\gdef\localbox#1{\epsfscale=800\vtop to1.5in{\null\vskip
-1in\epsfbox{Ans#1.eps}\vss}}
\long\def\hide #1\endhide{}

\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\def\whichexam{Exam I}
\def\examdate{February 5, 2002}
\def\numberpages{10}
\def\numberofproblems{15}
\def\multiplechoicemax{6}
\def\partialcreditmax{10}
%%
\newif\ifprelim
\prelimfalse
\prelimtrue
%%
\gdef\f(#1){\smfrac{1}{\cos(#1)}}
\gdef\f(#1){\sec(#1)}
\gdef\smmfrac#1#2{{{#1}\over{#2}}}
%%
%%%%%  The macro \correctanswers
%%%%%  sets the number of partial credit problems automatically.
%%%%%  We assume that the first answer listed after the problem is
%%%%%  the correct one.
%%%%%  The permutation indicates in which slots to put the answers
%%%%%  (so the first number in the permutation locates the correct answer!) 
\def\correctanswers{%
\ll 1 13425
\ll 2 15423
\ll 3 32451
\ll 4 53142
\ll 5 35214
\ll 6 21345
\ll 7 34215
\ll 8 45132
\ll 9 42513
\ll 10 25143
\ll 11 52341
\ll 12 24153
}

\def\rhb#1{\noindent\hbox to\hsize{\hfil#1}}

\def\longans#1{\vtop{\noindent\hsize=6.25 true in #1}}
\def\arctan{{\rm arctan\>}}

\long\def\Ans#1�#2�#3�#4�#5�#6�#7\endp{
\ifcase#1\relax\or\one\or\two\or\three\or\four\or\five\fi
\AN{#2}{#3}{#4}{#5}{#6}\Pra}

\def\dxd{\displaystyle}

\input epsf.def
\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\newdimen\picV
\newdimen\picH
\def\px#1{\setbox0=\hbox{\noindent\epsfxsize=\picH \epsfbox{:problem09:#1}}%
\picV=\ht0 %
\vbox to 0pt{\vskip-20pt\box0 \vss}\vrule width 0pt height 0pt depth\picV%
}

\newcount\cnumberofproblems
\cnumberofproblems=\numberofproblems
\global\advance\cnumberofproblems by 1 

\font\ld=cmbx12
\def\endp{\relax}
\newcount\totalproblems\totalproblems=\numberofproblems
\newtoks\ta\newtoks\tb
\long\def\rightappenditem#1\to#2{%
\ta={\\{#1}}\tb=\expandafter{#2}%
\xdef#2{\the\tb\the\ta}}
\def\leftremove#1\to#2{\expandafter\lopoff#1\lopoff#1#2}
\long\gdef\lopoff\\#1#2\lopoff#3#4{\gdef#4{#1}\gdef#3{#2}}

\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}
\def\loop#1\repeat{\def\body{#1}\iterate}
\def\iterate{\body\let\next=\iterate\else\let\next=\relax\fi\next}
\let\Correctanswers=\correctanswers
\def\correctanswers{\global\numbermultiplechoice=0 \gdef\list{}\Correctanswers}
\font\small=cmr7
\def\newpage{\vfill\eject}
\newdimen\pspacewidth
\newcount\credit
\newcount\pspace\pspace=0
\newcount\pc\pc=1
\newcount\breakpc\breakpc=1
\hfuzz=5pt
\let\xxsum=\sum
\edef\sum{\displaystyle\xxsum}

\newdimen\picH\newdimen\picV
\def\picture #1 by #2 (#3 scaled #4){
\picV=#2 \divide\picV by1000 \multiply\picV by#4 
\picH=#1 \divide\picH by1000 \multiply\picH by#4
\vtop to \picV{
      \hrule width \picH height 0pt depth 0pt
      \vfill
      \special{picture #3 scaled #4}}}

\newcount\mc
\def\newchoice#1{\ifnum\breakpc<3 \newpage\else\ifprelim\else\newpage\fi\fi\null%
\vskip10pt\centerline{{\bf#1}}\mc=0 %
\ifnum\breakpc<3\credit=\multiplechoicemax \else\credit=\partialcreditmax \fi}
\newif\ifmultichoice
\def\problemspacing{\ifprelim\vskip10pt\headline{\hfill}\else%
\ifmultichoice%
\ifnum\mc>0 \ifodd\breakpc\relax\newpage\null\vskip10pt\else\vfill\fi\fi%
\else%
\ifnum\mc>0 \newpage\null\vskip10pt\fi\fi%
\mc=1 \fi}

