OCR MEI C4 June 2016 Unofficial Mark Scheme

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.

Just going to pop this into the Maths Exams forum

What did you get for the Q for the height of AC?

Have I missed out one? Refresh my memory mate

I've got pretty much the same answers but my height is divided by 10? Was the height always in m or cm?

Yeah I got 10.3cm

I've got pretty much the same answers but my height is divided by 10? Was the height always in m or cm?

It was the one with the hill, and the guy could see the top 20m of the turbine, and you had to find the height from the ground at the tangent to the hill.

Yeah I got 10.3cm

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.

Question 3 was 8pi/3 because it was rotated 180 degrees not 360

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.

snip

Hey there. Some peoples are disputing that the angle in (7iv) (Paper A) was 57 degrees on the other thread. I got the same as you, however, by taking the product of the cos(theta) equation and subtracting it from 180.

Any insight? Because it seemed the right thing to do at the time.

Hey there. Some peoples are disputing that the angle in (7iv) (Paper A) was 57 degrees on the other thread. I got the same as you, however, by taking the product of the cos(theta) equation and subtracting it from 180.

Any insight? Because it seemed the right thing to do at the time.

I found the angle between the normal to the plane and the line - giving me 123.something degrees

Wait what answer did you get?

I found the angle between the normal to the plane and the line - giving me 123.something degrees - I'm assuming you did the same?

I then sat there and tried to imagine what that 123 degrees meant. In my head, it seemed like it was the angle between the normal to the plane pointing directly downwards, and the line coming out of the top side of the shape. Get what I mean? If that is true, then it's 180 - 123, giving 57. However in past papers, it's very common to see 90 - theta in mark schemes.

A few of my (smart) friends also did the same thing, and it was the first thing we spoke about after the exam, so I had confidence in that answer.

Worst case scenario, we lose 2 marks, though, so no real issue.

For question 6 in the comp paper where it asked you to find the height of C above A, I got something in the 80s. I think 83...m or something?