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So the triangle inequality (the simplest version) states that: |w1| + |w2| >= |w1 + w2| Now let w1 = z1 + z2 and let w2 = -z2. Then we have: |z1 + z2| + |-z2| >= |z1 + z2 - z2| |z1 + z2| + |z2| >= |z1| |z1 + z2| >= |z1| - |z2| Similarly (just switching z2 and z1 around in our...

Yeah - It wasn't that hard compared to previous years. It was only the final projectile motion question that contained a tricky bit which caught out most people.

The proofs that are solely by induction (i.e. no calculus) are quite tricky and the two common ones which are listed on the wikipedia article have a few not so straightforward steps. It is easier to prove it using calculus and fiddling around with maxima and minima. Fortunately, they will...

but why is it obvious that a monic polynomial can't have a non-integer rational root? The only way I can think of proving that is by my method where you show that q has to be +-1?

undalay's proof doesn't work. What about if the roots were -1/2, 1/2, 1/2, 16? they are all rational but none of them are the ones you suggested. Of course the theorem still holds but its proof is somewhat different. First, we suppose that x = p/q (p, q are integers with no common...

I just read through the submission by Bill Pender and have read through the revised syllabus documents and I have to agree almost completely with him on the points he makes. The main problem is that the Board is pursuing a change for change sakes' position which is completely unnecessary in the...

Nicely done - unfortunately I misread the question and only put the ranges instead of the widths, and I didnt top internally so I doubt I'll be getting an invite.

Don't worry - I didn't see the widths part of the question so I only put in the interval. I haven't heard about anyone yet who hasn't lost some mark somewhere so you never know - top mark may be 83 (or 83.5 if they give out half marks)

Yep - they are the old mistakes in terry lee's answers that i see. Hopefully he will correct them soon. So that's Q3(b)(ii) should be x < 4 not x < 2. Q4(a)(iii) should be 0.32 not 0.68 Q7(b)(iv) should be 19.51 <= x <= 20 so range is 0.49 or 0.5 (if rounded to 1dp)

All right I think, numerically at least (and ur integration for 1e is correct) apart from the very last question. the greatest range in the secodn interval will be achieved when m = 1 so the range should be 19.51 <= x <= 20 and the width should be .49.

I reckon someone will get full marks this year - 8000 kids doing a subject makes it fairly likely. However the fact that there were a number of fairly tricky questions could stuff people up. Much of question 7 was fairly tricky i thought and I know of a few top 4u people who stuffed up Q7a...

My numerical solutions: Q1a 16 + 8rt5 b (10,1) c 4x^3/(1 + x^8) d 45 degrees = pi/4 e 2 Q2a proof b i sketch ii 0 <= y <= 2pi c a = -7, b = 10 d i 1.4 ii 284 Q3 a pi^2/12 b i x = 4, y = 1 ii sketch c proofs Q4 a i 0.01 = 1% ii 0.285 iii 0.32 b proof c proof Q5a i...

They are similar but I have not heard of that as a possible similarity proof. As far as I know, there are three main proofs of similarity: AA, SAS, SSS whereas there are 4 for conguency: SSS, AAS, SAS, RHS. However the lack of RHS in the similarity system is basically just an arbitrary...

Where's teh general thoughts thread gone? - or is it just me?

Nup. Technically you can so the probability is (12C3*12C3*6!)/(24P6) but it is exactly the same so it doesn't really matter. Alternatively you can do it the really bashy way: (12/24)*(11/23)*(10/22)*(12/21)*(11/20)*(10/19)*6C3 (The last term is there for the ordering of which there aer...

I'm pretty sure there is because 1/0 + 1/0 =/= 1 so (0,0) can't be in the locus of P. other tricky questions were the 1(e) which I'm not sure if you had to simplify the whole way down to arctan(3/4) or not but considering it's worth 4 marks then there is that possibility. Also Q5d leaves a...

For the geometrical description of P. The locus is a circle centred at (1,0) with radius 1 but with a hole at the origin (0,0). Equation is y^2 + (x-1)^2 = 1. The hole exists because 1/0 is undefined (not sure if they will penalise this though). For the raindrop question, I think thats why...

For the probability question: (i) = 0.36 Exact fraction is (12C3*12C3)/(24C6) (consider them all as different i.e. R_1, R_2... then we have 12C3 ways of choosing the red ones, 12C3 ways of choosing the yellow ways. In the whole sample space there are 24C6 ways of choosing 6 balls) (ii)...

Yeah I reckon 90 should make E4 - remember that last year people were saying the same thing about the exam being way too easy and everything yet it seems like the raw marks weren't spectacularly high. And you have to remember that often it is the top people who frequent bos more often because...

raindrops question is 7 (u take the terminal velocity) the last part of the last question is pi^2/2 (but I somehow forgot to square it and multiply by 2 so I only got pi/2)