AQA Level 2 Certificate in Further Mathematics UNOFFICIAL MARKSCHEME Paper 2

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(1-sin^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+13n-5 [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/x-2 + .... 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

The length of CB was 35

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 (x-2)

At D, x = 0 so y - 11 = 24(0-2)

y-11 =-48

y = -37

-5--37 = 32 so CD is 32

If you think your method is right, I might be wrong so please tell me

Hope this helped

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 2-3 marks for each question lol it just doesn't flow. I've got 3-4 answers in different working methods ffs.
This paper was still a mofo.

oh **** youre right; sorry i remembered it wrong.

Yeah, I wasn't sure about the plane:

I got AC = root (740) using Pythagoras' theorem

I then got CX = root (185) and did cos-1 (root(185)/17)= 36.9 degrees

I'm probably wrong ...

Please tell me your method

I think you're right I got the same

Also i am pretty sure that k was 1

Thanks! I hate things in 3D so I'm kind of surprised that I got it right
How did you find the paper overall?

I'm pretty sure it was -9. Do you remember what the question was

I'm pretty sure it was -9. Do you remember what the question was

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

I got 42.83 (2dp). Is that correct?

I'm so sorry, I was getting confused with the p = -9 question. Yeah, I got the same.

Thanks! I hate things in 3D so I'm kind of surprised that I got it right
How did you find the paper overall?

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?