I will try and remember everything. Please correct me if I'm wrong
1) 15 square units [3]
2) a) x=2 [1]
b) x=0.8, x=4.8 [2]
3) a) (c/a, 0) [1]
b) a/b [1]
I don't know the question number for these:
sin^2(x)  3 cos^2(x) = sin^2(x)  3(1sin^2(x)) = sin^2(x)  3 + 3sin^2(x)
= 4sin^2(x)  3 [2]
solutions of the above is equal to 0 are x = 60, 120, 240, 300 (all in degrees) [4]
y > 45 degrees for tan y [1]
sin y = (p+1)/(root (2p^2+2)) [4]
p = 9 for the coefficient is 23 [3]
For the pyramid, I got 36.9 degrees but I'm not sure [4]
The graph started in the left and went below the x axis on the right [3]
To prove y = 3x on the parallel lines, you had to do angles around a point make 360 degrees and then use supplementary angles [4]
x/y = 12 [3]
f(x+1)  f(x) = 1/(2x+1)(2x+3) [5]
The 2/5* root x equation was 25/4 [2]
x^3 = 5x^2 was x=0, x=5 [2]
The nth term was 2n+5 [2]
The quadratic nth term was 6n^2+13n5 [3]
The circle theorems was 37.5 degrees [4]
The matrix mapping the points was that they can be the same as long as a = 3 [4]
The matrix transformation was (3 0) so scale factor 3 [5]
0 3
The equation with 3/x2 + .... was something like x = 1.23, x= 2.77 [6]
The length of CB was 32 units [6]
The range of f(x) was 3 <= f(x) <= 6 [2]
The tick box question was left box, middle box, right box, middle box [4]
The quadratic graph and coordinates of P were (0,7) [4]
The coordinates of P with the line ratios were (9/2, 47/8) [3]
The simplify fraction was x+2/x+3 I think [2]
The simplify was something like w^3x^2y^2 (w^2+y) [2]
The make the subject of the formula was something like x = 8w/y+8 [4]
The coordinates of intersection were (1.4, 3.4) [3]
When you were given the gradient and had to find k, k was 1 [4]
I can't remember 2 marks, sorry. Please tell me if you can remember but I will try and remember myself!
Hope this helps
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Maths
>< Maths Exams

AQA Level 2 Certificate in Further Mathematics UNOFFICIAL MARKSCHEME Paper 2
Announcements  Posted on  

Would YOU be put off a uni with a high crime rate? First 50 to have their say get a £5 Amazon voucher!  27102016 

 Follow
 1
 24062016 12:13
Last edited by GCSESTUDENT5000; 24062016 at 13:20. 
 Follow
 2
 24062016 12:16
The length of CB was 35

