(3) Let the function $f(x)=\ln x, g(x)=a x+\frac{b}{x}$, and their graphs have a common tangent line at a common point on the $x$-axis. Then, when $x>1$, the relationship between $f(x)$ and $g(x)$ is ( ).
(A) $f(x)>g(x)$
(B) $f(x)<g(x)$
(C) $f(x)=g(x)$
(D) Uncertain
3 B Hint: The graphs of $f(x)$ and $g(x)$ have a unique common point $(1,0)$ on the $x$-axis, so $g(1)=0$, which means $a+b=0$. Since $f^{\prime}(x)=\frac{1}{x}, g^{\prime}(x)=a-\frac{b}{x^{2}}$, by the problem, $f^{\prime}(1)=g^{\prime}(1)=1$, which means $a-b=1$. Therefore, $a=\frac{1}{2}, b=-\frac{1}{2}$. Let
$$
F(x)=f(x)-g(x)=\ln x-\left(\frac{1}{2} x-\frac{1}{2 x}\right),
$$

then
$$
F^{\prime}(x)=\frac{1}{x}-\frac{1}{2}-\frac{1}{2 x^{2}}=-\frac{1}{2}\left(\frac{1}{x}-1\right)^{2} \leqslant 0 .
$$

So $F(x)$ is monotonically decreasing in $(1, \infty)$.
Since $F(1)=0$, for $x>1$, $F(x)<0$, which means $f(x)<g(x)$. Choose B.