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) -a/b [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

Does anyone remember the question for 'p=-9 for the coefficient is -23'?

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(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

Sorry.... I tend to react quite strongly to things! Have a good Summer too. I didn't find it too difficult I just get a little jumpy after exams. I seriously hope that I didn't come across as too aggressive when you were just stating your opinion. I made a silly mistake on the last question doing inverse sine of + and -3/4 instead of + and - root 3/4 (if that was what was right..) and have been a little annoyed with myself. By the way, do you think I have a chance of getting any marks for that? I rearranged the equation correctly and did the correct method to find other angles.... I know they are quite stingy with method marks though...

You'll probably get 2 - 3 marks

It's kinda annoying cos I got basically everything right on the cylinder question but I got 1/12 at the end and not 12. How many marks do u think I've lost?


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It's kinda annoying cos I got basically everything right on the cylinder question but I got 1/12 at the end and not 12. How many marks do u think I've lost?
Posted from TSR Mobile

Please can someone explain how to work out the question that was something like this:

tan y = p+1 / p-1
find the expression for sin y

It's driving me insane that I can't work this out! Any help is really appreciated.

Draw the triangle, right angle - make theta any of the other 2 angles.
Label the opposite side as 'p+1' and the adjacent side as 'p-1'.
Work out the hypotenuse by doing (p+1)^2 + (p-1)^2
= p^2 + 2p + 1 + p^2 - 2p + 1
2p^2 + 2
Square root that = $\sqrt {2} p + \sqrt 2$

sin y = opp / hyp
= p + 1 / $\sqrt {2p^2 + 2}$

Thank you so much! It didn't occur to me to set up a triangle; I was just messing around with triginometric identities!