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I recently asked a question that got a great answer, but the answer's effectiveness is fairly limited to the extent that it seems like I'd have to "eyeball" the proper vertical alignment each time I wanted to implement the solution. For example, consider this picture:

enter image description here

The code:

\documentclass{exam}
\usepackage{amsmath}
\usepackage{tabularx}

\begin{document}

\begin{figure}[ht]
\begin{tabularx}{\linewidth}{@{}l>{\hsize=0.6\hsize}X |
                                 >{\hsize=0.3\hsize}X 
                             @{} }
    \hline 
 (a) &   $\begin{aligned}[t]
    f(x) & = 6\biggl[1-\frac{x}{3} + \frac{x^2}{2!3^2}-\frac{x^3}{3!3^3}+
            \cdots+\frac{(-1)^n x^n}{n!3^n}+\cdots\biggl]   \\
         &= 6-2x+\frac{x^2}{3}-\frac{x^3}{27}+\cdots+\frac{6(-1)^nx^n}{n!3^n}+\cdots
        \end{aligned}$ 

    &   \raisebox{-0.8\baselineskip}{$3:\begin{cases}
        1 : & \text{two of}\ 6, -2x, \frac{x^2}{3}, -\frac{x^3}{27}\\
        1 : & \text{remaining terms}\\
        1 : & \text{general term}
        \end{cases}$}
        \vspace{1ex}

        $\langle -1\rangle$ missing factor of 6\\[2em]
\hline
(b) & $g(0)=0$ and $g'(x)=f(x)$, so\newline

$\begin{aligned}[t]
g(x)
&= 6\biggl[x-\frac{x^2}{6}+\frac{x^3}{3!3^2}-\frac{x^4}{4!3^3}+
\cdots+\frac{(-1)^n x^{n+1}}{(n+1)!3^n}+\cdots\biggl]\\ 
&= 6x-x^2+\frac{x^3}{9}-\frac{x^4}{4(27)}+\cdots+
\frac{6(-1)^nx^{n+1}}{(n+1)!3^n}+\cdots
        \end{aligned}$ 

    &   \raisebox{-1.4\baselineskip}{$3:\begin{cases}
        1 : & \text{two terms}\\
        1 : & \text{remaining terms}\\
        1 : & \text{general term}
        \end{cases}$}
        \vspace{1ex}

        $\langle -1\rangle$ missing factor of 6\\
\hline
(c) & $f'(x)=-2e^{-x/3}$, so $h(x)=-2ke^{-ax/3}$\newline
$h(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+\cdots=e^x$\newline
$-2ke^{-ax/3}=e^x$\newline
$\frac{-a}{3}=1$ and $-2k=1$\newline
$a=-3$ and $k=-\frac{1}{2}$

\begin{center}OR\end{center}

$f'(x)=-2+\frac{2}{3}x+\cdots$, so\newline
$h(x)=kf'(ax)=-2k+\frac{2}{3}akx+\cdots$\newline
$h(x)=1+x+\cdots$\newline
$-2k=1$ and $\frac{2}{3}ak=1$\newline
$k=-\frac{1}{2}$ and $a=-3$

    &   \raisebox{-2.3\baselineskip}{$3:\begin{cases}
        1 : & \text{two terms}\\
        1 : & \parbox[t]{.3\columnwidth}{recognizes 
$h(x)=e^x$,\\\null\quad or\\equations 2 series for $h(x)$}\\
        1 : & \text{general term}
        \end{cases}$}

\end{tabularx}
\end{figure}

\end{document}

My question: Is there a way to have LaTeX automatically calculate the "proper value" C when using \raisebox{-C\baselineskip} to make sure vertical alignment is as it should be? I do not want to have add an hrule each time and adjust the cases based on an eyeballing approach. Any suggestions on how to approach this in a principled manner?

1
  • Instead of raising one of the sides you could box both and align them to the top (or better top - 1\baselineskip). This can be done using \begin{adjustbox}{varwidth=<width>,valign=T} (adjustbox package). Dec 20 '17 at 7:53
2

enter image description here

\documentclass{exam}
\usepackage{amsmath}
\usepackage{tabularx}

\begin{document}

\begin{figure}[ht]
\begin{tabularx}{\linewidth}{@{}
                                 >{\hsize=0.05\hsize\mbox{}\endgraf}X
                                 >{\hsize=0.60\hsize\mbox{}\endgraf\vspace{-.5\baselineskip}}X |
                                 >{\hsize=0.35\hsize\mbox{}\endgraf\vspace{-.5\baselineskip}}X 
                             @{} }
    \hline 
 (a) &   $\begin{aligned}[t]
    f(x) & = 6\biggl[1-\frac{x}{3} + \frac{x^2}{2!3^2}-\frac{x^3}{3!3^3}+
            \cdots+\frac{(-1)^n x^n}{n!3^n}+\cdots\biggl]   \\
         &= 6-2x+\frac{x^2}{3}-\frac{x^3}{27}+\cdots+\frac{6(-1)^nx^n}{n!3^n}+\cdots
        \end{aligned}$ 

    &   $3:\begin{cases}
        1 : & \text{two of}\ 6, -2x, \frac{x^2}{3}, -\frac{x^3}{27}\\
        1 : & \text{remaining terms}\\
        1 : & \text{general term}
        \end{cases}$

        \vspace{1ex}

        $\langle -1\rangle$ missing factor of 6\\[2em]
\hline
(b) & $g(0)=0$ and $g'(x)=f(x)$, so\newline

$\begin{aligned}[t]
g(x)
&= 6\biggl[x-\frac{x^2}{6}+\frac{x^3}{3!3^2}-\frac{x^4}{4!3^3}+
\cdots+\frac{(-1)^n x^{n+1}}{(n+1)!3^n}+\cdots\biggl]\\ 
&= 6x-x^2+\frac{x^3}{9}-\frac{x^4}{4(27)}+\cdots+
\frac{6(-1)^nx^{n+1}}{(n+1)!3^n}+\cdots
        \end{aligned}$ 

    &   $3:\begin{cases}
        1 : & \text{two terms}\\
        1 : & \text{remaining terms}\\
        1 : & \text{general term}
        \end{cases}$

        \vspace{1ex}

        $\langle -1\rangle$ missing factor of 6\\
\hline
(c) & $f'(x)=-2e^{-x/3}$, so $h(x)=-2ke^{-ax/3}$\newline
$h(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+\cdots=e^x$\newline
$-2ke^{-ax/3}=e^x$\newline
$\frac{-a}{3}=1$ and $-2k=1$\newline
$a=-3$ and $k=-\frac{1}{2}$

\begin{center}OR\end{center}

$f'(x)=-2+\frac{2}{3}x+\cdots$, so\newline
$h(x)=kf'(ax)=-2k+\frac{2}{3}akx+\cdots$\newline
$h(x)=1+x+\cdots$\newline
$-2k=1$ and $\frac{2}{3}ak=1$\newline
$k=-\frac{1}{2}$ and $a=-3$

    &  $3:\begin{cases}
        1 : & \text{two terms}\\
        1 : & \parbox[t]{.3\columnwidth}{recognizes 
$h(x)=e^x$,\\\null\quad or\\equations 2 series for $h(x)$}\\
        1 : & \text{general term}
        \end{cases}$

\end{tabularx}
\end{figure}

\end{document}

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