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

#1
Q1: D (graph with maxima in top left quadrant)
Q2:
Q3:
Q4: Show that. Factor theorem on both, set equal to each other, rearrange
Q5:
Q6: a=2, b= 2root3
Q7a: show that, differentiate, stick in 0, differentiate again to prove not point of inflection
Q7b: (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:
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
Last edited by MagnumKoishi; 2 years ago
0
2 years ago
#2
I got 2021 for question 8c
1
2 years ago
#3
51 years after 1970 =2021
Last edited by ERMmaths; 2 years ago
0
#4
Oof, I got 53. We'll see what people get when more arrive
Evil Homer btw
4
2 years ago
#5
53
1
2 years ago
#6
maybe I put 53, I might have mixed up lol
0
2 years ago
#7
I got 53 too!
0
2 years ago
#8
I got 53
0
2 years ago
#9
There may be a range I’m not sure, but I used the rounded to 3sig values of the graph and got 53
0
#10
(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
0
2 years ago
#11
I got 52.3 years so in 2022
12
2 years ago
#12
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
0
2 years ago
#13
I used the raw (not rounded values), when inserting p and q back into the formula, rather than using the rounded values.
0
2 years ago
#14
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.
0
#15
(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
0
2 years ago
#16
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.
0
2 years ago
#17
(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.
0
2 years ago
#18
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
5
2 years ago
#19
(Original post by ERMmaths)
2019,you?
1
#20
(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
0
X

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