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"what are the chances of being caught for plagarism [sic] in english advanced?" Probably not very high, in general.

It's what you get from applying the product rule twice.

$\noindent Maybe quicker to just use $(fg)'' = f''g + 2f'g' + g''f$. One of the terms will automatically vanish because the second derivative of the linear factor is $0$.$

Remember, the general rule is if A is (m x n) and B is (n x p), then AB is (m x p). Since A is (3 x 5) and AB is (3 x 1), we must have that B is (5 x 1). (No. of rows of B must be 5 in order for B to be compatible with A in the product AB, and no. of columns of B must be 1 in order that the no...

He put the "k" with the x^3 part instead of with the 1/x part. (Remember, you can put it with either part.)

$\noindent It's just a notation for function composition. The notation $D\circ M$ refers to the composition of $D$ and $M$. In general, $f\circ g$ basically means the function whose values are $f(g(x))$ (assuming appropriate domains/codomains). You can find out more and see examples on the...

Well for 2(c) for example, you should be able to show it's not linear by giving a specific example (it's basically not linear because the "length" function is not linear). And for 4, make sure to be familiar with the definitions of injective, surjective, and invertible (and it'll help to know...

They're equally likely by symmetry.

Try and consider the die modulo 3.

$\noindent If $f(x)$ is that cubic and $(a,b)$ is the coordinates of a stationary point of it, then $f(a) = b\Rightarrow f(a)-b = 0$ (since the point lies on the graph) and $f'(a) = 0$ (slope $0$ at a stationary point). Hence defining the cubic $C(x):= f(x) -b = x^{3}-3x-b $, we have $C(a) = 0$...

Ah OK. Well in any case, the markers are probably looking for the answer ln(2)/3. You should probably simplify your answer down to this because if you leave it as a complex arctangent, one could argue this isn't fully simplified, especially since we're dealing with a real integral (so we should...

Like I said, your answer involving complex numbers is essentially correct, but you need knowledge of the complex arctangent function to simplify it down to the desired form of the answer (ln(2)/3). And in the HSC, you should probably not use the complex method because it isn't in the syllabus...

Your answer is essentially still correct but you'll need to know about the complex arctangent function to simplify it down to the desired form of the answer (ln(2)/3). See for example here: http://scipp.ucsc.edu/~haber/archives/physics116A10/arc_10.pdf . Knowledge of these matters is of course...

It's one of the standard Taylor series. You can obtain it by integrating the geometric series.

$\noindent For the first one, split up the series and use $\sum_{n=1}^{\infty}z^n = \frac{1}{1-z}-1$ and $\sum_{n=1}^{\infty}\frac{1}{n}z^{n} = -\log{(1-z)}$, noting the intervals of convergence. $

Yes. "Would the value ever become constant?" Yes, it asymptotically approaches a constant. $\noindent We have $X_{n+1} = cX_{n} + I$, where $I$ is constant and $c \equiv 1 - R$ (replacing $B$ with $X_{n+1}$ and $A$ with $X_{n}$ to reflect evolution in time). This is an inhomogeneous first...

You can try using integration by parts.

$\noindent You essentially have to become good at ``guessing" these sorts of criteria (maybe by analysing tons of past paper criteria and trying to spot patterns in them). Also I think some past criteria to a longer-response question said that you need to make sure you include no extraneous...

You can use point C to get the slope of the line AB (since we are told that AB is parallel to OC, so slope(AB) = slope(OC) = 1/2).

$\noindent That one is even, but not odd.$