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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Yes - my mistake.

Re: MX2 2015 Integration Marathon Wolfram Integrator confirms 1/√6

A quadratic is a quadratic. It doesn't matter that is has resulted from the solution of a quartic - it is still solved the same way. You haven't had practice at solving quadratics which give complex solutions?

See my post 5 before this one.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread If sin2x = 1, then 2x = 90 OR 270 (0 < 2x < 360) x = 45, 135 (0 < x < 180)

Perhaps it should be called Astigmatic Maths.

His explanation is not right. There is only one way of arranging the two groups. But doing it the way you are describing, the number of ways of arranging (n-k+1) things in a circle is (n-k)! (You have learned this rule?)

As I said below, if there is a group of k, then the other n-k also form a group. There isn't any freedom to arrange two groups around a circle - there is only one way. The k! and (n-k)! come from arranging within the groups. And it looks like your textbook got all three answers wrong. Which...

I didn't know "denominate" was a word. I would have thought that would have a different meaning: "I nominated him for captain, but he pissed me off so I denominated him".

It is easier than that. If there is a group of k, then the other n-k also form a group.

The answer to 2 is also wrong. It is just k! (n-k)!

The number of numbers without restriction is 5!/2! = 60. So the answer can't be 72. The correct calculation is 4! + 4!/2!.

Re: MX2 2015 Integration Marathon When doing that recurrence relation, it is a little bit slicker if you start with I_n + I_(n-2).

Re: MX2 2015 Integration Marathon Experimenting with latex - so this could be rubbish: - \sum_{k=1}^{9}\frac{\cot^{k}x}{k}

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Why don't you substitute to check? Also consider whether you have missed any solutions.

Re: HSC 2015 3U Marathon The question says "AT THE TOP OF ITS MOTION it touches the top vertex of the triangle."

Re: HSC 2015 3U Marathon As you said, it's a PARABOLA. The axis of symmetry is half way between the roots. That makes the triangle isosceles.

Re: HSC 2015 3U Marathon Since alpha MUST equal beta, why not write tan theta = 2 tan alpha ?

They are NOT the answers. Two answers are purely real. And all answers include a √2. The answers above don't work on substitution.

I don't see any need to use trig. Just write (z+1)²=2(z-1)² and (z+1)²=-2(z-1)². Then rearrange and solve each equation using the quadratic formula. What is more interesting is that the answers all lie on a circle centred at (3,0) with radius 2√2. I'm wondering if anyone can show why...