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# OCR (not-MEI) FP1 - Friday 20th May 2016 Watch

1. For the sum I got -1/k for the product 1 and the the second part 1/k^2

And for the difference method question I got the sum to infinity as 1/3 but others got 1/6 how about u?
2. I thought that was alright, at least compared to M2...
I got:
1)
1/2 * n(n-1)(2n+3)
2)
i) sqrt(3)-3i
ii) -1/49 + 4sqrt(3)/49 i
3)
i) alpha + beta = -1/k
alpha * beta = 1
ii) (alpha + 1/alpha) * (beta + 1/beta) 1/k^2
4)
i) (5a-3b 10 0)
ii) (6b-5)
iii) (6a 12 18)
(4a 8 12)
(-a -2 -3)
5) [inductive proof]
6)
i) |z-3-3i| = sqrt(5)
ii) [argand diagram]
iii) 2 + 5i, 4 + i
7)
i) Shear, x axis invariant, (0,1) mapped to (3,1)
ii) (0 -1)
(-1 0)
Reflection in y=-x
8)
i) [Show that]
ii) n / 3(2n+3)
iii) 1/ 2(2n+1)
9)
i) [Show that]
ii) The equations were consistent
10)
i) +/- (5 + 4i)
ii) [Show that]
iii) +/- (5/41 +/- 4/41 i)
3. I think I got everything bar the argand diagram question (how many marks was that?) Strange times when C1 is harder than FP1
4. (Original post by farryharnworth)
I thought that was alright, at least compared to M2...
I got:
1)
1/2 * n(n-1)(2n+3)
2)
i) sqrt(3)-3i
ii) -1/49 + 4sqrt(3)/49 i
3)
i) alpha + beta = -1/k
alpha * beta = 1
ii) (alpha + 1/alpha) * (beta + 1/beta) 1/k^2
4)
i) (5a-3b 10 0)
ii) (6b-5)
iii) (6a 12 18)
(4a 8 12)
(-a -2 -3)
5) [inductive proof]
6)
i) |z-3-3i| = sqrt(5)
ii) [argand diagram]
iii) 2 + 5i, 4 + i
7)
i) Shear, x axis invariant, (0,1) mapped to (3,1)
ii) (0 -1)
(-1 0)
Reflection in y=-x
8)
i) [Show that]
ii) n / 3(2n+3)
iii) 1/ 2(2n+1)
9)
i) [Show that]
ii) The equations were consistent
10)
i) +/- (5 + 4i)
ii) [Show that]
iii) +/- (5/41 +/- 4/41 i)
I think I got the same answers for the questions I remember so that's pretty good but am gutted I couldn't do the last question
5. (Original post by Bobby21231)
For the sum I got -1/k for the product 1 and the the second part 1/k^2

And for the difference method question I got the sum to infinity as 1/3 but others got 1/6 how about u?
I got the same as you for both of those.
6. I have a feeling it was 2/3 for the argand diagram and 3 for the pints of intersection but I could be wrong
(Original post by SGHD26716)
I think I got everything bar the argand diagram question (how many marks was that?) Strange times when C1 is harder than FP1
7. (Original post by farryharnworth)
I thought that was alright, at least compared to M2...
I got:
1)
1/2 * n(n-1)(2n+3)
2)
i) sqrt(3)-3i
ii) -1/49 + 4sqrt(3)/49 i
3)
i) alpha + beta = -1/k
alpha * beta = 1
ii) (alpha + 1/alpha) * (beta + 1/beta) 1/k^2
4)
i) (5a-3b 10 0)
ii) (6b-5)
iii) (6a 12 18)
(4a 8 12)
(-a -2 -3)
5) [inductive proof]
6)
i) |z-3-3i| = sqrt(5)
ii) [argand diagram]
iii) 2 + 5i, 4 + i
7)
i) Shear, x axis invariant, (0,1) mapped to (3,1)
ii) (0 -1)
(-1 0)
Reflection in y=-x
8)
i) [Show that]
ii) n / 3(2n+3)
iii) 1/ 2(2n+1)
9)
i) [Show that]
ii) The equations were consistent
10)
i) +/- (5 + 4i)
ii) [Show that]
iii) +/- (5/41 +/- 4/41 i)
I agree with all of these except 8iii, which IIRC was a sum to infinity, so shouldn't have an n anywhere in it. <- THIS IS WRONG I CAN'T ****ING READ
8. (Original post by sdhand)
I agree with all of these except 8iii, which IIRC was a sum to infinity, so shouldn't have an n anywhere in it.
I'm fairly sure it was a sum from n to infinity
9. (Original post by sdhand)
I agree with all of these except 8iii, which IIRC was a sum to infinity, so shouldn't have an n anywhere in it.
Between n and infinity so the sum from 1 to n-1 needs to be defined in terms of n

Posted from TSR Mobile
10. (Original post by farryharnworth)
I'm fairly sure it was a sum from n to infinity
****, if that's true I didn't read the question correctly. There goes some marks.
11. Was also 1/blah instead of 2/blah as shown in part i

Posted from TSR Mobile
12. (Original post by farryharnworth)
I'm fairly sure it was a sum from n to infinity
I think there was just a sum
To infinity as well as this
13. (Original post by sdhand)
I agree with all of these except 8iii, which IIRC was a sum to infinity, so shouldn't have an n anywhere in it.
Actually the lower limit was n, so you had to subtract the equation with n-1 substituted for n. Their answer is right.
14. (Original post by drandy76)
Was also 1/blah instead of 2/blah as shown in part i

Posted from TSR Mobile
Yeah I'd don't read the 2 as well
15. (Original post by Bobby21231)
Yeah I'd don't read the 2 as well
Same lol. Think I'll have got 0 for that one.
16. Was a pretty good paper, only thing i fear now is very high boundaries :s
17. (Original post by sdhand)
Same lol. Think I'll have got 0 for that one.
At least it weren't just me Do u know how many marks it was?
18. (Original post by Bobby21231)
At least it weren't just me Do u know how many marks it was?
Think between 3 and 5, don't know for sure though.
19. (Original post by sdhand)
Think between 3 and 5, don't know for sure though.
Okay thanks I'm hoping I will have only dropped 10 marks at most as I know I lost 4 on the last question and then some on here
20. (Original post by Bobby21231)
Okay thanks I'm hoping I will have only dropped 10 marks at most as I know I lost 4 on the last question and then some on here
I think I dropped about 10 as well. Should be a low to middling A if the previous years grade boundaries are anything to go by.

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