[Satsui no hado]

Good luck all for tomorrow morning!

[corpuscallosum]

So this is a bit last minute but... has anyone got the answers for the Jan 2012 paper?

[san_M]

(Original post by corpuscallosum)
So this is a bit last minute but... has anyone got the answers for the Jan 2012 paper?

here you go. Good luck for tommorow!

[corpuscallosum]

(Original post by san_M)
here you go. Good luck for tommorow!

Thank you so much!

[Killjoy-]

The decision between sleeping and doing more revision is always a hard one to make...

[Anon 17]

Good luck for tommorow everyone!

[ElMoro]

Good luck guys

May God help us all

[dknt]

Good luck guys. I remember doing this exam last year. I was terrified.

[Killjoy-]

Good luck and good night (morning? lol)

[Anon 17]

Last minute, but I've got a few minutes to kill.

Good luck everyone!

[Anon 17]

How did everyone find the exam? Unofficial mark scheme time?

Still kicking myself for leaving 3/4 in the direction vector of the line in 7iii)...

[bananadude]

Hard paper I thought. Question 7 was awful

[san_M]

It took me a while to think about Q7, But I managed to do it, I couldn't do question 8 :< multiplied out the R matrix and got a reflection matrix and thats all I did for Question 8. I probably lost 15 marks or so :<

[lilipop]

(Original post by bananadude)
Hard paper I thought. Question 7 was awful

It was horrible.

[SBarns]

Not a nice paper at all, bloody OCR.

[Sgt.Incontro]

Question 7 was awful.
Question 8 was also awful, could only do first part.

I hope the grade boundaries are really low, then I might just make an A.

[SK1965542]

How did it go with the sin cubed times cos squared bit. I couldn't get the -1/16 bit. Any help?

[Satsui no hado]

I thought the paper was ok, for the second order d.e where you get m(m+4) A lot of people said you get Ae^-4x+B, but I left the constant out because otherwise i couldn't get a particular solution for ii).

Also for the last part of 8 I think it was just rotations of a hexagon

Hoping to get A* in this paper but a few parts were quite iffy

[Ree69]

what was q1..?

[Matt G]

(Original post by SK1965542)
How did it go with the sin cubed times cos squared bit. I couldn't get the -1/16 bit. Any help?

It would have been (2isintheta)^3 x (2costheta)^2 so to reduce that into the form sintheta^3 x costheta^2 you would divide by the 2i^3 x 2^2 which equals -32i and since there was already a factor of 2i on the RHS of the equation you would get -1/16.