Hey there! Sign in to join this conversationNew here? Join for free

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

    Offline

    0
    ReputationRep:
    Does anyone remember the question for 'p=-9 for the coefficient is -23'?
    Offline

    8
    ReputationRep:
    I got the left-most box for the last multiple choice one.

    I'm pretty sure the q said b-c>0 not >1
    Offline

    0
    ReputationRep:
    What was the question for 'p=-9 for the coefficient is -23'?
    Offline

    2
    ReputationRep:
    (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) -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
    Got everything right, you, apart from Q 3)b) - it should have been b/a, as you probably misread the question: it asked you for the parallel to the line - not the gradient of the line itself. Also, for factorisation, it should have been (x-2)(x+3), but you probably got that one right and just remembered it wrong. Same for the w,y,z factorise question - I think you remembered the answer wrong (not sure). Anyways, great job. You remembered MUCH more than I did and even basically got everything right.
    Offline

    2
    ReputationRep:
    (Original post by 123(6))
    Does anyone remember the question for 'p=-9 for the coefficient is -23'?
    (3x+4)(x^2 + px + 5).
    Offline

    0
    ReputationRep:
    same, i got -9.
    Offline

    2
    ReputationRep:
    Could Someone explain the Matrices Question!! I honestly read that question and began planning to fake a seizure in the middle of the exam...

    I said that Yes, P could Equal Q as they could both equal (-3,2) but what was the proper working out??
    Offline

    0
    ReputationRep:
    (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) -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
    Did you include the question that was about a hemisphere and a cylinder and how the volumes were equal?
    Offline

    10
    ReputationRep:
    (Original post by Bulbasaur10)
    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
    Offline

    11
    ReputationRep:
    (Original post by Redcoats)
    You'll probably get 2 - 3 marks
    Hopefully...! Thank you.
    Offline

    11
    ReputationRep:
    What did everyone get for the point of intersection simultaneous equations question? I don't remember getting (1.4,3.4), but it probably is....
    Offline

    5
    ReputationRep:
    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
    Offline

    19
    ReputationRep:
    (Original post by loshaw20)
    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
    1 mark or 2 maybe 1 method 1 accuracy.
    I think only 1 tbh.
    Offline

    11
    ReputationRep:
    (Original post by loshaw20)
    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
    I got lots wrong and am hoping for an A*. Albeit i may be being too harsh. Dunno until results day but i am hoping for that A*
    Offline

    4
    ReputationRep:
    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.
    Offline

    19
    ReputationRep:
    (Original post by iwouldn'tworry)
    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}
    Offline

    4
    ReputationRep:
    (Original post by Chittesh14)
    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!
    • TSR Support Team
    Offline

    21
    ReputationRep:
    TSR Support Team
    (Original post by iwouldn'tworry)
    Thank you so much! It didn't occur to me to set up a triangle; I was just messing around with triginometric identities!
    Me neither! I didn't know what "diagram" it wanted at all, in the end I just guessed an answer that was completely wrong as i didn't want to just leave it blank haha
    Offline

    0
    ReputationRep:
    Sorry if this has already been answered, I didn't have time to go through all of the replies.

    the one you missed was

    7) (2/3 x^3y)^3 = 8/27 x^9y^3 [2]

    The one's I'm not sure about are:

    4) The coordinates of intersection were (1.4, 3.4) [3] (3.4,1.4)????

    10)a) The simplify fraction was x+2/x+3 I think [2] (x-2)/(x+3)????
    b) The simplify was something like w^3x^2y^2 (w^2+y) [2] w^2x^3y^2(w^2+x^3y)????

    22) For the pyramid, I got 36.9 degrees but I'm not sure [4] 42.8????
    Offline

    14
    ReputationRep:
    Can someone draw the graph with the point of inflection and the minimum point please?


    Posted from TSR Mobile
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brexit voters: Do you stand by your vote?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.