[Tex/LaTex] How to make two columns in two columns

Question

How can I make two columns in two columns same as in the picture?

Answer 1

Maybe you can give the tasks package (used to be part of the exsheets bundle) a try:

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{multicol}
\usepackage{tasks}[2013/04/07]

% renew the {tasks} environment to use bold labels
% and use two columns as default settings:
\RenewTasks[counter-format= tsk.,label-format=\bfseries]{tasks}(2)

\begin{document}

\begin{multicols}{2}
Lorem ipsum dolor sit amet, consetetur sadipscing elitr
\begin{tasks}
 \task foo
 \task bar
 \task baz
 \task foobar
 \task foo
 \task bar
 \task baz
 \task foobar
\end{tasks}

Lorem ipsum dolor sit amet, consetetur sadipscing elitr
\begin{tasks}
 \task foo
 \task bar
 \task baz
 \task foobar
 \task foo
 \task bar
 \task baz
 \task foobar
\end{tasks}

Lorem ipsum dolor sit amet, consetetur sadipscing elitr
\begin{tasks}
 \task foo
 \task bar
 \task baz
 \task foobar
 \task foo
 \task bar
 \task baz
 \task foobar
\end{tasks}

Lorem ipsum dolor sit amet, consetetur sadipscing elitr
\begin{tasks}
 \task foo
 \task bar
 \task baz
 \task foobar
 \task foo
 \task bar
 \task baz
 \task foobar
\end{tasks}
\end{multicols}

\end{document}

---

Here's an example of how the appearance could be further customized (thanks to g.kov for doing the typing!). Edit: the updated version needs v0.10 (2014/07/20) for the \task! syntax.

The items that should span a complete line (items 31. and 32. in the picture below) can be achieved by one of the following methods:

• \task! – this will force the specific item to start in a new line using the whole line.
• \task* – this will force the specific item to use the remaining space of the line. In this case this means it uses the whole line if it happens to be an item in the first column.
• \task*(<num>) – this means the item will span <num> columns provided there are enough columns left in the current line. Otherwise it will use as much columns as it can.

\documentclass{article}
\usepackage[left=1.5cm,right=1.5cm,top=1.5cm,bottom=1.5cm]{geometry}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{multicol}
\usepackage{amsmath}

% use tasks v0.10 2014/07/20:
\usepackage{tasks}[2014/07/20]

% declare custom tasks instance that has no stretchable space between
% rows of items:
\DeclareInstance{tasks}{custom}{default}{
  counter-format  = tsk. ,
  label-format    = \bfseries ,
  label-width     = 1.5em ,
  label-offset    = .3333em ,
  after-item-skip = 0pt
}

% renew {tasks} environment to use the new instance and resume the
% item counting:
\RenewTasks[resume,style=custom,label-align=right]{tasks}(2)

\begin{document}

\begin{multicols}{2}
Derivative calculations\par
In Exercises \ref{tsk:st1}--\ref{tsk:end1}, given $y=f(u)$ and $u=g(x)$, find
$dy/dx=f^\prime(g(x))g^\prime(x)$
\begin{tasks}
 \task $y=6u-9,\ u=1/2 x^4$ \label{tsk:st1}
 \task $y=2u^3,\ u=8x-1$
 \task $y=\sin u,\ u=3x+1$
 \task $y=\cos u,\ u=-x/3$
 \task $y=\cos u,\ u=\sin x$
 \task $y=\sin u,\ u=x-\cos x$
 \task $y = \tan u,\ u=10x-5$
 \task $y=-\sec u,\ u=x^2+7x $ \label{tsk:end1}
\end{tasks}

In Exersises  \ref{tsk:st2}--\ref{tsk:end2}, write the function in the form
$y=f(u)$ and $u=g(x)$. Then find $dy/dx$ as a function of $x$.
\begin{tasks}
 \task $y=\left( 2x+1 \right)^5$ \label{tsk:st2} 
 \task $y=\left( 4-3x \right)^9$
 \task $y=\left( 1-\dfrac{x}7 \right)^{-7}$
 \task $y=\left( \dfrac{x}2 -1 \right)^{-10}$
 \task $y=\left( \dfrac{x^2}8 +x -\dfrac1{x} \right)^{4}$
 \task $y=\sqrt{2x^2-4x+6}$
 \task $y=\sec(\tan x)$
 \task $y=\cot\left( \pi -\dfrac1{x} \right)$
 \task $y=\sin^3 x$
 \task $y=5\cos^{-4} x$
 \task $y=e^{-5x}$
 \task $y=e^{2x/3}$
 \task $y=e^{5-7x}$
 \task $y=e^{4\sqrt{x}-x^2}$ \label{tsk:end2}
\end{tasks}

Find the derivatives of the functions in Exercises \ref{eq:st3}--\ref{eq:end3}:
\begin{tasks}
 \task $p=\sqrt{3-t}$ \label{eq:st3}
 \task $q=\sqrt[3]{2r-r^2}$
 \task $s=\dfrac{4}{3\pi}\sin{3t}+\dfrac{4}{5\pi}\cos{5t}$
 \task $s=\sin\dfrac{3\pi t}{2}+\cos\dfrac{3\pi t}{2}$
 \task $r=\left( \csc\theta +\cot\theta \right)^{-1}$
 \task $r=6\left( \sec\theta -\tan\theta \right)^{3/2}$
 \task $y=x^2\sin4x+x\cos^{-2}x$
 \task $y=\dfrac1{x}\sin^{-5}x-\dfrac{x}{3}\cos^{3}x$
 \task! $y=\dfrac1{21}(3x-2)^7+\left( 4-\dfrac1{2x^2}  \right)^{-1}$
 \task! $y=(5-2x)^{-3}+\dfrac1{8}\left( \dfrac2{x}+1  \right)^{4}$
 \task $y=(4x+3)^4(x+1)^{-3}$
 \task $y=(2x-5)^{-1}(x^2-5x)^{6}$
 \task $y=x e^{-x}+e^{3x}$
 \task $y=(1+2x)e^{-2x}$
 \task $y=(x^2-2x+2)e^{5x/2}$
 \task $y=(9x^2-6x+2)e^{x^3}$
 \task $h(x)=x\tan\left( 2\sqrt{x} \right)+7$
 \task $k(x)=x^2\sec\left( \dfrac1x \right)$ \label{eq:end3}
\end{tasks}
\end{multicols}

\end{document}