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# AQA FP3 June 2015 Unofficial Mark Scheme watch

1. (Original post by morgan8002)
What sid you get for AB? Off the top of my head it was .You set this equal to to get a cubic in , which you solve for . One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.
Yeah did that just couldn't solve the cubic
2. (Original post by Lau14)
In 6b I think you're missing the end of the first bracket (should be after Bsin(lnx/2) maybe?), and for question 7 part (a) was 5 marks and (b)(i) 4 marks from what I remember. Can't believe how fast you put this together and how much you can remember, thanks!
Updated all that now, thanks!
3. (Original post by Stepidermis)
Yeah did that just couldn't solve the cubic
You had to notice that one of the roots will correspond to A, so will be . Then it's just a case of algebraic polynomial division to get the other quadratic factor. Solving that with the quadratic formula will give you the other answers.
4. This unofficial mark scheme is a great help, really reassuring to see so many of my answers up there, yay! Any chance that an unofficial question paper has been put together? Only on the parts I got wrong I can't quite seem to remember the questions, thanks (:
5. Wait you had to state the validity of the series? I completely missed that hahaha...
6. (Original post by morgan8002)
What sid you get for AB? Off the top of my head it was .
You set this equal to to get a cubic in , which you solve for . One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.
It should be r = 3/(4sin theta).
Carrying on give me sin theta = 1/2 and (-1 +/- √13)/4. I substituted them back into the AB equation but its not the right answer.
7. I don't think there was a question on the validity of the series.
8. It was in the Ln[(1+2x)(1-2x)] question, had to state the range of validity for values of x
9. so no one else got AB=1.44 for the last question then...
10. I would suggest a slight refinement - set sin theta = 3/4r

You then can change 1 + cos 2 theta into 2 - 2 (3/4r)^2

Rearranging you get 8r^3 -16r^2 + 9 = 0

The key point then is that you know r = 3/2 is a solution, so 2r - 3 is a factor. Use that to find the quadratic factor and then solve with the formula or cts.
11. (Original post by Lordvulk)
It was in the Ln[(1+2x)(1-2x)] question, had to state the range of validity for values of x
Oh Damn! How many marks was it?
12. (Original post by SilenceOrNoise)
Oh Damn! How many marks was it?
The validity will probably just be worth 1 mark, can't imagine it would be worth any more than that.
13. (Original post by eddie1221)
The validity will probably just be worth 1 mark, can't imagine it would be worth any more than that.
thanks.
14. (Original post by Criggers)
I would suggest a slight refinement - set sin theta = 3/4r

You then can change 1 + cos 2 theta into 2 - 2 (3/4r)^2

Rearranging you get 8r^3 -16r^2 + 9 = 0

The key point then is that you know r = 3/2 is a solution, so 2r - 3 is a factor. Use that to find the quadratic factor and then solve with the formula or cts.
Cheers mate it works.I should have thought about that earlier.
15. (Original post by onlinekute17)
It should be r = 3/(4sin theta).
Carrying on give me sin theta = 1/2 and (-1 +/- √13)/4. I substituted them back into the AB equation but its not the right answer.
I did say it was off the top of my head.

So you got the right answer to II?
I used the calculator to find the value of theta at B(the question said not exact value). I then used cosine rule to find the length of AB. There are probably a few ways to do it.
16. (Original post by morgan8002)
I did say it was off the top of my head.

So you got the right answer to II?
I used the calculator to find the value of theta at B(the question said not exact value). I then used cosine rule to find the length of AB. There are probably a few ways to do it.
It was the right answer but not in the right form so I think I may lose 1 mark for that.
I spent too much time doing ii so I ran out of time for the last one (
17. Well this is worrying is think I only got 48/75
18. I don't really remember 4b but I don't think that was my answer...what was the question again?
19. Anyone posted the questions yet?
20. There's some bits missing/not sure on and the wording isn't right but these are the questions as far as I know
Spoiler:
Show
1. f(x,y) = (y2 + x)/x
f(2) = 5
(a) Use Euler with h = 0.05 to find an approximation for y(2.05) [2]
(b) Use the formula yr+1 = yr-1 + 2hf(xr, yr) to obtain an approximation for y(2.1), giving your answer to three significant figures [3]

2. Use an integrating factor to solve dy/dx + (tanx)y = tan3xsecx, given that y = 2 when x = π/4. [9]

3. (a)(i) Write down the expansion of ln(1 + 2x) up to and including terms in x4 [1]
(a)(ii) Find the first two non-zero terms of ln[(1 + 2x)(1 – 2x)] and write down the range of validity for this expansion [3]
(b) Find the value of Limx>0 [(3x – x√(9 + x))/ln[(1 + 2x)(1 – 2x)]] [4]

4. (a) Why is the integral (range 0 – infinity) of (x-2)e-2x improper? [1]
(b) Find the value of the integral, showing the limiting process used [6]

5. (a) Find the general solution of d2y/dx2 + 6dy/dx + 9 = (-36?)sin2x [7]
(b)(i) Given that y = f(x) is a solution of the equation, y(0) = 0 and y’(0) = 0, show that y’’(0) = 0 [1]
(b)(ii) Find the first two non-zero terms of the expansion of f(x) [3]

6. x = et, y is a function of x.
(a) Show that the substitution x = et turns the differential equation (?) into d2y/dt2 -2dy/dt + 2y = 4t2 + 5e-t [7]
(b) Hence find the general solution of the equation (?) [10]

7. (a) Find the area of r = 1 + cos2Θ, range –π/2 ≤ Θ ≤ π/2 [5]
(b)(i) The curve 1 + sinΘ intersects the curve 1 + cos2Θ at the origin and point A. Find the polar coordinates of A [4]
(b)(ii) A horizontal line through A intersects the curve 1 + cos2Θ again at B. Show that the length of OB is (√13 + 1)/4 [6]
(b)(iii) Find the length of AB [3]

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