Did anyone else get 112 for the Sigma question? 😦
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C2 Maths AS aqa 2016 (unofficial mark scheme new) watch
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 25052016 14:28

HistoryStudent1
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 25052016 14:33
Pretty sure I got all the marks except 3(d) when I put k = 4, how many marks does that drop?

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 25052016 14:34
(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 = 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 ( 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 [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]
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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 25052016 14:34
(Original post by Alexcor)
Did anyone else get 112 for the Sigma question? 😦
The question told us that sum of second and third term was 50
First term was 28
50 + 28 = 78 
HistoryStudent1
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 25052016 14:40
How is 3d k=5.5?

Porkieee.ee
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 25052016 14:42
For all you dummies who got 90 on 4c watch this video and you'll realise it was 198 😂😂😂😂
https://youtu.be/kCbEsrVEBco 
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 47
 25052016 14:45
(Original post by HistoryStudent1)
How is 3d k=5.5? 
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 25052016 14:47
(Original post by tajtsracc)
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 
Eisobdxhsonw
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 25052016 14:47
Wasn't 8bii 16+9sin^2x/53cosx as opposed to 35cosx on the bottom???
Also wouldn't that make the p+ qcosx equal to 53cosx therefore the minimum point would be when cosx=1 therefore the minimum value would be 53(1) which is 8?? 
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 25052016 14:48
(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/kCbEsrVEBco 
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 51
 25052016 14:48
Hi can anyone tell me how you got 2 for 1b as I got 4/7 thanks

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 25052016 14:48
(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 = 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 ( 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 [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]
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]
Binomial second part was worth 5 and also series question had a greater total marks than that 
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 53
 25052016 14:53
(Original post by FJCX)
Hi can anyone tell me how you got 2 for 1b as I got 4/7 thanks
Make the definite integral equal to 16
(36(3)^1 + (9a)/2)  (36/1 + a/2) = 16
Then you get 12 + 9a/2 + 36  a/2 = 16
24 + 4a = 16
4a = 8
a = 2 
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 25052016 14:53
(Original post by Bosssman)
It was 168  78 = 90
The question told us that sum of second and third term was 50
First term was 28
50 + 28 = 78 
Eisobdxhsonw
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 25052016 14:54
[QUOTE=Eisobdxhsonw;65156665]
Wasn't 8bii 16+9sin^2x/53cosx as opposed to 35cosx on the bottom?????
Also wouldn't that make the p+ qcosx equal to 53cosx therefore the minimum point would be when cosx=1 therefore the minimum value would be 53(1) which is 8??
Also does anyone know the exact logs question for the last one? Because I got a different answer but I'm convinced my method was correct 
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 56
 25052016 14:56
[QUOTE=Eisobdxhsonw;65156859]
(Original post by Eisobdxhsonw)
Wasn't 8bii 16+9sin^2x/53cosx as opposed to 35cosx on the bottom?????
Also wouldn't that make the p+ qcosx equal to 53cosx therefore the minimum point would be when cosx=1 therefore the minimum value would be 53(1) which is 8??
Also does anyone know the exact logs question for the last one? Because I got a different answer but I'm convinced my method was correct
The question was
Show that log4(2x+3) + log4(2x+15) = 1 + log4(14x + 5)
Also the trig question, it resulted in 5 + 3cosx, and so when x = pi, 3cosx = 3 so minimum value is 2Last edited by Bosssman; 25052016 at 15:05. 
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 25052016 15:07

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 25052016 15:09
For the one about expressing y in terms of n and m wasn't it m/26n instead of m/2+6n.

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 59
 25052016 15:11
(Original post by Parhomus)
For the one about expressing y in terms of n and m wasn't it m/26n instead of m/2+6n. 
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 25052016 15:12
(Original post by Bosssman)
Yes it was
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