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

    12
    ReputationRep:
    (Original post by ComputerMaths97)
    UNOFFICIAL MARK SCHEME FOR PAPER AND PAPER B NO PROBS DUUUUDES

    PAPER A:
    Q1)
    cos(theta) - 3 sin(theta) = Rcos(theta + alpha)
    R = sqrt(10)
    tan(alpha) = 3 => alpha = 1.249 rad

    4/sqrt(10) > 1 therefore no solutions to cos(theta) - 3sin(theta) = 4

    Q2)
    p = 2
    q = -2
    valid for |x| < 2

    Q3) 180 for y = x^4 therefore x^2 = y^(1/2)
    Volume = 16pi/3

    Q4) theta = 90 and 26.6 degrees (for some reason I put theta = 0 for cos(theta) = 0...)

    Q5) i)
    For triangle ABC, cos(theta) = x/AC
    For triangle ACD, cos(theta) = AC/AD
    For triangle ADE, cos(theta) = AD/2x
    Therefore cos^3(theta) = (x/AC)(AC/AD)(AD/2x)
    =1/2
    Therefore sec^3(theta) = 1/(cos^3(theta)) = 1/(1/2) = 2 as required

    ii) Do tan(theta) = ... for each triangle
    Then equate first and last triangle
    Giving required result

    Q6)
    dy/dx = (dy/dt)/(dx/dt) = (-2/t^2)(2) = -1/t^2
    Therefore y = (-1/t^2)x + R
    Also goes through (Q,0)
    => Q = t^2(R)
    Also goes through (2t,2/t)
    => R = 4/t
    Area = 1/2(QR) = 1/2(t^2)R^2
    = 1/2(16t^2/t^2) = 1/2(16) = 8 therefore independent of t

    Q7) i)
    Magnitude AG = 5sqrt(2)
    ii)
    n is perpendicular to two of the vectors on the plane DFG therefore n is normal to the plane
    therefore equation of plane is 15x - 20y + 4z = 20 since it goes through (any point on DFG substituded it)
    iii)
    Line AG meets plane at Q, lambda = 0.4 therefore Q: (2.4 , 1.2 , 2)
    iv) Angle = 180 - 123 = 57 (can't remember in more detail than that, but was 57.something)

    Q8)
    i) and ii) were easy show-that's
    iii) constant of proportionality = 1/4
    iv) do e^(-t) = e^(-ln(2+x/2-x)) then solve to show it
    mass tended to 2mg
    v) integrate, and c = k then result follows
    vi) k = 0.8113

    PAPER B:

    1) 149.5-45 = 104.5 m
    2) Angle of elevation = 11.35
    3) h = 10.3 cm
    4) On A4 sheet, h = 10.4 cm therefore since 10.3 ~ 10.4, and visual distance is 51.4 still, angle of elevation is still 11.35
    5) alpha = 5.1, beta + gamma = 6.7 (i think), gamma = 1.6(i think) therefore beta = 5.1 also
    6) Lot's of confusion here, not a clue tbh.
    I did 66 people too large, 170 okay and rest too small by splitting it in 50mm & 60mm, 70mm & 80mm, then the rest.
    I think the percentages were 18%, 49% and 33% respectively. (My answer)
    Assumptions I could think of were same sample and assuming that people who preferred 70mm or 80mm would've thought 75 was just about right.
    I got so many different answers to this...
    I got p and q as 4 and -4 respectively
    I ****ed up the volumes of revolution, God knows why I cube rooted x
    I remember getting only one solution for the trig question
    I seem to remember getting lambda = -1/3 and not 0.4
    Some of the other vectors work looks very unfamiliar.
    The only thing that looks like 18/18 is the differential equation...**** me...
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Cadherin)
    I got so many different answers to this...
    I got p and q as 4 and -4 respectively
    I ****ed up the volumes of revolution, God knows why I cube rooted x
    I remember getting only one solution for the trig question
    I seem to remember getting lambda = -1/3 and not 0.4
    Some of the other vectors work looks very unfamiliar.
    The only thing that looks like 18/18 is the differential equation...**** me...
    If it means anything, these are just the solutions from someone hoping for full ums. It's called unofficial for a reason, I'd be more surprised if there's no mistakes in there tbh
    Offline

