t$ is a decreasing function of $x$. To make $f(x)$ an increasing function on $x \in[0, +\infty)$, $g(t)$ needs to be a decreasing function on $t \in(0,1]$. Hence, $\frac{3 a^{2}+1}{2} \geqslant 1$, which means $a^{2} \geqslant \frac{1}{3}$, so $\frac{\sqrt{3}}{3} \leqslant a<1$.
In summary, the correct choice is B.