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AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread Watch

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    (Original post by Dhaden)
    How you guys learning the vectors stuff except from doing past papers?
    Try to learn some basic shapes that may crop up e.g. Rhombus bla bla bla
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    thank youuuuuu! <3 (and I'm sorry if I disturbed you) hope you do well tomorrow! good luck!
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    Name:  Screen Shot 2013-06-09 at 18.46.31.png
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    could anyone help me, on the second part where they give you cos2theta = x, i know how to do all the rearranging but is there a way to tell whether you should use the 2cos^(2)x - 1 version or the 1-2sin^(2)x as cos2theta, or even - does it matter? it would be waste of time to use the wrong version of cos2x and have to redo all the working!
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    I've just done June 2011, Jan 2012 and Jan 2013, all of them i have got nearly full marks (Probably because I've done them all 3 times)! Just need to remember some formulas and i should hopefully be ready! Recapping some C3 stuff and learning some basic shapes
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    Done barely anything today. Going to look through a textbook and see if I've missed any memorisy bits, go through integration by parts & substitution from C3, and then try get an early night

    Good luck to everyone, I'm sure you'll all do great!
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    (Original post by georgiaaaxo)
    Name:  Screen Shot 2013-06-09 at 18.46.31.png
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    could anyone help me, on the second part where they give you cos2theta = x, i know how to do all the rearranging but is there a way to tell whether you should use the 2cos^(2)x - 1 version or the 1-2sin^(2)x as cos2theta, or even - does it matter? it would be waste of time to use the wrong version of cos2x and have to redo all the working!
    You should find out by the question, x = cos0

    So technically, you should be using the cos2x = 2cos^2x - 1
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    (Original post by A-New-Start)
    i was looking at this picture and my mind decided to pick up the pen from the screen. -_-
    HAHAHA Amazing - Revision will do that to ya
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    (Original post by Enginerd.)
    You should find out by the question, x = cos0

    So technically, you should be using the cos2x = 2cos^2x - 1
    oh ok, so whatever x is equal to- if its cos something use the cos version, if its sin use the sin version?
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    (Original post by Jordn)
    I Have always thought intergration by inspection was

    1/x = ln|x|+c so 1/2x = 2ln|2x| + c

    or 3/x = 1/3ln|x| +c

    but this mark scheme is telling me differently
    have i been taught wrong?
    Attachment 224946
    yes, if it is 3/x+1 it integrates to 3ln|x+1|
    If it's 1/3x+1 it integrates to 1/3ln|3x+1|
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    so max values of cos's are 1 and min are -1
    and for sin's it's max 1 and min 0
    right?
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    (Original post by georgiaaaxo)
    oh ok, so whatever x is equal to- if its cos something use the cos version, if its sin use the sin version?
    Yes, i normally do that, and it works XD
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    (Original post by georgiaaaxo)
    oh ok, so whatever x is equal to- if its cos something use the cos version, if its sin use the sin version?
    Yes, i normally do that, and it works XD

    (Original post by A-New-Start)
    so max values of cos's are 1 and min are -1
    and for sin's it's max 1 and min 0
    right?
    min for sin is also -1, you can tell the min and max by looking at the graphs for the sin/cos/tan graphs
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    (Original post by beautywithbrains)
    Can somebody please help me out on this question? It's jan08 qu3c)
    Don't understand what to with the 8 on the bottom?
    Attachment 224993


    Posted from TSR Mobile
    That 8 just comes up to the top so you get 8^-0.5(2+3x)^0.5 take the 2^0.5 out and then expand binomially.
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    I need help guys, for Jan 2013 paper Q3.

    CotxCos2x = 0
    Therefore their are no solutions within the range for the Cotx but Cos2x has two solutions. 45 Degrees and 135 Degrees. Where does the 90 Degrees come from?

    http://img560.imageshack.us/img560/3194/cotx.png
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    Hi everyone. I am really confused but with the June 2007 question, Q8, they ask you to integrate 1/(1+2y) on LHS and then they say the answer is k root 1+2y where I get k to be 2 and then the answer is 1/1+2y=-1/x but the 2 is not in the answer. Am I doing it wrong ?
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    (Original post by Enginerd.)
    I need help guys, for Jan 2013 paper Q3.

    CotxCos2x = 0
    Therefore their are no solutions within the range for the Cotx but Cos2x has two solutions. 45 Degrees and 135 Degrees. Where does the 90 Degrees come from?

    http://img560.imageshack.us/img560/3194/cotx.png
    Oh god this question
    Think of it this way
     tan(x) = {sin(x)\over cos(x)}

\Rightarrow cot(x) = {cos(x)\over sin(x)}

\

From the question given

\

\Rightarrow cos(x) = 0

\

x = 90

    It got me too.

    Edit: LaTex didn't make sense
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    could anyone clear something up for me, memory of C3 is a bit hazy: when do you know whether to use integration by parts or substitution? do you only use substitution if you can use the substitution for both parts of the product? (dont know if that makes any sense)
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    (Original post by Bord3r)
    Oh god this question
    Think of it this way
     tan(x) = {sin(x)\over cos(x)}

\Rightarrow cot(x) = {cos(x)\over sin(x)}

\

From the question given

\

0= \fract{sin(x)}{cos(x)}

\

\Rightarrow cos(x) = 0

\

x=90

    It got me too.
    Hmm.. what if we do it the 1/tanx way? We get it wrong lol.
    Sin x = 0 their are no solutions as sin x = 0 and 180?
    Cos x = 0 the solution(within the range) is 90?

    Thanks, for explaining it XD
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    (Original post by Enginerd.)
    Hmm.. what if we do it the 1/tanx way? We get it wrong lol.
    Sin x = 0 their are no solutions as sin x = 0 and 180?
    Cos x = 0 the solution(within the range) is 90?

    Thanks, for explaining it XD
    Yeah just about sums up the Jan 2013 paper tbh
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    (Original post by Bord3r)
    Yeah just about sums up the Jan 2013 paper tbh
    Yep, was a nasty paper.
 
 
 
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