Join TSR now and get all your revision questions answeredSign up now
    Offline

    0
    ReputationRep:
    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?
    Offline

    1
    ReputationRep:
    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
    Offline

    1
    ReputationRep:
    For 9ii) I put (3/2pi - a) + 3/2pi instead of 3pi - a, how many marks would I lose?
    Offline

    1
    ReputationRep:
    (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???

    Name:  ImageUploadedByStudent Room1432134606.003804.jpg
Views: 152
Size:  102.1 KB


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    So if you solve for 3.3 rad you don't take it away from 6 pi because it's just symmetry?
    Offline

    1
    ReputationRep:
    (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 \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 S_{2N} - S_{N-1} (just like the previous question where \displaystyle\sum_{n=10}^{20} u_n \equiv S_{20} - S_9 (because you'd be excluding u_{10} if you did S_{20} - S_{10}).

    So finding S_{2N} and simplifying:

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

    Finding S_{N-1} and simplifying:

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

    This gets you brackets to expand:

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

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

    \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:
    4.5N^2 - 6.5N - 2748 = 0

    Then I used the quadratic formula, only bothering to calculate -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.

    \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
    Offline

    1
    ReputationRep:
    (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?


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (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 )
    Offline

    3
    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
    Offline

    1
    ReputationRep:
    (Original post by Lil_timmi)
    Sorry for sounding stupid, but how did you get 6pi in the first place and not 2pi?


    Posted from TSR Mobile
    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
    Offline

    0
    ReputationRep:
    (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 )
    Haha I did this too, the panic of exams does some very strange things...
    Offline

    1
    ReputationRep:
    Name:  ImageUploadedByStudent Room1432135522.206627.jpg
Views: 403
Size:  108.0 KB

    This is how I answered 7. iii)

    Can someone tell me where I went wrong???

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (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 )
    I know yea haha, I just had no time left thank you anyway
    Offline

    2
    ReputationRep:
    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

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (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


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (Original post by Lil_timmi)
    Name:  ImageUploadedByStudent Room1432135522.206627.jpg
Views: 403
Size:  108.0 KB

    This is how I answered 7. iii)

    Can someone tell me where I went wrong???

    Posted from TSR Mobile
    You can't just square the trig functions on each side and expect them to still equal eachother.
    Offline

    2
    ReputationRep:
    (Original post by Lil_timmi)
    Name:  ImageUploadedByStudent Room1432135522.206627.jpg
Views: 403
Size:  108.0 KB

    This is how I answered 7. iii)

    Can someone tell me where I went wrong???

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


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    Will we lose all 6 marks for getting N = -44 or N= 125/3 ?
    Offline

    2
    ReputationRep:
    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.

    Posted from TSR Mobile
 
 
 
Poll
How are you feeling about GCSE Results Day?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.