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2016 May 25th Edexcel Core 2 Questions and answers. [Unofficial mark scheme] 2016 Watch

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    (Original post by medapplicant2016)
    That's what I thought, I did get 16.something though, and did work out f"(x) > 0, which usually gives you 2 marks (you don't need to sub)
    If you got 16.6 then i think you may only lose 1 mark then just for not subbing for the minimum value
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    (Original post by dididid)
    the answer was b= (3a+5)/9
    I did this

    log 3 (3b+1/a-2) = -1
    3^-1 = 3b+1/a-2
    1/3= 3b+1/a-2
    a-2=9b+3
    a-5=9b
    b=a-5/9
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    (Original post by tory88)
    Is question 6 written out correctly? I get \frac{11\pi}{30}or \frac{31\pi}{30}

    EDIT: Seen that there have been amendments since then. Shouldn't 8i) be \frac{a-5}{9}?

    I think you're correct, this was my workings.

    log_3(3b+1)-log_3(a-2)=-1

    log_3(3b+1)+1=log_3(a-2)

    log_3(3b+1)+log_33=log_3(a-2)

    log_33(3b+1)=log_3(a-2)

    3(3b+1)=a-2

    9b+3=a-2

    9b=a-5

    b=\frac{a-5}{9}
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    (Original post by bulletman54)
    If you got 16.6 then i think you may only lose 1 mark then just for not subbing for the minimum value
    Cool, thanks so much!
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    (Original post by UKoE Luna)
    I think you're correct, this was my workings.

    log_3(3b+1)-log_3(a-2)=-1

    log_3(3b+1)+1=log_3(a-2)

    log_3(3b+1)+log_33=log_3(a-2)

    log_33(3b+1)=log_3(a-2)

    3(3b+1)=a-2

    9b+3=a-2

    9b=a-5

    b=\frac{a-5}{9}
    At this point I give up, Yes you are right however the question in the exam may have been \log_3 (3b+1)-\log_3 (2a-2)=-1. It has been too long, I forget. I'll wait till I can see the paper again tomorow
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    (Original post by X_IDE_sidf)
    we are on to V4 now http://docdro.id/18BTG2C

    Thanks
    Shouldn't question 9c) be to prove that

    P=\frac{1000}{x}+\frac{x}{12}(4\  pi+36-3\sqrt{3})

    I forget how to LATEX... Shouldn't the fraction outside the brackets be \frac{x}{12}
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    (Original post by X_IDE_sidf)
    At this point I give up, Yes you are right however the question in the exam may have been \log_3 (3b+1)-\log_3 (2a-2)=-1. It has been too long, I forget. I'll wait till I can see the paper again tomorow
    Fair enough, probably the best shout, you can only get so far with memory, but I must thank you for the awesome mark scheme!
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    (Original post by X_IDE_sidf)
    At this point I give up, Yes you are right however the question in the exam may have been \log_3 (3b+1)-\log_3 (2a-2)=-1. It has been too long, I forget. I'll wait till I can see the paper again tomorow
    Please quote me in if you give any corrections to the questions.
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    how many marks would b 90 UMS
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    for question 6 b, are we supposed tto give answrs to 1 decimal place?
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    (Original post by tory88)
    Shouldn't question 9c) be to prove that

    P=\frac{1000}{x}+\frac{x}{12}(4\  pi+36-3\sqrt{3})

    I forget how to LATEX... Shouldn't the fraction outside the brackets be \frac{x}{12}
    yes, you're right
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    (Original post by dadawdawd)
    for question 6 b, are we supposed tto give answrs to 1 decimal place?
    Pretty sure it said to give answers to 1dp yes.
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    (Original post by X_IDE_sidf)
    The question didn't have limits at that point, part (a) was simply \int (3x-x^{\frac{3}{2}})dx
    Well then it's just +c
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    Can somebody type out the actual question 7b? I cant remember what it was and may help me to remember if i got 24.3
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    (Original post by X_IDE_sidf)
    u_n=ar^{n-1}
    \therefore |u_9-u_{10}| = |(64)(\frac{3}{4})^{9-1}-(64)(\frac{3}{4})^{10-1}| = 1.602 (3 d.p.)
    Hey guys I did exactly this in my exam, I remember writing it identically however I got the wrong answer (calculator error maybe?) Would I get the method marks but not the answer mark?
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    (Original post by Dieselblue)
    Hey guys I did exactly this in my exam, I remember writing it identically however I got the wrong answer (calculator error maybe?) Would I get the method marks but not the answer mark?
    I did u10 - u9 or is that wrong? You should get the method mark if it was only a calculating error.
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    (Original post by Brydon hurst)
    I did u10 - u9 or is that wrong? You should get the method mark if it was only a calculating error.
    I did ar^(9-1)-ar^(10-1) I got -0.025 though
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    (Original post by Dieselblue)
    I did ar^(9-1)-ar^(10-1) I got -0.025 though
    Yeah you will definitely get method marks.
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    (Original post by Brydon hurst)
    Yeah you will definitely get method marks.
    Ok thanks thats good to know
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    any ideas for grade boundaries? roughly how many marks for an a?
 
 
 
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