1.66 The figure below represents the solution set of the inequality $a x > 2 x$ on the number line. Then the value of the expression
$$
\sqrt{a+2 \sqrt{a-1}}+\sqrt{a-2 \sqrt{a-1}}
$$
is
(A) $2 \sqrt{a-1}$.
(B) 2 .
(C) $a$.
(D) 1
(5th "Jinyun Cup" Junior High School Mathematics Invitational, 1988)
[Solution] From $a x>2 x$ and $x<0$, we get $a<2$.
Also, for the square root to be defined, $a-1 \geqslant 0$, i.e., $a \geqslant 1$, so $1 \leqslant a<2$.
$$
\begin{aligned}
& \sqrt{a+2 \sqrt{a-1}}+\sqrt{a-2 \sqrt{a-1}} \\
= & \sqrt{(a-1)+2 \sqrt{a-1}+1}+\sqrt{(a-1)-2 \sqrt{a-1}+1} \\
= & |\sqrt{a-1}+1|+|\sqrt{a-1}-1| \\
= & \sqrt{a-1}+1+1-\sqrt{a-1}=2 .
\end{aligned}
$$

Therefore, the correct choice is (B).