几位慷慨的 tex.stacker 帮助我整理了下面的代码。（h/t 尤其是@alchemist）
仅使用后，enumitem我很困惑哪些代码行在控制枚举。
是否\paracol生成枚举？或者可能\globalcounter*？请解释此代码中的枚举如何工作。
对于最后 2 个“额外学分”问题，我需要切换到字母枚举。此图显示了所需的字母枚举样式。
非常感谢您的帮助。
请注意额外学分问题从 mwe 的第二页开始。
\documentclass[12pt]{exam}
\printanswers
% un-comment to print solutions.
\renewcommand{\solutiontitle}{}
\usepackage{amsmath}
\usepackage{cancel}
\usepackage{framed}
%\usepackage{multicol}
\usepackage{paracol}
\globalcounter*
\usepackage[table]{xcolor}
\usepackage[a4paper,margin=0.5in,include head]{geometry}
\pagestyle{head}
%Note: \& forces & to be literal character
\header{Algebra II: Unit 11 Quiz (Rational Expressions \& Equations)}
{}
{Due 04/07/23}
\newcommand{\pagetop}{%
\noindent
\bigskip
\vspace{0.5mm}
}
\begin{document}
%definition for bigskip = 1 line to replace all \bigskip
\def\bigskip{\vskip\bigskipamount}
\begin{questions}
\fullwidth{\textbf{1-2 Simplify each expression.}}
\begin{paracol}{2}
% Prob #1
\question $\dfrac{4x+12}{x^2+5x+6}$
\\
\begin{solutionorbox}[3cm]
$\dfrac{4(x+3)}
{(x+3)(x+2)}$
\bigskip
$\dfrac{4\cancel{(x+3)}}{\cancel{(x+3)}(x+2)} = \boldsymbol{\dfrac{4}{x+2}}$
\end{solutionorbox}
\switchcolumn
% Prob #2
\question $\dfrac{x^2+3x+2}{x^2-4}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{(x+2)(x+1)}{(x+2)(x-2)}$
\bigskip
$\dfrac{\cancel{(x+2)}(x+1)}{\cancel{(x+2)}(x-2)} = \boldsymbol{\dfrac{x+1}{x-2}}$
\bigskip
\end{solutionorbox}
%force new page
\switchcolumn*[\newpage]
\fullwidth{\textbf{EXTRA CREDIT}}
\bigskip
%Extra credit problems
Divide and simiplify. Show all work clearly.
\question
$\dfrac{x^2+3x+2}{x^2-4}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{(x+2)(x+1)}{(x+2)(x-2)}$
\bigskip
$\dfrac{\cancel{(x+2)}}{\cancel{(x+2)}(x-2)} = \boldsymbol{\dfrac{x+1}{x-2}}$
\bigskip
\end{solutionorbox}
\end{paracol}
\end{questions}
\end{document}
答案1
我在课程手册第 39-40 页找到了你的问题的答案exam。
您需要在之前输入以下几行额外积分问题：
\qformat{\textit{\thequestion}} % not really needed, but emphasizes the capital A
\setcounter{question}{0}
\renewcommand{\thequestion}{\Alph{question}}
下一个问题将以大写字母 A 开头。
答案2
（部分）对已编辑问题的回答
比较qformat{}以下代码中的条目。要编译，请将代码插入到您的示例中。
%force new page
\switchcolumn*[\newpage]
\parbox{\textwidth}{\textbf{EXTRA CREDIT:} \textit{Show all work clearly and completely to earn credits.}}
\bigskip
\switchcolumn
\switchcolumn*
\qformat{\textit{\thequestion}} % not really needed, but emphasizes the capital A
\setcounter{question}{0}
\renewcommand{\thequestion}{\Alph{question}}
%Extra credit problems
Divide and simiplify. Show all work clearly.
\question
$\dfrac{x^2+3x+2}{x^2-4}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{(x+2)(x+1)}{(x+2)(x-2)}$
\bigskip
$\dfrac{\cancel{(x+2)}(x+1)}{\cancel{(x+2)}(x-2)} = \boldsymbol{\dfrac{x+1}{x-2}}$
\bigskip
\end{solutionorbox}
\switchcolumn
\qformat{\textbf{\thequestion})\hspace{5mm}} % not really needed, but emphasizes the capital A
\setcounter{question}{0}
\renewcommand{\thequestion}{\Alph{question}}
%Extra credit problems
Divide and simiplify. Show all work clearly.
\question
$\dfrac{3x^2-13x-10}{x^2+2x-35} \div \dfrac{6x^2-11x-10}{6x-15}$
\bigskip
\begin{solutionorbox}[5cm]
$\dfrac{(x-5)(3x+2)}{(x-5)(x+7)}\div\dfrac{(2x-5)(3x+2)}{3(2x-5)}$
\bigskip
$\dfrac{\cancel{(x-5)}(3x+2)}{\cancel{(x-5)}(x+7)}\div\dfrac{\cancel{(2x-5)}(3x+2)}{3\cancel{(2x-5)}}$
\bigskip
$\dfrac{\cancel{(3x+2)}}{(x+7)} \times \dfrac{3}{\cancel{(3x+2)}} = \boldsymbol{\dfrac{3}{x+7}}$
\bigskip
\end{solutionorbox}
如您所见，可以使用 来格式化标签\qformat。但是，我还不知道为什么标签不再与问题对齐。它应该仍然如此。
用于\parbox使用整页，这也意味着您必须重新对齐列。因此需要额外的\switchcolumns。