 Follow
 3
 24062016 12:18
Also i am pretty sure that k was 1

 Follow
 4
 24062016 12:20
(Original post by GCSESTUDENT5000)
I will try and remember everything. Please correct me if I'm wrong
1) 15 square units [3]
2) a) x=2 [1]
b) x=0.8, x=4.8 [2]
3) a) (c/a, 0) [1]
b) b/a [1]
I don't know the question number for these:
sin^2(x)  3 cos^2(x) = sin^2(x)  3(1sin^2(x)) = sin^2(x)  3 + 3sin^2(x)
= 4sin^2(x)  3 [2]
solutions of the above is equal to 0 are x = 60, 120, 240, 300 (all in degrees) [4]
y > 45 degrees for tan y [1]
sin y = (p+1)/(root (2p^2+2)) [4]
p = 9 for the coefficient is 23 [3]
For the pyramid, I got 36.9 degrees but I'm not sure [4]
The graph started in the left and went below the x axis on the right [3]
To prove y = 3x on the parallel lines, you had to do angles around a point make 360 degrees and then use supplementary angles [4]
x/y = 12 [3]
f(x+1)  f(x) = 1/(2x+1)(2x+3) [5]
The 2/5* root x equation was 25/4 [2]
x^3 = 5x^2 was x=0, x=5 [2]
The nth term was 2n+5 [2]
The quadratic nth term was 6n^2+13n5 [3]
The circle theorems was 37.5 degrees [4]
The matrix mapping the points was that they can be the same as long as a = 3 [4]
The matrix transformation was (3 0) so scale factor 3 [5]
0 3
The equation with 3/x2 + .... was something like x = 1.23, x= 2.77 [6]
The length of CB was 32 units [6]
The range of f(x) was 3 <= f(x) <= 6 [2]
The tick box question was left box, middle box, right box, middle box [4]
The quadratic graph and coordinates of P were (0,7) [4]
The coordinates of P with the line ratios were (9/2, 47/8) [3]
The simplify fraction was x+2/x+3 I think [2]
The simplify was something like w^3x^2y^2 (w^2+y) [2]
The make the subject of the formula was something like x = 8w/y+8 [4]
The coordinates of intersection were (1.4, 3.4) [3]
When you were given the gradient and had to find k, k was 1 [4]
I can't remember 2 marks, sorry. Please tell me if you can remember but I will try and remember myself!
Hope this helps
Lol, according to that mark scheme  I destroyed the paper, but I thought I flopped it.
Anyway, let's see  my working out I lost 23 marks for each question lol it just doesn't flow. I've got 34 answers in different working methods ffs.
This paper was still a mofo. 
 Follow
 5
 24062016 12:21
(Original post by poohplop)
The length of CB was 35
y = 2x^3  5
at C, x = 0, so y = 5
dy/dx = 6x^2
At P, x = 2 and y = 11, so dy/dx = 6(2)^2 = 6x4 = 24
Therefore the equation of the tangent at P is y  11 = 24 (x2)
At D, x = 0 so y  11 = 24(02)
y11 =48
y = 37
537 = 32 so CD is 32
If you think your method is right, I might be wrong so please tell me
Hope this helped 
 Follow
 6
 24062016 12:22
(Original post by GCSESTUDENT5000)
Ok, thanks. I just thought it was 32 because:
y = 2x^3  5
at C, x = 0, so y = 5
dy/dx = 6x^2
At P, x = 2 and y = 11, so dy/dx = 6(2)^2 = 6x4 = 24
Therefore the equation of the tangent at P is y  11 = 24 (x2)
At D, x = 0 so y  11 = 24(02)
y11 =48
y = 37
537 = 32 so CD is 32
If you think your method is right, I might be wrong so please tell me
Hope this helped 
 Follow
 7
 24062016 12:23
(Original post by Chittesh14)
For the plane, I got x = 42.8 degrees I think, something like that.
Lol, according to that mark scheme  I destroyed the paper, but I thought I flopped it.
Anyway, let's see  my working out I lost 23 marks for each question lol it just doesn't flow. I've got 34 answers in different working methods ffs.
This paper was still a mofo.
I got AC = root (740) using Pythagoras' theorem
I then got CX = root (185) and did cos1 (root(185)/17)= 36.9 degrees
I'm probably wrong ...
Please tell me your method 
 Follow
 8
 24062016 12:24
(Original post by poohplop)
oh **** youre right; sorry i remembered it wrong.
I probably remembered the whole markscheme wrong 
 Follow
 9
 24062016 12:30
(Original post by GCSESTUDENT5000)
Yeah, I wasn't sure about the plane:
I got AC = root (740) using Pythagoras' theorem
I then got CX = root (185) and did cos1 (root(185)/17)= 36.9 degrees
I'm probably wrong ...
Please tell me your method 
 Follow
 10
 24062016 12:32
(Original post by MaxHSloan)
I think you're right I got the same
How did you find the paper overall? 
 Follow
 11
 24062016 12:36
(Original post by poohplop)
Also i am pretty sure that k was 1 
 Follow
 12
 24062016 12:37
Can anyone remember where the other two marks came from?

 Follow
 13
 24062016 12:39
(Original post by GCSESTUDENT5000)
Thanks! I hate things in 3D so I'm kind of surprised that I got it right
How did you find the paper overall? 
 Follow
 14
 24062016 12:39
(Original post by Redcoats)
I'm pretty sure it was 9. Do you remember what the question was
When x = 2, the gradient is 12.
so 3/2 (2)  k (2)^4 + k = 12
3 15k = 12
15k = 15
k=1
Not sure though... maybe you're right 
 Follow
 15
 24062016 12:40
(Original post by Redcoats)
I'm pretty sure it was 9. Do you remember what the question was 
 Follow
 16
 24062016 12:41
(Original post by GCSESTUDENT5000)
It was dy/dx = 3/2 x  kx^4 + k
When x = 2, the gradient is 12.
so 3/2 (2)  k (2)^4 + k = 12
3 15k = 12
15k = 15
k=1
Not sure though... maybe you're right 
 Follow
 17
 24062016 12:41
(Original post by Redcoats)
I got 42.83 (2dp). Is that correct? 
 Follow
 18
 24062016 12:42
(Original post by Redcoats)
I'm so sorry, I was getting confused with the p = 9 question. Yeah, I got the same. 
 Follow
 19
 24062016 12:43
(Original post by GCSESTUDENT5000)
Thanks! I hate things in 3D so I'm kind of surprised that I got it right
How did you find the paper overall? 
 Follow
 20
 24062016 12:45
(Original post by MaxHSloan)
Thought it was ok overall, hoping I've done enough to get the A^. What about you? What do you think the grade boundaries will be like?
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: July 9, 2016
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.