[Tex/LaTex] Math book: how to write Exercise and Answers

booksexercisesmathematics

EDIT
within \documentclass[12pt]{book}I want to create chapter-wise exercises and put all the solutions (with or without hints) at the end of the book.

I want to do this in simple and non-tedious way like: For the input of the questions, I just want to add \questionfor each question and similar for answer, but all the answer should come at the end of the book.

I need the Exercises No as 1.1, 1.2, …, 2.1, 2.2, ….

Problem Nos as 1, 2, 3, …

The answer should include page No of the exercise as given in the attached jpg file.

Exercise style:

Solution type:

A MWE is as follows (I also don't know how to write solution in chapter 2 as problems are written within \begin{minipage} \end{minipage}. Suppose all solutions are: This is a solution Ex No.)

    \documentclass[12pt,openany,twoside]{book} 
\usepackage{fouriernc}
\usepackage[thmmarks,standard,thref]{ntheorem}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{answers}

\theoremseparator{.}
\theorembodyfont{\upshape}
\newtheorem{pro}{Problem}[chapter]

\Newassociation{loigiai}{Answer}{loigiaichung}


\begin{document}
\chapter{Firist }
\thispagestyle{empty}
\Opensolutionfile{loigiaichung}[pro]
In problem 1-3, determine whether the given differential equation is separable 
\begin{pro}
$\frac{dy}{dx}-\sin{(x+y)}=0$
\begin{loigiai}
This is a solution of Ex 1
\end{loigiai}
\end{pro}

\begin{pro}
$\frac{dy}{dx}=4y^2-3y+1$
\begin{loigiai}
This is a solution of Ex 2
\end{loigiai}
\end{pro}


\begin{pro}
$\frac{ds}{dt}=t\ln{(s^{2t})}+8t^2$
\begin{loigiai}
This is a solution Ex 3
\end{loigiai}
\end{pro}

In problem 4-6, solve the equation 
\begin{pro}
$\frac{dx}{dt}=3xt^2$
\begin{loigiai}
This is a solution Ex 4
\end{loigiai}
\end{pro}

\begin{pro}
$y^{-1}dy+ye^{\cos{x}}\sin{x}dx=0$
\begin{loigiai}
This is a solution Ex 5
\end{loigiai}
\end{pro}


\begin{pro}
$(x+xy^2)dx+ye^{\cos{x}}\sin{x}dx=0$
\begin{loigiai}
This is a solution Ex 5
\end{loigiai}
\end{pro}

\begin{pro}
$\frac{dy}{dt} = \frac{y}{t+1} + 4t^2 +  4t$, $\quad$ $y(1) = 10$
\begin{loigiai}
This is a solution Ex 6
\end{loigiai}
\end{pro}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{Second }
\thispagestyle{empty}
\begin{pro}
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{minipage}{.25\textwidth}
$(i)$\begin{align*}
\frac{dx}{dt} & = -x \\
\frac{dy}{dt} & = y-1\end{align*}
\end{minipage}
\begin{minipage}{.25\textwidth}
$(ii)$\begin{align*}
\frac{dx}{dt} & = x^2 - 1 \\
\frac{dy}{dt} & = y\end{align*}
\end{minipage}
\begin{minipage}{.25\textwidth}
$(iii)$\begin{align*}
\frac{dx}{dt} & = x+2y \\
\frac{dy}{dt} & = -y\end{align*}
\end{minipage}
\begin{minipage}{.25\textwidth}
$(iv)$\begin{align*}
\frac{dx}{dt} & = 2x \\
\frac{dy}{dt} & =  y\end{align*}
\end{minipage}

\begin{minipage}{.25\textwidth}
$(v)$\begin{align*}
\frac{dx}{dt} & = x \\
\frac{dy}{dt}  & = 2y\end{align*}
\end{minipage}
\begin{minipage}{.25\textwidth}
$(vi)$\begin{align*}
\frac{dx}{dt} & = x-1 \\
\frac{dy}{dt} & = -y\end{align*}
\end{minipage}
\begin{minipage}{.25\textwidth}
$(vii)$\begin{align*}
\frac{dx}{dt} & = x^2-1 \\
\frac{dy}{dt} & = -y\end{align*}
\end{minipage}
\begin{minipage}{.25\textwidth}
$(viii)$\begin{align*}
\frac{dx}{dt} & = x- 2y \\
\frac{dy}{dt} & =  -y\end{align*}
\end{minipage}
\end{pro}
\chapter{Answer to all problems}
\thispagestyle{empty}
\Closesolutionfile{loigiaichung}
\section{This is solution}
\input{pro}

\end{document}

Best Answer

Here is a solution with exercise package.

