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What would be the proper method of dealing with a definite integral of 1/x that involves both the positive and negative sides, taking into account the discontinuity at x=0? For example, Wolfram Alpha tells me that \int_{-1}^1 \frac 1x \, dx does not converge (and I could see why...

The theorem involving angles at the centre and circumference is only true when these angles are subtended by the same arc. In this case, the angle \angle C at the circumference is subtended by the arc DAB . And the angle subtended by the same arc DAB at the centre is the reflex angle at...

There are so many variables that affect exam performance. Amount of study is just one of them.

Is there any evidence that it had significantly impacted your other subjects? If you truly enjoy maths then I think you should keep 4U for a bit longer at least - remember that the first assessment tasks will usually not contribute a large amount towards the HSC. I found the 4U content so much...

What does that mean? If you're not finding 3U very challenging then you should be fine in 4U. Have you had an assessment for 4U yet? Unless you have, it's still way too early to tell. You can definitely afford to give it a bit more time. To be sure, 4U is harder than 2U and it's going to...

Cambridge Enhanced > Coroneos > Fitzpatrick Never used Terry Lee so can't comment on it. Cambridge alone should be enough to get you through the course.

\frac{\frac{1}{5! \cdot 11!}}{\frac{1}{3! \cdot 13!}} = \frac{1}{5! \cdot 11!} \div \frac{1}{3! \cdot 13!} = \frac{1}{5! \cdot 11!} \times {3! \cdot 13!} Sometimes if you have nested fractions they can be very confusing. Another way to think about it is to multiply both the numerator and...

(1+x)^{16} = \sum_{k=0}^{16} \binom{16}{k} 1^{16-k}x^k We want to find the terms in x^{13} and x^{11} . These are: \binom{16}{13}x^{13} \quad \mathrm{and} \quad \binom{16}{11}x^{11}. Ratio: \frac{\binom{16}{13}}{\binom{16}{11}} = \frac{\frac{16!}{3! \cdot 13!}}{\frac{16!}{5! \cdot 11!}}...

It depends on the notation. f(x) = \sqrt x is a function that denotes taking the principal square root, which is always the positive one, for real x . However, x^2 = k for k > 0 always has two solutions.

The exam should be up on NESA's site in a few days. If you look at previous papers it was similar to that except I think this year had more algorithms in it.

Didn’t see that F was not a letter....

https://drive.google.com/open?id=1FC6K1N6H0TijGdInHPeI-uYAK8Lgr9gz Make sure you look at SGS notes, they're really good

Going off the 2009 exam: A 68/100 was a 90 after alignment. That would translate to around a 48/70 this year, if the difficulty was judged to be the same.

Any angle subtended at the circumference by the diameter is always 90 degrees. The converse is also true.

If you used the inverse trig functions to find \cos \theta , then radians would have been necessary.

Paper was easy for me until the 4 marker, which I barely managed to get in the last few minutes. Weirdly enough I couldn’t get the result in the preceding 1 mark question even though I checked my similar triangles 5 times. Probably made a few sillies elsewhere. Was the exam out of 70? I...

I hope they don't ask us to draw a system flowchart because that's the one diagram I'm not too familiar with. Algorithms would be really nice too.

Alternatively: Let (K, M) denote the result of a game where Kim spins K and Mel spins M . The possible results for a victory are: (5, 4), \,(5,3), \,(5,2), \,(5,1) (4, 3), \,(4,2), \, (4,1) (3,2),\, (3,1) (2,1) \sum = 10 Clearly, there are 5 \cdot 5 = 25...

m_{OQ} = ap^2/2ap = p/2 and, similarly, m_{OP} = aq^2/2aq = q/2. \therefore m_{OQ}\cdot m_{OP} = -1 \implies pq = -4 \quad (*) The x intercept of PQ can be found by setting y=0 in the equation of PQ . This gives: \begin{aligned} x &= -apq \\ &= -a(-4) \quad {\rm from\, (*)} \\ &=...

You probably(?) won't be penalised for it as long as you show what you're doing, but typically you should never need to use 4U methods in 3U exams, or 3U methods in 2U exams etc. These methods are often slower or more tedious - and the exams are designed that way so as to not give any unfair...