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I can get them before the security date. Too bad my school didn't use those trials :(

I personally liked the Coroneos ones... but that's just me.

Re: why is it only 'medicine' for newcastle while other universities are medicine/sur I've heard that Newcastle is the best for Medicine.

Should really say that THERE EXISTS a point c such that ...

Because 3pi/8 + pi/8 = pi/2 So the second term becomes cot (pi/2 -x) = tan x So you're integrating tanx -tanx ie. 0 so integral is 0.

I don't know. Depends on your teacher. In the HSC you probably wont get this type of question exactly though. There'll be a step before it or give you a hint...

True, using the result \int^b_a f(x) dx = \int^b_a f(a+b-x)dx

What are your limits, and what are you integrating with respect to?

Well the notation \int^{\infty}_a f(x) dx means \lim_{b \to \infty} \int^b_a f(x) dx Mathematicians tend to be notoriously lazy sometimes :P

I think I've read in the examiner's report that half marks were not awarded.

Standard 2U question: Solve for x: |x - 1| = -5

I liked the telescoping series in first year. Interesting stuff :P

Triangle PQR is equilateral iff |PQ|=|QR|=|RP| and arg (PR) - arg (PQ) = arg(QP) - arg(QR) iff |q-p| = |r-q| = |p-r| and arg(r-p)-arg(q-p)=arg(p-q)-arg(r-q) iff |q-p| = |r-q| = |p-r| and arg[(r-p)/(q-p)] = arg[(p-q)/(r-q)] iff (r-p)/(q-p)=(p-q)/(r-q) iff (r-p)(r-q)= -(p+q)^2 , q=/=p and...

I think Mr Goodwin is alluding to complex analysis methods... which wouldn't even work for the above questions since there are no limits in the integral.

Probably... Also, I think you can let u=2x and then proceed from there... But that's basically the same as using the subs t=tanx because then dx = dt/(1+t^2), whereas the the t=tan(x/2) subs yields dx = 2dt/(1+t^2)

\text{Use } \ddot{x}=\frac{d}{dx} \left( \frac{1}{2} v^2 \right)

I think 1st year maths is easier than 4U actually (you don't get impossible Q7/8's)... But the content is somewhat different isn't it? eg. Matrices, vector geometry etc...

That would be second year maths.

In complex analysis there are two types of logarithmic functions: 1) Normal logarithmic function (ie. log) 2) Principal logarithmic function (ie. Log) By definition, log z = ln|z| + i arg z and Log z = ln|z| + iArg z Arg z is the principal argument, ie. -pi < Arg z <= pi arg z = Arg z + 2...

They probably wont. 1st year maths (at any university) is quite standard stuff though...