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\newif\ifsolution
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\begin{document}

\title{Mathematics Preparedness Self-Assessment}
\author{Jacobs University}
\date{Fall Semester 2015}
\maketitle

\emph{Instructions}: Set aside two hours of quality time where you can
work undisturbed.  Try to answer as many of these question as possible
``closed book'', i.e., without looking up any information in books or
online.  

Solutions will be posted on Sunday, September 6, after 18:00 together
with instructions on how to self-evaluate your results, on
\begin{quote}
\url{http://math.jacobs-university.de/oliver/}
\end{quote}
\bigskip
\ifsolution \begin{center}
\textsc{\large Scoring}
\end{center}

To self-grade your exam, please assign a score between 0
and 5 points to each problem as follows:
\begin{itemize}
\item[0] No substantial work done.
\item[1] Some correct idea, but little progress toward a full
solution.
\item[2] Some progress, but still far from a correct solution.
\item[3] Solution substantially given, but major error or omission.
\item[4] Solution almost there, but minor error or omission.
\item[5] Fully complete and correct solution.
\end{itemize}
(Note: these are essentially the criteria by which your homework and
quizzes are graded.)

You may then enter your scores \emph{anonymously} on
\begin{center}
\small\url{https://docs.google.com/spreadsheets/d/1bVRztVndxqTWtImmVyiS08rq6VcbGG52BAlFWeGA0GA/edit?usp=sharing}
\end{center}
Since everybody can edit this sheet, please treat this self-evaluation
with respect and enter only your data, leaving other's data
untouched.  In the first column, we appreciate if you indicate your
current enrollment (A=Applied Calculus, C=Calculus, E=Elements of
Analysis). \newpage
\fi

\begin{center}
\textsc{\large Problems}
\end{center}

\begin{enumerate}

\item Compute
\[
  \frac23 - \frac16 \,.
\]

\ifsolution
\item[\textit{Solution}:] 

\[
  \frac23 - \frac16 = \frac46-\frac16 = \frac36 = \frac12 \,.
\]
\fi

\item Solve the quadratic equation
\[
  2 x^2 + x - 1 = 0 \,.
\]

\ifsolution
\item[\textit{Solution}:] 
In this case, it is easy to spot the factorization
\[
  2 x^2 + x - 1 = (2x-1)(x+1) \,,
\]
so the two solutions are $x=\tfrac12, -1$.  You may, of course, also
use the standard solution formula for quadratic equations.
\fi

\item Solve
\[
  \frac{kT-1}T = S
\]
for $T$.
\ifsolution
\item[\textit{Solution}:] Multiplying by $T$, we obtain
\[
  kT -1 = TS
  \implies
  kT - TS = 1
  \implies
  T(k-S) = 1 
  \implies
  T = \frac1{k-S} \,.
\]
\fi
\item Solve the simultaneous equations
\begin{gather*}
  3x+2y = 1 \,, \\ 6x-4y = 3 \,.
\end{gather*}
\ifsolution
\item[\textit{Solution}:] Adding twice the first equation to the
second equation, we find $12x=5$, so $x=\tfrac5{12}$.  Resubstituion
of this value into the second equation gives $\tfrac52-4y=3$, so that
$4y=\tfrac52-\tfrac62$ and finally $y=-\tfrac18$.
\fi
\item Simplify the expression
\[
  \frac{(\e^b)^{1+\ln a}}{\e^b} \,.
\]
\ifsolution
\item[\textit{Solution}:] 
\[
  \frac{(\e^b)^{1+\ln a}}{\e^b}
  = \frac{\e^{b (1+\ln a)}}{\e^b}
  = \e^{b (1+\ln a) - b}
  = \e^{b \, \ln a}
  = a^b \,.
\]


\item[\textit{Comment}:] If you have difficulties in problems 1--5,
you need to look for self-study material on ``College Algebra'' and
revise your basic maths skills very seriously.  Although we will
review the basic concepts very briefly, all of the university-level
methods classes assume prior familiarity.
\fi
\item Sketch the graph of the function
\[
  y = 3 \, \cos 2x 
\]
including appropriate labeling of the coordinate axes.
\ifsolution
\item[\textit{Solution}:] See extra sheet.
\fi


