You are Here: Home

Pure 2 watch

1. There's a couple of questions on the EDEXCEL JUNE 2003 PURE 2 paper I cannot do. Does anyone happen to know where I could find solutions for this paper? Thanks a lot.
2. Why not post the problem here? The forum has many superb maths brains about. And I'd give it a go myself.
3. (Original post by XTinaA)
Why not post the problem here? The forum has many superb maths brains about. And I'd give it a go myself.
you do m2? check my m2 post! cheers !
4. (Original post by kikzen)
you do m2? check my m2 post! cheers !
Err this is P2 and I did M1, I chose S2 instead of M2 for my further maths AS...

I'll have a look.
5. Umm, well it's one with a graph in...so I can't really show it. But if you go to http://www.mrhughes.net/Maths/Hidden%20Pages/login.htm username-alevel password-mrhughes. The paper's on there.

I need help with 6 and 7. Thanks for your time!
6. (Original post by AATTMM)
Umm, well it's one with a graph in...so I can't really show it. But if you go to http://www.mrhughes.net/Maths/Hidden%20Pages/login.htm username-alevel password-mrhughes. The paper's on there.

I need help with 6 and 7. Thanks for your time!
I've printed the questions you asked for and am looking at them, mind if I get it done tomorrow? I'll have a go now anyway...
7. Sure ^_^.
8. 6

a) dy/dx = -c/x²

when x = p, dy/dx = -4 => 4 = c/p² => c = 4p²

b) 5 = 1 + (4p²)/p => 4=4p => p=1 => c=4.

c) V = pi.integral[1,2] (1+4/x)² dx

=pi.integral[1,2](1 + 8/x + 16/x²)dx

=pi.[x + 8ln(x) - 16/x][1,2]

=pi.[2-1 +8ln(2) - 8 + 16]

=pi.(9 + 8 ln(2))

=> k = 9, q = 8.
9. 7a) At M:
x=0, y=7
dy/dx = 2e^x = 2 so the gradient of the normal is -0.5
=> y-7=-0.5(x-0)
2y-14 = -x
2y+x=14

7b) At N, y=0 so 2y+x=14 => x=14

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 19, 2004
Today on TSR

Edexcel C4 Maths Unofficial Markscheme

Find out how you've done here

2,272

students online now

Exam discussions