練習問題パッケージの一部として、短い回答と長い回答の両方を用意することは可能ですか?
たとえば、次の例をご覧ください。数学の本 - 演習と解答の書き方 - Stack Exchange
答えだけを記した短答セクションと、完全に解答された長答セクションの両方を設けることは可能でしょうか?
MWE:
\documentclass[11pt]{article}
\usepackage[margin=2cm,includefoot,bottom=2.55cm,top=2.025cm,headsep=0.5cm,footskip=0.65cm]{geometry}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{multicol}
\usepackage{ifthen}
\newboolean{firstanswerofthesection}
\usepackage{xcolor}
\definecolor{e}{RGB}{0,40,120}
\usepackage{chngcntr}
\usepackage{stackengine}
\usepackage{tasks}
\newlength{\longestlabel}
\settowidth{\longestlabel}{(m)}
\settasks{label-format=\color{e}, counter-format={(tsk[a])}, label-width=\longestlabel,
item-indent=0pt, label-offset=2pt, column-sep={10pt}}
\usepackage[lastexercise,answerdelayed]{exercise}
\counterwithin{Exercise}{section}
\counterwithin{Answer}{section}
\renewcounter{Exercise}[section]
\newcommand{\QuestionNB}{\color{e}\bfseries\arabic{Question}.\,}
\renewcommand{\ExerciseName}{EXERCISE}
\renewcommand{\ExerciseHeader}{\noindent\def\stackalignment{l}% code from https://tex.stackexchange.com/a/195118/101651
\stackunder[0pt]{\colorbox{e}{\textcolor{white}{\textbf{\large\ExerciseName\;\large\ExerciseHeaderNB}}}}{\textcolor{e}{\rule{\linewidth}{2pt}}}\medskip}
\renewcommand{\AnswerName}{Exercises}
\renewcommand{\AnswerHeader}{\ifthenelse{\boolean{firstanswerofthesection}}%
{\bigskip\noindent\textcolor{e}{\textbf{SECTION \thesection}}\newline\newline%
\noindent\bfseries\emph{\textcolor{e}{\AnswerName\ \ExerciseHeaderNB, page %
\pageref{\AnswerRef}}}\smallskip}
{\noindent\bfseries\emph{\textcolor{e}{\AnswerName\ \ExerciseHeaderNB, page \pageref{\AnswerRef}}}\smallskip}}
\setlength{\QuestionIndent}{16pt}
\begin{document}
\section{First}
\begin{Exercise}\label{EX11}
\vspace{-\baselineskip}% <-- You don't need this line of code if there's some text here
\Question In problem \ref{EX11-1-i}-\ref{EX11-1-iii}, determine whether the given differential equation is separable
\begin{tasks}(2)
\task\label{EX11-1-i} $\frac{dy}{dx}-\sin{(x+y)}=0$
\task $\frac{dy}{dx}=4y^2-3y+1$
\task\label{EX11-1-iii} $\frac{ds}{dt}=t\ln{(s^{2t})}+8t^2$
\end{tasks}
\Question In problem \ref{EX11-2-iv}-\ref{EX11-2-viii}, solve the equation
\begin{tasks}[resume=true](2)
\task\label{EX11-2-iv} $\frac{dx}{dt}=3xt^2$
\task $y^{-1}dy+ye^{\cos{x}}\sin{x}dx=0$
\task $(x+xy^2)dx+ye^{\cos{x}}\sin{x}dx=0$
\task\label{EX11-2-viii} $\frac{dy}{dt} = \frac{y}{t+1} + 4t^2 + 4t$, $\quad$ $y(1) = 10$
\end{tasks}
\end{Exercise}
\setboolean{firstanswerofthesection}{true}
\begin{multicols}{2}
\begin{Answer}[ref={EX11}]
\Question
\begin{tasks}(3)
\task 45
\task 32
\task 32
\end{tasks}
\Question
\begin{tasks}[resume=true]
\task This is a solution of Ex 4
\task This is a solution of Ex 5
\task This is a solution of Ex 6
\task This is a solution of Ex 7
\end{tasks}
\end{Answer}
\end{multicols}
\setboolean{firstanswerofthesection}{false}
\begin{Exercise}\label{EX12}
Another exercise.