Its option answerdelayed allows you to print the answer where you want with \shipoutAnswer.

lastexercise means that if you don't put anything as answer reference, the Answer environment is referred to the previous exercise. However, I have also put \begin{Answer}[ref={...}], in this way you can move the code for the answers where you want, but you have to keep the order. For how I have created the code for the first answer of the chapter (to print CHAPTER ... in the final list), you also have to write it within the relative chapter.

Moreover, To have CHAPTER ... printed in the final answer list, you have to set \setboolean{firstanswerofthechapter}{true} at the beginning and \setboolean{firstanswerofthechapter}{false} at the end of the first answer of a chapter (maybe there's a smarter way to do it but it's above my level of knowledge).

Eventually, if you need horizontal lists you have to use tasks environment, otherwise, the ordinary \Question is enough I put an example of both in my MWE.

Edit: I've changed the numbering as required by the OP in their comment.

If you want the roman numbers to begin from i. in the second question, just remove the [resume=true] option.

Second edit: To have the list of the second exercise correctly aligned, you have to use aligned instead of align*. Since these are systems, maybe it's better to put a curly bracket before, like here in Ex. 2.2:

Third edit: In the original post I wrote counter-format={tsk[r].} as an option of the tasks package. But, since October 2019, that option is deprecated. Hence, the current version has label=\roman*..
Thanks to EthanAlvaree for pointing it out.

\documentclass{book}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{multicol}

\usepackage{ifthen}
\newboolean{firstanswerofthechapter}  

\usepackage{xcolor}
\colorlet{lightcyan}{cyan!40!white}

\usepackage{chngcntr}
\usepackage{stackengine}

\usepackage{tasks}
\newlength{\longestlabel}
\settowidth{\longestlabel}{\bfseries viii.}
\settasks{label=\roman*., label-format={\bfseries}, label-width=\longestlabel,
item-indent=0pt, label-offset=2pt, column-sep={10pt}}

\usepackage[lastexercise,answerdelayed]{exercise}
\counterwithin{Exercise}{chapter}
\counterwithin{Answer}{chapter}
\renewcounter{Exercise}[chapter]
\newcommand{\QuestionNB}{\bfseries\arabic{Question}.\ }
\renewcommand{\ExerciseName}{EXERCISES}
\renewcommand{\ExerciseHeader}{\noindent\def\stackalignment{l}% code from https://tex.stackexchange.com/a/195118/101651
    \stackunder[0pt]{\colorbox{cyan}{\textcolor{white}{\textbf{\LARGE\ExerciseHeaderNB\;\large\ExerciseName}}}}{\textcolor{lightcyan}{\rule{\linewidth}{2pt}}}\medskip}
\renewcommand{\AnswerName}{Exercises}
\renewcommand{\AnswerHeader}{\ifthenelse{\boolean{firstanswerofthechapter}}%
    {\bigskip\noindent\textcolor{cyan}{\textbf{CHAPTER \thechapter}}\newline\newline%
        \noindent\bfseries\emph{\textcolor{cyan}{\AnswerName\ \ExerciseHeaderNB, page %
                \pageref{\AnswerRef}}}\smallskip}
    {\noindent\bfseries\emph{\textcolor{cyan}{\AnswerName\ \ExerciseHeaderNB, page \pageref{\AnswerRef}}}\smallskip}}
\setlength{\QuestionIndent}{16pt}

\begin{document}
    \chapter{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{firstanswerofthechapter}{true}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX11}]
            \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 
            \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{firstanswerofthechapter}{false}
    
    \begin{Exercise}\label{EX12}
        Another exercise. 
        \Question If you don't need a horizontal list, you can simply use \verb|\Question|
    \end{Exercise}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX12}]
            \Question This is a solution of Ex 1
        \end{Answer}
    \end{multicols}
    
    \chapter{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{firstanswerofthechapter}{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{firstanswerofthechapter}{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}
    
    \chapter{Answer to all problems}
    
    \begin{multicols}{2}\raggedcolumns
        \shipoutAnswer
    \end{multicols}
    
\end{document}

Related Solutions

[Tex/LaTex] Solutions of exercises produced with tcolorbox

It is possible to prepend the \chapter command to write some information about the chapter into the record file. This will fail for the first chapter since the recording has not started then.