\long\def\Newp#1.#2�#3a)#4\endp{
\newp#1.#2
\Xan#3
a)#4\endp}


\long\def\newp#1.#2\endp{\problemspacing%
\noindent\vbox{\noindent{\bf\number\pc.}(\number\credit\ pts.)
\global\advance\pc by1 %
\global\advance\breakpc by1 %
#2\endp}\vfil}


\newbox\Aa\newbox\Ab\newbox\Ac
\newbox\Ad\newbox\Ae
\newbox\aa\newbox\ab\newbox\ac
\newbox\ad\newbox\ae

\newdimen\Aspace
\newcount\Ax
\newcount\tmpPC
\def\rememberPC{\global\tmpPC=\pc\global\advance\tmpPC by-1 }

\hfuzz=20pt

\def\five{\global\Ax=5 \global\Aspace=.2\hsize\relax\global\advance\Aspace by-20pt}
\def\four{\global\Ax=4 \global\Aspace=.25\hsize\relax\global\advance\Aspace by-20pt}
\def\three{\global\Ax=3 \global\Aspace=.3\hsize\relax\global\advance\Aspace by-20pt}
\def\two{\global\Ax=2 \global\Aspace=.5\hsize\relax\global\advance\Aspace by-20pt}
\def\one{\global\Ax=1 \global\Aspace=\hsize\relax\global\advance\Aspace by-20pt}

\long\def\AN#1#2#3#4#5{%
\setbox\Aa=\hbox to\Aspace{#1\hfil}%
\setbox\Ab=\hbox to\Aspace{#2\hfil}\setbox\Ac=\hbox to\Aspace{#3\hfil}%
\setbox\Ad=\hbox to\Aspace{#4\hfil}\setbox\Ae=\hbox to\Aspace{#5\hfil}}

\newdimen\Ls\Ls=10pt

\def\prafive{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac(d)\quad\box\Ad%
(e)\quad\box\Ae}

\def\prafour{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac(d)\quad\box\Ad\vskip\Ls\noindent%
(e)\quad\box\Ae}

\def\prathree{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac\vskip\Ls\noindent(d)\quad\box\Ad%
(e)\quad\box\Ae}

\def\pratwo{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
\vskip\Ls\noindent(c)\quad\box\Ac(d)\quad\box\Ad%
\vskip\Ls\noindent(e)\quad\box\Ae}

\def\praone{\vskip\Ls\noindent(a)\quad\box\Aa\vskip\Ls\noindent(b)\quad\box\Ab%
\vskip\Ls\noindent(c)\quad\box\Ac\vskip\Ls\noindent(d)\quad\box\Ad%
\vskip\Ls\noindent(e)\quad\box\Ae}

\def\resetboxes{\setbox\Aa=\box\aa\setbox\Ab=\box\ab\setbox\Ac=\box\ac%
\setbox\Ad=\box\ad\setbox\Ae=\box\ae}
\def\switchbox#1#2{\ifcase#1 \relax\or\setbox\aa=\box#2\or\setbox\ab=\box#2\or%
\setbox\ac=\box#2\or\setbox\ad=\box#2\or\setbox\ae=\box#2 \fi}

\newcount\pcc
\def\mooseAntlers#1#2#3#4#5@{\gdef\pA{#1}\gdef\pB{#2}\gdef\pC{#3}\gdef\pD{#4}\gdef\pE{#5}}
\def\Pra{\leftremove\list\to\moose%
\expandafter\mooseAntlers\moose@%
\switchbox{\pA}{\Aa}\switchbox{\pB}{\Ab}%
\switchbox{\pC}{\Ac}\switchbox{\pD}{\Ad}\switchbox{\pE}{\Ae}%
\resetboxes%
\ifcase\Ax \or\praone\or\pratwo\or\prathree\or\prafour\or\prafive\fi%
\pcc=\pc\advance\pcc by -1 %
\immediate\write16{Problem \number\pcc - Correct Answer \number\pA}%
}


\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\def\Lim{\displaystyle\lim}
\def\Int{\displaystyle\int}
\def\arcsec{\rm arcsec}

\def\frac#1#2{{\displaystyle{{\textstyle#1}\over{\textstyle#2}}}}

\long\def\boxit#1#2#3{\vbox{\hrule\hbox{\vrule\kern#2
\vbox{\hsize=#3\kern#2#1\kern#2}\kern#2\vrule}\hrule}}
\newcount\pc
\def\pr{\vskip10pt\noindent\number\pc.\global\advance\pc by 1}
\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}

