For those aiming for full ums or close to it. Lets hope the grade boundaries for this paper are close to June 2013
IMO an A will probably be 6062 with full UMS being 73/75
June 2015: 64/75 for an A
75 = 100
74 = 98
73 = 96
72 = 95
71 = 93
70 = 91
June 2014: 62/75 for an A
75 = 100
74 = 100
73 = 98
72 = 97
71 = 95
70 = 93
June 2013: 59/75 for an A
75 = 100
74 = 100
73 = 100
72 = 100
71 = 100
70 = 98
Jan 2013: 64/75 for an A
June 2012: 58/75 for an A
75 = 100
74 = 100
73 = 100
72 = 100
71 = 99
70 = 97
Jan 2012: 61/75 for an A
75 = 100
74 = 100
73 = 100
72 = 98
71 = 97
70 = 95
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C1 Maths AS aqa 2016 (unofficial mark scheme new)
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 201
 18052016 20:10
Last edited by Pentaquark; 18052016 at 20:11. 
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 202
 18052016 20:19
Oh thank god, i found the area of triangle to be 9 too. And the area of the shaded region was 299=20, is that right?
Posted from TSR Mobile 
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 203
 18052016 20:22
(Original post by Jin99)
Oh thank god, i found the area of triangle to be 9 too. And the area of the shaded region was 299=20, is that right?
Posted from TSR Mobile 
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 204
 18052016 20:35
For question 3 c) Translation of (1/2 , 41/4), I got 1/2 , could someone explain why I'm wrong please, thanks

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 205
 18052016 20:38
(Original post by Igzzy__)
For question 3 c) Translation of (1/2 , 41/4), I got 1/2 , could someone explain why I'm wrong please, thanks
y=(x7/2)^2  41/4 > y=(x4)^2
So the vertex of the first is (7/2, 41/4)
So to move from that to the vertex of the other curve (4,0)
We need to move from 3.5 > 4 therefore a translation of 0.5 in the xdirection
and upwards of 41/4 from 41/4. 
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 206
 18052016 20:52
(Original post by Pentaquark)
These are the two lines I think...
y=(x7/2)^2  41/4 > y=(x4)^2
So the vertex of the first is (7/2, 41/4)
So to move from that to the vertex of the other curve (4,0)
We need to move from 3.5 > 4 therefore a translation of 0.5 in the xdirection
and upwards of 41/4 from 41/4. 
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 207
 18052016 20:58
For the complete the square questions i used decimals, would i still get full marks?

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 208
 18052016 21:01
So here is a mark scheme which has the equations any everything. Anyone can edit it  so if you wanna colaberate on it just jump in and add the marks you know. We're missing 7. a) and b)
https://docs.google.com/document/d/1...it?usp=sharing 
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 209
 18052016 21:07
(Original post by timtjtim)
Okay, I've added a little bit.
1) a)Asked to work out gradient of a parallel line, m=  3/2
b) Asked to find coordinates of B, B(  3,4)
c) Asked to find K, K=  30
2) a) simplify (3√5)^2 = 45
b) Put ((3√5)^2 + √5) / (7 + 3 √5) into m + n√5= 75  32√5
3) a) y=(x7/2)^2  41/4
b) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3  5x^2 8x + 48. p(3) = 3^3  5 * 3^2  8 * 3 + 48 = 0. As p(3) = 0, (x+3) is a factor of p(x).
b) Three linear factors goes to (x+3)(x4)(x4)
c) find the remainder when x^3  5x^2 8x + 48 was divided by (x+2) R=20
d) factorise p(x) using your answer to part c) as R = (x2)(x^23x+14) + 20
5) a) asked to find equation of circle (x5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find coordinates of B (AB is diameter) B ( 12, 7)
c) asked to find the equation of the tangent at A 7x4y+18=0
d) asked to find the length of CT = 9
6) a)
7) a) I think there were 4 parts to this question I can't remember part a and b
b)
c) Definite integral (from 1 to 2) = 81/4 (20.25)
d) Area of shaded region (integral  triangle area) = 45/4 (11.25)
8) a) can't remember anything except the answers were k>6 and k<3/2 (NOT 6 > k > 3/2 )
Answers that need a question to be assigned to
 Q(5/4), y=32x40, upside down positive graph passing through y axis at 8
 k=4 and k=20
 x = 1±√5
 d^2y/dx^2=  2x  9x^2 sub in xcoord of p to get 45, 45>0 therefore minimum 
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 210
 18052016 21:07
(Original post by Rit101)
Nope pretty sure the top students in my Sixth Form got root65 as the radius 
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 211
 18052016 21:12
(Original post by Igzzy__)
yh so if you move from 3.5 > 4 won't it be 0.5 because you're moving to the right in the xdirection and because its the x value you have to give it in negative
It is simply a translation of 1/2 across and 41/4 up.
