1. Let $a>0, f(x)=\frac{x}{\sqrt{a}}-\sqrt{a(x+a)}$ $(x \geqslant-a)$, denote the minimum value of $f(x)$ as $m(a)$. If $m(a)=-2006 \sqrt{a}$, then $a$ equals ( ).
(A) 8020
(B) 8016
(C) 2006
(D) 2005
-.1.A.
Let $\sqrt{x+a}=t(t \geqslant 0)$, then $x=t^{2}-a$. Hence $f(t)=\frac{t^{2}-a}{\sqrt{a}}-\sqrt{a} t=\frac{1}{\sqrt{a}}\left(t^{2}-a t-a\right)$ $=\frac{1}{\sqrt{a}}\left[\left(t-\frac{a}{2}\right)^{2}-a-\frac{a^{2}}{4}\right]$.

Obviously, when $t=\frac{a}{2}>0$, $f(x)$ reaches its minimum value $m(a)=-\frac{1}{\sqrt{a}}\left(a+\frac{a^{2}}{4}\right)=-\sqrt{a}\left(1+\frac{a}{4}\right)$. Also, $m(a)=-2006 \sqrt{a}$, then $1+\frac{a}{4}=2006$. Solving for $a$ gives $a=4 \times 2005=8020$.