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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Note that $n^{\frac{2}{3}} = \left(n^{\frac{1}{3}}\right)^2$. So $n^{\frac{2}{3}} = 9\Longleftrightarrow \left(n^{\frac{1}{3}}\right)^2 = 9$, so $n^{\frac{1}{3}} = \pm 3$, so upon cubing, we have $n = \pm 27$.$

You could try starting assessments early to avoid all-nighters.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The LHS would be $1+2+2^2$. The RHS would be $2^3 -1$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent For $n=1$, the LHS should be $1+2$, since the general expression of the LHS starts with a 1 (as it is $1+2+2^2 + \cdots + 2^n$).$

Re: HSC 2016 4U Marathon These polynomials are called palindromic polynomials. The curious HSC 4U student may want to read about some of their properties here: https://en.wikipedia.org/wiki/Reciprocal_polynomial#Palindromic_and_antipalindromic_polynomials

Re: HSC 2016 4U Marathon $\noindent We shouldn't do it like this (without first proving the roots lie on the unit circle, or unless we are told to assume it without proof). Rather, sub. $y = x+\frac{1}{x}$ and observe we will obtain a quadratic in $y$ $\Big{(}$using manipulations like $x^2 +...

Here is another way of going about it. $\noindent The answer is $\binom{9}{2}\cdot \binom{7}{2} \cdot \binom{5}{2}\cdot 3!$ (which evaluates to $45360$ as Paradoxica had). The reasoning is that we first choose spots for $1$ and $2$ (note that once we choose the spots, we have automatically...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread What is the question?

Be prepared for a lot of rote learning.

You can try comparing these marks to the people you talked to who had 30 ATAR. Also, try putting estimated marks into some ATAR Calculators (there are many available online).

\noindent \text{A) Let} \ P(x) = x^{100}. \ \text{Write} \ P(x)=x^{100} \equiv Q(x) (x-1)(x-2) + ax+b \ \text{(division transformation)}, \text{where ax + b is the remainder and Q is a polynomial of degree 98. Now,} \ P(1) = 1^{100} = 1 \Rightarrow 1 = 0 + a\cdot 1 +b \ \text{(from division...

While a lot of 99+ ATAR students do get coaching, it is neither necessary nor sufficient in order to obtain such an ATAR (or even 99.9+ ATAR, as asked on another thread). It can be very helpful for some people though, of course.

$\noindent They can be proved in many different ways. One way is to draw a right-angled triangle (this does it for acute angles), and then use symmetries to derive it for angles in the other quadrants.$

Re: HSC 2016 3U Marathon $\noindent Not entirely sure what you did here with the $a$ etc., but to show that $\binom{n}{0}=1$ for non-negative integers $n$, recall that $0!=1$, so $\binom{n}{0} =\frac{n!}{0!(n-0)!} =\frac{1}{0!} = 1$.$

$\noindent The first is $\sin X$. The second is $\cot \theta$, assuming this is defined.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread What is the question?

It is indeed possible to get 99.9+ without doing 4U maths. Doing 4U is neither necessary nor sufficient for getting 99.9+ ATAR (although it can be helpful). Also, I doubt that 99% of people with 99.9+ ATAR did 4U maths. That seems too big a proportion. For example, in 2015, 92 people got 99.9+...

Do you need to do two related texts for this? Or can you just do one? (Haven't been following this thread sorry, so don't know if you've already answered this.) If you can get away with doing just one and this essay isn't worth any marks, you could just do one and go to sleep now. :)

Some HSC subjects have fewer than 10 people in the state doing them, so probably only one person does them at some schools.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread This is due to the HSC Circle Geometry theorem "line from centre of circle to midpoint of chord is perpendicular to chord". This can be easily proved by proving that those triangles are congruent (use SSS, due to equal radii...