It will also fail if the chapter number is reset.

Since the prepending occurs before the chapter number is increased, I've decided to trick \c@chapter using a number increased by one within a TeX group, so the counter value is not really changed but \thechapter will use the tricked value for while and apply the correct format as defined for \thechapter. This is better than using something like \arabic{chapter} etc, in my point of view.

Change the \solutionchapterformat macro at will to produce any nice/fancy heading of the solution-chapter-group (but leave the \bgroup... \egroup)

\documentclass{book}

\usepackage[most]{tcolorbox}

\usepackage{xpatch}

% Formatting command as a 'headline' of the solutions of chapter X
\NewDocumentCommand{\solutionchapterformat}{m}{%
  \noindent \bgroup\bfseries Solutions of the exercises of the chapter #1\egroup%
}
\makeatletter


\xpretocmd{\chapter}{%
  \begingroup
  \ifnum\value{chapter}>0\relax
  \tcbrecord{\string\clearpage}% Write a clearpage after the first chapter for each new chapter
  \fi
  \c@chapter \numexpr\c@chapter+1% Increase the count register `\@chapter` by one to trick `\thechapter` using the 'correct' chapter number
  \tcbrecord{%
    \solutionchapterformat{\thechapter}}%
  \endgroup
}{}{}

\NewDocumentCommand{\extrasolutioncontent}{+m}{%
  \tcbrecord{Extra solution stuff\par}% Remove this later on!
  \tcbrecord{\detokenize{#1}}%
}

\newcommand{\fetchsolutions}{%
%For the first chapter
\begingroup
\c@chapter1%
\solutionchapterformat{\thechapter}%
\endgroup% Now get the rest of the stuff
\tcbinputrecords
}
\makeatother

%\tcbuselibrary{skins,breakable}

%\usepackage{polyglossia}
%\setmainlanguage{english}
%\usepackage{fontspec}
%\setmainfont{Times New Roman}
%\setsansfont{Arial}
%\setmonofont{Courier New}


\NewTColorBox[auto counter,number within=chapter]{exercise}{m+O{}}{%
    enhanced,
    colframe=green!20!black,
    colback=yellow!10!white,
    coltitle=green!40!black,
    fonttitle=\bfseries,
    underlay={\begin{tcbclipinterior}
        \shade[inner color=green!80!yellow,outer color=yellow!10!white]
            (interior.north west) circle (2cm);
        \draw[help lines,step=5mm,yellow!80!black,shift={(interior.north west)}]
            (interior.south west) grid (interior.north east);
        \end{tcbclipinterior}},
    title={Exercise~ \thetcbcounter:},
    label={exercise:#1},
    attach title to upper=\quad,
    after upper={\par\hfill\textcolor{green!40!black}%
        {\itshape Solution on page~\pageref{solution:#1}}},
    lowerbox=ignored,
    savelowerto=solutions/exercise-\thetcbcounter.tex,
    record={\string\solution{#1}{solutions/exercise-\thetcbcounter.tex}},
    #2
}

\NewTotalTColorBox{\solution}{mm}{%
    enhanced,
    colframe=red!20!black,
    colback=yellow!10!white,
    coltitle=red!40!black,
    fonttitle=\bfseries,
    underlay={\begin{tcbclipinterior}
        \shade[inner color=red!50!yellow,outer color=yellow!10!white]
            (interior.north west) circle (2cm);
        \draw[help lines,step=5mm,yellow!80!black,shift={(interior.north west)}]
            (interior.south west) grid (interior.north east);
        \end{tcbclipinterior}},
    title={Solution of Exercise~\ref{exercise:#1} on page~\pageref{exercise:#1}:},
    phantomlabel={solution:#1},
    attach title to upper=\par,
}{\input{#2}}

\tcbset{no solution/.style={no recording,after upper=}}

\begin{document}
\chapter{The first chapter}

\tcbstartrecording
\begin{exercise}{Ex1}[coltitle=cyan!80!black]
Compute the derivative of the following function:
\begin{equation*}
f(x)=\sin((\sin x)^2)
\end{equation*}
\tcblower
The derivative is:
\begin{align*}
f’(x) &= \left( \sin((\sin x)^2) \right)’
=\cos((\sin x)^2) 2\sin x \cos x.
\end{align*}
\end{exercise}
\vspace*{1cm}
The solution of the exercise \ref{exercise:Ex1} is in the page \pageref{solution:Ex1}
\vspace*{1cm}
%\tcbstoprecording
%\tcbinputrecords
\chapter{The second chapter}

