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 (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=(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
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) [3]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [5]

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

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

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

Predicted Grade Boundaries

A - 61
B - 56
C -

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

Think the answer for 4c was 90

4)c) Is S21 - S3, which = 90

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]