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OCR (not MEI) C2 - Wednesday 20th May 2015

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Reply 420
I got 3.3 rad but then took it away from 6 pi (graph is stretched by SF 3 on x-axis) and got 12 point something rad?
You know from the graph in the previous question that the point of intersection does not occur when sinx=0. Therefore you can divide by it as you know it can't be 0
For 9ii) I put (3/2pi - a) + 3/2pi instead of 3pi - a, how many marks would I lose?
(edited 8 years ago)
Original post by ayl21
I got 3.3 rad but then took it away from 6 pi (graph is stretched by SF 3 on x-axis) and got 12 point something rad?


But the two graphs didn't intersect at the same point so you can't take it away from 6 pi???

ImageUploadedByStudent Room1432134606.003804.jpg


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Reply 424
So if you solve for 3.3 rad you don't take it away from 6 pi because it's just symmetry?
Original post by DominoMatrix227
What did you get for the 2N thing? I got -44 and 125/3 which I think are both wrong.


I got 24

It was something like ∑n=N2Nun=2750\displaystyle\sum_{n=N}^{2N} u_n = 2750 iirc. The sigma notation is indicating the sum of all the terms from n=N to n=2N which is the same as saying S2N−SN−1S_{2N} - S_{N-1} (just like the previous question where ∑n=1020un≡S20−S9\displaystyle\sum_{n=10}^{20} u_n \equiv S_{20} - S_9 (because you'd be excluding u10u_{10} if you did S20−S10S_{20} - S_{10}).

So finding S2NS_{2N} and simplifying:

S2N=2N2[2×5+3(2N−1)]=N(6N+7)=6N2+7N\displaystyle S_{2N} = \frac{2N}{2}\left[2 \times 5 + 3(2N-1)\right] = N(6N+7) = 6N^2 + 7N

Finding SN−1S_{N-1} and simplifying:

SN−1=N−12[10+3(N−2)]\displaystyle S_{N-1} = \frac{N-1}{2}\left[10 + 3(N-2)\right]

This gets you brackets to expand:

=(N−1)(3N+4)2=3N2+N−42=1.5N2+0.5N−2\displaystyle = \frac{(N-1)(3N+4)}{2} = \frac{3N^2 + N - 4}{2} = 1.5N^2 + 0.5N - 2

very fiddly :tongue: but the question depends on you being able to exploit the algebra

S2N−SN−1=(6N2+7N)−(1.5N2+0.5N−2)=4.5N2+6.5N+2=2750\displaystyle S_{2N} - S_{N-1} = (6N^2 + 7N) - (1.5N^2 + 0.5N - 2) = 4.5N^2 + 6.5N + 2 = 2750

We can form a quadratic from this:
4.5N2−6.5N−2748=04.5N^2 - 6.5N - 2748 = 0

Then I used the quadratic formula, only bothering to calculate −b+b2−4ac-b + \sqrt{b^2 - 4ac} in the numerator because it was clear that using ±\pm would mean that I would get a negative solution as well as a positive one, which wouldn't make sense in the context of sequences.

−6.5+6.52−(4×4.5×−2748)9=−6.5+49506.259=−6.5+222.59=2169=24\displaystyle \frac{-6.5 + \sqrt{6.5^2 - (4 \times 4.5 \times -2748)}}{9} = \frac{-6.5 + \sqrt{49506.25}}{9} = \frac{-6.5 + 222.5}{9} = \frac{216}{9} = 24
(edited 8 years ago)
Original post by ayl21
So if you solve for 3.3 rad you don't take it away from 6 pi because it's just symmetry?


Sorry for sounding stupid, but how did you get 6pi in the first place and not 2pi?


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Original post by jamesplay73
For 9ii) I put (3/2pi - a) + 3/2pi instead of 3pi - a, how many marks would I lose?


That's literally exactly the same as 3pi-a... Did it not cross your mind to just add the 3/2 pi together? (For the record, you'll lose no marks but that's such a weird thing to write :tongue:)
Could someone draw out the graph question drawing sin1/3 second last part on 9 . And draw out how they did the last question ? So confused would really appreciate it !!!! Xx
Original post by Lil_timmi
Sorry for sounding stupid, but how did you get 6pi in the first place and not 2pi?


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its a 1/3 x graph, hence stretch by SF3 in x direction. as it showed a full wave of cos on the diagram we know it must be 6pi not 2pi
Original post by nurav11
That's literally exactly the same as 3pi-a... Did it not cross your mind to just add the 3/2 pi together? (For the record, you'll lose no marks but that's such a weird thing to write :tongue:)


Haha I did this too, the panic of exams does some very strange things...
ImageUploadedByStudent Room1432135522.206627.jpg

This is how I answered 7. iii)

Can someone tell me where I went wrong???

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(edited 8 years ago)
Original post by nurav11
That's literally exactly the same as 3pi-a... Did it not cross your mind to just add the 3/2 pi together? (For the record, you'll lose no marks but that's such a weird thing to write :tongue:)


I know yea haha, I just had no time left thank you anyway :smile:
That might be necessary. I remeber completing it think what I have written doesnt look like 4 marks. But if you did that you would get the next answer right. When they say simplist form I am sure they mean without fractions. Everybody may have lost a mark if you are correct

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Original post by andygeor
its a 1/3 x graph, hence stretch by SF3 in x direction. as it showed a full wave of cos on the diagram we know it must be 6pi not 2pi


Ahhhh **** I got that wrong. Thanks anyway :smile:


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Original post by nurav11
That would be a godsend honestly, I got screwed over badly by the 2N question and could easily have lost 4 maybe even all 6 marks there. Apart from that though, thought it went very well, if boundary is 53 or something, then where do you reckon 100 UMS will be? 67/68? How'd you find it in general aswell?


Used my fp1 knowledge for that question :P. I thought it was a paper that required you to be careful, but otherwise was quite straight forward. I know i've lost all my marks on the last question cus of radians smh
Original post by Lil_timmi
ImageUploadedByStudent Room1432135522.206627.jpg

This is how I answered 7. iii)

Can someone tell me where I went wrong???

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You can't just square the trig functions on each side and expect them to still equal eachother.
Original post by Lil_timmi
ImageUploadedByStudent Room1432135522.206627.jpg

This is how I answered 7. iii)

Can someone tell me where I went wrong???

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Make it in terms of tanx dont square them, then remember the limits are larger because its 1/3


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Will we lose all 6 marks for getting N = -44 or N= 125/3 ?
I was considering this but when you are working to find solutions dont you do 360+ and 180- to get a set of solutions then whatever is combined with the x on the other side gets factored in to the solutions giving you the real set of solutions. The 1/3 may have only become relevant in the x intercept of the x axis for cos whatever it was. For example 2x=cos^-1(1/2), 2x=60, from here you would do 720-60 and 360+60.

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