AQA Level 2 Certificate in Further Mathematics UNOFFICIAL MARKSCHEME Paper 2

Maybe... I'm not sure. I don't think there has been a decisive majority of what the answer for this question is; 2 people have got 42.8 and 2 have got 36.9 so I'm not sure yet

How did you find the paper?

For the 3D pythagoras:

sqrt(16² + 22²) = 2sqrt(185)

Divide this by 2 to get the value of the center of the rectangle to the point A (length of AX)

Therefore, AX is sqrt(185)

Therefore the height of the pyramid (VX) is sqrt(17² - sqrt(185)²) = 2 sqrt(26)

Then find the midpoint of AB = 8

Find the length of V to the midpoint of AB = sqrt(17² - 8²) = 15

Therefore angle of plane is sin^-1(2 sqrt(26) / 15) = 42.83 (2dp)

Really easy. For the non calculator paper, I apparently got 70 / 70 according to the unofficial mark scheme and for this paper I got all the same answers apart from the 3D pythagoras so anything between 100 - 105 / 105

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

I think i got all of these apart from pyramid angle i got 48 or something

Posted from TSR Mobile

The length of CB was 35

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

Was the gradient -a/b or -b/a?

It was:
ax + by = c
by = c - ax
y = c/b - ax/b

So, isn't that -a/b times by x, which gives the gradient?

I got pretty much the same, but a few of my answers were different. I may have done these completely wrong, but I'm pretty sure I got the left-most box for the last 'tick' question. I also got completely different x values for the last trig identity question, and different co-ordinates of intersection. Also, I'm not sure whether it was -b/a. I thought it was b/a, as the equation was ax+by = c, so it was the same as by = -ax+c, and y = -a/bx+c. I must have misread it because I thought you had to work out the gradient of a line perpendicular to it, and did the negative reciprocal of -a/b, which is b/a. (It must have said parallel.... ) Sorry for the long post.... I'm probably 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 got pretty much the same, but a few of my answers were different. I may have done these completely wrong, but I'm pretty sure I got the left-most box for the last 'tick' question. I also got completely different x values for the last trig identity question, and different co-ordinates of intersection. Also, I'm not sure whether it was -b/a. I thought it was b/a, as the equation was ax+by = c, so it was the same as by = -ax+c, and y = -a/bx+c. I must have misread it because I thought you had to work out the gradient of a line perpendicular to it, and did the negative reciprocal of -a/b, which is b/a. (It must have said parallel.... ) Sorry for the long post.... I'm probably wrong.

I also thought you had to find the perpendicular gradient, I got the same as you

The left tick box I think was always true

It said b>1 and -1<c<1
The statement was b-c>1
So, if b = 2 and c = -0.5, b-c=2.5
If b = 1.2 and c = 0.6, b-c = 0.6
Therefore, it is only sometimes true I think

I might have remembered some of my answers wrong so you could be right, but, even if you're wrong, you still might get method marks as long as you have shown a valid method

Hope this helps

I also thought you had to find the perpendicular gradient, I got the same as you

x

Ah..... I thought it said more than 0!!! Thanks.

x