    8
    ReputationRep:
    (Original post by Groodydavis)
    I know the other guy who's predicted A* in my class got 30 something, I think people found the angle between the normal an the line, I'm gonna ask my maths teacher on Monday though
    Hello,
    have you asked your teacher yet??
    Offline

    0
    ReputationRep:
    (Original post by ComputerMaths97)
    UNOFFICIAL MARK SCHEME FOR PAPER AND PAPER B NO PROBS DUUUUDES

    PAPER A:
    Q1)
    cos(theta) - 3 sin(theta) = Rcos(theta + alpha)
    R = sqrt(10)
    tan(alpha) = 3 => alpha = 1.249 rad

    4/sqrt(10) > 1 therefore no solutions to cos(theta) - 3sin(theta) = 4

    Q2)
    p = 2
    q = -2
    valid for |x| < 2

    Q3) 180 for y = x^4 therefore x^2 = y^(1/2)
    Volume = 16pi/3

    Q4) theta = 90 and 26.6 degrees (for some reason I put theta = 0 for cos(theta) = 0...)

    Q5) i)
    For triangle ABC, cos(theta) = x/AC
    For triangle ACD, cos(theta) = AC/AD
    For triangle ADE, cos(theta) = AD/2x
    Therefore cos^3(theta) = (x/AC)(AC/AD)(AD/2x)
    =1/2
    Therefore sec^3(theta) = 1/(cos^3(theta)) = 1/(1/2) = 2 as required

    ii) Do tan(theta) = ... for each triangle
    Then equate first and last triangle
    Giving required result

    Q6)
    dy/dx = (dy/dt)/(dx/dt) = (-2/t^2)(2) = -1/t^2
    Therefore y = (-1/t^2)x + R
    Also goes through (Q,0)
    => Q = t^2(R)
    Also goes through (2t,2/t)
    => R = 4/t
    Area = 1/2(QR) = 1/2(t^2)R^2
    = 1/2(16t^2/t^2) = 1/2(16) = 8 therefore independent of t

    Q7) i)
    Magnitude AG = 5sqrt(2)
    ii)
    n is perpendicular to two of the vectors on the plane DFG therefore n is normal to the plane
    therefore equation of plane is 15x - 20y + 4z = 20 since it goes through (any point on DFG substituded it)
    iii)
    Line AG meets plane at Q, lambda = 0.4 therefore Q: (2.4 , 1.2 , 2)
    iv) Angle = 180 - 123 = 57 (can't remember in more detail than that, but was 57.something)

    Q8)
    i) and ii) were easy show-that's
    iii) constant of proportionality = 1/4
    iv) do e^(-t) = e^(-ln(2+x/2-x)) then solve to show it
    mass tended to 2mg
    v) integrate, and c = k then result follows
    vi) k = 0.8113

    PAPER B:

    1) 149.5-45 = 104.5 m
    2) Angle of elevation = 11.35
    3) h = 10.3 cm
    4) On A4 sheet, h = 10.4 cm therefore since 10.3 ~ 10.4, and visual distance is 51.4 still, angle of elevation is still 11.35
    5) alpha = 5.1, beta + gamma = 6.7 (i think), gamma = 1.6(i think) therefore beta = 5.1 also
    6) Lot's of confusion here, not a clue tbh.
    I did 66 people too large, 170 okay and rest too small by splitting it in 50mm & 60mm, 70mm & 80mm, then the rest.
    I think the percentages were 18%, 49% and 33% respectively. (My answer)
    Assumptions I could think of were same sample and assuming that people who preferred 70mm or 80mm would've thought 75 was just about right.
    do you still remember what are all the questions to this paper? I want to discuss it with my friends but I can't remember much of it.
 
 
 
  • 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
    What's your favourite Christmas sweets?
  • 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

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