# C2 Maths AS aqa 2016 (unofficial mark scheme new)Watch

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3 years ago
#41
Did anyone else get 112 for the Sigma question? 😦
0
3 years ago
#42
Pretty sure I got all the marks except 3(d) when I put k = 4, how many marks does that drop?
0
3 years ago
#43
(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 + ax2 /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 co-ordinate 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
re-arrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
perimeter of shaded = (9-r) + (8-r) + 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=(x2 +9)1/2 onto y = 5 + (x2 +9)1/2
Translation by (0, 5) [2]
c) describe transformation of graph y=(x2 +9)1/2 onto y=3(x2 +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 x10 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) re-arrange 16+9sin2x / 3 - 5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

9)re-arrange something into the form (c)1/2 / d^2 to get 3y 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]
hi I got 4/7 for 1b please can you tell me how you got -2
0
3 years ago
#44
(Original post by Alexcor)
Did anyone else get 112 for the Sigma question? 😦
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
0
3 years ago
#45
How is 3d k=5.5?
0
3 years ago
#46
For all you dummies who got 90 on 4c watch this video and you'll realise it was 198 😂😂😂😂

https://youtu.be/kCbEsrVEBco
0
3 years ago
#47
(Original post by HistoryStudent1)
How is 3d k=5.5?
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
0
3 years ago
#48
(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
That's what I wrote.
0
3 years ago
#49
Wasn't 8bii 16+9sin^2x/5-3cosx as opposed to 3-5cosx on the bottom???

Also wouldn't that make the p+ qcosx equal to 5-3cosx therefore the minimum point would be when cosx=-1 therefore the minimum value would be 5-3(-1) which is -8??
0
3 years ago
#50
(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
It was definitely 90
2
3 years ago
#51
Hi can anyone tell me how you got -2 for 1b as I got 4/7 thanks
0
3 years ago
#52
(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 + ax2 /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 co-ordinate 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
re-arrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
perimeter of shaded = (9-r) + (8-r) + 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=(x2 +9)1/2 onto y = 5 + (x2 +9)1/2
Translation by (0, 5) [2]
c) describe transformation of graph y=(x2 +9)1/2 onto y=3(x2 +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 x10 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) re-arrange 16+9sin2x / 3 - 5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

9)re-arrange something into the form (c)1/2 / d^2 to get 3y 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
0
3 years ago
#53
(Original post by FJCX)
Hi can anyone tell me how you got -2 for 1b as I got 4/7 thanks
If you integrated correctly for 1a then the integral should have been -36x^-1 + (ax^2)/2
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
1
3 years ago
#54
(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
Yes i got 168 and planned to take away (50 + U1). I might have typed something worng i guess 😀 fs thanks anyway
0
3 years ago
#55
[QUOTE=Eisobdxhsonw;65156665]

Wasn't 8bii 16+9sin^2x/5-3cosx as opposed to 3-5cosx on the bottom?????

Also wouldn't that make the p+ qcosx equal to 5-3cosx therefore the minimum point would be when cosx=-1 therefore the minimum value would be 5-3(-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
0
3 years ago
#56
[QUOTE=Eisobdxhsonw;65156859]
(Original post by Eisobdxhsonw)

Wasn't 8bii 16+9sin^2x/5-3cosx as opposed to 3-5cosx on the bottom?????

Also wouldn't that make the p+ qcosx equal to 5-3cosx therefore the minimum point would be when cosx=-1 therefore the minimum value would be 5-3(-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 answer to the log question was definitely 5/2

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 2
0
3 years ago
#57
(Original post by Porkieee.ee)
18/2(56+17(-2))
9(56-34)
9(22)
=198
It would be 90 because it said the sum of the first 21 was 168, and you know that the value of a is 28 from the previous question working out and that the sum of u2 and u3 is 50 therefore you do 168-(50+28)=90
0
3 years ago
#58
For the one about expressing y in terms of n and m wasn't it m/2-6n instead of m/2+6n.
0
3 years ago
#59
(Original post by Parhomus)
For the one about expressing y in terms of n and m wasn't it m/2-6n instead of m/2+6n.
Yes it was
1
3 years ago
#60
(Original post by Bosssman)
Yes it was
Thanks; I'm just annoyed that I didn't notice the one about the stretch in sf 1/3 for x and the fact that I drew 0.2^x as a normal exponential graph instead of it being reflected in the y axis despite getting the third part of the question correct for going from 0.2^x to 5^x. And messing up the adding up of the coefficients; besides that I'm fine :P.
0
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