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\Question If you don't need a horizontal list, you can simply use
\end{Exercise}
\begin{multicols}{2}
\begin{Answer}[ref={EX12}]
\Question This is a solution of Ex 1
\end{Answer}
\end{multicols}
\section{Second}
\begin{Exercise}\label{EX21}
\vspace{-\baselineskip}% <-- You don't need this line of code if there's some text here
\Question Eight systems of differential equations and five direction fields are given below. Determine the system that corresponds to each direction field and sketch the solution curves that correspond to the initial conditions $(x_0, y_0) = (0,1)$ and $(x_0, y_0) = (1,-1)$.
\begin{tasks}(3)
\task $\begin{aligned}
\frac{dx}{dt} & = -x \\
\frac{dy}{dt} & = y-1
\end{aligned}$
\task $\begin{aligned}
\frac{dx}{dt} & = x^2 - 1 \\
\frac{dy}{dt} & = y
\end{aligned}$
\task $\begin{aligned}
\frac{dx}{dt} & = x+2y \\
\frac{dy}{dt} & = -y
\end{aligned}$
\task $\begin{aligned}
\frac{dx}{dt} & = 2x \\
\frac{dy}{dt} & = y
\end{aligned}$
\task $\begin{aligned}
\frac{dx}{dt} & = x \\
\frac{dy}{dt} & = 2y
\end{aligned}$
\task$\begin{aligned}
\frac{dx}{dt} & = x-1 \\
\frac{dy}{dt} & = -y
\end{aligned}$
\task$\begin{aligned}
\frac{dx}{dt} & = x^2-1 \\
\frac{dy}{dt} & = -y
\end{aligned}$
\task $\begin{aligned}
\frac{dx}{dt} & = x- 2y \\
\frac{dy}{dt} & = -y
\end{aligned}$
\end{tasks}
\end{Exercise}
\setboolean{firstanswerofthesection}{true}
\begin{multicols}{2}
\begin{Answer}[ref={EX21}]
\Question
\begin{tasks}
\task This is a solution of Ex 1
\task This is a solution of Ex 2
\task This is a solution of Ex 3
\task This is a solution of Ex 4
\task This is a solution of Ex 5
\task This is a solution of Ex 6
\task This is a solution of Ex 7
\task This is a solution of Ex 8
\end{tasks}
\end{Answer}
\end{multicols}
\setboolean{firstanswerofthesection}{false}
\newpage
\begin{Exercise}\label{EX22}
Since these are systems, maybe it's better to put the \verb|aligned| enviroment within \verb|\left\{| and \verb|\right.|:
\Question Eight systems of differential equations and five direction fields are given below. Determine the system that corresponds to each direction field and sketch the solution curves that correspond to the initial conditions $(x_0, y_0) = (0,1)$ and $(x_0, y_0) = (1,-1)$.
\begin{tasks}(3)
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = -x \\
\frac{dy}{dt} & = y-1
\end{aligned}\right.$
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = x^2 - 1 \\
\frac{dy}{dt} & = y
\end{aligned}\right.$
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = x+2y \\
\frac{dy}{dt} & = -y
\end{aligned}\right.$
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = 2x \\
\frac{dy}{dt} & = y
\end{aligned}\right.$
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = x \\
\frac{dy}{dt} & = 2y
\end{aligned}\right.$
\task$\left\{\begin{aligned}
\frac{dx}{dt} & = x-1 \\
\frac{dy}{dt} & = -y
\end{aligned}\right.$
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = x^2-1 \\
\frac{dy}{dt} & = -y
\end{aligned}\right.$
\task $\left\{\begin{aligned}
\frac{dx}{dt} & = x- 2y \\
\frac{dy}{dt} & = -y
\end{aligned}\right.$
\end{tasks}
\end{Exercise}
\begin{multicols}{2}
\begin{Answer}[ref={EX22}]
\Question
\begin{tasks}
\task This is a solution of Ex 1
\task This is a solution of Ex 2
\task This is a solution of Ex 3
\task This is a solution of Ex 4
\task This is a solution of Ex 5
\task This is a solution of Ex 6
\task This is a solution of Ex 7
\task This is a solution of Ex 8
\end{tasks}
\end{Answer}
\end{multicols}
\newpage
\section{Answer to all problems}
\begin{multicols}{2}\raggedcolumns
\shipoutAnswer
\end{multicols}
\end{document}