For the M inverse question, did u need to explicitly write it out? I wrote this:(Original post by decombatwombat)
1)i) 1  x^2 + x^4 (3)
ii) x  1/3 x^3 + 1/5 x^5 +c but c = 0 (3)
iii) pi/6 (4)
iv) Draw the graph, sort of like a loop but starting at pi/4 and ending on the horizontal axis (2)
As the angles tends to 0 r tends to infinity (1)
v) a^2 ln(2 root2) or 1/2 a^2 ln 8 (4)
2)i) 1z (3)
ii) Show that C + jS thing (8)
iii) Modulus of the cube roots was root2 the angles were pi/18 13pi/18 25pi/18 (7)
3)i) Eigenvalues were 1/6, 1 eigenvectors were (1 1) and (3 4) I think, cant quite remember the order (8)
ii) M^n tended towards one seventh of (3 4) or something like that order might be different (6)
................................ ......................(3 4)
iii) Not too sure on this, I think it didnt tend to a limit (maybe infinity), as the elements of the matrix were greater than 1 but some were negative so it could have positive or negative infinity (4)
4)i) Show that arcosh thing (5)
ii) ln((3+root5)/2) and ln((3root5)/2) (5)
iii) The graph sort of looked like x^2 but started at y=2
The area bound by the curve was 5root5 / 2
The area bound by the line y = 5 was 5ln((7+3root5)/2) There are other ways of writing this for example 10ln(...) (8)
Then take one away from the other.
P* ((6)^n 0)* P^1
.....(0 1)
As (6)^n is unbounded, (M^1)^n doesn't tend to a finite limit.
Will i drop a mark ?
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OCR MEI FP2 Thread  AM 27th June 2016
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 181
 27062016 13:11
Last edited by Gifted; 27062016 at 13:12. 
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 182
 27062016 13:11
(Original post by decombatwombat)
1)i) 1  x^2 + x^4 (3)
ii) x  1/3 x^3 + 1/5 x^5 +c but c = 0 (3)
iii) pi/6 (4)
iv) Draw the graph, sort of like a loop but starting at pi/4 and ending on the horizontal axis (2)
As the angles tends to 0 r tends to infinity (1)
v) a^2 ln(2 root2) or 1/2 a^2 ln 8 (4)
2)i) 1z (3)
ii) Show that C + jS thing (8)
iii) Modulus of the cube roots was root2 the angles were pi/18 13pi/18 25pi/18 (7)
3)i) Eigenvalues were 1/6, 1 eigenvectors were (1 1) and (3 4) I think, cant quite remember the order (8)
ii) M^n tended towards one seventh of (3 4) or something like that order might be different (6)
................................ ......................(3 4)
iii) Not too sure on this, I think it didnt tend to a limit (maybe infinity), as the elements of the matrix were greater than 1 but some were negative so it could have positive or negative infinity (4)
4)i) Show that arcosh thing (5)
ii) ln((3+root5)/2) and ln((3root5)/2) (5)
iii) The graph sort of looked like x^2 but started at y=2
The area bound by the curve was 5root5 / 2
The area bound by the line y = 5 was 5ln((7+3root5)/2) There are other ways of writing this for example 10ln(...) (8)
Then take one away from the other. 
 Follow
 183
 27062016 13:12
(Original post by Gifted)
For the M inverse question, did u need to explicitly write it out? I wrote this:
P* ((6)^n 0)* P^1
(0 1)
As (6)^n is unbounded, (M^1)^n doesn't tend to a finite limit.
Will i drop a mark ? 
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 184
 27062016 13:13
(Original post by ComputerMaths97)
I lost so many fricking marks.
1  I couldn't do the c/s thing because I turned both c and s in terms of theta instead of half theta for some retarded reason.
For the area bounded by y=5 and curve, didn't read the question correct so just got area under curve
Didn't bother with the +c for question 1. Not sure why, clearly just forgot.
Only had the positive root for the ln(3+root(5)/2) one
Only drew the graph for positive x values because I thought cosh(x) doesn't exist for negative x.
And here was me hoping for an A :/ Defo missed my cam offer now, 3 huge basic mistakes and a few more mistakes just ruined everything looool. I even got full UMS in S2 near abouts, but still no A* in FM, great
what subject did you apply for? which college? how about your other A2 FM exams?Last edited by mrk1357; 27062016 at 13:30. 
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 27062016 13:13
(Original post by Sopranos)
for part 2i how can it be z1. didn't you have to take out a factor of 1/j since the cos part had a j in it and the sin part did not 
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 27062016 13:15
(Original post by decombatwombat)
This was probably the question I found the hardest, so not entirely sure. I am sure writing it like that is fine however, and at most if it were not to be okay, you would lose a mark. 
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 27062016 13:16
(Original post by decombatwombat)
I put 1z? I expanded the brackets, the right hand side simplified to jsin(x) and the left part could be written as 1 cos^2...... etc.. 
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 27062016 13:17
(Original post by decombatwombat)
I put 1z? I expanded the brackets, the right hand side simplified to jsin(x) and the left part could be written as 1 cos^2...... etc.. 
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 189
 27062016 13:17
(Original post by decombatwombat)
I put 1z? I expanded the brackets, the right hand side simplified to jsin(x) and the left part could be written as 1 cos^2...... etc.. 
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 27062016 13:17
(Original post by Bunderwump)
If people can remember the questions I can provide unofficial mark schemes for the ones people are wondering about (except last part of hyperbolics). I solved the C + jS question but I can't remember the questions. 