\item Compare the functions
\[
  f(x) = \frac{3x}{(x-1)(x+2)}
  \qquad \text{and} \qquad
  g(x) = \frac{1+x}{1-x^2} - \frac{2}{2+x} \,.
\]
Are they the same, or do they differ?
\ifsolution
\item[\textit{Solution}:]

Proceeding formally, we find
\begin{align*}
  \frac{1+x}{1-x^2} - \frac{2}{2+x}
  & = \frac{1+x}{(1-x)(1+x)} - \frac{2}{2+x} \\
  & = \frac{1+x}{(1-x)(1+x)} - \frac{2}{2+x} \\
  & = \frac{-1}{x-1} - \frac{2}{2+x} \\
  & = \frac{-2-x}{(x-1)(x+2)} - \frac{2 (x-1)}{(x-1)(x+2)} \\
  & = \frac{3x}{(x-1)(x+2)} \,.
\end{align*}
So the expressions coincide whenever the two sides are well-defined.

Note, however, that the domains of the two functions differ: $f$ is
undefined for $x = 1, -2$, while $g$ is undefined for $x=1,-1,-2$.

\item[\textit{Comment}:] This question tests if you can execute the
rules for manipulating fractions and are able to recognize
factorizations of polynomial expressions in a moderately complicated
context.  Congratulations if you managed to work it out without a
hitch.  Very likely, you overlooked the different domains of
definition.  Don't worry, but for Elements of Analysis, you need to
appreciate that this is indeed an issue.
\fi

\item Find all local minima and maxima of the function
\[
  f(x) = x^3 - 3 x + 1 \,.
\]
\ifsolution
\item[\textit{Solution}:] We test the necessary condition for
existence of local extrema of a smooth function on the reals, namely
$f'(x)=0$.  Here, 
\[
  f'(x) = 3x^2 - 3
\]
with roots $x=\pm 1$.  Since $f'(x) >0$ for $\lvert x \rvert >1$ and
$f'(x) < 0$ for $\lvert x \rvert < 1$, $f$ has a local maximum at
$x=-1$ and a local minimum at $x=1$.

\fi
\item Consider the function
\[
  f(x) = x \, \e^{-x} \,.
\]
Find its horizontal and vertical asymptotes, local minima, local
maxima, and points of inflection.  Identify the regions where the
graph of $f$ is concave upward or concave downward.  Draw a sketch of
the graph.
\ifsolution
\item[\textit{Solution}:]  

The function is defined on the entire real line, so cannot have
vertical asymptotes.  For the horizontal asymptotes, note that
\[
  \lim_{x\to-\infty} f(x) = -\infty \,, \qquad
  \lim_{x\to\infty} f(x) = 0 \,.
\]
So $f$ has a horizontal asymptote as $x\to\infty$ at $y=0$.

To find local extrema, we first look at the critical points
$f'(x)=0$.  We compute
\[
  f'(x) = \e^{-x} - x \, \e^{-x} = (1-x) \, \e^{-x} 
\]
which has a single root at $x=1$.  Clearly, $f'(x)<0$ for $x<1$ and
$f'(x)>0$ for $x>1$, so $f$ has a local maximum at $x=1$ where it
takes the value $f(1)=\e^{-1}$.  There are no other local or global
extrema.

To find the points of inflection, we look at the roots of 
\[
  f''(x) = - \e^{-x} - (1-x) \, \e^{-x} = (x-2) \, \e^{-x}
\]
which has a single root at $x=2$.  Clearly, $f''(x) < 0$ for $x<2$ and
$f''(x)>0$ for $x>2$, so $f$ is concave down to the left, has a point
of inflection at $x=2$ where it takes the value $f(2)=2 \, \e^{-2}$,
and is concave up to the right.

For a sketch, please see extra sheet.
\fi
\item Compute
\[
  \int_0^{2\pi} x \, \sin x \, \d x \,.
\]
\ifsolution
\item[\textit{Solution}:] We integrate by parts:
\[
 \int_0^{2\pi} x \, \sin x \, \d x
  = x \, (-\cos x) \biggr|_0^{2\pi}
    - \int_0^{2\pi} (-\cos x) \, \d x 
  = - 2 \pi \,,
\]
where we note that the remaining integral of the cosine function in
the center expression is over a complete period, thus vanishes by
symmetry.

