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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by Student403)
    Stop spreading incorrect information.

    Why don't you try it on your calculator and tell me what you get?

    Try the total of [(1/3)lnx] between 3 and 2. Then try the total of [(1/3)ln(3x)] between 3 and 2.

    You will find that in both cases you get 0.135

    you're right my bad, never realised that could happen, why is that mathematically possible?
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    (Original post by Dayzh)
    guys, in the june 2015 paper, question 6a: http://qualifications.pearson.com/co...e_20150616.pdf

    where did they obtain the factor of 2cos(theta) in the mark scheme: http://qualifications.pearson.com/co...c_20150812.pdf ??

    Need help, thanks
    They do dx/dtheeta = 2costheeta
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    (Original post by jtlmao)
    How do you know when to do long division before partial fractions?
    If it is top heavy e.g Name:  Screen Shot 2016-06-23 at 18.54.45.png
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    9x^2 over x^2
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    13 1 20 8 42 Is this even C4?

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    (Original post by Nick-F007)
    you're right my bad, never realised that could happen, why is that mathematically possible?
    Because when you integrate it to one version, you get (1/3)lnx + C. If you integrate it to the other version, you get (1/3)ln3x + D. Where C and D are some constants. Taking the second version and applying log rules, we can expand

    (1/3)ln3x + D

    to (via logarithm product rule)

    (1/3)lnx + (1/3)ln3 + D

    In the first version we saw that we got (1/3)lnx + C.

    So by comparing the two, C = (1/3)ln3 + D.

    As you can see C and D are not equal, so basically when you take the indefinite integral, you get a different constant depending on how you integrate it.


    Now when you take the definite integral, the constants always cancel each other out anyway, so it doesn't make a difference which version you use.
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    (Original post by Don Pedro K.)
    13 1 20 8 42 Is this even C4?

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    Isn't that one of TeeEm's hard/very hard papers
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    (Original post by Don Pedro K.)
    13 1 20 8 42 Is this even C4?

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    That is definitely C4. It can be done with further methods, but C4 teaches you all the information you need to solve that
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    (Original post by 13 1 20 8 42)
    Isn't that one of TeeEm's hard/very hard papers
    Yeah paper s.

    (Original post by Student403)
    That is definitely C4. It can be done with further methods, but C4 teaches you all the information you need to solve that
    Ah okay...well I guess I'll have a got at it :s! Never seen anything like it before though!
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    (Original post by Don Pedro K.)
    Yeah paper s.



    Ah okay...well I guess I'll have a got at it :s! Never seen anything like it before though!
    Good luck. The solution's pretty simple if you take it step by step
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    (Original post by Don Pedro K.)
    Yeah paper s.



    Ah okay...well I guess I'll have a got at it :s! Never seen anything like it before though!
    I wouldn't worry about S and T. But that question is one of the easier ones I think. Just think about what you know about minimums and variables..
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    question 7a
    http://qualifications.pearson.com/co...e_20150616.pdf
    it doesnt say whether to do radians or in degrees unless being stupid but can some one help
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    (Original post by Nick-F007)
    you're right my bad, never realised that could happen, why is that mathematically possible?
    Ln 3x and ln x both differentiate to 1/x
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    (Original post by NotoriousS)
    Attachment 555713

    Can someone please help me on part b?

    it tells you that x<8/9.

    So you can take out 1000 from under the cube root sign leaving you with
    10 cube root 7.1

    Then 7.1= 8-9x
    Solve for x=0.1
    Sub 0.1 into your previous expansion and times by 10

    In summary you need your x to be small, so you make that possible using the method above. P.S the mark scheme for this question has an error
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    (Original post by shamk123)
    question 7a
    http://qualifications.pearson.com/co...e_20150616.pdf
    it doesnt say whether to do radians or in degrees unless being stupid but can some one help
    In calculus you always use radians
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    (Original post by Student403)
    Good luck. The solution's pretty simple if you take it step by step
    What is the solution, I'm not sure how to do this?

    Thank you
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    i know this sounds stupid but ive got a mind blank on the easiest question
    Can someone please differentiate 5/2t for my stupid ar*e
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    (Original post by taysc)
    What is the solution, I'm not sure how to do this?

    Thank you
    The shortest line joining two lines, l1 and l2, lies perpendicular to both l1 and l2. Try to take it from here
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    (Original post by coolguy123456)
    phew
    But then again this is edextrick.. They gave a messed up C4 q in fp2
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    𝐬𝐢𝐧𝟓 𝒙 𝒄𝒐𝒔 𝒙 How do you integrate this?
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    (Original post by celessi)
    𝐬𝐢𝐧𝟓 𝒙 𝒄𝒐𝒔 𝒙 How do you integrate this?
    sin5xcosx or sin5xcosx?
 
 
 
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