1. Given $\sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a}$. Then $\sqrt{4 x+x^{2}}=$
(A) $a-\frac{1}{a}$
(B) $\frac{1}{a}-a$
(C) $a+\frac{1}{a}$
(D) cannot be determined
-1. B.
$$
\begin{array}{l}
\text { Given } \sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a} \Rightarrow x=\frac{1}{a}+a-2 \\
\Rightarrow x+2=\frac{1}{a}+a .
\end{array}
$$

Then $\sqrt{4 x+x^{2}}=\sqrt{(x+2)^{2}-4}$
$$
\begin{array}{l}
=\sqrt{\left(\frac{1}{a}+a\right)^{2}-4}=\sqrt{\left(\frac{1}{a}-a\right)^{2}} \\
=\left|\frac{1}{a}-a\right| . \\
\text { Also } \frac{1}{\sqrt{a}}-\sqrt{a} \geqslant 0 \Rightarrow \frac{1}{a} \geqslant a .
\end{array}
$$

Therefore, $\sqrt{4 x+x^{2}}=\frac{1}{a}-a$.