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

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2 years ago
#21
Isn't the perimeter of the circle 13.8cm?
1
2 years ago
#22
(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
0
2 years ago
#23
9a and b were both worth 4 marks
1
2 years ago
#24
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
0
2 years ago
#25
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
0
#26
(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
0
2 years ago
#27
(Original post by Roxyfreeman)
Isn't the perimeter of the circle 13.8cm?
"Circle"
0
2 years ago
#28
(Original post by Khalidak)
"Circle"
I meant sector as you probably know
0
2 years ago
#29
I got 88 for 4c? Anyone else get that?

I did the sum of 12 - sum of 3
0
2 years ago
#30
(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
0
2 years ago
#31
(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.
0
2 years ago
#32
(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 😂😂😂😂
0
2 years ago
#33
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]
0
2 years ago
#34
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!!
0
2 years ago
#35
(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?
0
2 years ago
#36
(Original post by Porkieee.ee)
I'm pretty sure Perimeter of shaded area was 13.8 cm
Yeh I got that as well
0
2 years ago
#37
(Original post by Harryg123456)
Yeh I got that as well
Was it to 3 sig fig?
0
2 years ago
#38
(Original post by Eliottooooo)
Was it to 3 sig fig?
Yeah it was
0
2 years ago
#39
(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
1
2 years ago
#40
(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
0
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