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C2 Maths AS aqa 2016 (unofficial mark scheme new)

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1. Isn't the perimeter of the circle 13.8cm?
2. (Original post by Porkieee.ee)
18/2(56+17(-2))
9(56-34)
9(22)
=198
That sort of series question doesn't work like that.
The series started from 4 and ends at 21, so the final answer will be S21 - S3, so you get
21/2(2(28)+20(-2)) - 3/2(2(28)+2(-2))
= 168 - 78
= 90
3. 9a and b were both worth 4 marks
4. You assumed a was 28 from the parts before, but it wasn't when doing the sigma part. As you were starting from the 4th term for this last part, it gave a (first value aka U4), as 22

Therefore 9(44-34)
9(10)
=90
5. Wasn't it 13.8 for perimeter? Some are saying its 13.6 but pretty sure it was 13.8 if you didn't round before the final answer
6. (Original post by Excuse Me!)
4B

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?
i did that but i ended up with a+2d=0 lol gg marks
7. (Original post by Roxyfreeman)
Isn't the perimeter of the circle 13.8cm?
"Circle"
8. (Original post by Khalidak)
"Circle"
I meant sector as you probably know
9. I got 88 for 4c? Anyone else get that?

I did the sum of 12 - sum of 3
10. (Original post by Grakata)
That sort of series question doesn't work like that.
The series started from 4 and ends at 21, so the final answer will be S21 - S3, so you get
21/2(2(28)+20(-2)) - 3/2(2(28)+2(-2))
= 168 - 78
= 90
ah crap i done that and crossed it out and done it the other way that gave the wrong answer, awesome
11. (Original post by Bosssman)
Wasn't it 13.8 for perimeter? Some are saying its 13.6 but pretty sure it was 13.8 if you didn't round before the final answer
I can't remember if my answer was 13.8 or 13.6, though I know I did absolutely no rounding before the final answer.
12. (Original post by Grakata)
That sort of series question doesn't work like that.
The series started from 4 and ends at 21, so the final answer will be S21 - S3, so you get
21/2(2(28)+20(-2)) - 3/2(2(28)+2(-2))
= 168 - 78
= 90
Ok son, I will award you 1 mark for effort 😂😂😂😂
13. Made a few changes and sorted the marks!

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 [3]
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 [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 [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 that’s what I wrote) [2]

7)a) expand (2+x)5 work out p q r, i think p was 10 q was +40 and r was -80) [4]
b)find the coefficient of x10 in the expansion of (2+x)5 * (something else)7
it was something like -1648 [5]

8) find the value of tan(x) tan(x) = -5/4 [2]
bi) values of tan(x) between 0 and 360 degrees (45, 129, 225, 309) all in degrees [3]
bii) 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]
14. Where are the 7marks that aren't included! Counted up the marks from the inoffical mark scheme and it was only 68 but the paper was out of 75!!
15. (Original post by voltz)
I got 88 for 4c? Anyone else get that?

I did the sum of 12 - sum of 3
I'm not sure what you did if you meant you did S21 - S3 and you had a typo in your post, though I don't really know what you did, though I'm very sure that the final answer was 90.

Maybe you dropped a number somewhere?
16. (Original post by Porkieee.ee)
I'm pretty sure Perimeter of shaded area was 13.8 cm
Yeh I got that as well
17. (Original post by Harryg123456)
Yeh I got that as well
Was it to 3 sig fig?
18. (Original post by Eliottooooo)
Was it to 3 sig fig?
Yeah it was
19. (Original post by Harryg123456)
Yeah it was
For god's sake, I was too buzzed in the moment with knowing how to do it, left it as 13.76
20. (Original post by Grakata)
I'm not sure what you did if you meant you did S21 - S3 and you had a typo in your post, though I don't really know what you did, though I'm very sure that the final answer was 90.

Maybe you dropped a number somewhere?
Yeah I meant 21, I got 88 as the final answer? I did sum to 21 which was 168 - 80 which I got for the sum to 3

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