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

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**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 co-ordinate of the maximum point M

**M(9,6) [3]**

c) equation of normal to curve at P, P was (4,5)

**y=-2x+13 [3]**

d) the normal to curve at P is translated by (k,0) find value of k,

**k = 5.5 [3]**

4)a) show that

**a+10d=8 [3]**

b)sum to 2nd and 3rd term is 50, work out the 12th term of the series

**D=-2 and A=28, u12 = 6 [4]**

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 [3]**

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 5.829**

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=(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 (1+2x)

^{5}work out p q r, i think p = -10, q= 40 r = - 80

**(correct me if im wrong) [3]**

b)find the coefficient of x

^{10}in the expansion of (1+2x)^5 * (2-x)^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+9sin

^{2}x / 5 - 3cosx 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]**

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#2

Predicted Grade Boundaries

A - 61

B - 56

C - 51

Was easier than last year but not the easiest paper there's been.

A - 61

B - 56

C - 51

Was easier than last year but not the easiest paper there's been.

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i dont think this paper was too hard, to me it seems harder than last year but at the same time easier lol

MATHEMATICS UNIT MPC2 2015 last year was A 58 B 50 C 42 D 35 E 28

i think this year it will be A 63 B 55 C 47 D 40 E 33

MATHEMATICS UNIT MPC2 2015 last year was A 58 B 50 C 42 D 35 E 28

i think this year it will be A 63 B 55 C 47 D 40 E 33

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#6

(Original post by

1)a) integrate something to get

b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1,

2)a)

b) use logarithms to solve (0.2)

c) describe transformation of (0.2)

3)a) differentiate something to get

b) find y co-ordinate of the maximum point M

c) equation of normal to curve at P i think P was (5,6)

d) the normal to curve at P is translated by (0,k) find value of k,

4)a) show that

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

c) work out the sum of the 21st term minus sum of 3rd term (i dont know the answer)

5)a) show that thetha (in radians) was

b) work out area of triangle

c) work out perimeter of shaded area, first piece of info given ->

you had to realise that the

re-arrange to get 2(area of sector) = 19.9 and solve for

perimeter of shaded = (9-r) + (8-r) + 5 + (length of arc = r*theta)

6)a)trapezium rule to get

b) describe transformation of graph y=(x

c) describe transformation of graph y=(x

7)a) expand (2+x)

b)find the coefficient of x

it was something like

8) find the value of tan(x)

b) re-arrange 16+9sin

9)re-arrange something into the form (c)

b) find the one and only solution of something,

**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 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 (i dont know the answer)

**[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 = 13.6 [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**b) re-arrange 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)re-arrange something into the form (c)

^{1/2}/ d^2 to get 3^{y }**y = 1/2 ( m + 12n) [3]**b) find the one and only solution of something,

**x = 5/2**( you should get two equal roots i think)**[5]**
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(Original post by

3a) Was something like 6x^(1/2) -x

6c) Is that not a stretch along the y-axis SF 3?

**jake4198**)3a) Was something like 6x^(1/2) -x

6c) Is that not a stretch along the y-axis SF 3?

6c i dont think so because think of it like this

y=f(x)

if i applied y=f(1/3 x)

that means i stretch it by sf 3 because it would be 1/(1/3) or 1/a

hence its not the same as stretching in y because you only stretch all x values by sf 3 on the graph but in the equation it is written as 1/3

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#8

Hardest c2 I've seen , I'd say grade boundaries the same or lower than last year. I hope around 58-60 for an A and 50-52 for a B

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#10

(Original post by

when u differeentiate u should get what i got in the mark scheme

6c i dont think so because think of it like this

y=f(x)

if i applied y=f(1/3 x)

that means i stretch it by sf 3 because it would be 1/(1/3) or 1/a

hence its not the same as stretching in y because you only stretch all x values by sf 3 on the graph but in the equation it is written as 1/3

**beanigger**)when u differeentiate u should get what i got in the mark scheme

6c i dont think so because think of it like this

y=f(x)

if i applied y=f(1/3 x)

that means i stretch it by sf 3 because it would be 1/(1/3) or 1/a

hence its not the same as stretching in y because you only stretch all x values by sf 3 on the graph but in the equation it is written as 1/3

Either way for 6c, 2 marks shouldn't be too detrimental. As long as the boundaries are at most 60 for an A I should be alright.

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#11

(Original post by

Predicted Grade Boundaries

A - 61

B - 56

C -

Was easier than last year but not the easiest paper there's been.

**jake4198**)Predicted Grade Boundaries

A - 61

B - 56

C -

Was easier than last year but not the easiest paper there's been.

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it was easier than last years but it was very tricky at the same time

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#16

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#17

Would you get atleast a mark for showing this?

I wrote 1/2r²θ = 19.9 - 1/2r²θ

**area of sector = area of triangle - area of sector**I wrote 1/2r²θ = 19.9 - 1/2r²θ

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#18

(Original post by

4)a) show that

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

c) work out the sum of the 21st term minus sum of 3rd term

5)a) show that thetha (in radians) was

b) work out area of triangle

c) work out perimeter of shaded area, first piece of info given ->

you had to realise that the

re-arrange to get 2(area of sector) = 19.9 and solve for

perimeter of shaded = (9-r) + (8-r) + 5 + (length of arc = r*theta)

**beanigger**)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 = 13.6 [6]****4**B

From part a) you know that a + 10d = 8.

Sum of U3 and U2 = 50 so you wrote equations for each (U3 = a + d....) and add them together to get 2a + 3d = 50 (think those were the numbers).

Then you solve simultaneously with equation from part a) to get your a and d values.

For 5c I think it was 13.8?

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#19

http://www.thestudentroom.co.uk/show....php?t=4117603 Made a poll for the AQA C2 exam

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