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### AEA Maths Questions

Posted: Tue Jun 24, 2008 11:14 am UTC
I am sitting the Advanced Extension Award maths exam tomorrow. It's similar to the STEP but by most accounts harder (I do not intend defend this against anyone who believes otherwise, It's just what people have told me). I am not expected to pass this exam. To put it into context only two people from my FE college have passed in the last decade. However I am determined to try anyway.

To the point:
I was working through past paper questions when I came to this question.
7c

The curve C has the equation:
y=e-x sin x x≥0

The terms of the sequence A1, A2, ..., An,... represent areas between C and the x-axis for successive portions of C where y is positive.

Find an expression for An in terms of n and π.

Part b of the question asked you to show that the integral of the function was: -1/2 e-x (sin x + cos x) + c. (by using integration by parts twice and rearranging. )

I managed to derive the result: An = 1/2 (e-(2n-1)π/4 + e-nπ/2)

My question is: is this the result they are asking for, or is there some way to get rid of the exponents?
(Eulers formular: e=-1 is not on the syllabus)

Any help or advise on this question or anything else to do with this exam would be much appreciated.

Thanks,
Frimble

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 12:54 pm UTC
You can factor out exp(-n pi /2), but its' not really any better.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 2:25 pm UTC
Thanks TABD. I thought there wasn't. It seemed strange to me asking for a variable in terms of a constant anyway. They probably just mentioned pi to make it clear they wanted the answer in radians.

More taxing questions to follow I suspect.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 4:00 pm UTC
Just to cheer you up: AEA is much easier than STEP, except possibly STEP I which I didn't take.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 4:15 pm UTC
Nope AEA is easier that STEP I as well (I took it yesterday and will rue it forever that I didn't take III given that I could solve 4 of their questions)

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 4:53 pm UTC
Hey, I don't need to pass. My offer for Oxford is AAA, and I already have AA. This is just for fun. In fact I had to negotiate with the head of department to let me sit it, I would have sat the STEP papers too, but that would have been pushing it a bit. Next question if anyone has time:

Given that:

d/dx (u x0.5) = du/dx * d(x0.5)/dx 0<x<0.5

where u is a function of x, and that u=4 when x=3/8, find u in terms of x.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 5:19 pm UTC
Frimble wrote:Hey, I don't need to pass. My offer for Oxford is AAA, and I already have AA. This is just for fun. in fact I had to negotiate with the head of department to let me sit it, I would have sat the STEP papers too, but that would have been pushing it a bit. ]

Try explaining wanting to take STEP and AEA early before even getting an offer. You missed out by not taking STEP, it would have been much better preparation for Oxford than the AEA (which is basically revision of the A level core)

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 5:26 pm UTC
I did take the AEA Physics early. (clearly out scienced)

Are you sitting this exam tomorrow too?

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 5:31 pm UTC
Indeed, although I'd much rather be taking STEP II, but still, better than a kick in the teeth.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 5:41 pm UTC
Have you done the June 2005 paper?

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:00 pm UTC
Which one was that?

(I'm fairly sure I've done them all)

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:07 pm UTC
The one from which I took the differential equation above. ^
(I don't get a unique solution when I try to solve it.)

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:17 pm UTC
Out of interest, how did you go about solving it (I have a single solution)

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:26 pm UTC
I started by using product rule to get:

du/dx x0.5 + d(x0.5)/dx u = du/dx * d(x0.5)/dx

Is this appropriate/correct?

From here I can't decide whether to multiply throughout by dx.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:31 pm UTC
I'd carry out all the differentiation on the x's first.

BTW, have you been taught that its acceptable to "multiply through by dx"

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:34 pm UTC
Already trying that. Yes, much nicer.

Yes, thats how separating variables works isn't it?

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:44 pm UTC
Thats not "why" it works, but its one way of implementing it, although in general multiplying through by dx isn't allowed

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 6:49 pm UTC
Oh? Is it not? Why is it valid when separating variables then?

Spoiler:
u=1/(1-2x) ?

Ed. let dy/dx=dz/dx.

dy=dz

y=z+c

Is that valid? I have been trying to find a good explanation of this sort of thing for some time.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 7:05 pm UTC
Your value for u looks familiar. (Not guaranteeing anything since I haven't done the question in ages)

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 7:21 pm UTC
I had to promise not to look at the questions until I had sat my A levels which is why I'm still working on them now.

### Re: AEA Maths Questions

Posted: Tue Jun 24, 2008 7:26 pm UTC
Hmm.. I would probably say that that was a bad idea. STEP and AEA do require some work.

