%Taylor
\def\whichexam{Final}
\def\examdate{May 7, 2002}

\newcount\n\newcount\na\newcount\nb

%% List of problem number at top of each page with last entry = number of problems +1
%% Begin with the problem number at the top of the second page since problem 1 is
%% always at the top of the first page of problems.
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\\{21}\\{23}\\{25}\\{26}}

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%%\gdef\boldlines{\\{1}\\{7}\\{12}\\{18}\\{34}}


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

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\centerline{\bf Math 126, \whichexam}
\centerline{\examdate}
\vskip10pt

\itemitem{$\bullet$}
The Honor Code is in effect for this examination.
All work is to be your own.
\itemitem{$\bullet$} Be sure that you have all \numberpages\ pages of the test. 
\itemitem{$\bullet$} No calculators are to be used.
\itemitem{$\bullet$} The exam lasts for two hours.
\itemitem{$\bullet$} You are to hand in just the front page.
\vskip5pt
\centerline{Good Luck!}
\vskip10pt
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\long\def\Exam{%
\coversheet{ }%

\vfill\eject\headline{\hfil}
\five
\newp 1. 1@
Let $f(x)= e^x-1$ and
let $f^{-1}$ denote the inverse function.
Then $(f^{-1})^\prime(e^2-1)=\ ?$
\ans
$e^{-2}$�
$e^{-1}$�
$\frac1{e^2-1}$�
$e^2$�
$e$�
\endp 1.

\five
\newp 2. 1@
$\Int_{\tan(1)}^{\tan(2)}\frac{\arctan^5(x)}{1+x^2}\>dx =\ ?$
\ans
$\frac{21}{2}$�
$\frac{\tan(3)}{6}$�
$\frac{32}{3}$�
$\frac{11}{2}$�
Diverges�
\endp 2.

\two
\newp 3. 1@
$\Int_{1}^{2}\arctan(x)\>dx =\ ?$
\ans
$2\arctan(2)-\arctan(1)+\frac{\ln 2-\ln 5}{2}$�
$2\arctan(1)-\arctan(2)+\frac{\ln 5-\ln2}{2}$�
$\frac{\pi}{4}-\ln(5/2)$�
$\ln\bigl(\arctan(2)\bigr)-\arctan\bigl(\ln(5)\bigr)$�
$\sqrt{\frac\pi2\, }\cdot\frac32$�
\endp 3.
\five

\newp 4. 2@
$\Lim_{x\to 0^+}\ (\cos x)^{^\smfrac 1{x^2}}=$
\medskip
\ans
$e^{^{-\smfrac {1}{2}}}$�
$1$�
$e$�
$\infty$�
Does not exist�
\endp 4.

\three
\newp 5.  1@
The moment about the $y${--}axis of 
the plane region above the parabola $y=x^2$ and below the line
$y=9$ is given by the integral
\ans
$\Int_{-3}^3x(9-x^2)\ dx$�
$\Int_{-3}^{3}\frac{(9-x^2)^2}{2}\>dx$�
$4\Int_{0}^{3}x(9-x^2)\ dx$�
$\Int_0^9 \frac{y^2}{2}\>dy$�
$\Int_0^9 y(9-y)\>dy$�
\endp 5.

\three
\newp 6. 1@
The solution to the initial value problem
$$x\frac{dy}{dx}+2y = e^{x^2}\qquad y(1) = 0$$
is
\def\xs{-{{\vrule width 0pt height 0pt depth 2pt\,\scriptstyle x^{\hbox to 0pt{$\scriptscriptstyle2$\hss}}}%
\over{\vrule width 0pt height 6pt depth 0pt\scriptstyle2}}}
\ans
$y=\frac{e^{x^2}-e}{2x^2}$�
$y=\frac{e^{x}-e}{2x}$�
$y=\frac{e^{x^2}-e}{2x}$�
$y=\frac{e^{x}-e}{2x^2}$�
$y=xe^x-e$�
\endp 6.

\two
\newp 7. 1@
The solution to the initial value problem
$$y^\prime=x\cos y\hskip 1in y(2)=0$$
satisfies the implicit equation
\ans
$\ln\vert\sec(y)+\tan(y)\vert=\frac{x^2}{2}-2$�
$\frac{ey}{2}=e^{\cos x}-e^{\cos 2}$�
$\cos y=x-1$�
$\cos(y)=x+\cos(2)$�
$e^{2y+1}=\arcsin(x-2)+e$�
\endp 7.