I can't see why it would be 0.5 since that means it will move to the left in the negative direction?
e.g. moving y=x^2 across by +4 units will give the equation y=(x4)^2
This is a translation of [4] across. Not 4.
Its describing a movement of one graph to another.
The new equation of (x7/2)41/4 moving to (x4)^2
will be (x3.50.5)41/4 + 41/2 => (x4)^2
If it moves +0.5 then inside the bracket it will be 0.5 
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 212
 18052016 21:18
(Original post by timtjtim)
So here is a mark scheme which has the equations any everything. Anyone can edit it  so if you wanna colaberate on it just jump in and add the marks you know. We're missing 7. a) and b)
https://docs.google.com/document/d/1...it?usp=sharing
https://docs.google.com/document/d/1...it?usp=sharing
If you wish to "chat" with each other and me, go to https://www.yourworldoftext.com/aqac12016msLast edited by timtjtim; 18052016 at 21:25. 
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 213
 18052016 21:20
sorry could someone post the most updated mark scheme so i can update it? im kind of busy to search around :P
thanks 
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 214
 18052016 21:30
(Original post by beanigger)
Unoffical Mark scheme for C1 AQA 2016
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m=  5/3 [2]
b)Asked to find coordinates of B B(  3,4) [3]
c)Asked to find K K=  30 [2]
2) a) simplify (3√5)^2 = 45 [1]
b) i cant remember the question but the answer was75  32√5 [4]
3) a) y=(x7/2)^2  41/4
b) min value = 41/4
c) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3  5x^2 8x + 48, p(3) = 0, (x+3) is a factor of p(x).
b) three linear factors of p(x) = (x+3)(x4)(x4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x2)(x^23x14) + 20 [3]
5) a) asked to find equation of circle (x5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find coordinates of B (AB is diameter) B ( 12, 7) [2]
c) asked to find equation of tangent at A 7x4y+18=0 [5]
d) asked to find length of CT = 9 [2]
6) a) y=32x40
b)Q(5/4) [1]
c) upside down positive graph passing through y axis at 8 [2]
d) x = 1±√5
7) a) i think there was 4 parts to this question i cant remember part a and b
b)
c) definite integral = = 81 /4 [5]
d) area of shaded region = = 45/4 [3]
8) a) cant remember anything except the answers were k>6 and k<3/2
Answers that need a question to be assigned to
 k=4 and k=20
 d^2y/dx^2=  2x  9x^2 sub in xcoord of p to get 45, 45>0 therefore minimum 
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 215
 18052016 21:34
http://prntscr.com/b5o7n1
Do you need to put "+c" for the integration? I was taught you don't have to if you find the integral between x and y. :'( 
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 216
 18052016 21:35
Calculating coordinate Q was in Q7 I am sure because you had to use that to find the area of the triangle.
I dont understand why it would be in 6 (a) & 6 (b) ?
(a) You had to find the derivative of the equation 4x^23x^3 using point P(2,24)
dy/dx= (2x9x^2) so subbing in 2 gives 32.
y24=32(x+2)
y=32x64+24
y=32x40
(b) then you'd use that to find coordinate Q by subbing in y=0
into y=32x40
0=32x40 to get
x=5/4
(c)? Then you had to find that integral.
(d)?
To find the area of the triangle you use Coordinate Q 5/4
You had to subtract 5/4 from the limit (2)
So 5/4   2 = 3/4
Base of the triangle is then 3/4
So the area of the triangle is 0.5 * 0.75 * 24 = 9Last edited by Pentaquark; 18052016 at 21:36. 
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 217
 18052016 21:42
(Original post by Pentaquark)
Calculating coordinate Q was in Q7 I am sure because you had to use that to find the area of the triangle.
I dont understand why it would be in 6 (a) & 6 (b) ?
(a) You had to find the derivative of the equation 4x^23x^3 using point P(2,24)
dy/dx= (2x9x^2) so subbing in 2 gives 32.
y24=32(x+2)
y=32x64+24
y=32x40
(b) then you'd use that to find coordinate Q by subbing in y=0
into y=32x40
0=32x40 to get
x=5/4
(c)? Then you had to find that integral.
(d)?
To find the area of the triangle you use Coordinate Q 5/4
You had to subtract 5/4 from the limit (2)
So 5/4   2 = 3/4
Base of the triangle is then 3/4
So the area of the triangle is 0.5 * 0.75 * 24 = 9
i remember that because there was so much god dam space for question 7 and to work out x=5/4 i ran out of space to write that and had to get extra paper so there is no way that was in question 7 
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 218
 18052016 21:42
Posted from TSR Mobile
It definitely said 3 linear factors. But one was a repeated root. So I guess as long as you put (x4)^2 I guess you'll get the mark 
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 219
 18052016 21:42
(Original post by Jaspal3131)
what was the working for the last question on 1 
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 220
 18052016 21:43
Anyone got a hard copy?
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Updated: May 27, 2016
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