%\tcbstartrecording
\begin{exercise}{Ex2}[coltitle=cyan!80!black]
Compute the derivative of the following function:
\begin{equation*}
f(x)=(x^2+1) \sqrt{x^4+1}
\end{equation*}
\tcblower
The derivative is:
\begin{align*}
f’(x) &= \left( (x^2+1) \sqrt{x^4+1} \right)’
= 2x\sqrt{x^4+1} + \frac{2x^3(x^2+1)}{\sqrt{x^4+1}}.
\end{align*}
\end{exercise}
\vspace*{1cm}
The solution of the exercise \ref{exercise:Ex2} is in the page \pageref{solution:Ex2}
\vspace*{1cm}

\chapter{The third chapter}

\begin{exercise}{Ex3}[coltitle=cyan!80!black]
Compute the derivative of the following function:
\begin{equation*}
f(x)=(x^2+1) \sqrt{x^4+1}
\end{equation*}
\tcblower
The derivative is:
\begin{align*}
f’(x) &= \left( (x^2+1) \sqrt{x^4+1} \right)’
= 2x\sqrt{x^4+1} + \frac{2x^3(x^2+1)}{\sqrt{x^4+1}}.
\end{align*}
\end{exercise}
\vspace*{1cm}
The solution of the exercise \ref{exercise:Ex3} is in the page \pageref{solution:Ex3}
\vspace*{1cm}


\extrasolutioncontent{$\textcolor{blue}{E=mc^2}$}

\begin{exercise}{Ex31}[coltitle=cyan!80!black]
Compute the derivative of the following function:
\begin{equation*}
f(x)=(x^2+1) \sqrt{x^4+1}
\end{equation*}
\tcblower
The derivative is:
\begin{align*}
f’(x) &= \left( (x^2+1) \sqrt{x^4+1} \right)’
= 2x\sqrt{x^4+1} + \frac{2x^3(x^2+1)}{\sqrt{x^4+1}}.
\end{align*}
\end{exercise}
\vspace*{1cm}
The solution of the exercise \ref{exercise:Ex31} is in the page \pageref{solution:Ex31}
\vspace*{1cm}


\chapter{The fourth chapter}

\begin{exercise}{Ex4}[coltitle=cyan!80!black]
Compute the derivative of the following function:
\begin{equation*}
f(x)=(x^2+1) \sqrt{x^4+1}
\end{equation*}
\tcblower
The derivative is:
\begin{align*}
f’(x) &= \left( (x^2+1) \sqrt{x^4+1} \right)’
= 2x\sqrt{x^4+1} + \frac{2x^3(x^2+1)}{\sqrt{x^4+1}}.
\end{align*}
\end{exercise}
\vspace*{1cm}
The solution of the exercise \ref{exercise:Ex4} is in the page \pageref{solution:Ex3}
\vspace*{1cm}




\tcbstoprecording
\newpage
\section{Solutions of the exercices}
\fetchsolutions
\end{document}

[Tex/LaTex] Math book: Exercise – Solution environment

I suggest to you to read exercise package documentation.

There are the package options:

lastexercise if no references is given for an answer, then the answer is supposed to refer to the last exercise (see section 2.1).

and

answerdelayed save the answers instead of typeset them. The answers can be included later in the document with the command \shipoutAnswer.

and the Exercise environment option:

difficulty The difficulty of the exercise (a number). It will be available later with the counter \ExerciseDifficulty.

You can start with this simple example and then make your customization. As you can see, if you write your Answer environment immediately after the corresponding Exercise, the answer is automatically joined to the previous exercise.

\documentclass[11pt]{book}
\usepackage[lastexercise,answerdelayed]{exercise}

\usepackage{lipsum}

\begin{document}

    \chapter{Test 1}

    \lipsum[1]

    \section{Problems}

    \begin{Exercise}[difficulty={1}]
        My easy exercise.
    \end{Exercise}
    \begin{Answer}
        My answer to my easy exercise.
    \end{Answer}

    \chapter{Test 2}

    \lipsum[1]

    \section{Problems}
    \begin{Exercise}[difficulty={2}]
        My difficult exercise.
    \end{Exercise}
    \begin{Answer}
        My answer to my difficult exercise.
    \end{Answer}

    \chapter{Solutions}

    \shipoutAnswer

\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Solutions of exercises produced with tcolorbox

[Tex/LaTex] Math book: Exercise – Solution environment

Related Question