AEA Official Thread Summer 2016

Thought it was a very nice paper, personally. Made a silly mistake at the last minute on 2 by putting pi + pi/4 instead of pi - pi/4 but hopefully that's just one mark. I had no problems with the rest of the paper.

Reply 62

Original post by A Slice of Pi

Thought it was a very nice paper, personally. Made a silly mistake at the last minute on 2 by putting pi + pi/4 instead of pi - pi/4 but hopefully that's just one mark. I had no problems with the rest of the paper.

how did you end up with pi-pi/4 anyway? I ended up with pi/4+pi/2???

Reply 63

Original post by Aph

how did you end up with pi-pi/4 anyway? I ended up with pi/4+pi/2???

Year that's also correct. They both give 3pi/4. It depends how you tackle the problem but I put them all in terms of arctan and ended up with arctan(-1) which I erroneously said was pi/4 + pi (omitting the minus sign in front of the pi/4). There'll be a few ways of doing the problem though.

Reply 64

Original post by A Slice of Pi

Year that's also correct. They both give 3pi/4. It depends how you tackle the problem but I put them all in terms of arctan and ended up with arctan(-1) which I erroneously said was pi/4 + pi (omitting the minus sign in front of the pi/4). There'll be a few ways of doing the problem though.

yeah I know that they are both correct... but why would you put

arccos(\frac{1}{\sqrt{3}})

in terms of tan when you can solve as it is? just seems a bit weird to me.

Reply 65

Original post by Aph

yeah I know that they are both correct... but why would you put

arccos(\frac{1}{\sqrt{3}})

in terms of tan when you can solve as it is? just seems a bit weird to me.

Yeah, I'm aware of that, but it works out quite nicely if you get everything in terms of arctan and then use the arctan sum identity to get everything as a single arctan term. That's the method that came quickest to me under exam conditions.

Reply 66

Original post by A Slice of Pi

Yeah, I'm aware of that, but it works out quite nicely if you get everything in terms of arctan and then use the arctan sum identity to get everything as a single arctan term. That's the method that came quickest to me under exam conditions.

ahhh okay. that makes sense now.

Reply 67

username1424648

2

Struggled with Q2 and Q5… Hopefully I can still get the Merit that I need.

Reply 68

Bunderwump

How do you get started with Q2?

Edit: and also, finding E for the vectors?

(edited 7 years ago)

Reply 69

Original post by Bunderwump

How do you get started with Q2?

arccos(\frac{1}{\sqrt{3}})+arcsin(\frac{1}{3})+2arctan(\frac{1}{\sqrt{2}})

?

Reply 70

Bunderwump

Original post by Aph

arccos(\frac{1}{\sqrt{3}})+arcsin(\frac{1}{3})+2arctan(\frac{1}{\sqrt{2}})

?

Yep.

Reply 71

Original post by Bunderwump

Yep.

okay I started:

arccos(\frac{1}{\sqrt{3}})+arcsin(\frac{1}{3})+2arctan(\frac{1}{\sqrt{2}})

arccos(\frac{1}{\sqrt{3}})=\frac{\pi}{4}

arcsin(\frac{1}{3})=\theta \Rightarrow sin(\theta)=\frac{1}{3} \Rightarrow cos(\theta)=\frac{2\sqrt{2}}{3} \therefore tan(\theta)=\frac{1}{2\sqrt{2}}

2arctan(\frac{1}{\sqrt{2}})= \theta \Rightarrow tan(\frac{\theta}{2})=\frac{1}{ \sqrt{2}}

using the double angle formula I could then say

tan(\theta)=2\sqrt{2}

can walk you through some more but it should be fairly obvious from here.

Reply 72

Bunderwump

Assuming I got full marks on the questions I think I got right (can someone remember the marks for each question?)

Q1: 7/7
Q4: 11/11
Q6: 18/18
Q7: 16/22

Gives 52. I probably got one or two method marks for 2 and 3, and quite a few for 5 and 7(a). But I've probably also made mistakes. Maybe I can scrape a merit.

Reply 73

Original post by kingaaran

Managed to do 2, 5a but couldn't do 5b either, as I found it difficult to sum rx^(-r), but hopefully I should get a mark for taking out the 1/X constant

Posted from TSR Mobile

The summation follows directly from (b). Carry out differentiation and then multiply though by x. Then substitute x = 2. A similar question was asked in the specimen paper.

(edited 7 years ago)

Reply 74

student0042

This paper wasn't too difficult in my opinion. The only parts I couldn't do was 5b and 5c. I divided by x for 5b, so I may get a mark there, but I'm pretty sure I'll get a distinction unless I've mucked up my mental calculations everywhere.

Reply 75

ChanE

1

Glad it wasn't part of my offer (just doing it for a bit of fun) but think I might have scraped a merit. I did struggle a lot though so I'm not sure. Still i enjoyed doing harder than A level maths

Reply 76

Well for the summations (part b) I thought I had been really cleaver and went and said that It was a GP

a=x^{-2}, r=2^{log_{2}(2)}*x^{-1}

Because I was thinking that to get a 1 unit increase each term you should use some form of adding logs bye because it was a GP you had to times them together... In the exam it made perfect sense but now it really doesn't make sense at all. How were you meant to do that one?

Reply 77

username1162972

18

Original post by Aph

Well for the summations (part b) I thought I had been really cleaver and went and said that It was a GP

a=x^{-2}, r=2^{log_{2}(2)}*x^{-1}

Because I was thinking that to get a 1 unit increase each term you should use some form of adding logs bye because it was a GP you had to times them together... In the exam it made perfect sense but now it really doesn't make sense at all. How were you meant to do that one?

Differentiate it

Posted from TSR Mobile

Reply 78

Original post by physicsmaths

Differentiate it

Posted from TSR Mobile

With respect to r? How's that work?

Reply 79