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# AQA 2016 C1 Unofficial Mark Scheme

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1. (Original post by BamSrown)
Anyone got any idea what grade boundaries will be like?
I assume they'll be lower than last year
2. Can someone please put the marks next to every answer much appreciated

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3. Should be around 60 for an A. I hope.
4. (Original post by Katie904)
Can someone please put the marks next to every answer much appreciated

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Don't want to put ones i don't know. Sorry.
5. [QUOTE=Iamz;64916099]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)
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.]

Can you fill in the marks please
6. [QUOTE=Tjcmcfc;64939097]
(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) (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)
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.]

Can you fill in the marks please
Translation by (1/2 , -41/4) you replace x with x-1/2 and y with -(-41/4) which gives you y+41/4 but because they wanted it in the form y= you subtract the 41/4 and it becomes y=(x-1/2)^-41/4

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