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AQA A2 MFP2 Further Pure 2 – 24th June [Exam Discussion Thread]

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Original post by C0balt
What was 3b about that everyone's talking about?


I believe it was the evaluation of s (the arc length).
For anyone wondering how it was done, you could simply decompose into partial fractions or do what I did and add and subtract 2 in the numerator.
Original post by IrrationalRoot
I believe it was the evaluation of s (the arc length).
For anyone wondering how it was done, you could simply decompose into partial fractions or do what I did and add and subtract 2 in the numerator.

Ah right, I was about to panic but I did partial fraction in the end haha
That was a pretty fun paper, must admit. :biggrin:
1) A=4,
50/609

2) beta = 1-2i,
alpha*beta = 5,
gamma = 1/3,
P=-7, Q=-5

3) -3/4 + ln7

4) root3*pi / 18

5) Modulus = 8,
Locus is a circle centred at (-4root3, 4) radius 4, arg(w) = 2pi/3
z = 2e^i*theta where theta = -7pi/18, 5pi/18, 17pi/18

6) ln(2/3)
3arcosh(2) + 6root3 + 5

7)
Original post by IrrationalRoot
If anyone wants to know how the induction went:(1+p)1≥1+p(1+p)^1 \geq 1+p obviously, so base case established.Assume (1+p)k≥1+kp(1+p)^k \geq 1+kp. Clearly we need to multiply by 1+p1+p and we can do so without changing the sign of the inequality because p≥−1⇒1+p≥0p\geq -1\Rightarrow 1+p \geq 0.Therefore (1+p)k+1≥(1+kp)(1+p)=1+p+kp+kp2=1+(k+1)p+kp2≥1+(k+1)p(1+p)^{k+1} \geq (1+kp)(1+p)=1+p+kp+kp^2=1+(k+1)p+kp^2 \geq 1+(k+1)p, since p2≥0p^2 \geq 0, k≥1⇒kp2≥0k \geq 1 \Rightarrow kp^2 \geq 0.Therefore result holds for n=k+1n=k+1 and by induction we have result.


8) itan(3pi/8), itan(5pi/8), itan(7pi/8)
tan2(pi/8)tan2(3pi/8) = 1
tan2(pi/8) + tan2(3pi/8) = 6
What UMS would 55 marks get me?

Fairly easy paper generally with quite a few impossible questions mixed in, I thought.
Fair paper but I didnt answer like 14 mark; no proof, missed out the last 8 marks and couldnt integrate the fraction with x squared fully. Probs got a nice B
Reply 186
Avatar for jjsnyder
jjsnyder
OP
10
Original post by bencurnow
1) A=4,
50/609

2) beta = 1-2i,
alpha*beta = 5,
gamma = 1/3,
P=-7, Q=-5

3) -3/4 + ln7

4) root3*pi / 18

5) Modulus = 8,
Locus is a circle centred at (-4root3, 4) radius 4, arg(w) = 2pi/3
z = 2e^i*theta where theta = -7pi/18, 5pi/18, 17pi/18

6) ln(2/3)
3arcosh(2) + 6root3 + 5

7)

8) itan(3pi/8), itan(5pi/8), itan(7pi/8)
tan2(pi/8)tan2(3pi/8) = 1
tan2(pi/8) + tan2(3pi/8) = 6


There's an unofficial MS out btw :smile:
http://www.thestudentroom.co.uk/showthread.php?t=4184065
(edited 7 years ago)
how many marks was the proof by induction?
Original post by fffffg1g1
how many marks was the proof by induction?

Six
was x=ln(2/3) definitely the root?
What UMS would 67 give me?
Original post by C0balt
What UMS would 67 give me?


93-97 probably.
Original post by student0042
93-97 probably.

Hopefully 90s...
Original post by C0balt
Hopefully 90s...


Well there's no point trying to over analyse your exam when you have others left to do.
Original post by C0balt
What UMS would 67 give me?


Last year that was the 100 cap and imo this paper was harder and more people will have lost larger chunks of marks
Original post by MrMagic12345
Last year that was the 100 cap and imo this paper was harder and more people will have lost larger chunks of marks

Well I found last year's harder apart from induction but I was less prepared when I did the mock so maybe you're right. Oh well. Onto M2 now haha
Reply 196
Avatar for Jm098
Jm098
2
can anyone remember the limits for the 1+x^2/1-x^2? thanks
Original post by Jm098
can anyone remember the limits for the 1+x^2/1-x^2? thanks


0 and 3/4 I think
Thought it was a very nice paper, surprisingly. I agree on all the posted answers.
Original post by A Slice of Pi
Thought it was a very nice paper, surprisingly. I agree on all the posted answers.


What'd you think o 'the induction? (broken keyboard)

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