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 191
 27062016 13:18
(Original post by decombatwombat)
1)i) 1  x^2 + x^4
ii) x  1/3 x^3 + 1/5 x^5 +c but c = 0
iii) pi/6
iv) Draw the graph, sort of like a loop but starting at pi/4 and ending on the horizontal axis
As the angles tends to 0 r tends to infinity
v) a^2 ln(2 root2) or 1/2 a^2 ln 8
2)i) 1z
ii) Show that C + jS thing
iii) Modulus of the cube roots was root2 the angles were pi/18 13pi/18 25pi/18
3)i) Eigenvalues were 1/6, 1 eigenvectors were (1 1) and (3 4) I think, cant quite remember the order
ii) M^n tended towards one seventh of (3 4) or something like that order might be different
................................ ......................(3 4)
iii) Not too sure on this, I think it didnt tend to a limit (maybe infinity), as the elements of the matrix were greater than 1 but some were negative so it could have positive or negative infinity
4)i) Show that arcosh thing
ii) ln((3+root5)/2) and ln((3root5)/2)
iii) The graph sort of looked like x^2 but started at y=2
The area bound by the curve was 5root5 / 2
The area bound by the line y = 5 was 5ln((7+3root5)/2) There are other ways of writing this for example 10ln(...)
Then take one away from the other.
i) Correct
ii) Correct
iii) Correct
iv) Correct
v) Correct
2)
i) Correct
ii) Got C+js then got as far as c and s = .... in terms of theta/2, then for some reason decided to turn them in terms of theta, so was then stuck (3)
iii) Correct
3)
i) Got eigenvalues correct, 1 of the eigenvectors wrong, the other correct. (2)
ii) Got M^n tended to some matrix, hopefully got it right in terms of the incorrect eigenvector (either full marks or 0 lol)
iii) I got that it fluctuates between positive and negative infinity as (6)^n for even n tends to infinity but as n increases as 1, the becomes negative infinity, so does not tend to a finite limit. Hoping I got e.c.f there. (Either full marks or 0 there, again xD)
4)
i) Correct
ii) Didn't have the negative root (2)
iii) Only had the graph for positive x values as I thought cosh(x) had no values for negative x (2)
Only did area under curve, for the one side I had drawn it, so maybe 3 or 4?
Therefore 72  3  2  2  2  4 = 59/72 best case scenario. Dammit. Defo not an A let's be realistic :/
Thanks tho 
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 192
 27062016 13:18
(Original post by Gifted)
I did something similar. I had (1z)^n . Instead of 1z i wrote down the result in part i and raised it to the power of n. But this wasn't the same as the show that. But i multiplied by j^4 which is 1, and I got what was required. This is a valid technique right ? 
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 27062016 13:20
(Original post by decombatwombat)
I dont see why not, and yeah you had to manipulate it slightly to get what they wanted, I did a similar thing by multiplying by j etc... 
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 27062016 13:21
(Original post by ComputerMaths97)
1)
i) Correct
ii) Correct
iii) Correct
iv) Correct
v) Correct
2)
i) Correct
ii) Got C+js then got as far as c and s = .... in terms of theta/2, then for some reason decided to turn them in terms of theta, so was then stuck (3)
iii) Correct
3)
i) Got eigenvalues correct, 1 of the eigenvectors wrong, the other correct. (2)
ii) Got M^n tended to some matrix, hopefully got it right in terms of the incorrect eigenvector (either full marks or 0 lol)
iii) I got that it fluctuates between positive and negative infinity as (6)^n for even n tends to infinity but as n increases as 1, the becomes negative infinity, so does not tend to a finite limit. Hoping I got e.c.f there. (Either full marks or 0 there, again xD)
4)
i) Correct
ii) Didn't have the negative root (2)
iii) Only had the graph for positive x values as I thought cosh(x) had no values for negative x (2)
Only did area under curve, for the one side I had drawn it, so maybe 3 or 4?
Therefore 72  3  2  2  2  4 = 59/72 best case scenario. Dammit. Defo not an A let's be realistic :/
Thanks tho 
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 195
 27062016 13:22
(Original post by Gifted)
Part i had to multiply by j^2 and ended up with z1. But because i multiplied by 1, had to divide by 1 again, to get the final 1z. It was quite a fiddly question :/ 
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 196
 27062016 13:22
(Original post by Gifted)
Part i had to multiply by j^2 and ended up with z1. But because i multiplied by 1, had to divide by 1 again, to get the final 1z. It was quite a fiddly question :/ 
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 197
 27062016 13:23
(Original post by decombatwombat)
I'm sure everyone, including myself, who took the exam found it hard, meaning that the boundaries will be lower than expected. No point getting hung up about the exam now, get Physics out of the way, and who knows, come results day, I am sure you will surprise yourself with atleast an A in FP2 and an A* overall, especially considering how well you have done in the other modules. 
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 198
 27062016 13:23
Does anybody remember the exact question for the C + jS sum? I'll provide full working for it.

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 199
 27062016 13:24
(Original post by Bunderwump)
Does anybody remember the exact question for the C + jS sum? I'll provide full working for it.
S = 0  nC1 sinx + nC2 sin2x  ... + nCn sin nx
Then show that C/S = cot(1/2 x)
Then you had to get 2sin(x/2)(sin(x/2)  jcos(x/2)) into 1zLast edited by decombatwombat; 27062016 at 13:28. 
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 200
 27062016 13:29
(Original post by decombatwombat)
C = 1  nC1 cosx + nC2 cos2x  ... + nCn cos nx
S = 0  nC1 sinx + nC2 sin2x  ... + nCn sin nx
Then show that C/S = cot(1/2 x)
Then you had to get 2sin(x/2)(sin(x/2)  jcos(x/2)) into 1z
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Updated: August 13, 2016
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