C2 Maths AS aqa 2016 (unofficial mark scheme new)

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 it crosses 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 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]

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

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 it crosses 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 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]

For all you dummies who got 90 on 4c watch this video and you'll realise it was 198 😂😂😂😂

https://youtu.be/kCbEsrVEBco

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

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 it crosses 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 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]

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.

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 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!?

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 it crosses 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 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]

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 ?

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?

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 it crosses 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 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]