OCR MEI C4 Mathematics June 2016 - Official Thread

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?

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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.

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

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

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?

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No you dont if you look at the c4 specification it says "the angle between the normals of two planes is the same as the angle between the planes

Me! I found Paper A fairly easy overall (with Part B about similar difficulty to Part A), but the comprehension was dire for me

we werent asked to find the angle of two normals or two planes. we were asked to find angle between line and plane

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Oh dear, I got p and q as 4 and -4...

I did the same as you! For what little reassurance that is.

Same hopefully we will get a couple of method marks, still don't know where I've gone

Can someone make an unofficial markscheme 😅