Turn on thread page Beta
    Offline

    0
    ReputationRep:
    (Original post by tomiswellcool)
    On question 4)i) Did the solution have to be simplified, because i lieterally put everything into the quotient rule, then just left it. Will i be docked marks for this?
    I did that too, i don't think you should lose marks there because there really wasn't much to simplify
    Offline

    10
    ReputationRep:
    (Original post by crazedmonkey09)
    Did anyone solve the Cosec^2 - Cot=1 by
    1/sin^2 - CosSin/Sin^2 =1
    cosSin/Sin^2 = 1
    Cos/Sin=1
    cot=1

    This working is completely wrong because i negated the 1 (i only just realised now ) but i still got 2 solutions, do you think i will still get the method marks?
    thats would get one of the anwers but not the one you got.

    you should've got cos(2x) = 0 from that
    Offline

    0
    ReputationRep:
    this paper was bloody hard compared to the one june 09...i did it as mock and got 80% on that with no revision at all....this time i would be happy to get half that :eek: :mad:

    oh well there is always a retake possible...:confused:
    Offline

    0
    ReputationRep:
    Looked through the markscheme...and was just wondering about the cosec^2x - cotx = 1

    I got the cotx(cotx - 1) = 0 ...(or whatever that turned out to be)

    ..but surely cotx can't = 0, as rewritten that would be 1/tanx = 0 ... multiply both sides by tanx, and that leaves you with 1 = 0 which is incorrect? And tanx can never be 0, because 1/0 is impossible..

    Have I got that wrong?
    Offline

    0
    ReputationRep:
    (Original post by Anonymous0155)
    thats would get one of the anwers but not the one you got.

    you should've got cos(2x) = 0 from that
    hows that? from this crazy working i got the solutions 22.5, 112.5 its a sin squared on the bottom and a single sin on top so i get cot but why =0?
    Offline

    10
    ReputationRep:
    (Original post by crazedmonkey09)
    Did anyone solve the Cosec^2 - Cot=1 by
    1/sin^2 - CosSin/Sin^2 =1
    1 - cos/Sin/Sin^2 = sin^2
    1 - Cos/Sin^3 = sin^2

    This working is completely wrong because i negated the 1 (i only just realised now ) but i still got 2 solutions, do you think i will still get the method marks?
    actually, thats just kinda wrong; but you got the right answer somehow.

    the bold is how it should've gone, coz if you multiply you gotta do it both sides.
    Offline

    0
    ReputationRep:
    (Original post by Smeh)
    That was my plan too!
    I wish I was born year earlier...None of this A* business.
    I should really be revising chemistry right now but instead i'm moping :emo:
    Well after spending several hours looking on the bright side only to have my retinas burnt by more incorrect answers :mad: the grade boundaries should be quite low and if they are as low as 55 for example we could be looking at 93+.
    Offline

    0
    ReputationRep:
    (Original post by crazedmonkey09)
    I did that too, i don't think you should lose marks there because there really wasn't much to simplify

    Yeah, the way i looked at it this was: They asked for dy/dx of f(x), and i gave them dy/dx of f(x). They didn't ask for it simplified so i didn't :p:
    Offline

    0
    ReputationRep:
    So for question 8 there was meant to be four solutions?  cot2x = 0

     \frac{cos2x}{sin2x} = 0 multiply through by sin2x

     cos2x = 0 which equals zero at  \frac{\pi}{2} , \frac{3\pi}{2}
    Offline

    0
    ReputationRep:
    (Original post by Anonymous0155)
    actually, thats just kinda wrong; but you got the right answer somehow.

    the bold is how it should've gone, coz if you multiply you gotta do it both sides.
    i didn't multiply the whole thing by sin i just wrote the fraction differently

    cos2x/sin2x
    as
    Cos2xSin2x/Sin^2 2x

    those fractions are identical because sin can be canceled. My wrong working was where it should become 1-Cos2xSin2x/Sin^2 but don't include the 1- and i get the right answer
    Offline

