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# AQA Level 2 Certificate in Further Mathematics UNOFFICIAL MARKSCHEME Paper 2

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1. (Original post by GCSESTUDENT5000)
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
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)
2. (Original post by GCSESTUDENT5000)
How did you find the paper?
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
3. (Original post by Redcoats)
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)
Thanks. I wasn't particularly sure.
4. (Original post by Redcoats)
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
Haha, that's literally exactly the same as me but I probably got 0 on the 3D question and I kept rubbing out my graph so it might have been a bit messy

Well done
5. (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(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

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6. (Original post by ErinMei)
I think i got all of these apart from pyramid angle i got 48 or something

Posted from TSR Mobile
Well done!

Well, everyone seems to be getting different answers on this pyramid question so you might be right!
7. (Original post by poohplop)
The length of CB was 35
No it was 32
8. 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?
9. (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(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 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.
10. (Original post by Ishan_2000)
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?
It is -a/b
Sorry, that's my bad memory - I'll edit the markscheme
11. (Original post by Bulbasaur10)
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.
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
12. (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 cos-1 (root(185)/17)= 36.9 degrees

I'm probably wrong ...

Tbh, I don't even remember lol.
But it was the plane, there was 1 side which equals to 16 cm.
I found the midpoint connected it to the midpoint of the plane (ABCD or whatever it was) and then connected that line to the vertex or whatever it was V.
So, I think the midpoint of the base was X.
I labelled the midpoint half way through the 16 cm side as D.
So XD was 11 cm because it was 1/2 AB which was 22 cm.
I forgot what it was but I think VB = VC and they both equal to 17 cm.

So, VB or whatever it was.. VB^2 - VD^2 (8 cm) = one of the side = 15 cm.

So, angle theta = Cos -1 (A/H) = Cos -1 (11/15) = 42.8...
Surprised how I remember it lmfao.

I also worked out an extra side just incase to cross-check and it was O/A then so tan-1 (10.2 or something/11) = 42.8 also

and then I also done sin lool sin-1 (10.2/15) = 42.8
13. (Original post by Bulbasaur10)
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
14. (Original post by MaxHSloan)
I also thought you had to find the perpendicular gradient, I got the same as you
Lol, I got -b/a and b/a lol from y=mx+c and ax + by = c
I just went with b/a at the end..
15. (Original post by GCSESTUDENT5000)
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
Ah..... I thought it said b-c was more than 0!!! Thanks.
16. So what is the answer to the 3D Pythagoras question
17. (Original post by MaxHSloan)
I also thought you had to find the perpendicular gradient, I got the same as you
Phew... maybe it is correct if other people thought it said that too? Thank you!
18. (Original post by GCSESTUDENT5000)
x
What was the 12, x/y.
How did you get that, did you also get x = 12y and do 12y/y = 12?
19. (Original post by Bulbasaur10)
Ah..... I thought it said more than 0!!! Thanks.
So is the answer Sometimes True?
20. (Original post by GCSESTUDENT5000)
x
You forgot the question where it told you to draw the graph of the stationary and point of inflection [3].
So, you've got 1 question or something where you've wrote an extra mark lol.

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