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AQA A2 MFP2 Further Pure 2 – 24th June [Exam Discussion Thread]

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Reply 140

JackHKeynes

Original post by Jm098

What's the expression for alpha^3 + beta^3 + gamma^3?

(a+b+g)^3 - 2(a+b+g)(ab+bg+ga)
At least I think so...

(edited 7 years ago)

Reply 141

mazsaz

Can anyone provide me with a quick step by step method of how to attack proof by induction questions please? Really struggle with them :frown:

Reply 142

sarcastic-sal

Original post by C0balt

I will accept my fate as soon as I see 9 questions with hard-core DM question as 9th.

I'm still holding out a bit of hope that this could be nice because surely it'll just be compensation for how awful FP3 and M3 were.

That hope will promptly disappear as soon as I can read the first question and AQA suddenly decide that the gamma function is on the syllabus or some other curveball.

Reply 143

2014_GCSE

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-QP-JUN11.PDF

On 7b when it says "the three other roots of the equation", isn't there an unlimited amount. Like theta = 6pi/5, 7pi5, 8pi/5, etc.

ALSO, on part c), why do you reject -root5.

(edited 7 years ago)

Reply 144

Buntar

I'm having trouble understanding last year's partial fractions question, could someone please show their solution

Screen Shot 2016-06-23 at 22.46.59.png

Cheers

Reply 145

TheLifelessRobot

Original post by Buntar

I'm having trouble understanding last year's partial fractions question, could someone please show their solution

Screen Shot 2016-06-23 at 22.46.59.png

Cheers

I posted an explanation on page 5. I hope it helps!

Reply 146

Buntar

Original post by TheLifelessRobot

I posted an explanation on page 5. I hope it helps!

Perfect, exactly what I was after!

Thanks a lot

Reply 147

Ano123

Original post by 2014_GCSE

For part b, you're right there are infinite solutions for theta, but they give the same value as the first 4,

\tan(\pi/5)=\tan(11 \pi/2)

, so its theta mod

\pi

.
You reject

-\sqrt 5

for c as you know that

\tan(2\pi/5)

and

\tan(\pi/5)

are both positive and so cannot multiply to give an negative value.

(edited 7 years ago)

Reply 148

2014_GCSE

On 6c) how do you get costheta from root(4-sin^2(theta))

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-MS-JUN14.PDF

Reply 149

IrrationalRoot

Original post by 2014_GCSE

On 6c) how do you get costheta from root(4-sin^2(theta))

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-MS-JUN14.PDF

I think what you mean is how do you get

\cos\theta

from

\sqrt{4-(2\sin\theta)^2}

since the substitution is

x=2\sin\theta

.

In which case that is worrying to hear right before the FP2 exam, but nonetheless you get it from this:

\sqrt{4-(2\sin\theta)^2} =\sqrt{4-4\sin^2\theta}=\sqrt{4(1-\sin^2\theta)}=2\sqrt{\cos^2 \theta}

.

Now you can be very hand-wavy (just like the mark scheme is here) and just say this is equal to

2\cos\theta

.

(If you want to be technically correct, then you would actually write

2\sqrt{\cos^2\theta}=2|\cos \theta|

and we can actually let

\theta \in [-\frac{\pi}{2},\frac{\pi}{2}]

since this gives the full range of

x=\sin\theta

(so the substitution is still valid). Now

\cos\theta \geq 0

on this interval and so

|\cos\theta|=\cos\theta

and we can then carry on as normal.
Notice that the restriction we placed on

\theta

also allows us to take arcsines.)

Hopefully the reason you were confused was that you were actually trying to get

\cos\theta

from

\sqrt{4-\sin^2\theta}

, which is impossible.

(edited 7 years ago)

Reply 150

IrrationalRoot

Original post by mazsaz

Can anyone provide me with a quick step by step method of how to attack proof by induction questions please? Really struggle with them :frown:

This is probably too late, but if you have a statement P(n) which you want to prove for all positive integers n:

Prove P(1) (usually easy).
Assume that P(k) is true. Use that to prove that P(k+1) is true.
Then make some statement like 'P(1) is true and P(k)

\Rightarrow

P(k+1), so P(n) holds for all positive integers n by induction.'

I must warn you that it is much more important to understand how induction works rather than memorising the method, since the method could vary (as an example, you might only have to prove P(n) for all even positive integers in which case you'd prove P(2) and then assume that P(k) is true (where k is an even positive integer) and use that to prove that P(k+2) is true). Usually the way in which it'll vary will be that maybe you have to prove P(0) or P(2) as your base case.