\item[\textit{Comment}:] The last three questions cover standard
Calculus material.  For Elements of Analysis, you should be able to do
these problems without difficulty, though the techniques will be
reviewed in brief.  In (Applied) Calculus, such problems will be
covered in detail, but be aware that these are elementary examples, so
even if you can do them now, do not assume that you know it all.
\fi
\item Consider the Fibonacci sequence
\begin{gather*}
  a_{n+1} = a_{n} + a_{n-1} \qquad \text{for } n \geq 1 \,, \\
  a_0 = a_1 = 1 \,.
\end{gather*}
Show that this sequence $\{a_n\}$ is unbounded.
\ifsolution
\item[\textit{Solution}:]

Clearly, all $a_n$ are non-negative.  Thus $a_{n+1} \leq a_n + 1$ so
that, proceeding inductively, $a_{n} \geq n+1$.  Letting $n \to
\infty$, we see that $a_n \to \infty$.  In particular, $\{a_n\}$ is
unbounded. 

\item[\textit{Comment}:] The task here is to find precise mathematical
language to make an ``obvious'' statement rigorous; the proof given
here is just one of many different possibilities.  If you can follow
this reasoning, you are a candidate for \emph{Elements of Analysis}.
\fi

\item How many roots (zeros) does the function
\[
  f(x) = x^7 +x^5 + x^3 + x
\]
have?
\ifsolution
\item[\textit{Solution}:]

Of course you can factor
\[
  f(x) = x \, (x^6 + x^4 + x^2 + 1) \,,
\]
so $x=1$ is a root.  Since the second factor is the sum of
non-negative terms and one, there cannot be another root.  

An even simpler argument is to look at the original form of of the
function $f$: it is the sum of terms each of which increases
monotonically from $-\infty$ to $+\infty$.  Thus, $f$ must have at
least one zero.  Since the sum of monotonic functions is monotonic, it
cannot have more than one zero, either.

\item[\textit{Comment}:] Although the more direct solution through
factorization is available here, it may not be possible for other,
very similar problems.  So Mathematicians will try to abstract the
argument to the minimal set of features necessary to answer the
question.  This leads to the second argument given.  This quest for
abstraction will play a prominent role in Elements of Analysis (but is
present, at some level, in the other courses as well).
\fi
\item A triangle has edges of length $3$, $4$, and $5$.  Is it a
right-angled triangle?
\ifsolution
\item[\textit{Solution}:] Note that three numbers satisfy the
Pythagorean relation
\[
  3^2 + 4^2 = 5^2 \,.
\]
By the law of cosines, this is a sufficient condition for the
occurrence of a right angle.

\item[\textit{Comment}:] The answer requires the logical converse to
how the Pythagorean theorem is usually stated.  This needs to be
appreciated and a first-principles argument (such as via the law of
cosines) needs to be given.
\fi

\item Show that for arbitrary real numbers $a$ and $b$,
\[
  a \, b \leq \tfrac12 \, a^2 + \tfrac12 \, b^2 \,.
\]

\ifsolution

\item[\textit{Solution}:] 
\[
  \tfrac12 \, a^2 + \tfrac12 \, b^2 - a \, b
  = \tfrac12 \, (a^2 - 2 \, a \, b + b^2)
  = \tfrac12 \, (a-b)^2 
  \geq 0 \,.
\]
\fi
\item Find the indefinite integral
\[
  \int \sqrt{\frac{1-x}{1+x}} \, \d x \,.
\]
\ifsolution
\item[\textit{Comment}:] This is a hard problem which requires good
algebraic manipulation skills and some experience or stamina.  A
solution is given on p.~4 of 
\begin{center}
  \url{http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-Strategy-Integra_Stu.pdf}
\end{center}
If you cannot do this (easily), don't worry.  If you can do it, you
should be very fit for higher maths.

\fi

\end{enumerate}

\end{document}