\def\rhs#1#2{\par\noindent\hbox to\hsize{\hfill\hbox to#1{#2\hfill}}}
\def\ul#1{\vrule width#1 height -3pt depth3.1pt}
\def\UL#1#2{\hbox to0pt{\qquad #2\hss}\ul{#1}}
\newif\ifnotmarkanswers\notmarkanswerstrue
\newcount\ansmark\newcount\numbermultiplechoice\newcount\numberpartialcredit
\newdimen\AA\AA=20pt
\def\ll#1 #2#3#4#5#6{\tabalign {#1.}&
\global\ansmark=#2 \relax%
\ifnotmarkanswers \global\ansmark=0 \fi%
\ifnum\ansmark=1 \relax($\bullet$)\else(a)\fi&
\ifnum\ansmark=2 \relax($\bullet$)\else(b)\fi&
\ifnum\ansmark=3 \relax($\bullet$)\else(c)\fi&
\ifnum\ansmark=4 \relax($\bullet$)\else(d)\fi&
\ifnum\ansmark=5 \relax($\bullet$)\else(e)\fi&
\hskip\AA\cr\global\advance\numbermultiplechoice by1 
\rightappenditem{#2#3#4#5#6}\to\list%
}

\def\marked{\pageno=1 %
\global\notmarkanswersfalse
\headline{\vbox{\rhs{3in}{Name:\ \UL{3in}{ANSWERS}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.3in}{ANSWERS}\hss}%
}}
\answersheet}

\long\def\xan#1a.#2b.#3c.#4d.#5e.#6\endp{%
\Ans#1�
#2�
#3�
#4�
#5�
#6�
\endp}
\let\\=\relax
\long\def\Xan#1a)#2b)#3c)#4d)#5e)#6\endp{%
\Ans#1�
#2�
#3�
#4�
#5�
#6�
\endp}

%%%%%%%
\font\sfm=cmr8

\def\eat#1{}
\def\exam#1{\ifprelim\Exam{Bullwinkle}\else\expandafter\EExam#1,@\fi}
\def\EExam#1,#2@{\Exam{#1}\def\xx{#2}%
\ifx\empty\xx\let\next=\eat\else\let\next=\EExam\fi\next#2@}

\def\+{\tabalign}
\long\def\Exam#1{%
\pc=1 \pageno=1 \pspace=0 %
\headline{\vbox{\rhs{3in}{Name:\ \ul{3in}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.3in}{#1}\hss}%
\rhs{3in}{\hskip 1cm\sfm If Cao please indicate 10:40 or 11:45}%
}}
\long\gdef\answersheet{
\null\vskip10pt
\centerline{\whichexam}
\centerline{\examdate}
\vskip10pt

\itemitem{$\bullet$}
The Honor Code is in effect for this examination.
All work is to be your own.
\itemitem{$\bullet$} No calculators.
\itemitem{$\bullet$} The exam lasts for one hour.
\itemitem{$\bullet$} Be sure that your name is on every page in case
pages become detached. 
\itemitem{$\bullet$} Be sure that you have all \numberpages\ pages of the test.
\vskip5pt
\centerline{Good Luck!}
\vskip4pt
\def\HH{\hskip.75in}
\settabs\+9.\qquad &\HH&\HH&\HH&\HH&\HH&\cr
\hskip.5 in\boxit{\hbox to 3.8in{\vbox{\obeylines\baselineskip20pt
\noindent
PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!
\correctanswers
}\hss}}{10pt}{5in}
%%\vskip10pt

\global\numberpartialcredit=\numbermultiplechoice
\global\advance\numberpartialcredit by 1
\settabs\+Total multiple choice: &\hskip 2in\cr
\hfill{{\vbox to0pt{\boxit{\vbox{\obeylines\hsize=3in\baselineskip26pt
DO NOT WRITE IN THIS BOX!
\tabalign Total multiple choice: &\quad\ul{1in}\cr
\loop\ifnum\numberpartialcredit< \cnumberofproblems
\tabalign\hfill\number\numberpartialcredit.&\quad\ul{1in}\cr\global\advance\numberpartialcredit by 1
\repeat

\tabalign\hfil {\bf Total:}&\quad\ul{1in}\cr
}}{10pt}{1in}\vss}}}}
\answersheet
\multichoicetrue
\newchoice{Multiple Choice}

\def\kg{{\rm kg}}
\def\log{{\rm log}}
\credit=5
\Newp 1.
Solve the following equation for $x$.
$$\ln(x+6)+\ln(x-3)=\ln5+\ln2$$
�3
a)$x=4$\\
b)$x=5$, $x=2$\\
c)$x=5$, $x=-7$\\
d)$x=4$, $x=-5$\\
e)$x=2$\\
\endp

\Newp 2.
 The derivative of $x^{\ln x}$ is equal to\\
�3
a)$\frac{2\ln x }{x}\ x^{\ln x}$\\
b)$0$\\
c)$\frac{x^{\ln x }}{x\ln x }$\\
d)$(\ln x ) (x^{\ln x-1})$\\
e)$x^{\ln x }\ln(\ln x )$\\
\endp

