Unoffical Mark scheme for C2 AQA 2016
It would help if you could correct the amount of marks in each question if it is wrong please
Questions:
1)a) integrate something to get 36x^{1} + ax^{2 }/2 [3]
b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a =  2 [2]
2)a) Draw the graph of (0.2)^{x }and state where itcrosses axis (0,1)[2]
b) use logarithms to solve (0.2)^{x} = 4 , x = 0.861 [2]
c) describe transformation of (0.2)^{x } onto (5)^{x} reflection in y axis [1]
3)a) differentiate something to get 3x^{1/2 }1 [2]
b) find y coordinate of the maximum point M M(9,6) [3]
c) equation of normal to curve at P, P was (4,5) y=2x+13 [3]
d) the normal to curve at P is translated by (k,0) find value of k, k = 5.5 [3]
4)a) show that a+10d=8 [3]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series D=2 and A=28, u12 = 6 [4]
c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]
5)a) show that thetha (in radians) was 0.568 [3]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given > area of sector = area of shaded area
you had to realise that the area of sector = area of triangle  area of sector
rearrange to get 2(area of sector) = 19.9 and solve for r to get 5.829
perimeter of shaded = (9r) + (8r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.8 [6]
6)a)trapezium rule to get 65.6 [4]
b) describe transformation of graph y=(x^{2} +9)^{1/2} onto y = 5 + (x^{2} +9)^{1/2}
Translation by (0, 5) [2]
c) describe transformation of graph y=(x^{2} +9)^{1/2} onto y=3(x^{2} +1)^{1/2} Stretch by scale factor 1/3 in x direction [2]
7)a) expand (1+2x)^{5} work out p q r, i think p = 10, q= 40 r =  80 (correct me if im wrong) [3]
b)find the coefficient of x^{10} in the expansion of (1+2x)^5 * (2x)^7
it was something like 1648 [5]
8)a)find the value of tan(x) tan(x) = 5/4 [2]
b) values of tanx in the range of 0 to 360 degrees 45, 129, 225, 309 [3]
b) rearrange 16+9sin^{2}x / 5  3cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]
9)rearrange something into the form (c)^{1/2} / d^2 to get 3^{y }y = 1/2 ( m  12n) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4]
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C2 Maths AS aqa 2016 (unofficial mark scheme new)
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 1
 25052016 12:40
Last edited by username2592579; 29052016 at 12:56.Post rating:12 
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 2
 25052016 12:45
Predicted Grade Boundaries
A  61
B  56
C  51
Was easier than last year but not the easiest paper there's been.Post rating:3 
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 3
 25052016 12:47
Would do you guys predict an A will be? More than or less than 60?
Post rating:1 
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 4
 25052016 12:47
Think the answer for 4c was 90
Post rating:1 
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 5
 25052016 12:49
i dont think this paper was too hard, to me it seems harder than last year but at the same time easier lol
MATHEMATICS UNIT MPC2 2015 last year was A 58 B 50 C 42 D 35 E 28
i think this year it will be A 63 B 55 C 47 D 40 E 33 
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 6
 25052016 12:49
(Original post by beanigger)
Unoffical Mark scheme for C2 AQA 2016
It would help if you could correct the amount of marks in each question if it is wrong please
Questions:
1)a) integrate something to get 36x^{1} + ax^{2 }/2 [3]
b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a =  2 [2]
2)a) Draw the graph of (0.2)^{x }and state where itcrosses axis (0,1)[2]
b) use logarithms to solve (0.2)^{x} = 4 , x = 0.861 [2]
c) describe transformation of (0.2)^{x } onto (5)^{x} reflection in y axis [1]
3)a) differentiate something to get 3x^{1/2 }1 [2]
b) find y coordinate of the maximum point M M(9,6) [3]
c) equation of normal to curve at P i think P was (5,6) y=2x+13 [2]
d) the normal to curve at P is translated by (0,k) find value of k, k = 5.5 [3]
4)a) show that a+10d=8 [2]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series (if someone could explain how to do this it would be nice i got D=1 and A=2) the answer everyone seems to be getting is D=2 and A=28, u12 = 6 [3]
c) work out the sum of the 21st term minus sum of 3rd term (i dont know the answer) [3]
5)a) show that thetha (in radians) was 0.568 [2]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given > area of sector = area of shaded area
you had to realise that the area of sector = area of triangle  area of sector
rearrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
perimeter of shaded = (9r) + (8r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.6 [6]
6)a)trapezium rule to get 65.6 [4]
b) describe transformation of graph y=(x^{2} +9)^{1/2} onto y = 5 + (x^{2} +9)^{1/2}
Translation by (0, 5) [2]
c) describe transformation of graph y=(x^{2} +9)^{1/2} onto y=3(x^{2} +1)^{1/2} Stretch by scale factor 1/3 in x direction ( not 100% sure about this but thats what i wrote) [2]
7)a) expand (2+x)^{5} work out p q r, i think p was 10 q was 80 and r was 80 (correct me if im wrong) [4]
b)find the coefficient of x^{10} in the expansion of (2+x)^{5} * (something else)^{7}
it was something like 1648 [4]
8) find the value of tan(x) tan(x) = 5/4
b) rearrange 16+9sin^{2}x / 3  5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]
9)rearrange something into the form (c)^{1/2} / d^2 to get 3^{y } y = 1/2 ( m + 12n) [3]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [5]Last edited by jake4198; 25052016 at 12:51.Post rating:3 
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 7
 25052016 12:52
(Original post by jake4198)
3a) Was something like 6x^(1/2) x
6c) Is that not a stretch along the yaxis SF 3?
