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AQA 2016 C1 Unofficial Mark Scheme

AQA 2016 UNOFFICIAL MARK SCHEME

1)
a. Work out the gradient of a tangent, m = -5/3 (2marks)
b. Find the co-ordinates of B, B(-3,4) (3marks)
c. Work out k from (3+5k, 4-3k) ?, k = -30 (2marks)

2)
a. Simplify (3√5)^2 = 45 (2marks)
b. ((3√5)^2+√5) / (7+3√5) = 75 - 32√5 (5marks)

3)
a. Put x^2-7x-2, y= (x - 7/2)^2 - 41/4
bi. minimum value, - 41/4 (1mark)
c. What is the geometrical transformation from y=(...) onto y=(x-4)^2, Translation by (1/2 , 41/4)

4)
a. Show that (x+1) was a factor, subbed in to = 0 (2marks)
b. Give p(x) as three linear factors, (x+3)(x-4)(x-4) (3marks)
c. Find remainder when p(x) is divided by (..), r = 20 (2marks)
d. Divide p(x) by (x-2) and leave in the form (x-2)(x^2-bx+c)+n, (x-2)(x^2-3x+14) + 20 (3marks)

5)
ai. Find equation of circle, (x-5)^2+(y+3)^2 = 65 (2marks)
ii. radius = √65 (1mark)
b. Find the co-ordinates of B when AB is the diameter, B(12,-7)
c. Find the length CT, 81 = 9
d. Find the equation of the tangent at A, 7x-4y+18=0

6)
a. Find the equation of the tangent, y = -32x - 40
ii. Find the value of x (something), x = -1±√5
b. Sketch the graph of 8-4x-2x^2, n-shaped parabola with y-intercept = 8 & x axis intercepts at -1+√5 & -1-√5 (3marks)
c. k=4 and k=20

7)
a.
b. Find the value of Q, Q(-5/4)
c. Find the definite integral between -2 & 1, 81/4 (5marks)
d. Find the shaded area ( definite integral - area of triangle), 45/4 (3marks)

8)
ai. Find d^2y / dx^2, -2x - 9x^2 (2marks)
ii. d^2y/dx^2= - 2x - 9x^2 sub in x-coordinate of p to get 45, 45>0 therefore it is a minimum value
bi. Y= k(4x + 1) & y= ...... , You had to make the y's equal each other then rearrange to form an equation given
bii. Solve the inequality (9k^2.....), k < -3/2 & k > 6 (4marks)

Hope this helped you in any way.
(edited 7 years ago)

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Reply 1
Original post by Iamz
Hoping we can put together the questions and answers to this paper. If there are any mistakes let me know.
1a.
b.
c. k =30
2a. (3√5)^2 = 45
b. ((3√5)^2+√5) / (7+3√5)
= 75 - 3√5
Circle Question = (x
Radius = √65
Integration Question = 81/4
Shaded Area = 45/4
Geometrical Transformation Question = Translation by [1/2- 41/4]


2b = 75-32rt5

http://www.thestudentroom.co.uk/showthread.php?t=4095033&page=7&p=64916141&posted=1#post64916141
Reply 2
Original post by Iamz


4a. Show that (x+1) was a factor, subbed in to = 0
b. Give p(x) as three linear factors (x+3)(x-4)(x-4)
c. Find remainder when p(x) is divided by (..), r = 20
d. Divide p(x) by (x-2) and leave in the form (x-2)(x^2-bx+c)+n, (x-2)(x^2-bx+c)+52



shouldn't 4(d) be +20 at the end because that was the remainder? that's what I got and it seemed to work.
Reply 3
Original post by sean.17
shouldn't 4(d) be +20 at the end because that was the remainder? that's what I got and it seemed to work.


Good point
Original post by Iamz
Unofficial 2016 AQA mark scheme.
Hoping we can put together the questions and answers to this paper. If there are any mistakes let me know.

