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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    And since we have only 2 weeks left reckon doing most Solomon papers all gold papers and all papers from the last 2 years including Jan ial should be enough


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    (Original post by Supermanxxxxxx)
    Attachment 540855
    Spent so long on a still couldn't do it


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    Did you divide through by (X-6)(X-3) then do partial fractions to put it as A/(x-6) + B/(X-3) ?
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    (Original post by Middriver)
    Did you divide through by (X-6)(X-3) then do partial fractions to put it as A/(x-6) + B/(X-3) ?
    Nope I didn't even think of that, thanks


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    [QUOTE=Supermanxxxxxx;65322473]Attachment 540855
    Spent so long on a still couldn't do it


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    Edit:nevermind
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    (Original post by Clovers)
    Hi,
    Can someone please explain why, when solving a differential equation, you don't put constants of integration on both sides? Thanks in advance

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    hi look at this video to understand it http://www.examsolutions.net/maths-r...utorial-1a.php
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    (Original post by Supermanxxxxxx)
    Nope I didn't even think of that, thanks


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    yh use partial fractions
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    Anybody know how I should go about learning vectors, I think I understand the topic but can't do any pp questions!
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    (Original post by metrize)
    Can someone help on jan 13 c4 question 8?

    https://googledrive.com/host/0B1ZiqB...%20Edexcel.pdf

    This is my method:
    -ln(3-theta all over A)=t/125
    So 3-theta over A=e^(-t/125)
    So 3-theta = Ae^(-t/125)
    So theta = 3-Ae^(-t/125)


    But the answer is meant to be theta = 3+Ae^(-t/125)
    Not sure why you start with -ln(3-theta all over A) I thought the A would be a constant on the dt side, so it would start -ln(3-theta)= t/125 +c ?
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    (Original post by Middriver)
    Not sure why you start with -ln(3-theta all over A) I thought the A would be a constant on the dt side, so it would start -ln(3-theta)= t/125 +c ?
    Yeah i called the constant ln A so then when i took it away and with logs rules i made it -ln( 3-theta all over A)
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    (Original post by metrize)
    Yeah i called the constant ln A so then when i took it away and with logs rules i made it -ln( 3-theta all over A)
    Oh right, did you manage to get it right?
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    guys for c3, when drawing modulus graphs and equating it to eachother how would you know if they intercept twice or not?

    https://07a69ccf283966549a9350d1a669...%20Edexcel.pdf

    heres an example, question 7
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    (Original post by Middriver)
    Oh right, did you manage to get it right?
    I did the steps I had said at first but fot the signs wrong in the end so im not sure where i went wrong
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    (Original post by metrize)
    I did the steps I had said at first but fot the signs wrong in the end so im not sure where i went wrong
    Hmh shall I send my workings might be able to see where sign change went wrong? I didn't use lnA though so don't want to confuse you or anything.
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    (Original post by Middriver)
    Hmh shall I send my workings might be able to see where sign change went wrong? I didn't use lnA though so don't want to confuse you or anything.
    Yeah please do, thanks
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    (Original post by metrize)
    Can someone help on jan 13 c4 question 8?

    https://googledrive.com/host/0B1ZiqB...%20Edexcel.pdf

    This is my method:
    -ln(3-theta all over A)=t/125
    So 3-theta over A=e^(-t/125)
    So 3-theta = Ae^(-t/125)
    So theta = 3-Ae^(-t/125)


    But the answer is meant to be theta = 3+Ae^(-t/125)
    hello, your first step is wrong.. why are you dividing by A? thats not how its done.. the constant youre meant to be Adding is C and not A. Heres how i did it:

    1) Dtheta/dt = (3-theta) / 125
    2) 1/(3-theta) with respect to dtheta = 1/125 with respect to dt
    3) -ln(3-theta) = t/125 + 'C'
    4) ln(3-theta) = -t/125 - C
    5) (3-theta) = e-0.008t-c
    6) theta= 3 - e-0.008t-c
    7) theta= 3 - (e-0.008t e-c )
    8) theta = -e-c e-0.008t +3
    9) theta =Ae-0.008t +3
    where A= -e-c it says A is a constant and -e-c is a constant
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    (Original post by jovdawesome)
    hello, your first step is wrong.. why are you dividing by A? thats not how its done.. the constant youre meant to be Adding is C and not A. Heres how i did it:

    1) Dtheta/dt = (3-theta) / 125
    2) 1/(3-theta) with respect to dtheta = 1/125 with respect to dt
    3) -ln(3-theta) = t/125 + 'C'
    4) ln(3-theta) = -t/125 - C
    5) (3-theta) = e-0.008t-c
    6) theta= 3 - e-0.008t-c
    7) theta= 3 - (e-0.008t e-c )
    8) theta = -e-c e-0.008t +3
    9) theta =Ae-0.008t +3
    where A= -e-c it says A is a constant and -e-c is a constant
    I did that because at the beginning i said the constant is ln A and so i took away ln A and with log rules it divides by A inside the log, its the way that our teacher had taught it us, will learn how to do it with c, thanks for posting the working out
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    (Original post by ryandaniels2015)
    Anybody know how I should go about learning vectors, I think I understand the topic but can't do any pp questions!
    any help?
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    Nvm he's done it^
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    (Original post by metrize)
    I did that because at the beginning i said the constant is ln A and so i took away ln A and with log rules it divides by A inside the log, its the way that our teacher had taught it us, will learn how to do it with c, thanks for posting the working out
    When adding the constant, we must use 'C' in most cases. We must only add lnC if it makes it easier for us to solve it. For example if a question was ln(x-2)= ln5.. we can add lnC here because it will be easier for us to solve as we can cancel the ln on both sides. However in all questions ur simply meant to add C and NOT lnC..
    I recommend you go over videos on examsolutions and learn this topic ASAP as we have just above 2 weeks left for the exam .. Good luck
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    (Original post by jovdawesome)
    When adding the constant, we must use 'C' in most cases. We must only add lnC if it makes it easier for us to solve it. For example if a question was ln(x-2)= ln5.. we can add lnC here because it will be easier for us to solve as we can cancel the ln on both sides. However in all questions ur simply meant to add C and NOT lnC..
    I recommend you go over videos on examsolutions and learn this topic ASAP as we have just above 2 weeks left for the exam .. Good luck
    Oh wait i think my A is different to the A in the actuak exam question and that i might have gotten confused with the letters... so my -A is a constant which is just A as theyre both still constants and so can i change the sign of it at the end?
 
 
 
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