6c i dont think so because think of it like this
y=f(x)
if i applied y=f(1/3 x)
that means i stretch it by sf 3 because it would be 1/(1/3) or 1/a
hence its not the same as stretching in y because you only stretch all x values by sf 3 on the graph but in the equation it is written as 1/3Post rating:1 
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 8
 25052016 12:54
Hardest c2 I've seen , I'd say grade boundaries the same or lower than last year. I hope around 5860 for an A and 5052 for a B
Last edited by Rectum_resizer; 25052016 at 12:55.Post rating:4 
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 9
 25052016 12:54
The answer for 4c was 198
Post rating:1 
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 10
 25052016 12:57
(Original post by beanigger)
when u differeentiate u should get what i got in the mark scheme
6c i dont think so because think of it like this
y=f(x)
if i applied y=f(1/3 x)
that means i stretch it by sf 3 because it would be 1/(1/3) or 1/a
hence its not the same as stretching in y because you only stretch all x values by sf 3 on the graph but in the equation it is written as 1/3
Either way for 6c, 2 marks shouldn't be too detrimental. As long as the boundaries are at most 60 for an A I should be alright. 
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 11
 25052016 12:57
(Original post by jake4198)
Predicted Grade Boundaries
A  61
B  56
C 
Was easier than last year but not the easiest paper there's been. 
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 12
 25052016 12:58
(Original post by Harryg123456)
Think the answer for 4c was 90Post rating:1 
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 13
 25052016 12:58
4)c) Is S21  S3, which = 90
Post rating:3 
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 14
 25052016 12:58
it was easier than last years but it was very tricky at the same time

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 15
 25052016 13:00
I'm pretty sure Perimeter of shaded area was 13.8 cm
Post rating:3 
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 16
 25052016 13:02
(Original post by Grakata)
4)c) Is S21  S3, which = 90
9(5634)
9(22)
=198 
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 25052016 13:02
Would you get atleast a mark for showing this?
area of sector = area of triangle  area of sector
I wrote 1/2r²θ = 19.9  1/2r²θ 
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 18
 25052016 13:05
(Original post by beanigger)
4)a) show that a+10d=8 [2]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series (if someone could explain how to do this it would be nice i got D=1 and A=2) the answer everyone seems to be getting is D=2 and A=28, u12 = 6 [3]
c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]
5)a) show that thetha (in radians) was 0.568 [2]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given > area of sector = area of shaded area
you had to realise that the area of sector = area of triangle  area of sector
rearrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
perimeter of shaded = (9r) + (8r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.6 [6]
From part a) you know that a + 10d = 8.
Sum of U3 and U2 = 50 so you wrote equations for each (U3 = a + d....) and add them together to get 2a + 3d = 50 (think those were the numbers).
Then you solve simultaneously with equation from part a) to get your a and d values.
For 5c I think it was 13.8? 
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 19
 25052016 13:06
http://www.thestudentroom.co.uk/show....php?t=4117603 Made a poll for the AQA C2 exam
Post rating:2 
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 20
 25052016 13:07
6061 for an A imo.
Full ums probably 73?
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Updated: June 4, 2016
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