I did most of the one with Psinx + Qsin2x + Rsin3x as well, but that worked out really horribly in the end and I probably got it wrong.
I wonder how many marks you lose for getting most of an answer right, and then just writing the final thing down incorrectly? I think I wrote the wrong answer for Q4) - Do'h! ...I hate exam pressure :).
The last part of Q4) turns out to be incredibly fidgety (or maybe that was just my method). Wasn't sure how to answer it. Any ideas?
I fidgeted a lot for question 4, but it's really not that bad.
Find the roots of the equation as 2nx=mp etc. without worring about the restriction on x. Then find the maximum value of m for each form of the roots can take in terms of n for even and odd n bearing in mind that x < p. Then say that one's bigger than that one so that's the maximum root. If that made no sense, post back and I'll do it properly.Yes, for the last part of Q4, use the previous part.
In that case I made Q4 unnecessarily complicated.
Hi everyone.
Does anyone know what the final answers to 4 actually are? I got
pi/4 for n even and I kinda guessed pi/6 for n odd!
I put pi/4 for even Nikhil, but when I came out of the exam I
started thinking about it, and I don't think it's right. It's
probably something like m pi /2n, where m is the largest positive
integer such that m/n < 1, ie m = n - 1. This is just out of
memory now, so it's probably not right (I haven't attempted the
question again).
The thing that got me when I had finished the question was the
fact that it said 'Find an expression ' which made me
realise I'd probably messed it all up. But by then I was too
tired with that question to change everything, so I just left it
:)
My method for the last part of question 4 was:
g(x) = sin2n x + sin4n x - sin6n x g(x) = sin2n x + sin2(2n x) - sin3(2n x) g(x) = sin2n x (2 + 2cos2n x - 4cos2 2n x) g(x) = 0 Þ sin2n x = 0, cos2n x = -1/2, or cos2n x = 1 2n x = mp, where m is an integer 2n x = (2m+1) + /3 < par/ > 2n x = (2m+1)p- p/3 2n x = 2m p So x = mp/2n x = (m + 2/3)p/n x = (m + 1/3)p/n x = mp/n Now for even n let n = 2k and choose m in terms of k so that m is an integer and x < p/2 but x is as large as possible. So x = (2k - 1)p/4k x = (k - 1 + 2/3)p/2k x = (k - 1 + 1/3)p/2k x = (k - 1)p/2k Now choose the largest value of x and rewrite it in terms of n and that's the answer. Repeat for odd n, n = 2k + 1, noting that (k + 2/3)p/(2k + 1) > p/2