STEP 2001 Paper 2 Question 4

This thread is an excerpt from a discussion of the STEP papers immediately after they took place.

By Olof Sisask on Thursday, June 28, 2001 - 01:07 pm :

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.

By Olof Sisask on Thursday, June 28, 2001 - 01:40 pm :

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 :).

By Olof Sisask on Thursday, June 28, 2001 - 01:47 pm :

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?

By Tom Hardcastle on Thursday, June 28, 2001 - 02:04 pm :

I fidgeted a lot for question 4, but it's really not that bad.

Find the roots of the equation as

2 nx = m π

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 < π

. 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.

By Michael Doré on Thursday, June 28, 2001 - 02:18 pm :

Yes, for the last part of Q4, use the previous part.

By Olof Sisask on Thursday, June 28, 2001 - 03:24 pm :

In that case I made Q4 unnecessarily complicated.

By Nikhil Shah on Thursday, June 28, 2001 - 04:07 pm :

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!

By Olof Sisask on Thursday, June 28, 2001 - 04:18 pm :

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 :)

By Tom Hardcastle on Thursday, June 28, 2001 - 05:09 pm :

My method for the last part of question 4 was:

g (x) = \sin 2 n x + \sin 4 n x - \sin 6 n x

$g (x) = \sin 2 n x + \sin 2 (2 n x) - \sin 3 (2 n x)$

$g (x) = \sin 2 n x (2 + 2 \cos 2 n x - 4 \cos 22 n x)$

$g (x) = 0$ $\Rightarrow \sin 2 n x = 0$ , $\cos 2 n x = - 1 / 2$ , or $\cos 2 n x = 1$

$2 n x = m π$ , where $m$ is an integer

$2 n x =$ (2m+1) + /3 $< par / >$ 2n x = $(2 m + 1) π - π / 3$

$2 n x = 2 m π$

$x = m π / 2 n$

$x = (m + 2 / 3) π / n$

$x = (m + 1 / 3) π / n$

$x = m π / n$

Now for even $n$ let $n = 2 k$ and choose $m$ in terms of $k$ so that $m$ is an integer and $x < π / 2$ but $x$ is as large as possible. So

$x = (2 k - 1) π / 4 k$

$x = (k - 1 + 2 / 3) π / 2 k$

$x = (k - 1 + 1 / 3) π / 2 k$

$x = (k - 1) π / 2 k$

Now choose the largest value of $x$ and rewrite it in terms of $n$ and that's the answer.

Repeat for odd $n$ , $n = 2 k + 1$ , noting that

$(k + 2 / 3) π / (2 k + 1) > π / 2$

Tom.