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In my opinion a tutor is only needed if you are not bothered to study or your teacher is awful. But that is another debate. Doing homework and practising regularly on NEW questions is more important than relearning content learnt at school. Ask the teacher for help if you do not understand a...

Re: HSC 2013 4U Marathon I'll add it in the latex so people don't miss it

Re: HSC 2013 4U Marathon DONE: I'm not bothered to post working out, basically let x = that root, then find x^6, x^4 and x^2 using binomial theorem, then manipulation then: 6x^6-36x^4+54x^2-245 = 0 I better not of made a silly mistake :/ ============== E(x)=\sum_{r=0}^{n} \frac{x^{r}}{r!}...

Eh

About 5 seconds

\pi \neq \sqrt{6(\sum_{k=1}^{\infty} \frac{1}{k^2})} So pi cannot be constant if this is the case, and we are merely representing pi as a variable.

I am theorizing that either 1. Trolling 2. Doesn't understand differentiation well 3. Is concerned about the constancy of irrational numbers. - > In this case, it is reasonable to see where the confusion aroused. Pi being the irrational value of 3 + 0.1 + 0.04 + 0.001 + 0.0005 + ... And...

That would be greatly appreciated And no problem

Two valuable points to consider when doing 4U. 1. Since 4U makes you not sit the 2U exam, Extension 2 Maths becomes 2 Units worth, Extension 1 becomes 2 units worth. This then essentially doubles your 3U mark, so if you perform well enough in 3U which by itself has incredible scaling (best...

Re: Chess thread? http://lichess.org/zdk2ig1t Play me (anyone will do)

I think he is referring to the polynomial in x (the actual question)

The reason why we do not get zeroes from k=\pm i is when we are solving for k in C in this particular case. I am guessing it is similar to the reason why some solutions do not work when we are solving for multiple absolute values: For instance: |x+a|+ |x+b| = c When we solve for this we can...

We get -20i

Re: HSC 2013 4U Marathon Yes I think there are no all three real solutions. This comes from the fact that, if we have a polynomial roots x, y, z with the restrictions in the question. Then the polynomial equation will be: \beta = \alpha^3 + d Where it is a polynomial in alpha, and d is any...

There is a mistake in the 5th line of your working for the first part. ========================= Anyway, for the question: 1+\tan \theta 1+\frac{2t}{1-t^2} \\ \\ = \frac{t^2-2t-1}{t^2-1} That is the furthest simplification without going back to the start, unless you want to do something...

Express the polynomial in terms of its quotient, remainder and divisor: x^4-2x^3+x^2+ax+b=(x^2+x-1) A(x) + R(x) For it to be exactly divisible, R(x)=0 x^4-2x^3+x^2+ax+b=(x^2+x-1)A(x) Notice how A(x) will have to be some quadratic equation: x^4-2x^3+x^2+ax+b=(x^2+x-1)(px^2+qx+r) Note how...

Related to the Quadratic function: $A quadratic function$ f(x) $ is given. Find a general form of the parabola such that $ \\ \\ f(f'(x))=f'(f(x))

Re: HSC 2013 4U Marathon Well yes, if I prove it to be true for complex numbers, then it must be true for the real numbers. But I was just pointing out that the question can be extended to non-zero complex numbers.

Re: The p00n thread http://en.lichess.org/my7dzi5flxbf

Re: The p00n thread http://lichess.org/my7dzi5f Play me