[Ree69]

(Original post by Anon 17)
Also, in the case of 7A question 5, they ask you to use a previously worked out expansion of sin8theta when the actual way to solve the question is by using a different expansion of it...

Exercise 7A of FP3 is horrid, probably the hardest exercise in the entire book. I wouldn't worry too much about it - the questions are (considerably) harder than what comes up in past papers. To be honest, the OCR textbooks in general have a tendency to do this. Although question 7 does remind me of a harder variation to the last part of the last question in the Jan 12 paper.

With regards to question 5, the trick is to realise that only four solutions are given, and (any) 'expansion' of will have 8 solutions for . It's given in that particular form to help you get rid of some of the roots. (You need to get rid of four, so you should be aiming to end up with some quartic equation).

[Anon 17]

Alright thanks, glad I'm not the only one who thought so.

My teacher attempted question 5 and 9, and got stuck at the same part as me in both cases. I've decided just to skip 7A and move on...

I agree though, the OCR books have some ridiculously hard questions in them at some points.

EDIT - On groups now, if anyone knows why |a-b| has no identity (even though 0 works to my knowledge as |a-0| = a, and |0-a| = |-a| = a) or how to know which function to find for 9C question 3 it'd be appreciated. In the case of 9C question 3 from finding inverses I find one of the two functions but I have no idea how you'd try to find the second one since it's self inverse.

[Ree69]

(Original post by Anon 17)
On groups now, if anyone knows why |a-b| has no identity (even though 0 works to my knowledge as |a-0| = a, and |0-a| = |-a| = a)

The identity axoim doesn't exist there because there's no unique single identity element for the whole set. 2a is just as much of an identity element here as 0 is.

[lilipop]

Don't like some of those groups questions, but still can do them…

[Anon 17]

Ah, knew there had to be some reason, thanks =)

Groups don't seem too bad this time round, although most of the proofs for two groups to be isomorphic seem to kind of ignore the one-one clause, although I'm guessing having a mapping function between two infinite groups (in which you've proved the identity of G maps to the identity of H, and that f(ab) = f(a)f(b) for a, b belonging to G) is enough to show this?

Sorry for all the questions by the way, I'll happily answer anything here that I'm able to if anyone else has questions.

[zozzie94]

I just did the jan o8 paper and got it correct except question 3 where i lost almost all the marks!! Can someone please explain q3 ii and iii to me? Thanks

[Satsui no hado]

I'm comfortable on differential equations, trig, and most of groups, need to brush up a bit on remembering all the vector rules though :P

(Original post by zozzie94)
I just did the jan o8 paper and got it correct except question 3 where i lost almost all the marks!! Can someone please explain q3 ii and iii to me? Thanks

From what i remember doing this paper you use a.b=modaxmodbxsin(theta) nhat
and where (r-a)=AP and (r-b) = BP, which is at an angle of pi to each other and as sin(pi)=0, you can say a.b or (r-a)(r-b)=0

for iii) this is an equation for a plane so the vector in front of r is perpendicular to the plane (so it's pependicular to BA) and that it passes the origin because the constant at the end of the equation is 0.

Sorry if my explanations hard to follow i'm not sure if it's completely correct either but I hope it helped

[Anon 17]

(Original post by zozzie94)
I just did the jan o8 paper and got it correct except question 3 where i lost almost all the marks!! Can someone please explain q3 ii and iii to me? Thanks

Part 2: P is a point on the line AB, so the position vector r therefore lies on the line between a and b. As (r-a) and (r-b) give two vectors in the same direction (from A to P and B to P, both on the same line) then the angle between them will be 0 (or 180) degrees. As the vector product is equal to |a||b|sin(x)n, if x is 0/180 degrees then sin(x) is 0 and thus the vector product is 0.

Part 3: The vector product of r and (a-b) being zero means that r must have an angle of 0/180 degrees with the line AB. Thus the locus of r is a line, in the direction of AB. I'm unsure why it passes the origin; I thought the equation of a plane was (r-a) . n (as it's r.n = a.n), not the vector product?

June 2009, question eight parts one and four:

I understand the maths, but I'm unsure how you'd determine that was the correct method to approach the problem. Is there any set list or at least guidelines on what kind of methods you'd use for groups? Like, is there a process as with the differentials (e.g. check if it's 1st or 2nd order, linear, can seperate the variables etc...) or is it just spotting the right properties to use?

[ElMoro]

Messed this exam up in january because I focused on other exams and didn't practice enough? Guess what I've been doing this time 'round?

[Anon 17]

^ Same, how are you finding it this time round? I'm finding it much easier, although the Jan '12 paper was a lot harder than the past papers (imo).

[Ree69]

(Original post by Anon 17)
June 2009, question eight parts one and four:

I understand the maths, but I'm unsure how you'd determine that was the correct method to approach the problem. Is there any set list or at least guidelines on what kind of methods you'd use for groups? Like, is there a process as with the differentials (e.g. check if it's 1st or 2nd order, linear, can seperate the variables etc...) or is it just spotting the right properties to use?

Lol.

[ElMoro]

Jan 10 questions
2) iv) Why is 5 being prime significant?

[Ree69]

(Original post by ElMoro)
Jan 10 questions
2) iv) Why is 5 being prime significant?

It's not really.

It's just that prime numbers have the property that their squareroot is always irrational.

[Killjoy-]

(Original post by ElMoro)
Jan 10 questions
2) iv) Why is 5 being prime significant?

a would have to be irrational for an inverse not to exist, but by definition a is rational.
Square root of all primes is irrational or I think you could just say five is not a perfect square.

[Satsui no hado]

One quick thing, I'm quite good at vectors now - the only thing I struggle with is finding the perpendicular distance between a point and a line/plane. Can anyone help in explaining how to go about these styles of questions?

[Ree69]

(Original post by Satsui no hado)
One quick thing, I'm quite good at vectors now - the only thing I struggle with is finding the perpendicular distance between a point and a line/plane. Can anyone help in explaining how to go about these styles of questions?

Fear not, the equation for the perpular distance between a point and a plane is given in the formula booklet.

[Satsui no hado]

(Original post by Ree69)
Fear not, the equation for the perpular distance between a point and a plane is given in the formula booklet.

Thanks now what about between a point and a line? xD

[ElMoro]

(Original post by Killjoy-)
a would have to be irrational for an inverse not to exist, but by definition a is rational.
Square root of all primes is irrational or I think you could just say five is not a perfect square.

(Original post by Ree69)
It's not really.

It's just that prime numbers have the property that their squareroot is always irrational.

Thanks guys - makes sense

[ElMoro]

For the june 09 paper, question (7)(ii)(b)

How do you justify whether is or ?

Thanks

[ElMoro]

Think groups and complex numbers&trig are the hardest topics

Groups because it's just so out there and some of those trig questions really need you to think out the box imo

Think vectors and differential equations are fairly standard.

What do you guys think?