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# OCR MEI C4 June 2016 Unofficial Mark Scheme

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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...
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
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Updated: June 27, 2016
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