Expecting full marks, all of my answers seem identical to the ones here. Still, not the easiest paper!
~60 for an A I reckon.
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C2 Maths AS aqa 2016 (unofficial mark scheme new)
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 61
 25052016 14:13

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 62
 25052016 14:14
(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 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 = 13.8 [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 (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 [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) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4] 
 Follow
 63
 25052016 14:19
(Original post by Bosssman)
You had to find the x value of the normal line at y = 6, I think it was 3.5, then the x value at M was 9, so 9  3.5 = 5.5 
 Follow
 64
 25052016 14:22
(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 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 = 13.8 [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 (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 [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 / 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) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4] 
 Follow
 65
 25052016 14:28
(Original post by Porkieee.ee)
For all you dummies who got 90 on 4c watch this video and you'll realise it was 198 😂😂😂😂
https://youtu.be/kCbEsrVEBcoPost rating:1 
 Follow
 66
 25052016 14:28
I got 2688 or something for the binomial😂😂 damn as I thought you had to find out x^5 for each term and then times them as you add the powers. Lol I failed

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 67
 25052016 14:32
(Original post by bethey180)
I got 2688 or something for the binomial😂😂 damn as I thought you had to find out x^5 for each term and then times them as you add the powers. Lol I failed
So you were missing two of them to finish it off. 
 Follow
 68
 25052016 14:32
Will I get a mark for not saying (0, 1) on 2a? I did put 1 where it crossed the yaxis and zero on the graph but didn't state the coordinates.

 Follow
 69
 25052016 14:34
I can't remember if I did the normal right! I'm doing it now though and I can't get an answer so are the numbers in the mark scheme correct!?

 Follow
 70
 25052016 14:35
(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 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 = 13.8 [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 (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 [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 / 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) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4] 
 Follow
 71
 25052016 14:36
(Original post by jake4198)
Will I get a mark for not saying (0, 1) on 2a? I did put 1 where it crossed the yaxis and zero on the graph but didn't state the coordinates. 
 Follow
 72
 25052016 14:36
(Original post by jake4198)
Will I get a mark for not saying (0, 1) on 2a? I did put 1 where it crossed the yaxis and zero on the graph but didn't state the coordinates. 
 Follow
 73
 25052016 14:37
(Original post by Hbassett26)
I can't remember if I did the normal right! I'm doing it now though and I can't get an answer so are the numbers in the mark scheme correct!? 
 Follow
 74
 25052016 14:38
(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 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 = 13.8 [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 (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 [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 / 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) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4] 
 Follow
 75
 25052016 14:44
(Original post by Bosssman)
Yes, they generally say in the mark scheme allow yintercept to be marked on the graph as a 1
but the sketch was correct,
and above the sketch i said 'when x=0 , y=1'
but i put the corordinate the other way around
will i still get the full marks ? 
 Follow
 76
 25052016 14:49
yup, I most definitely failed. with dignity x)
Post rating:1 
 Follow
 77
 25052016 14:53
(Original post by moneyforall)
i put (1,0) by mistake
but the sketch was correct,
and above the sketch i said 'when x=0 , y=1'
but i put the corordinate the other way around
will i still get the full marks ?Post rating:1 
 Follow
 78
 25052016 14:54
QUESTION 8B:
(Copied from first post):
8) find the value of tan(x) tan(x) = 5/4 [2]
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]
For part B, wouldnt x= 0, 2pi?
As it was 5  3cos(theta)
If theta = pi then its:
5  (3*1) = 8
hence theta should be 0 or 2pi as this gives:
5  (3*1) =2 = Minimum value?
Also i think the question had it as plural??
Or am i missing something here? 
 Follow
 79
 25052016 14:59
(Original post by IsThisReallyLife)
QUESTION 8B:
(Copied from first post):
8) find the value of tan(x) tan(x) = 5/4 [2]
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]
For part B, wouldnt x= 0, 2pi?
As it was 5  3cos(theta)
If theta = pi then its:
5  (3*1) = 8
hence theta should be 0 or 2pi as this gives:
5  (3*1) =2 = Minimum value?
Also i think the question had it as plural??
Or am i missing something here?
Also the question had it as 'value', not plural 
 Follow
 80
 25052016 15:01
(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 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 = 13.8 [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 (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 [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 / 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) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4]
but then I thoughts
you can't just "say" that you can't put logs at negative so +5/2 would be the answer so I crossed it out and used b^24ac= 0 on the quadratic that it formed  will I still get the marks???? FMLLLLLLLLast edited by alkaline.; 25052016 at 15:03.
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Updated: June 4, 2016
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