### Re: AEA Maths Questions

Posted: Wed Jun 25, 2008 2:53 am UTC
Frimble wrote:let dy/dx=dz/dx.

dy=dz

y=z+c

Is that valid? I have been trying to find a good explanation of this sort of thing for some time.

In the case that I came across the expression

dy/dx = dz/dx

I would just integrate both sides with respect to x, and you would get the same result.

### Re: AEA Maths Questions

Posted: Wed Jun 25, 2008 8:58 am UTC
Yes, when I say multiply throughout by dx really I mean integrating everything with respect to x. It's annouther one of those cases where functions,... no sorry, linear operators can be treated as numbers.

### Re: AEA Maths Questions

Posted: Wed Jun 25, 2008 4:11 pm UTC
Since the paper is online, you can all chat about it.

I myself am of to visit some unis!

http://eiewebvip.edexcel.org.uk/Reports/Confidential%20Documents/0806/9801_01_que_20080625.pdf

### Re: AEA Maths Questions

Posted: Wed Jun 25, 2008 5:07 pm UTC
I doubt I'll pass. I left unanswered/partially unanswered 29 marks worth of questions.

Note that the character in Q5 is needs to have a western name because she has made a mistake.

Q3 is easy but a bit open ended, you can prove it however you like!

Q7 was nasty. I haven't done any serious vector geometry for a year now.

### Re: AEA Maths Questions

Posted: Wed Jun 25, 2008 9:57 pm UTC
I had nothing better to do.

Spoiler:
1) 8100. Formula for sum of an arithmetic series (not hard to deduce if you can't remember it) and simple arithmetic.

2)a) P = (-4,-3). Knowing what the tangent is and plugging things into an equation.
2)b) [imath]y = \frac{\displaystyle 8(x+2)^2}{\displaystyle x+1}[/imath]. Separate variables, integrate, simplify.
3)a) Half-angle formula for tan, and knowing cos 60 and sin 60. Seriously, degrees?
3)b) 15,165,195,345. An exercise in connecting two parts of a question (and addition formulae, of course).

4)a) B = (arccos(e-1),e-1). I love it when everything cancels at once.
4)b) A pretty tricky integration, but it works out. I'm wondering if there's an easier method that I'm missing.

5)i) p = 9, q = 9/8. Odd numbering on this question, but basic log stuff.
5)ii) x = 12. Again, prove yourself better than Anna at remembering facts about logs.

6)a) [imath]f^{-1}(x) = \frac{\displaystyle 2x-b}{\displaystyle a-x}[/imath]. Note f(x) = a is impossible unless a = b/2, in which case f-1(a) = 0.
6)b) a = -2. Barely even worth 2 marks.
6)c)i) Asymptotes: x = -2, y = -2. Intersection points: (b/2,0) , (0,b/2).
6)c)ii) Asymptotes: x=0, y=0. Intersection points: none. Note graph is [imath]y = \frac{\displaystyle 4+b}{\displaystyle x}[/imath].
6)d) 4x+51, if I haven't made a stupid mistake. Don't you just love it when they tell you how to answer? This is definitely not worth 5.

7)a) 1/3. Best to transform so B is the origin, then take a dot product.
7)b) Use the above transform - if you make A and B the same length, you can just average them, then transform back.
7)c) AB = OB-OA.
7)d) [imath]\left(\frac{\displaystyle 3}{\displaystyle 2},2,-\frac{\displaystyle 9}{\displaystyle 2}\right)[/imath]. The centre of an incircle is the average of the vertices (each weighted by the proportion of the triangle's perimeter taken up by the opposite side).
7)e) [imath]\frac{\displaystyle 3\sqrt{2}}{\displaystyle 2}[/imath]. Twice the triangle's area divided by the perimeter.

### Re: AEA Maths Questions

Posted: Thu Jun 26, 2008 5:49 pm UTC
Q2b - answer is (9/4)(x+2)^2/(x+1)

davet

### Re: AEA Maths Questions

Posted: Thu Jun 26, 2008 9:31 pm UTC
Yeah, apparently I can't tell which way round x and y are in coordinates.

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 12:28 am UTC
What was your method for question 6d? I mucked it up in the exam, but have since worked out that the answer should be 51, which tallies with yours. You say it's definitely not worth 5 marks, but my method (after the exam) was lengthy: get the transformed function, differentiate it, equate derivative to -1/gradient of tangent, get equations of normals, transform function back again, suss which normal has equation y=4x-39, and get the value of k from the other one. What shorter method am I missing???

