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AQA A level Maths Paper 2 Unofficial Marks Scheme 12th Jun

Q1: D (graph with maxima in top left quadrant)
Q2: a815a^{\frac{8}{15}}
Q3: x2x^2
Q4: Show that. Factor theorem on both, set equal to each other, rearrange
Q5: t2=−2x(ln⁡x+1)+6t^2 = \frac{-2}{x}(\ln{x} + 1) + 6
Q6: a=2, b= 2root3
Q7a: show that, differentiate, stick in 0, differentiate again to prove not point of inflection
Q7b: −4p3<q<0-4p^3<q<0 (turning points must be one above and one below x axis. Minimum when is x = 0, maximum when x=-2p)
Q8b: p=7940, q=1.08
Q8c: 2021/2022/2023. It depends on whether you use rounded or exact values from part b. It didnt specify which to use- so like physics, these will all probably be accepted
Q9b: between plus/minus root2
Q9c: 0.39467
Q9d: invalid as the value is outside the range of a small angle approximation (as derived from maclaurin series)
Q10: B (deceleration)
Q11: 1000N
Q12: - 400
Q13b: teacher is correct- resistive force should make it lower than expected; it's actually greater so machine must be faulty
Q14b: Moments, n=3
Q15a: show trapezium- show that two sides are parallel, the other 2 are not (eg show multiple, use dot product methods etc)
Q15b: 17m/s
Q16a: 11.5
Q16b: s=11.71t+58445e−0.9t−0.1e0.3t−115990s=11.71t + \frac{584}{45} e^{-0.9t} - 0.1e^{0.3t} - \frac{1159}{90}
Q16c: the model is good, as if you sub in earlier values, you get very close to 100
Q17a: show that. N2L and resolving
Q17b: parts a and b rely on the limiting case of friction as nonzero velocity- as Lizzy and her sledge are not moving in the new case, friction is likely to be less than μR
(edited 4 years ago)

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Reply 1
Avatar for ERMmaths
ERMmaths
3
I got 2021 for question 8c
Reply 2
Avatar for ERMmaths
ERMmaths
3
51 years after 1970 =2021
(edited 4 years ago)
Oof, I got 53. We'll see what people get when more arrive
@Evil Homer btw
53
Reply 5
Avatar for ERMmaths
ERMmaths
3
maybe I put 53, I might have mixed up lol
I got 53 too!
Reply 7
Avatar for Ariyaa
Ariyaa
11
I got 53
There may be a range I’m not sure, but I used the rounded to 3sig values of the graph and got 53
Original post by Edexcel Are mugs
There may be a range I’m not sure, but I used the rounded to 3sig values of the graph and got 53


There'll be a range for the values of p and q, but the range won't be big enough to allow anything other than one year
Reply 10
Avatar for 123mjr
123mjr
4
I got 52.3 years so in 2022
Reply 11
Avatar for _zoe
_zoe
6
I got 0<q<14p^3 for the cubic one and for the year question I think it depends on wether you used the unrounded values of p and q so hopefully the mark scheme will have a range
I used the raw (not rounded values), when inserting p and q back into the formula, rather than using the rounded values.
Just checked with pretty much the correct values and was still 2023
Original post by ERMmaths
I used the raw (not rounded values), when inserting p and q back into the formula, rather than using the rounded values.
Original post by _zoe
I got 0<q<14p^3 for the cubic one and for the year question I think it depends on wether you used the unrounded values of p and q so hopefully the mark scheme will have a range


I'm almost sure q has to be negative, but I may indeed be wrong. We'll see what others get
Oh Thank god! I did the same. That’s what I saw on the mark scheme for a similar past paper question.
Original post by ERMmaths
I used the raw (not rounded values), when inserting p and q back into the formula, rather than using the rounded values.
Original post by quantumgen
Oh Thank god! I did the same. That’s what I saw on the mark scheme for a similar past paper question.

what answer did you get?
Wasn’t q12 -390?
The y part was 0. X part was 400. And the speed was 10m/s.
Original post by MagnumKoishi
Q1: D (graph with maxima in top left quadrant)
Q2: a^(8/15)
Q3: x^2
Q4: Show that. Factor theorem on both, set equal to each other, rearrange
Q5: t^2 = -2/x(lnx 1) 6
Q6: a=2, b= 2root3
Q7a: show that, differentiate, stick in 0, differentiate again to prove not point of inflection
Q7b: -4p^3<q<0
Q8b: p=7940, q=1.08
Q8c: 2023
Q9b: between - 2root2
Q10: B (declaration)
Q11: 1000N
Q12: - 400
Q13b: correct as greater than expected
Q14b: 17
Q15b: 17
Q16a: 11.5
Q16b: 11.71t 584/45e^-0.9t - 0.1e^0.3t - 1159/90
Q17: show that. N2L and resolving
Original post by ERMmaths
what answer did you get?


2019,you?
Original post by quantumgen
Wasn’t q12 -390?
The y part was 0. X part was 400. And the speed was 10m/s.


No. Constant speed means no resultant force

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