\Newp 3.
Evaluate $\displaystyle\int_0^{\pi/4}\tan x\  dx$\\
�5
a)$\ln(\sqrt{2})$\\

b)$\ln(\pi/4)$\\

c)$\ln\bigl(\sqrt{3}/2\bigr)$\\

d)$0$\\

e)$\ln(4)$\
\endp

\Newp 4.
 Evaluate $\displaystyle\int_{0}^{1}3^{x^2}x\ dx$\\
�5
a)$\frac{1}{\ln 3}$\\

b)$0$\\

c)$\frac{2}{\ln 3}$\\

d)$2$\\

e)$\frac{\ln 6}{\ln 3}$\\
\endp

\Newp 5.
Find $\displaystyle\lim_{x\rightarrow 0^+}\frac{\ln x}{x}$\\
�5
a)$-\infty$\\

b)$0$\\

c)$1$\\

d)$\infty$\\

e)$2$\\
\endp

\Newp 6.
Evaluate $\displaystyle\int \frac{e^x}{\sqrt{1-e^{2x}}}\>
dx$\\ 
�3
a)$\arcsin(e^x)+C$\\
b)$\arctan(e^x)+C$\\
c)$\arcsec(e^x)+C$\\
d)$\ln(\sqrt{1-e^{2x}})+C$\\
e)$\ln(1-e^{2x})+C$\\
\endp

\Newp 7.
Find $\displaystyle\lim_{x \rightarrow \infty}\frac{x^2}{e^x}$\\
�5
a)$0$\\
b)$e^2$\\
c)$\infty$\\

d)$\frac{2}{e}$\\

e)Does not exist\\
\endp

\Newp 8.
Evaluate $\displaystyle\int_{1}^{e}{\frac{\ln x }{x}}dx$\\
�5
a)$\frac{1}{2}$\\

b)$\frac{e}{2}$\\

c)$2$\\

d)$\sqrt{e}$\\

e)$\ln(e-1)$\\
\endp

\Newp 9.
The function $x^3-2x$ is increasing for $x\geq
1$. What is the value for the derivative of $f^{-1}$ for $x=4$?\\
�5
a)$\frac{1}{10}$\\

b)$\frac{1}{40}$\\

c)$\frac{1}{4}$\\

d)$\frac{-1}{2}$\\

e)$\frac{1}{54}$\\
\endp

\Newp 10.
Evaluate $\displaystyle\int_{1}^{e} x^2 \ln x\ dx$\\
�5
a)$\frac{1}{9}(2e^3+1)$\\

b)$\frac{e}{3}$\\

c)$\frac{1}{2}(e^3-1)$\\

d)$\frac{1}{6}(e^3+2)$\\

e)$0$\\
\endp

\Newp 11.
 Evaluate $\displaystyle\int_{-1}^{1} \arcsin x\  dx$\\
�5
a)$0$\\

b)$\frac{\pi}{2}$\\

c)$1-\frac{\pi}{4}$\\

d)$2 \pi$\\

e)$\pi$\\
\endp

\Newp 12.
Find the derivative of $(\arcsec\> x)^3$\\
�2
a)$\frac{3 (\arcsec\> x)^2}{x\sqrt{x^2-1}}$\\

b)$3 (\arcsec\> x )^2\bigl((\arcsec\> x)(\arctan\> x) \bigr)$\\

c)$\frac{-3 \tan\> x }{(sec\> x )^3}$\\

d)$\frac{3 (\arcsec\> x )^2}{\sqrt{1-x^2}}$\\

e)$\frac{3 (\arcsec\> x)^2}{1+x^2}$\\
\endp

\multichoicefalse
\def\vvbv{\vbox{\hbox to 6in{\hfill Partial Credit\hfill}%
\hbox to 6in{\rm%
\hfill You must show your work on the partial credit problems 
to receive credit!\hfill}%
\vskip10pt}}
\newchoice{\vvbv}
\credit=12
\newp 13.
Find the area under the region
bounded by the curves $y=2^x$ and $y=5^x$, between $x=0$ and
$x=1$.\\

\endp

\newp 14.
Evaluate the limit:\\

$$\displaystyle\lim_{x \rightarrow 0^+}(\sin x )^{\tan x}$$\\
\endp

\credit=16
\newp 15.
Use logarithmic differentiation to
find the derivative of 
$$y= \frac{(x^2+1)^{2.2}(x+1)^{1.2}}{(x^3-1)^{0.3}}$$
\endp

\endp


\vfill\eject}


\exam{}\marked{}
\end
