C2 Maths AS aqa 2016 (unofficial mark scheme new)

think it was p=-10 q-40 r--80 as the other 80 in the expansion was already showed in the question

It was 90😭😭

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]

In x isnt it the opposite, so moving right it a negative value?

Either way I'm sure it was (k,0) not (0,k) like in the OP

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]

Haha here's hoping! Best of luck to you too!

could you link me to the physics thread? This is for the new specification right?

Yep, new spec:
http://www.thestudentroom.co.uk/showthread.php?t=4114965

This is for the Breadth, Im not sure if there is one for the Depth one yet

i think i failed c2 tbh, maybe a D ? if im lucky

lets say if i get;
'C' in c1
'D' in c2
what will i have to get in s1 to get a B (and is it even still possible for me to get a B ? LMAO )

I mean B as an overall grade in AS maths

i think i failed c2 tbh, maybe a D ? if im lucky

lets say if i get;
'C' in c1
'D' in c2
what will i have to get in s1 to get a B (and is it even still possible for me to get a B ? LMAO )

I mean B as an overall grade in AS maths

hhmmm idk probably a high a but mayb you did nt do as bad as you think

Thanks Gabby the Greek !

i like your motivation , im going to vote to stay in EU just for you !
long like Greece and the UK !!

Couldn't do 8b! How did you get it into the form p+qcosx