1a. Work out the gradient of a tangent, m = -5/3
b. Find the co-ordinates of B, B(-3,4)
c. Work out k from (3+5k, 4-3k) ?, k = -30

2a. Simplify (3√5)^2 = 45
b. ((3√5)^2+√5) / (7+3√5) = 75 - 32√5

3a. Put x^2-7x-2, y= (x - 7/2)^2 - 41/4
bi. minimum value = - 41/4
c. Translation of (1/2 , 41/4)

4a. Show that (x+1) was a factor, subbed in to = 0
b. Give p(x) as three linear factors (x+3)(x-4)(x-4)
c. Find remainder when p(x) is divided by (..), r = 20
d. Divide p(x) by (x-2) and leave in the form (x-2)(x^2-bx+c)+n, (x-2)(x^2-3x+14) + 20

5ai. Find equation of circle, (x-5)^2+(y+3)^2 = 65
ii. radius = √65
b. Find the co-ordinates of B when AB is the diameter, B(12,-7)
c. Find the equation of the tangent at A, 7x-4y+18=0
d. Find the length CT = 81 = 9

6a. Find the equation of the tangent, y = -32x - 40
bi. Find the value of Q, Q(-5/4)
ii. Sketch the graph of 8-4x-2x^2, n-shaped parabola with y-intercept = 8
c. k=4 and k=20

7a. Find d^2y / dx^2,
b.
c. Find the definite integral between -2 & 1, 81/4
d. Find the shaded area ( definite integral - area of triangle), 45/4

8a.
b.
8bii. Solve the inequality (9k^2.....), -3/2< k & k<6

Side note
I remember a question with -2(x+1)^2+10
and you get this from solving for x, x = -1±√5


How many marks was the circle question worth altogether ?
Reply 5
Original post by ALI7861105555555
How many marks was the circle question worth altogether ?


If i was to guess, I'd say 12 marks in total for circles section.
First post here, hello!

Some of your solutions for six were for question seven if I recall correctly - finding the tangent and the coordinates of Q were specifically to do with the integral if I'm not mistaken?

Question six had you solve the quadratic for -1 ± 5, sketch the graph, do a 1 mark 'show that' question and then solve the inequality, I think it just needs to be rearranged a bit.
Question 5) c) and d) are the wrong way round :tongue:
Reply 8
Original post by Iamz
Unofficial 2016 AQA mark scheme.
Hoping we can put together the questions and answers to this paper. If there are any mistakes let me know.

1a. Work out the gradient of a tangent, m = -5/3 (2marks)
b. Find the co-ordinates of B, B(-3,4) (2marks)
c. Work out k from (3+5k, 4-3k) ?, k = -30 (3marks)

2a. Simplify (3√5)^2 = 45 (2marks)
b. ((3√5)^2+√5) / (7+3√5) = 75 - 32√5 (4marks)

3a. Put x^2-7x-2, y= (x - 7/2)^2 - 41/4
bi. minimum value = - 41/4 (1mark)
c. Translation of (1/2 , 41/4)

4a. Show that (x+1) was a factor, subbed in to = 0 (2marks)
b. Give p(x) as three linear factors (x+3)(x-4)(x-4) (3marks)
c. Find remainder when p(x) is divided by (..), r = 20 (2marks)
d. Divide p(x) by (x-2) and leave in the form (x-2)(x^2-bx+c)+n, (x-2)(x^2-3x+14) + 20 (3marks)

5ai. Find equation of circle, (x-5)^2+(y+3)^2 = 65 (2marks)
ii. radius = √65 (1mark)
b. Find the co-ordinates of B when AB is the diameter, B(12,-7)
c. Find the length CT = 81 = 9
d. Find the equation of the tangent at A, 7x-4y+18=0

6a. Find the equation of the tangent, y = -32x - 40
ii. Find the value of x (something), x = -1±√5
bi. Find the value of Q, Q(-5/4)
ii. Sketch the graph of 8-4x-2x^2, n-shaped parabola with y-intercept = 8 & x axis intercepts at -1+√5 & -1-√5 (3marks)
c. k=4 and k=20

7a.
b.
c. Find the definite integral between -2 & 1, 81/4
d. Find the shaded area ( definite integral - area of triangle), 45/4

8ai. Find d^2y / dx^2, -2x - 9x^2 (2marks)
ii. d^2y/dx^2= - 2x - 9x^2 sub in x-coordinate of p to get 45, 45>0 therefore it is a minimum value
bi. Y= k(4x + 1) & y= ...... , You had to make the y's equal each other then rearrange to form an equation given
8bii. Solve the inequality (9k^2.....), -3/2 < k & k > 6 (4marks)


for 8bii shouldn't k<-3/2 and k>6
Reply 9
Original post by Astro_Joe
First post here, hello!

