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

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1. (Original post by Rit101)
This means that the gradient of AB was -3/5
Hence the parallel line should have the same gradient correct?
That's what I did
2. (Original post by Ywna097)
Thanks
It may be wrong
My friends said something like 4 - 8x - 2x^2 but I feel like that was for another question Argh I normally remember nearly all the questions but today was just a blur tbf.
3. (Original post by timtjtim)
Unofficial Mark scheme for C1 AQA 2016

It would help if you could link answers to questions as I can't remember them

Questions:

1)
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) 75-32√5

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 were 4 parts to this question I can't remember part a and b
b)
c) Definite integral of = 81 /4 [5]
d) Area of shaded region = = 45/4 [3]

8) a) can't 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-coordinate of p to get 45, 45>0 therefore minimum

Updated again.

R = (x-2)(x^2-3x+14) + 20
For this question, shouldnt it be -14
So that when you multiply it out you'd get +28+20 which would be +48 which was the original number in the equation.
4. How do you work out ques 1. It was find the coordinate of B ... lies on diameter???

Posted from TSR Mobile
5. (Original post by Rit101)
Why was it not 6>k>-3/2
because it was equation>0, so it was all the values above the x axis. Meaning it was anything under -3/2 (x<-3/2) and anything above 6 (x>6)
6. For 6d
I used the quadratic formula, I got -1+-squareroot|80

How did you get -1+-squareroot|5 ?
Sorry for poor quality post, I'm on my **** phone.
7. Would error carried forward be possible Q7 for the triangle area and co ordinate of Q if 7a was done wrong to get the line.
8. smh the only marks i'm counting on are method marks.
9. (Original post by kiiten)
How do you work out ques 1. It was find the coordinate of B ... lies on diameter???

Posted from TSR Mobile
You know that C is the midpoint of the line AB.

To get from A to C, you go down by 4 and across by 7. So from C to B, you do the same. C(5,-3) so B(5+7,-3-4) = B(12,-7)
10. messed up the tangent to circle question (5marks) think I used different coordinates to what the point actually was. If I did the working out correct but did this how many marks do you think ill gain?
11. (Original post by Pavansanga01)
For 6d
I used the quadratic formula, I got -1+-squareroot|80

How did you get -1+-squareroot|5 ?
Sorry for poor quality post, I'm on my **** phone.
It worked out as (-4 +- sqrt80)/(4) which turns into (-4 +- sqrt16 * sqrt5)/(4) which turns into (-4 +- 4*sqrt5)/(4) which turns into -1 +- sqrt5
12. (Original post by vyyzx)
What was the question for k=4 and k=20?
i think it was the one about if line blank is a tangent to the circle what is k and you had to do a discriminant or some shi
13. (Original post by timtjtim)
Unofficial Mark scheme for C1 AQA 2016

It would help if you could link answers to questions as I can't remember them

Questions:

1)
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) 75-32√5

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 were 4 parts to this question I can't remember part a and b
b)
c) Definite integral of = 81 /4 [5]
d) Area of shaded region = = 45/4 [3]

8) a) can't 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-coordinate of p to get 45, 45>0 therefore minimum

Updated again.
Thanks this is really useful! Was 5d definitely only 2 marks?? I completely missed it and only saw it when I was closing the paper and checking through and I think I remember it being 3 marks? Would rather it was 2 obviously aha
14. (x-5)^2-25+(y+3)^2-9=0
= (x-5)^2+(y+3)^2=34 therefore radius=Square root of 34
15. (Original post by Pavansanga01)
For 6d
I used the quadratic formula, I got -1+-squareroot|80

How did you get -1+-squareroot|5 ?
Sorry for poor quality post, I'm on my **** phone.
thats what i did :'(
16. (Original post by beanigger)

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
How many marks was 6d??? 3/4?
17. (Original post by Applebananaaaa)
(x-5)^2-25+(y+3)^2-9=0
= (x-5)^2+(y+3)^2=34 therefore radius=Square root of 34
Thats what I got!
18. for Q4 d) i swear the answer is (x-2)(x^2-3x+28) + 20

because 28+20= 48 ...
19. (Original post by money-for-all)
for Q4 d) i swear the answer is (x-2)(x^2-3x+28) + 20

because 28+20= 48 ...
I think when you times the whole thing out you had to get +48
In your case when you times it all out then you'd get -56+20
20. (Original post by timtjtim)
You know that C is the midpoint of the line AB.

To get from A to C, you go down by 4 and across by 7. So from C to B, you do the same. C(5,-3) so B(5+7,-3-4) = B(12,-7)
this is like the only question i agree with you tim tim my man

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