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Part b) is a famous elementary proof of the Harmonic series' divergence: S = 1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15) + ... > 1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + ... = 1...

a) Actually follows immediately from the p-series test (here, it's summing 1/n^p, where p = 1/2 < 1 ==> divergence). $\noindent To show it the way they want, we can use induction. Note that $s_{2} = 1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ (since the LHS is greater than $1.5$, since...

You need to enclose your code in tex tags: [.tex] [./tex] (without the full stops). You can practise it here for instance: http://community.boredofstudies.org/3/non-school/345040/latex-practice.html .

$\noindent (1) The motion is $\ddot{x} = - \omega^{2} x$, where $\omega^{2} = \frac{\pi\cdot 40^{2}g}{m}$. Hence the period is $T = \frac{2\pi}{\omega} = 2\pi \times \sqrt{\frac{m}{\pi \cdot 40^{2}g}}$. We are told the period is $2.5$ seconds, so equate the expression for $T$ to 2.5 and solve...

Re: MX2 2016 Integration Marathon Correct!

Re: MX2 2016 Integration Marathon $\noindent Find $\int _{0}^{\pi} \frac{1 + 2\cos x}{5 + 4 \cos x}\, \mathrm{d}x$.$

Re: HSC 2016 4U Marathon $\noindent Note that the set of values of $\pm \frac{\pi}{2} + 2k\pi$ as $k$ runs through the integers is the same as that of $\frac{\pi}{2} +k \pi$ as $k$ runs through the integers, since $\frac{\pi}{2}$ and $-\frac{\pi}{2}$ are distance $\pi$ apart.$

$\noindent (2) (a) Separate variables again and we obtain $\ln y = \lambda \ln x + c$. So $y = Ax^{\lambda}$, where $A$ is an arbitrary constant.$ $\noindent (b) Notice that $\ln y$ is linearly related to $\ln x$. So we could plot the data using a \textsl{log}-\textsl{log plot}. You could use...

$\noindent (1) (a) Just separate variables and integrate, so $\ln y = k \left(t + \frac{a}{2\pi}\sin (2\pi t)\right) + c \Rightarrow y = A \exp \left(k \left(t + \frac{a}{2\pi} \sin (2\pi t)\right)\right)$, where $A$ is an arbitrary constant.$ $\noindent (b) The growth rate is $k(t) = 0.02 -...

$\noindent (3) Letting $D$ be the differentiation operator, we formally have $L = aD^2 + bD + c$, and $L^2 = \left(aD^2 + bD + c\right)^2$ (remember these operators work just like polynomials; you could expand this out to find $L^2$ in terms of powers of $D$ up to $D^4$). So the characteristic...

$\noindent (2) By inspection, the solution is $c_1 \cos \omega t + c_2 \sin \omega t$, if $\omega >0$. (Standard SHM solution.)$ $\noindent To derive this via the proposed substitution (though it's faster to just use the standard method of solving second-order ODE's with constant coefficients...

$\noindent (1) Let $v = y'$. Then the ODE is $2v v' = 1+v^2$. So $v' = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{1+v^2}{2v}$. Thus $\ln \left(1+v^2\right) = t + c$. So $v^2 = Ae^t -1$ for some positive constant $A \equiv e^c$. So $v = y' = \pm \sqrt{A}\sqrt{e^t -B^2}$, where $B= \sqrt{A^{-1}}$...

$\noindent Your answer would be right if (and only if) the polynomial was monic (had leading coefficient $1$). But the leading coefficient is $2$, so you need to put a $2$ in front of your answer. You can then obtain Terry Lee's answer by multiplying the $2$ into the $\left(x -...

$\noindent Of course, for part (ii), we are having $x > \frac{n}{n-1}$ (which is greater than $1$), and we are going to use part (i) by saying $\left(1 + (x-1)\right)^n \geq 1 + n\left(x-1\right)$. So it suffices to just prove Bernoulli's inequality for $x > 0$ say for part (i) in order to use...

$\noindent Oh, also realised, Q14 (a) of the Ext. 2 paper, it's saying let $x$ be a real number and show that $\left(1+x\right)^{n} \geq 1 + nx$ ($n\geq 2$ an integer). But it should say something like let $x \geq -1$, since that inequality doesn't hold in general if $x$ is allowed to be less...

Yeah I figured out the intended meaning of the Q. pretty quickly because it wouldn't make sense otherwise. (Iirc as it was written, it syntactically meant a "for all" rather than "there exists". Like if we say "Let n be a positive integer. Show n has a prime factorisation.", it means show this...

The answer is ln|sec(x)| + c, since tan(x) = sin(x)/cos(x). Since sin(x) is the negative of the derivative of cos(x), the primitive of tan(x) is -ln|cos(x)| + c = ln|sec(x)| + c. Unnecessarily complicated to use u = sin(x) as a substitution, but you can do it as shown by trecex1 below.

For the Ext. 2 paper, it seems like the second last sub-part of the last part of Question 16 (part (c) (iii)) doesn't have its mark allocation written. (Luckily all the other parts of Question 16 have their mark allocations written, and these sum to 12, so part (c) (iii) should be worth 3 marks...

Also Question 11(a) (4U paper) is dodgily worded too. Obviously if g(x) can by any function continuous at alpha (which is what the wording technically means and hence what I thought it meant on first reading), then f(x) need not be identical to (x-alpha)^2 * g(x). E.g. take g(x) to be the zero...

$\noindent You're right in that if the pipettes did something like always move 10 mL and 5 mL back and forth, we wouldn't reach an equilibrium (well we'd get A end up empty and B end up full and). So I'm going to assume the following is how the transferring works: if a beaker has height $h$ of...