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# C1 Maths AS aqa 2016 (unofficial mark scheme new)

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1. 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 60-62 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
2. Oh thank god, i found the area of triangle to be 9 too. And the area of the shaded region was 29-9=20, is that right?

Posted from TSR Mobile
3. (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 29-9=20, is that right?

Posted from TSR Mobile
I didn't get the area of triangle, but area under curve was 20.25, so area shaded region was 20.25-9=11.25 from what I've seen on here.
4. For question 3 c) Translation of (1/2 , 41/4), I got -1/2 , could someone explain why I'm wrong please, thanks
5. (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
These are the two lines I think...
y=(x-7/2)^2 - 41/4 ---> y=(x-4)^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 x-direction
and upwards of 41/4 from -41/4.
6. (Original post by Pentaquark)
These are the two lines I think...
y=(x-7/2)^2 - 41/4 ---> y=(x-4)^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 x-direction
and upwards of 41/4 from -41/4.
yh so if you move from 3.5 -> 4 won't it be -0.5 because you're moving to the right in the x-direction and because its the x value you have to give it in negative
7. For the complete the square questions i used decimals, would i still get full marks?
8. 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)

9. (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 co-ordinates 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 - 325

3) a) y=(x-7/2)^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)(x-4)(x-4)
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 = (x-2)(x^2-3x+14) + 20

5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
c) asked to find the equation of the tangent at A 7x-4y+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=-32x-40, 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 x-coord of p to get 45, 45>0 therefore minimum
Why did you change 1a? and how many marks was each question worth?
10. (Original post by Rit101)
Nope pretty sure the top students in my Sixth Form got root65 as the radius
I got the radius as 7? Then the distance between C and T as root65?
11. (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 x-direction and because its the x value you have to give it in negative
If you plot the vertex of equation 1 and the vertex of equation 2.
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=(x-4)^2
This is a translation of [4] across. Not -4.
Its describing a movement of one graph to another.

The new equation of (x-7/2)-41/4 moving to (x-4)^2
will be (x-3.5-0.5)-41/4 + 41/2 => (x-4)^2
If it moves +0.5 then inside the bracket it will be -0.5
12. (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)

Due to spam, we are locking the document. It is Viewable only (commenting off).

If you wish to "chat" with each other and me, go to https://www.yourworldoftext.com/aqac12016ms
13. sorry could someone post the most updated mark scheme so i can update it? im kind of busy to search around :P
thanks
14. (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 co-ordinates 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=(x-7/2)^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)(x-4)(x-4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x-2)(x^2-3x-14) + 20 [3]

5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7) [2]
c) asked to find equation of tangent at A 7x-4y+18=0 [5]
d) asked to find length of CT = 9 [2]

6) a) y=-32x-40
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 x-coord of p to get 45, 45>0 therefore minimum
what was the working for the last question on 1
15. 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. :'(
16. 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 4-x^2-3x^3 using point P(-2,24)
dy/dx= (-2x-9x^2) so subbing in -2 gives -32.
y-24=-32(x+2)
y=-32x-64+24
y=-32x-40

(b) then you'd use that to find coordinate Q by subbing in y=0
into y=-32x-40
0=-32x-40 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
17. (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 4-x^2-3x^3 using point P(-2,24)
dy/dx= (-2x-9x^2) so subbing in -2 gives -32.
y-24=-32(x+2)
y=-32x-64+24
y=-32x-40

(b) then you'd use that to find coordinate Q by subbing in y=0
into y=-32x-40
0=-32x-40 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
the part c and d you are talking about was definitely question 7 but not part a and b
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
18. Posted from TSR Mobile

It definitely said 3 linear factors. But one was a repeated root. So I guess as long as you put (x-4)^2 I guess you'll get the mark
19. (Original post by Jaspal3131)
what was the working for the last question on 1
im not sure
20. Anyone got a hard copy?

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