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# AQA A2 MFP3 Further Pure 3 – 18th May 2016 [Exam Discussion Thread] watch

1. (Original post by JDMason)
1. Ae^5x/2 - 2x - 4/5

2. 5/3, 8/5

3. Approximation: 2.081, 2.080

4. c=36, d=20, traslate 4 across

5. 0.5x^2 +3x +ln(x+1) +1

6. 1/36

7. I got my constants wrong but I believe this is correct: y= 2cos2x - 1/2(e^pi)sin2x + 1/2(e^4x) + xSin2x - 2xcos2x

8. (1/2)ln3 - 1/2

roast me
Got all the same answers except for p and q, but I think I got that wrong. I think you should post this as unofficial ms. 😃
2. I noticed that AQA have completely changed the design of the front cover of the paper, dk why.
3. (Original post by IrrationalRoot)
I noticed that AQA have completely changed the design of the front cover of the paper, dk why.
They're changing everything because they ****ing hate us.
4. The unnoficial answers posted seem all correct, I will append them to the first post later
I made some silly errors but nothing content wise, just arithmetic slips :/

Posted from TSR Mobile
5. For the very last question, the area one:

Was the asnwer the integral of (1+tanX) between pi/3 and the value of theta at B, minus the area of the triangle (3+2√3)/6
??

Posted from TSR Mobile
6. I got something along the lines of y= 0.5x^2 +3x -ln(X+1) +cx -(2+C)ln(X+1) +2cln(X+1) + B is this along the right lines, I had noticed a similar answer on the thread but without the two constants? Were there constraints that I failed to spot?
7. (Original post by Barbecuetime123)
I got something along the lines of y= 0.5x^2 +3x -ln(X+1) +cx -(2+C)ln(X+1) +2cln(X+1) + B is this along the right lines, I had noticed a similar answer on the thread but without the two constants? Were there constraints that I failed to spot?
8. What was question number 5?
9. (Original post by Adam_98)
For the very last question, the area one:

Was the asnwer the integral of (1+tanX) between pi/3 and the value of theta at B, minus the area of the triangle (3+2√3)/6
??

Posted from TSR Mobile
This is what I did, got something like .5ln(3) -.5
10. (Original post by Adam_98)
For the very last question, the area one:

Was the asnwer the integral of (1+tanX) between pi/3 and the value of theta at B, minus the area of the triangle (3+2√3)/6
??

Posted from TSR Mobile
(1+tanX)^2 mate, you gotta square it in polar. also half it

but otherwise I think so yeah
11. (Original post by Xzielon)
5 I think, where u = y' abd you had to make a subs.
12. (Original post by Barbecuetime123)
5 I think, where u = y' abd you had to make a subs.
I didn't know what to do there either. I differentiated u and tried plugging in the partial fractions but that's as far as I went.
13. (Original post by Xzielon)
I didn't know what to do there either. I differentiated u and tried plugging in the partial fractions but that's as far as I went.
Once I made the substitution, I used an integrating factor and then integrated both sides as normal for u, then integrated to get y. Not sure why my answer was so messed up, I had to do a bit of nasty manipulating to be able to integrate too.
14. (Original post by Barbecuetime123)
Once I made the substitution, I used an integrating factor and then integrated both sides as normal for u, then differentiated to get y. Not sure why my answer was so messed up, I had to do a bit of nasty manipulating to be able to integrate too.
I wanted to use the integrating factor but I wasn't sure if I should. I thought that's a way to go but it would make it messy so I kept thinking about an easier way and time run out as I wanted to try it for the sake of it.
15. (Original post by Xzielon)
I didn't know what to do there either. I differentiated u and tried plugging in the partial fractions but that's as far as I went.
basically what I did:

sub in u and du/dx for dy/dx and d2y/dx2

get integrating factor using partial fractions integral (I got the reciprocal of the RHS bit)

integrate 1 and solve how you would an integrating factor one

sub u back out so you get dy/dx = a load of crap

then you have some horrible shite to integrate and you had to rewrite it as x+3+1/(x+1) I think it was, something like that

then you have equation, sub in 0 to find constant value

I got what everyone else got, y = x^2/2 + 3x + ln(x+1) +1
16. Let's just hope the grade boundaries are like, 20 marks for an A :'(
Flopped too hard, I got all positives for question 1, question 2 can't remember, done half of question 4 and flopped 6. Missed out the 4 marker for question 8 :'(
17. (Original post by Mop3476)
basically what I did:

sub in u and du/dx for dy/dx and d2y/dx2

get integrating factor using partial fractions integral (I got the reciprocal of the RHS bit)

integrate 1 and solve how you would an integrating factor one

sub u back out so you get dy/dx = a load of crap

then you have some horrible shite to integrate and you had to rewrite it as x+3+1/(x+1) I think it was, something like that

then you have equation, sub in 0 to find constant value

I got what everyone else got, y = x^2/2 + 3x + ln(x+1) +1
Can you remember the original equation? I wanna try what I planned to do
18. (Original post by Adam_98)
For the very last question, the area one:

Was the asnwer the integral of (1+tanX) between pi/3 and the value of theta at B, minus the area of the triangle (3+2√3)/6
??

Posted from TSR Mobile
what value of theta at b did you get?
19. (Original post by IrrationalRoot)
I noticed that AQA have completely changed the design of the front cover of the paper, dk why.
That **** didn't even tell us how many Qs there were...
20. anyone know the allocation of marks for all parts of the questions if i remember right it was something like
1a. 3
b. 3

2a. 1
b. 4

3a. 6
b. 3
c. 3
ci. 1

4. 5

5a. 1
b. 11

6a. 3
b. 4

7. 10

8a. 3
b. 4
c. 3
d. 7

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