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

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2 years ago
#61
Expecting full marks, all of my answers seem identical to the ones here. Still, not the easiest paper!
~60 for an A I reckon.
0
2 years ago
#62
(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 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 [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 [4]
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]
7a was the question about the expansion of (1-2x)^5 and p was -10, q was 40 and r was -80 i think.
0
2 years ago
#63
(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
Dammit, I forgot to take it away from 9 -.-
0
2 years ago
#64
(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 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 [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 [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) 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]
I got 3^(-m+4n) for 9a and 9/10 for b? Also, -1552 for 7b
0
2 years ago
#65
(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 90😭😭
1
2 years ago
#66
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
0
2 years ago
#67
(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
What you did is correct; you just forgot that x^7 x x^3=x^10 and that x^4 x x^6=x^10.
So you were missing two of them to finish it off.
0
2 years ago
#68
Will I get a mark for not saying (0, 1) on 2a? I did put 1 where it crossed the y-axis and zero on the graph but didn't state the coordinates.
0
2 years ago
#69
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!?
0
2 years ago
#70
(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 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 [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 [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) 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]
thanks for this
0
2 years ago
#71
(Original post by jake4198)
Will I get a mark for not saying (0, 1) on 2a? I did put 1 where it crossed the y-axis and zero on the graph but didn't state the coordinates.
Yes, they generally say in the mark scheme allow y-intercept to be marked on the graph as a 1
0
2 years ago
#72
(Original post by jake4198)
Will I get a mark for not saying (0, 1) on 2a? I did put 1 where it crossed the y-axis and zero on the graph but didn't state the coordinates.
I think that's fine; but in doubt I would probably put 0,1.
0
2 years ago
#73
(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!?
Mtangent was=1/2 therefore Mnormal=-2. Therefore you just write y-5=-2(x-4) or whatever. I think he just expanded and rearranged it which is fine too.
0
2 years ago
#74
(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 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 [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 [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) 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]
I got R (for perimeter) as 5.919 I did 0.5r^2theta = 9.95 then rearranged it for R and got 5.919? Anyone else
0
2 years ago
#75
(Original post by Bosssman)
Yes, they generally say in the mark scheme allow y-intercept to be marked on the graph as a 1
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 cor-ordinate the other way around

will i still get the full marks ?
0
2 years ago
#76
yup, I most definitely failed. with dignity x)
1
2 years ago
#77
(Original post by money-for-all)
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 cor-ordinate the other way around

will i still get the full marks ?
Potentially, mark schemes generally disallow marks if there is a contradiction, and in your case there is a contradiction, so really it depends on how observant the marker is!
1
2 years ago
#78
QUESTION 8B:

(Copied from first post):

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]

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?
0
2 years ago
#79
(Original post by IsThisReallyLife)
QUESTION 8B:

(Copied from first post):

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]

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?
It was 5 + 3cosx
Also the question had it as 'value', not plural
0
2 years ago
#80
(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 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 [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 [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) 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]
can I just say I did that for 9b at first and got plus/minus 5/2
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^2-4ac= 0 on the quadratic that it formed - will I still get the marks???? FMLLLLLLL
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