    0
    ReputationRep:
    (Original post by Robbyv)
    ^ so much this.
    Having 8 exams this month and having sat 4 of them and finding them piss, i thought this would be another cakewalk. First question stumped me for a bit. the X=Tany, i got the the dy/dx= 1/sec^2(y) and got stuck (i'm really annoyed i did now :mad:) The graph of ln|x| was fail, i've only drawn lnx for +ve values (another thing :mad:) and then there was that 9a)ii). I figured it out, literally when they said pens down, so i couldn't write it :mad:. Oh and now theres a debate whether cot2x could give you solutions for x. well June here i come
    When i got to 1/sec^2(y) i changed it to 1/1+tan^2(y) and x=tany
    so 1/1+tan^2(y) = 1/1+x i hope thats right =S
    Offline

    0
    ReputationRep:
    But that could also be written as:

    1/tan2x = 0

    ...meaning tan2x couldn't = 0 ?
    Offline

    10
    ReputationRep:
    (Original post by crazedmonkey09)
    i didn't multiply the whole thing by sin i just wrote the fraction differently

    cos2x/sin2x
    as
    Cos2xSin2x/Sin^2 2x

    those fractions are identical because sin can be canceled. My wrong working was where it should become 1-Cos2xSin2x/Sin^2 but don't include the 1- and i get the right answer
    Yes you had

    (1-cos2xsin2x)/sin^2(2x) = 1

    but you can't just make it 1 - cos2xsin2x = 1

    1 - cos2xsin2x = sin^2(2x) (you need a common denominator for the whole equation in order to cancel it out; not just one side.
    Offline

    0
    ReputationRep:
    (Original post by Riten Patel)
    When i got to 1/sec^2(y) i changed it to 1/1+tan^2(y) and x=tany
    so 1/1+tan^2(y) = 1/1+x i hope thats right =S
    thats correct.
    yo, do u go to a catholic 6th form?
    Offline

    0
    ReputationRep:
    (Original post by Anonymous0155)
    Yes you had

    (1-cos2xsin2x)/sin^2(2x) = 1

    but you can't just make it 1 - cos2xsin2x = 1

    1 - cos2xsin2x = sin^2(2x) (you need a common denominator for the whole equation in order to cancel it out; not just one side.
    I never did make it that
    1 stays on the other side of the equals sign i never canceled the 1's i just forgot about the other 1 which should have been on top of the fraction, thats how i managed cot2x=1. call it serendipity if you will but i doubt i'll get the marks because my working is badly wrong

    from your working you multiplied the whole thing by sin^2 but somehow divided it as well, i can't understand what you've done there
    Offline

    10
    Cot 2x can equal zero; look at the graph of cot x. It gives 2 extra solutions, 45, 135 which ARE valid in the original equation. I know it is confusing but it has been explained it detail earlier. If you didn't include them you've lost 2 marks.
    Offline

    0
    ReputationRep:
    can someone tell me what page i can download the unofficial mark scheme from

    thank you
    Offline

    0
    ReputationRep:
    easiest paper every i think i got 70/75
    Offline

    10
    (Original post by Catherine91)
    Looked through the markscheme...and was just wondering about the cosec^2x - cotx = 1

    I got the cotx(cotx - 1) = 0 ...(or whatever that turned out to be)

    ..but surely cotx can't = 0, as rewritten that would be 1/tanx = 0 ... multiply both sides by tanx, and that leaves you with 1 = 0 which is incorrect? And tanx can never be 0, because 1/0 is impossible..

    Have I got that wrong?
    you've kind of proved why the solutions work. If the logic is that the solutions are not valid then because 1/0 is impossible tanx can never be zero; but it can, an infinite number of times in fact.
    Offline

    10
    (Original post by aswat300)
    can someone tell me what page i can download the unofficial mark scheme from

    thank you
    33 I think
 
 
 
Turn on thread page Beta
Updated: June 14, 2010

University open days

  • University of Warwick
    Undergraduate Open Days Undergraduate
    Sat, 20 Oct '18
  • University of Sheffield
    Undergraduate Open Days Undergraduate
    Sat, 20 Oct '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 20 Oct '18
Poll
Who is most responsible for your success at university

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

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