I reckon I got a mark of 75-76.

Someone just told me on question 6d that there is indeed a shorter method: use the fact that the transformed function is odd, get from y=4x-45 to y=4x+45 for the second normal, and transform back to get y=4x+51.

davet

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 12:33 am UTC
frimble, I don't know how things went on the questions you did answer, but the grade boundary for a distinction will be around 68-70, to judge by previous years, and don't forget to take into account the S marks. Hard not to get the 1 available for Q1 if you worked out the right answer really! (There are 2 available on each of the later questions, although they only tot up S marks from the 3 questions where you got the most).

davet

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 5:36 am UTC
davet wrote:Someone just told me on question 6d that there is indeed a shorter method: use the fact that the transformed function is odd, get from y=4x-45 to y=4x+45 for the second normal, and transform back to get y=4x+51.

That was exactly it. I guess it's only easy if you can see how.

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 10:31 am UTC
Yep - I guess doing it that way would suggest say 3-4 marks, but if the longer way were the only way, it would be worth maybe 6-7!! I wonder what proportion of candidates who got the answer used the respective methods? A good question really, because students should have asked themselves what the point of the transform was. STEP-like in a sense, what with the relation between the two parts.

Q5b is terrible though - it's not clear whether x=2 is in the solution set or not.

x=8/3 would involve log(0); undefined; so x=8/3 is definitely not a solution.
x=12 definitely is a solution.
With x=2, the position is unclear. It involves logs of negative numbers, which are complex. The original LHS and RHS then come out as equal.

So is x=2 allowed? It depends on whether you allow the use of complex numbers to get the original LHS and RHS. One view is that this is banned. But why? Nowhere do they suggest this. It's a very different matter from requiring the log of 0. What's a candidate supposed to do? Many who sit AEA are already aware of complex numbers, even though they're not on the A level syllabus.

Oops! We should bring this to Edexcel's attention!

davej

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 10:58 am UTC
x=2 is not a solution even if you use a complex log, as far as I calculate it. The LHS is log3(12)/log3(2), which is real, but the RHS is 0.5 + log3(-2), which is not real.

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 8:14 pm UTC
Remember log is multivalued. I think this can be used to equate LHS and RHS although not sure exactly how.

davej

### Re: AEA Maths Questions

Posted: Fri Jun 27, 2008 10:57 pm UTC
The real parts are equal, but the complex part of the RHS is not a multiple of 2πi, so multivaluing doesn't work.

### Re: AEA Maths Questions

Posted: Sat Jun 28, 2008 10:10 am UTC
I thought it could work by choosing complex logs on numerator and denominator of the LHS as well as on the RHS, but you are right!! There is no way integer multiples of 2πi can be chosen to equate the real parts, except if they are all zero, which as you say leaves different imaginary parts. Thanks for this!

dave

### Re: AEA Maths Questions

Posted: Sat Jun 28, 2008 3:10 pm UTC
Q7d - centre of incircle is at (3/2, 1, -9/2) unless I'm going mad... ### Re: AEA Maths Questions

Posted: Sun Jun 29, 2008 11:00 am UTC
I think I answered questions, 2, 3 and perhaps 5 fine, so that should be 38 marks. I made an embarrassing error on question 1 but much of my working was correct. I have no idea what I'll get for the other questions though, theres no mark breakdown for this paper. When I sat the Physics AEA last year I answered every question and go 66% so I don't work that accurately it seems.

I use a different method for 3a to token:

I first proved that tan 45 = 1 and that tan 60 = 3^0.5 by drawing appropriate triangles.

I then stated that as the formula tan (A-B) = (tan A)-(tan B)/1+(tan A)(tan B) was in the formula booklet I was going to use this as a lemma and proceeded from there.

Question 3b, 15,165,195,345? Thats not in the range 0<θ<360, has he just added a multiple of 360?

I agree though: Why degrees?

PS. Complex numbers are not on the syllabus. Using them might be easier but it will certainly not be necessary.

### Re: AEA Maths Questions

Posted: Sun Jun 29, 2008 12:55 pm UTC
Frimble wrote:Question 3b, 15,165,195,345? Thats not in the range 0<θ<360, has he just added a multiple of 360?

θ=15, θ=165, θ=195 and θ=345 are all solutions.

### Re: AEA Maths Questions

Posted: Sun Jun 29, 2008 1:04 pm UTC
Oh right, you meant,
15, 165, 195, 345 then. I thought your answer was 15165195345. davet: I never get S marks. 