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

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

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    Well done!

    Well, everyone seems to be getting different answers on this pyramid question so you might be right!
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    (Original post by poohplop)
    The length of CB was 35
    No it was 32
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    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?
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    (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.
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    (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
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    (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
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    (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 ...

    Please tell me your method
    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
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    (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
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    (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..
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    (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.
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    So what is the answer to the 3D Pythagoras question
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    (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!
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    (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?
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    (Original post by Bulbasaur10)
    Ah..... I thought it said more than 0!!! Thanks.
    So is the answer Sometimes True?
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    (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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