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# OCR MEI C4 June 2016 Unofficial Mark Scheme Watch

1. (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
=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...
2. (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
3. (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,
4. (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
=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.

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