\one
\newp 8.  2@
Which integral below gives the arclength of the curve
$x=1-2\cos t$, $y=\sin^2\bigl(t/2\bigr)$, $0\leq t\leq 2\pi$?
\noindent
\ans
$\Int_0^{2\pi}\sqrt{4\sin^2 t+\sin^2(t/2)\cos^2(t/2)\ }\ dt$�
$\Int_0^{2\pi}\sqrt{1-2\cos(t)+\cos^2(t)+\sin^4(t/2)\ }\ dt$�
$\Int_0^{2\pi}\sqrt{1-2\cos(t)+\cos^2(t)+\sin^2(t/2)\cos^2(t/2)\ }\ dt$�
$\Int_0^{2\pi}\sqrt{4\sin^2 t+\sin^4(t/2)\ }\ dt$�
$\Int_0^{2\pi}\sqrt{\sin^2 (t/2)-2\sin^2(t/2)\cos(t)\ }\ dt$�
\endp 8.

\five
\newp 9. 2@
$\Int^{2}_0\frac{dx}{1-x\>}=$ 
\ans
Diverges�
$0$�
$\ln2$�
$\frac\pi{\sqrt{2\ }}$�
$\frac{\pi}{6}$�
\endp 9.

\five
\newp 10. 2@
$\Int_0^{\pi/2}x\cos(x)dx=$ 
\ans
$\frac\pi2 - 1$�
$1-\frac\pi2$�
$0$�
$\frac{\sqrt{3\,}}{2}-\frac{\sqrt{2\,}}{2}$�
Diverges�
\endp 10.

\five
\newp 11. 2@
You begin an experiment at 9am with a sample of 1000 bacteria.
An hour later your population has doubled. 
Assuming exponential growth, what is the population at noon?
\ans
$8000$�
$1000e^3$�
$4000$�
$1000e^{-3}$�
$32000$�
\endp 11.

\three
\newp 12. 2@
If you expand $\frac{2x+1}{x^3+x}$ as a partial fraction, which
expression below would you get?
\ans
$\frac{1}{x}+\frac{-x+2}{x^2+1}$�
$\frac{2}{x}+\frac{1}{x^2+1}$�
$\frac{-1}{x}+\frac{x}{x^2+1}$�
$\frac{-1}{x^2}+\frac{1}{x+1}$�
$\frac{-2}{x}+\frac{1}{x^2+1}$�
\endp 12.


\four
\newp 13. 4@
The point $\brace{bigg}{2,\frac{11\pi}3}$ in polar coordinates corresponds to 
which point below in Cartesian coordinates?
\ans
$(1,-\sqrt 3\>)$�
$(-\sqrt 3, 1)$�
$(-1,\sqrt 3\>)$�
$(\sqrt 3, -1)$�
Since $\frac{11\pi}3>2\pi$, there is no such point\hss�
\endp 13.

\three
\newp 14.  3@
Which integral below gives the surface area of the surface
of revolution obtained by rotating the polar curve
$r=\cos\theta$, $0\leq\theta\leq\pi$ about the $y${--}axis?
\par\noindent{\bf Hint:} A polar curve is also a parameterized curve.
\ans
$2\pi\Int_0^{\pi} \cos^2\theta\> d\theta$�
$2\pi\Int_0^{\pi}\sin^2\theta\ \cos\theta\> d\theta$�
$2\pi\Int_0^{\pi}\sin^2\theta\> d\theta$�
$2\pi\Int_0^{\pi}\sin\theta\ \cos\theta\> d\theta$�
$\frac{\pi}2\Int_0^{\pi}\sin^2\theta\ \cos\theta\> d\theta$�
\endp 14.

\five
\newp 15. 4@
Find the area inside the cardioid $r=1+\sin\theta$.
\HideDisplacementBoxes
\par\pictboxdim=120pt
\pictboxstrut\ForceWidth{\pictboxdim}\vskip-100pt\BoxedEPSF{cardioid.eps}
\par
\ans
$\frac{3\pi}{2}$�
$2\pi$�
$\frac{3}{2}$�
$2$�
$3\pi+\ln 4$�
\endp 15.


\four
\newp 16. 3@
Which point below is a focus for the hyperbola
$\frac{x^2}{9}-\frac{y^2}{16}=1$?
\ans
$(-5,0)$�
$(0,5)$�
$(3,0)$�
$(0,4)$�
$(4,0)$�
\endp 16.

\five
\newp 17.  3@
Find 
$\Sum n1 \frac{2^{2n}}{3\cdot5^{n-1}}$ 
\ans
$\frac{20}{3}$�
$\frac{4}{15}$�
$\frac{5}{4}$�
$\frac{5}{3}$�
$\frac{5}{12}$�
\endp 17.

\one
\newp 18. 3@
Which of the following series converge conditionally?

