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ou-entry.tex
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ou-entry.tex
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https://mcs-notes2.open.ac.uk/files/m337dq.pdf
\includegraphics[width=400pt]{img/ou-entry-2236.png}
\begin{enumerate}
\item
\begin{align*}
3(2ac + 2bc + 4) - 6(ab + ac + 2)
&= 6ac + 6bc + 12 - 6ab -6ac -12 \\
&= 6bc - 6ac \\
&= 6c(b-a)
\end{align*}
\item (b)
\item
\begin{align*}
\frac{1}{\sqrt{2} + 1}
&= \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} \\
&= \sqrt{2} - 1
\end{align*}
\item (b)
\item
\begin{align*}
x(x^2 - 5x + 6) = x(x-2)(x-3)
\end{align*}
\item \begin{align*}
y &= x(x - 3) \\
&= x^2 - 3x
\end{align*}
Second derivative is positive.
$y > 0$ for $x \in (-\infty, 0) \cup (3, \infty)$.
\item \begin{align*}
\frac{1}{x(x - 3)} &= \frac{A}{x} + \frac{B}{x-3} \\
1 &= Ax - 3A + Bx \\
A &= -1/3 \\
B &= 1/3 \\
\frac{1}{x(x - 3)} &= \frac{-1}{3x} + \frac{1}{3(x - 3)}
\end{align*}
Check:
\begin{align*}
\frac{-1}{3x} + \frac{1}{3(x - 3)}
&= \frac{-3(x - 3) + 3x}{9x(x - 3)} \\
&= \frac{9}{9x(x - 3)} \checkmark
\end{align*}
\includegraphics[width=400pt]{img/ou-entry-9439.png}
\begin{enumerate}
\item
\begin{table}[h!]
\centering
\begin{tabular}{c|ccccccccc}
$\theta$ & 0 & $\pi/6$ & $\pi/4$ & $\pi/3$ & $\pi/2$ & $2\pi/3$ & $3\pi/4$ & $5\pi/6$ & $\pi$ \\
\hline
$\sin \theta$ & 0 & 1/2 & $1/\sqrt{2}$ & $\sqrt{3}/2$ & 1 & $\sqrt{3}/2$ & $1/\sqrt{2}$ & 1/2 & 0\\
$\cos \theta$ & 1 & $\sqrt{3}/2$ & $1/\sqrt{2}$ & 1/2 & 0 & -1/2 & $-1/\sqrt{2}$ & $-\sqrt{3}/2$ & -1\\
$\tan \theta$ & 0 & $1/\sqrt{3}$ & 1 & $\sqrt{3}$ & $\infty$ & $-\sqrt{3}$ & -1 & $-1/\sqrt{3}$ & 0
\end{tabular}
\end{table}
\end{enumerate}
\item \begin{align*}
\sin^2(2\theta)
&= (2\sin\theta\cos\theta)^2 \\
&= 4\cos^2(1-\cos^2\theta)
\end{align*}
\item $\cos(2\theta) = -1/2$ implies $2\theta = 4\pi/6$ or $2\theta = 8\pi/6$, i.e. $\theta \in \{\pi/3, 2\pi/3 \}$.
\item
\begin{enumerate}[label=(\alph*)]
\item $e$
\item \begin{align*}
\frac{e^{2x}e^{3x+1}}{e^x} = e^xe^{3x+1}
\end{align*}
\item \begin{align*}
\frac{1}{3} \log_e 8 - \frac{1}{2} \log_e 4
&= 3 \cdot \frac{1}{3} \log_e 2 - 2 \cdot \frac{1}{2} \log_e 2 \\
&= 0
\end{align*}
\item \begin{align*}
e^{-\log_e 2} = 1/2
\end{align*}
\end{enumerate}
\item \begin{align*}
e^{2x} + 2e^x - 3 = 0
\end{align*}
Let $y = e^x$. Then we have
\begin{align*}
y^2 + 2y - 3 &= 0 \\
(y + 3)(y - 1) &= 0
\end{align*}
hence $e^x = -3$ or $e^x = 1$,
hence the only real solution is $x = \ln 1$.
\end{enumerate}
\includegraphics[width=400pt]{img/ou-entry-cdc6.png}
\includegraphics[width=400pt]{img/ou-entry-ce7c.png}
\begin{enumerate}
\item \begin{align*}
\sum_{n=0}^\infty \Big(\frac{1}{5}\Big)^n
\end{align*}
Geometric series formula...how to derive?
\item
\begin{enumerate}[item=(\alph*)]
\item Ratio test:
\begin{align*}
\lim_{n\to\infty} \frac{1/(n+1)}{1/n}
&= \lim_{n\to\infty} \frac{n}{n+1} \\
&= \lim_{n\to\infty} \frac{1}{1 + 1/n} \\
&= 1
\end{align*}
Inconclusive by the ratio test.
This is the harmonic series and it diverges.
\item \begin{align*}
\frac{(-1)^{n+2}}{n+1} / \frac{(-1)^{n+1}}{n}
&= \frac{(-1)^{n+2}n}{(-1)^{n+1}(n+1)} \\
&= (1 + n)(-1) < 0
\end{align*}
Converges by Alternating Series test.
\end{enumerate}
\item Ratio test:
\begin{align*}
\lim_{n\to\infty} \frac{1/(n+1)^2}{1/n^2}
&= \lim_{n\to\infty} \frac{n^2}{n^2 + 2n + 1} \\
&= \lim_{n\to\infty} \frac{1}{1 + 2/n + 1/n^2} \\
&= 1
\end{align*}
Inconclusive. This converges; every term is smaller than harmonic series.
\end{enumerate}
\includegraphics[width=400pt]{img/ou-entry-a82d.png}
\includegraphics[width=400pt]{img/ou-entry-efb5.png}
\begin{enumerate}
\item False. E.g. the maximum of $f(x) = x$ occurs at $b$ and yet $f'(b) = 1$.
\item False. This is only true if $f$ is continuous. Counter-example
\begin{align*}
f(x) =
\begin{cases}
-1 &~~~\text{if}~~~ x <= 0\\
1 &~~~\text{if}~~~ x > 0
\end{cases}
\end{align*}
\item
\end{enumerate}