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    I also got 90-56 like someone mentioned. 56 is the angle between the normal and the line no ? If you want the angle between the plane and the line like in the question you'd 90-56.
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    That's also what I did. 56 degrees would be the angle between the plane vector and the line, so you need to take it away from 90 to find the angle between the actual plane and the line.
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    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 = 103 m
    4) On A4 sheet, h = 104 m therefore since 103 ~ 104, 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.
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    It said show that the area OQR is independent of t, so I just showed that it was independent of t instead of working out the area itself
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    I couldn't get my head around how much easier section B was than section A!
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    (Original post by ComputerMaths97)
    I will continue to edit this as people remind me of the questions and the order they were in!!! (I am constantly editing and saving so that you always have most up to date version as I work through it, and people can help with what the questions were)

    My unofficial mark scheme.

    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) ?? Remind me of question please

    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
    For the angle, do you not take it off 90 degrees. because you worked out the angle of line to normal. it asked for angle of line and plane so dont you have to take it off 90?

    Posted from TSR Mobile
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    (Original post by HFancy1997)
    For the angle, do you not take it off 90 degrees. because you worked out the angle of line to normal. it asked for angle of line and plane so dont you have to take it off 90?

    Posted from TSR Mobile
    This is just what I put in the exam. If you imagine it geometrically, subtracting it from 180 makes more sense, at least to me
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    (Original post by WhiteBison)
    I personally felt that while the comprehension paper was quite easy as they come (again, just moi), and Part B of the core paper was very standard, Part A seemed a fair bit more difficult than in previous years.
    For the independence part of the Part A parametric (Q6 I think), I did the tangent thing and even found the intercepts, so the answer was right there, but I then elected to find the area in terms of t after doing all the working and I don't know why... 5/7 OCR pls? Apollo where art thou. However, many of my friends didn't get that far. In fact, the Q8 parametric felt quite easy in comparison, though if you approached it by finding dx/dt instead of 1/(dt)/dx), you'd have a bad time.

    Aside from Q6, I've also probably lost a couple of marks by fudging the last step of the last part of the stacked triangles question. T'was a mighty fudge. So idk. 85/90 would be nice.
    From what I can tell the rest of it was fine. I've got the same answers as HFancy1997 if y'all need any more verification.

    I'd be interested to know if anyone found Part A easier than part B.
    Me! I found Paper A fairly easy overall (with Part B about similar difficulty to Part A), but the comprehension was dire for me
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    (Original post by ComputerMaths97)
    UNOFFICIAL MARK SCHEME FOR PAPER A (DOING PAPER B NOW)

    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
    Oh dear, I got p and q as 4 and -4...
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    (Original post by ComputerMaths97)
    This is just what I put in the exam. If you imagine it geometrically, subtracting it from 180 makes more sense, at least to me
    I did the same as you! For what little reassurance that is.
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    (Original post by HFancy1997)
    For the angle, do you not take it off 90 degrees. because you worked out the angle of line to normal. it asked for angle of line and plane so dont you have to take it off 90?

    Posted from TSR Mobile
    No you dont if you look at the c4 specification it says "the angle between the normals of two planes is Name:  Screenshot_2016-06-24-14-31-26-373.jpeg
Views: 203
Size:  476.5 KBthe same as the angle between the planes
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    (Original post by Maliha01)
    No you dont if you look at the c4 specification it says "the angle between the normals of two planes is Name:  Screenshot_2016-06-24-14-31-26-373.jpeg
Views: 203
Size:  476.5 KBthe same as the angle between the planes
    we werent asked to find the angle of two normals or two planes. we were asked to find angle between line and plane

    Posted from TSR Mobile
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    (Original post by Cadherin)
    Me! I found Paper A fairly easy overall (with Part B about similar difficulty to Part A), but the comprehension was dire for me
    Am I the only one who found that much harder than normal?
    Im really concerned because I found it pretty damn hard, and everyone seems to be okay with it.
    Q5 & Q6 I just could not do.
    messed up with K as 4 instead of 1/4 and got the ratio wrong too.
    Luckily I think I aced the comprehension as I realised that was my last chance to give myself a possibility of an A*.
    Think I dropped about 17 marks, which is so awful coming from someone who did every c4 paper and got 80+ on pretty much all of them.
    Probably ****ed my chances of an A*.
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    (Original post by HFancy1997)
    we werent asked to find the angle of two normals or two planes. we were asked to find angle between line and plane

    Posted from TSR Mobile
    Oh yeah i forgot
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    (Original post by Cadherin)
    Oh dear, I got p and q as 4 and -4...
    Same hopefully we will get a couple of method marks, still don't know where I've gone
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    Dudes incase you missed it, tell you friends your family even your pet rock...

    That the unofficial mark scheme for C4 mei ocr june 2016 has been made (both papers)

    http://www.thestudentroom.co.uk/show....php?t=4184549
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    (Original post by WhiteBison)
    I did the same as you! For what little reassurance that is.
    I did the same I'm sure there was a similar question on a past paper
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    (Original post by Crozzer24)
    Same hopefully we will get a couple of method marks, still don't know where I've gone
    Perhaps we're right?
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    (Original post by Fhhafhhs)
    Can someone make an unofficial markscheme 😅
    http://www.thestudentroom.co.uk/show....php?t=4184549
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    To illustrateName:  Untitled.jpg
Views: 306
Size:  36.4 KB
    the reason you take it from 90.
 
 
 
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