\noindent\null\hfil(1) $\Sum n0 \frac{\sin(2n)}{n!}$\hfil(2) $\Sum n2 \frac{(-1)^n\>n}{(\ln n)^2}$\hfil
(3) $\Sum n1(-1)^n\frac{n^2}{n^{3}+1}$\hfil\null
\ans
(3) converges conditionally, (1) and (2) do not converge conditionally�
(1) and (2) converge conditionally, (3) does not converge conditionally�
(1) and (3) converge conditionally, (2) does not converge conditionally�
(2) and (3) converge conditionally, (1) does not converge conditionally�
(2) converges conditionally, (1) and (3) do not converge conditionally�
\endp 18.

\five
\newp 19.  3@
Find the interval of convergence for 
$$\Sum n1 \frac{x^n}{\root 3\of{n^2+2}}$$
\par\noindent{\bf Remark:} 
$\frac{1}{\root3\of{n^2+2}}$ is decreasing for $n>0$.
\ans
$-1\leq x< 1$�
$-1<x < 1$�
$-1< x \leq  1$�
$-1\leq x\leq 1$�
$-1\leq x\leq 0$�
\endp 19.

\two
\newp 20. 4@
If $f(x)=\Sum n0 \frac{(-1)^n\,(x-2)^{n}}{(2n+1)!}$,
find the power series centered at $2$ for the function
$\Int_2^x f(t)\>dt$.
\ans
$\Sum n0 \frac{(-1)^n\,(x-2)^{n+1}}{(n+1)(2n+1)!}$�
$\Sum n0 \frac{(-1)^n\,(x-2)^{n+1}}{(n^2)(2n+1)!}$�
$\Sum n0 \frac{(-1)^n\,(x-2)^{2n+1}}{(n+1)(2n)!}$�
$\Sum n0 \frac{(-1)^n\,(x-2)^{n+1}}{(n+1)!}$�
The given function can not be represented by a power
series centered at $2$.\hss�
\endp 20.

\three
\newp 21.  4@
Which series below is the MacLaurin series (Taylor series centered at $0$)
 for $\frac{x^2}{1+x}$?
\ans
$\Sum n0 (-1)^n\ x^{n+2}$�
$\Sum n0  x^{2n+2}$�
$\Sum n0 \frac{x^{n+2}}{n+2}$�
$\Sum n2 \frac{(-1)^nx^{2n-2}}{n!}$�
$\Sum n0 (-1)^nx^{2n}$�
\endp 21.

\three
\newp 22.  4@
Find the degree 3  MacLaurin polynomial (Taylor polynomial centered at $0$)
for the function
$$\frac{e^x}{1-x^2}$$
\ans
$1+x+\frac{3x^2}{2}+\frac{7x^3}{6}$�
$1+x-\frac{x^3}{6}$�
$1-\frac{x^2}{2}+\frac{x^3}{5}$�
$1+x+\frac{x^2}{6}+0\>x^3$�
$1+x-\frac{5x^3}{3}$�
\endp 22.

\five
\newp 23. 4@
$\Lim_{x\to 0}\ \frac{\sin(x^3)-x^3}{x^9}=$
\vskip10pt\noindent{\bf Hint:}
Without MacLaurin series this may be a long problem.
\ans
$-\frac16$�
$\infty$�
$0$�
$\frac{9}{7}$�
$\frac{7}{9}$�
\endp 23.


\four
\newp 24.  3@
Which series below is a power series for $\cos(\sqrt{ x })\ $?
\ans
$\Sum n0 \frac{(-1)^nx^n}{(2n)!}$�
$\Sum n0 \frac{(-1)^n\sqrt{x}^{\,n}}{(2n)!}$�
$\Sum n0 \frac{(-1)^nx^{2n-\smfrac{1}{2}}}{(2n)!}$�
$\Sum n0 \frac{(-1)^nx^n}{(2n+1)!}$�
$\Sum n0 \frac{(-1)^nx^n}{n^2+1}$�
\endp 24.


\five
\newp 25. 4@
What is the radius of convergence for the binomial series
for the function $\frac{1}{\root 5\of{3-x}}\ $?
\ans
$3$�
$\frac{2}{3}$�
$1$�
$\root 5\of 3$�
$\frac{3}{5}$�
\endp 25.


}

\Exam{}
\vtop to 0pt{\ifprofver
\vskip20pt\noindent\vrule width\hsize depth0pt height.1pt

Exam 1 = \number\eA\quad
Exam 2 = \number\eB\quad
Exam 3 = \number\eC\quad
Exam 4 = \number\eD
\fi\vss}
\vfill\eject
\marked{}
\end
