AQA Mathematics GCE, A2, 24th January 2011 MPC4

MU ?

The greek letter. I'm talking about the vector question. Mu=1, or 5/7? I went for 5/7, as the shape was a trapezium, not a parallelogram.

http://en.wikipedia.org/wiki/Mu_(letter)

The greek letter. I'm talking about the vector question. Mu=1, or 5/7? I went for 5/7, as the shape was a trapezium, not a parallelogram.

http://en.wikipedia.org/wiki/Mu_(letter)

I got 54 for that question as well, it came out as 53.15 which I think you round up? The wording of the question confused me.

hmm I cant remember sorry. did m0/100=m0*2^(-t/8) and solved for t i think.

What did people get for the partial fractions/binomial expansion question?

Yeah, that's what I got! Nice to know that I have three marks then...

I didn't mind the wording of it so much, but some questions on papers in the past have been horrendously-put, to the extent that I've not had a clue if they were asking this or that. They should bear in mind that maths people don't tend to be any good with words

Just realised tom, i did all that but did 1/10 not 1/100 so got 26 DAMN IT! Thats 3 marks aswell what a waste

A=3 B=-6

And I think I mucked up the expansion. Can anyone remember the exact question?

The greek letter. I'm talking about the vector question. Mu=1, or 5/7? I went for 5/7, as the shape was a trapezium, not a parallelogram.

http://en.wikipedia.org/wiki/Mu_(letter)

The greek letter. I'm talking about the vector question. Mu=1, or 5/7? I went for 5/7, as the shape was a trapezium, not a parallelogram.

I got this too!

I can't remember what I got exactly, but there was a -23/50 in there somewhere, and I thought it was valid between -3/5 and 3/5.

I got this too!

I can't remember what I got exactly, but there was a -23/50 in there somewhere, and I thought it was valid between -3/5 and 3/5.

Ah crap I think I put 5/3

(a + x)^n is valid provided -a < x < a, so (3 + 5x)^-1 is valid provided -3 < 5x < 3, which is -3/5 < x <3/5. If you did it in stages, you may well get the first mark; if you just put 5/3, I think that may well be two marks gone

I got this, after a ridiculous amount of working, and got C as (64/7, -18/7, 65/7) or something like that.
The only thing I definitely got wrong is using degrees instead of radians after the differential equation (stupid) and maybe that last vectors one isn't right, we'll see

1) R=root29, a=28.1?

2)(3x+1)(3x-1)(x+2)=f(x)
3x[3x-1]^-1
p=-9

3)3(1+x) -6(3+5x)
1+x/3 -23x/9
-0.6<x<0.6

4)2e^t/3 + 2/3e^3t so 4/3
3y + 12 =4x
K=9

5)about 3
d=32
53.16 => n=54

6)...
X = tan^-1(+-root3) so 60,120
->2sin^2x + 2sinx -1=0
sin x isn't <-1 so 0.5(root3 -1)=Sinx

7) x=(2-cos0.5t)^2
9metres
3.6seconds at 5metres

8) AB=(6,0,3)-(3,-2,4)=(3,2,-1)
85.9deg
D3,-2,4)+2(2,-1,3)=(7,-4,10) so l2=(7,-4,10) + MU(3,2,-1)
And lastly:
|AD|=root56=|BC|
so (1+3MU)^2 + (2MU-4)^2 + (7-MU)^2 = 56
so MU=1 or 5/7.
MU =\ 1 because |DC|=\|AB| so MU=5/7
thus C64/7,-18/7,65/7)

Yep, aside from a few typos, I think I agree.