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 showthat's
iii) constant of proportionality = 1/4
iv) do e^(t) = e^(ln(2+x/2x)) then solve to show it
mass tended to 2mg
v) integrate, and c = k then result follows
vi) k = 0.8113
PAPER B:
1) 149.545 = 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.
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OCR MEI C4 June 2016 Unofficial Mark Scheme
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 1
 24062016 14:39
Last edited by ComputerMaths97; 24062016 at 15:04.Post rating:4 
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 2
 24062016 14:47
(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
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 showthat's
iii) constant of proportionality = 1/4
iv) do e^(t) = e^(ln(2+x/2x)) then solve to show it
mass tended to 2mg
v) integrate, and c = k then result follows
vi) k = 0.8113
PAPER B:
1) 149.545 = 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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 3
 24062016 14:48
Just going to pop this into the Maths Exams forum

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 4
 24062016 14:49
(Original post by Fox Corner)
Just going to pop this into the Maths Exams forum 
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 5
 24062016 14:50
(Original post by decombatwombat)
What did you get for the Q for the height of AC? 
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 6
 24062016 14:51
(Original post by ComputerMaths97)
Have I missed out one? Refresh my memory mate 
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 7
 24062016 14:54
I've got pretty much the same answers but my height is divided by 10? Was the height always in m or cm?

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 8
 24062016 14:55
(Original post by Crozzer24)
I've got pretty much the same answers but my height is divided by 10? Was the height always in m or cm? 
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 9
 24062016 14:56
(Original post by decombatwombat)
Yeah I got 10.3cmPost rating:1 
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 10
 24062016 15:03
(Original post by Crozzer24)
I've got pretty much the same answers but my height is divided by 10? Was the height always in m or cm?(Original post by decombatwombat)
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.(Original post by decombatwombat)
Yeah I got 10.3cm 
 Follow
 11
 24062016 15:10
(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
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 showthat's
iii) constant of proportionality = 1/4
iv) do e^(t) = e^(ln(2+x/2x)) then solve to show it
mass tended to 2mg
v) integrate, and c = k then result follows
vi) k = 0.8113
PAPER B:
1) 149.545 = 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. 
 Follow
 12
 24062016 15:11
(Original post by mattpotter)
Question 3 was 8pi/3 because it was rotated 180 degrees not 360Post rating:1 
 Follow
 13
 24062016 15:18
(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
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 showthat's
iii) constant of proportionality = 1/4
iv) do e^(t) = e^(ln(2+x/2x)) then solve to show it
mass tended to 2mg
v) integrate, and c = k then result follows
vi) k = 0.8113
PAPER B:
1) 149.545 = 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.
Did they ask what assumptions were made in the last question for the comprehension paper? 
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 14
 24062016 15:19
(Original post by ComputerMaths97)
snip
Any insight? Because it seemed the right thing to do at the time. 
 Follow
 15
 24062016 15:22
(Original post by WhiteBison)
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. 
 Follow
 16
 24062016 15:22
(Original post by WhiteBison)
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 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. 
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 17
 24062016 15:33
(Original post by ComputerMaths97)
I found the angle between the normal to the plane and the line  giving me 123.something degreesPost rating:1 
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 18
 24062016 15:35
(Original post by 12284)
Wait what answer did you get?
(Original post by ComputerMaths97)
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.
*Having just checked*
January 2013, (7i): You calculate theta and then subtract it from 180 to get the correct answer. This is to work out the angle between a vector and the horizontal plane, which is just a very specific variant of the question in our paper (the normal vector is, after all, just another vector).
Idk. I'm not saying that we were definitely correct, but from what you've said, and from this in turn, I think we might be? 
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 19
 24062016 15:59
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?

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 20
 24062016 16:13
(Original post by ktait98)
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?
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Updated: June 27, 2016
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