7.83 The graph of the function $y=f(x)$ is $c$. The graph $c_{1}$ is the reflection of $c$ about the line $x=1$, and the graph $c_{2}$ is obtained by shifting $c_{1}$ one unit to the left. Then the function corresponding to $c_{2}$ is
(A) $y=f(-x)$.
(B) $y=f(1-x)$.
(C) $y=f(2-x)$.
(D) $y=f(3-x)$.
(China Beijing High School Mathematics Competition, 1992)
[Solution] Let the function corresponding to $c_{1}$ be $f_{1}(x)$. Given that $c_{1}$ is symmetric to $c$ with respect to the line $x=1$, we have $f(1-x)=f_{1}(1+x)$,
then
that is
$$
\begin{array}{c}
f[1-(x-1)]=f_{1}[1+(x-1)], \\
f_{1}(x)=f(2-x) .
\end{array}
$$

Therefore, the expression for $c_{1}$ is $\quad y=f(2-x)$.
Since $c_{2}$ is obtained by shifting $c_{1}$ one unit to the left, the function corresponding to $c_{2}$ is
$$
y=f_{1}(x+1)=f(1-x) \text {. }
$$

Hence, the correct choice is $(B)$.