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AEA Prep Thread - 25th June 2015 (Edexcel)

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oh okay thanks then. no matter how i look at the last question, i can't help but think that the trigonometric answer can't be any less horrendous than the one i gave. Have you got the answer to that one ?

Yeah, you have to use the same substitution as part (a), it then gave you cot^4øcosecø, you then split that up into (cot^3ø)(cotøcosecø), and integrated cotøcosecø and differentiate cot^3ø which then got you an integral which you could split up into part b and the thing you are trying to integrate so you had 4I = {-cosecøcot^3ø} between the limits -3xpart b I believe.

Reply 421

tiny hobbit

Original post by physicsmaths

Tinyhobbit and teeem. What do you think of the paper? Hardest questions ? I reckon the trig one was the hardest!

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Like TeeEm, I'll need to try it, which I'm not going to do until I've got a legible version tomorrow or so.

Reply 422

dhee

Original post by fatart123

Cos(2x + 78) = cos(90-x)

I went the long way around with it (i think)
and i got to sin(2x-12) = sinx
and i got x=4
were there any other solutions?

Reply 423

dhee

Original post by SH0405

I got sqrt(30).

yup that's right (or at least what i got)

Reply 424

username1162972

Original post by dhee

I went the long way around with it (i think)
and i got to sin(2x-12) = sinx
and i got x=4
were there any other solutions?

Yh it did it your way well except 12-2x and look at my solution there 124,244 and 192 aswell.

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Reply 425

dhee

Original post by physicsmaths

Yh it did it your way well except 12-2x and look at my solution there 124,244 and 192 aswell.]

yeah i saw your stuff
i was going about that way but thought nah they wouldn't do that
it means about 4 more marks thrown away there fml

Reply 426

username1162972

Original post by TeeEm

I am currently writing solutions to AQA, C1, June 2014 paper. I am about to start Q6. Any good?

In the vectors question I made a mistake in d leading me to the wrong answer in e. How many marks would I lose in e?

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Reply 427

TeeEm

Original post by physicsmaths

In the vectors question I made a mistake in d leading me to the wrong answer in e. How many marks would I lose in e?

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I am not familiar with AEA marking but if it was to be marked as the standard papers you will lose 2 marks from e and 1 mark from f

Reply 428

DFranklin

Original post by physicsmaths

Did you get 4,124,244,192?I've been reading this and similar comments without seeing the question and thinking:

"Wow, the answer was 4 billion, 124 million, 244 thousand, 192? That looks scary!"

:smile:

Reply 429

DFranklin

physicsmaths

Regarding the integral everyone seems to have struggled with: can you say what the exponent was (either of the integral or the curve)? (It's not legible on the attachment you posted).

Reply 430

username1162972

Original post by DFranklin

I've been reading this and similar comments without seeing the question and thinking:

"Wow, the answer was 4 billion, 124 million, 244 thousand, 192? That looks scary!"

:smile:

LOL. Hahahahaha. I thought it was an amazing question where people thought the 78 came out of nowhere.

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Reply 431

DFranklin

Somewhat guessing as I can't read the question but the following might have helped:

Define

\displaystyle I_n= \int \dfrac{1}{(x^2-1)^{n/2}}\,dx

then IBP with dv = 1, gives us

\displaystyle I_n = x \dfrac{1}{(x^2-1)^{n/2}} - \int x \dfrac{d}{dx}\dfrac{1}{(x^2-1)^{n/2}} \,dx

\displaystyle = x \dfrac{1}{(x^2-1)^{n/2}} - \int 2x^2 \frac{n}{2} \dfrac{1}{(x^2-1)^{(n+2)/2}} \,dx

\displaystyle = x \dfrac{1}{(x^2-1)^{n/2}} - n \int x^2 \dfrac{1}{(x^2-1)^{(n+2)/2}} \,dx

\displaystyle = x \dfrac{1}{(x^2-1)^{n/2}} - n \int ((x^2-1) + 1) \dfrac{1}{(x^2-1)^{(n+2)/2}} \,dx

\displaystyle = x \dfrac{1}{(x^2-1)^{n/2}} - n \int \dfrac{1}{(x^2-1)^{n/2}} + \dfrac{1}{(x^2-1)^{(n+2)/2}} \,dx

\displaystyle = x \dfrac{1}{(x^2-1)^{n/2}} - nI_n - nI_{n+2}

\implies n I_{n+2} = x \dfrac{1}{(x^2-1)^{n/2}} -(n+1) I_n

You can use this reduction formula to get to an "easier" version of the integral.

(N.B. I've ignored arbitrary constants 'cos I'm lazy, but for the actual question same argument with definite integrals is probably easiest).

Edit: This is perhaps slightly more complied than it needs to be - I thought the power was something like 5/2 but it's actually 3.

(edited 8 years ago)

Reply 432

username1162972

Original post by DFranklin

Regarding the integral everyone seems to have struggled with: can you say what the exponent was (either of the integral or the curve)? (It's not legible on the attachment you posted).

Curve is y=1/(x^2-1)^3/2. Rather nice integral after doing step for 6 months. Shame i didnt have time to finish that part couldve got 95/100. Oh well I shall live haha.

Reply 433

DFranklin

Original post by physicsmaths

Curve is y=1/(x^2-1)^3/2. Rather nice integral after doing step for 6 months. Shame i didnt have time to finish that part couldve got 95/100. Oh well I shall live haha.

So actual integral was "just"

\displaystyle \int \dfrac{1}{(x^2-1)^3} \,dx

?

That should have been nice for everyone who did STEP III Q1 this year...

Reply 434

username1162972

last year the official MS was posted rather quickly. A few days i thinkz could the same happen this year?

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Reply 435

username1162972

Original post by DFranklin

So actual integral was "just"

\displaystyle \int \dfrac{1}{(x^2-1)^3} \,dx

?

That should have been nice for everyone who did STEP III Q1 this year...

Yep. Lost 8 on that since didnt have time. Sorry i repeat this but it is annoying lol. Shud get 88 ish so still not too bad.
There are a few ways to do it but would be interesting which way they wanted.

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Reply 436

DFranklin

Original post by physicsmaths

Yep. Lost 8 on that since didnt have time. Sorry i repeat this but it is annoying lol. Shud get 88 ish so still not too bad. Probably more than I'd have got. I hate AEA papers.