Some of your solutions for six were for question seven if I recall correctly - finding the tangent and the coordinates of Q were specifically to do with the integral if I'm not mistaken?

Question six had you solve the quadratic for -1 ± 5, sketch the graph, do a 1 mark 'show that' question and then solve the inequality, I think it just needs to be rearranged a bit.


Thanks
Reply 10
Original post by RoadtoSuccess
Question 5) c) and d) are the wrong way round :tongue:


Original post by mohs112
for 8bii shouldn't k<-3/2 and k>6


Ty
Anyone know what was the integral ?
I think it was (4 - x^2 - 3x^3) dx with limits 1 and -2.
Reply 13
Original post by Iamz
AQA 2016 UNOFFICIAL MARK SCHEME

1)
a. Work out the gradient of a tangent, m = -5/3 (2marks)
b. Find the co-ordinates of B, B(-3,4) (2marks)
c. Work out k from (3+5k, 4-3k) ?, k = -30 (3marks)

2)
a. Simplify (3√5)^2 = 45 (2marks)
b. ((3√5)^2+√5) / (7+3√5) = 75 - 32√5 (4marks)

3)
a. Put x^2-7x-2, y= (x - 7/2)^2 - 41/4
bi. minimum value, - 41/4 (1mark)
c. What is the geometrical transformation from y=(...) onto y=(x-4)^2, Translation by (1/2 , 41/4)

4)
a. Show that (x+1) was a factor, subbed in to = 0 (2marks)
b. Give p(x) as three linear factors, (x+3)(x-4)(x-4) (3marks)
c. Find remainder when p(x) is divided by (..), r = 20 (2marks)
d. Divide p(x) by (x-2) and leave in the form (x-2)(x^2-bx+c)+n, (x-2)(x^2-3x+14) + 20 (3marks)

5)
ai. Find equation of circle, (x-5)^2+(y+3)^2 = 65 (2marks)
ii. radius = √65 (1mark)
b. Find the co-ordinates of B when AB is the diameter, B(12,-7)
c. Find the length CT, 81 = 9
d. Find the equation of the tangent at A, 7x-4y+18=0

6)
a. Find the equation of the tangent, y = -32x - 40
ii. Find the value of x (something), x = -1±√5
b. Sketch the graph of 8-4x-2x^2, n-shaped parabola with y-intercept = 8 & x axis intercepts at -1+√5 & -1-√5 (3marks)
c. k=4 and k=20

7)
a.
b. Find the value of Q, Q(-5/4)
c. Find the definite integral between -2 & 1, 81/4
d. Find the shaded area ( definite integral - area of triangle), 45/4

8)
ai. Find d^2y / dx^2, -2x - 9x^2 (2marks)
ii. d^2y/dx^2= - 2x - 9x^2 sub in x-coordinate of p to get 45, 45>0 therefore it is a minimum value
bi. Y= k(4x + 1) & y= ...... , You had to make the y's equal each other then rearrange to form an equation given
bii. Solve the inequality (9k^2.....), k < -3/2 & k > 6 (4marks)

Hope this helped you in any way.


2b) was 5 marks and the marks for 1B) and 1C) need to be switched as 1b was the 3 mark simultaneous equations question
(edited 7 years ago)
How many marks was 7c and 7d?
Anyone got any idea what grade boundaries will be like?
Reply 16
Original post by BamSrown
Anyone got any idea what grade boundaries will be like?


Last year it was 64/75 for an A which is high considering it was a pretty hard paper but I think this years was slightly harder so I'd say maybe 60, possibly less depending on how well people across the whole country done..
Reply 17
Original post by Waweegee
How many marks was 7c and 7d?

(c) was 5 marks
(d) was 3 marks :smile:
Reply 18
Hilter Reacts to AQA Core 1... https://www.youtube.com/watch?v=JGdwNwi4G80
Reply 19
K sorted marks out now

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