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AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread] watch

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    (Original post by aaaae)
    Anyone know how to draw the graph for y=f(|x|), in comparison to y= |f(x)|
    y=f(|x|) will be symmetrical about y axis because x=-1 will give same y value as x=1 etc
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    http://www.mrbartonmaths.com/resourc...3-QP-Jun13.pdf

    question 6 . can anyone explain to me how the second graph works...? i thought it would be a reflection in the x axis and a movement of pi in the y direction. But in the mark scheme they've done what looks like just a reflection in the y axis.

    Mark scheme
    http://www.mrbartonmaths.com/resourc...-MS-June13.pdf
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    Name:  c3 june 2005 q1.png
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Size:  166.6 KB Hi, can anyone help with question 1. I have differentiated and I know how to integrate for part b)ii) doing by parts but it says hence and the mark scheme gets straight to the answer after one step, is there a shorter way of integrating it?
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    (Original post by C0balt)
    do the values not have to be satisfied by the ranges of both parent functions?
    (i'm just asking in general not this particular question)
    i believe they don't but the composite function has a new domain which this case is (x*less than or equal to* 4
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    (Original post by aaaae)
    Anyone know how to draw the graph for y=f(|x|), in comparison to y= |f(x)|
    literally just reject in y axis
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    (Original post by FluorescentM)
    Name:  c3 june 2005 q1.png
Views: 428
Size:  166.6 KB Hi, can anyone help with question 1. I have differentiated and I know how to integrate for part b)ii) doing by parts but it says hence and the mark scheme gets straight to the answer after one step, is there a shorter way of integrating it?
    Are you familiar with integration by recognition/reverse chain rule?
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    Why on June 2014 question 5B do you reject the inverse function that is minus (x+4)^0.5
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    (Original post by Sirelis)
    http://www.mrbartonmaths.com/resourc...3-QP-Jun13.pdf

    question 6 . can anyone explain to me how the second graph works...? i thought it would be a reflection in the x axis and a movement of pi in the y direction. But in the mark scheme they've done what looks like just a reflection in the y axis.

    Mark scheme
    http://www.mrbartonmaths.com/resourc...-MS-June13.pdf
    The reflection in the x axis followed by a translation by pi in the y direction is, in this case, a reflection in the y axis.
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    (Original post by Fudge2)
    Are you familiar with integration by recognition/reverse chain rule?
    I think so, I tried doing that so I put the 8 back in as the coefficient of x and then took 1/8 out of the integral to account for this, but then what happens to the 8x?
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    (Original post by Dapperblook22)
    The reflection in the x axis followed by a translation by pi in the y direction is, in this case, a reflection in the y axis.
    Oh cheers :P i see what I did now, I moved it by - pi instead of pi
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    (Original post by FluorescentM)
    I think so, I tried doing that so I put the 8 back in as the coefficient of x and then took 1/8 out of the integral to account for this, but then what happens to the 8x?
    Way I think about it is that you must have had (x^2-6)^4 somewhere in order to differentiate and get back to it cubed, and if you differentiate that you get 4(x^2-6)^3 * 2x which gives you 8x(x^2-6)^3. You want what differentiates to give you just x(x^2-6)^3, so you want 1/8(x^2-6)^4 to get that. It's kinda hard to explain so https://www.youtube.com/watch?v=Fag7U3NtUmw might be a better bet if that doesn't make sense
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    (Original post by Fudge2)
    Way I think about it is that you must have had (x^2-6)^4 somewhere in order to differentiate and get back to it cubed, and if you differentiate that you get 4(x^2-6)^3 * 2x which gives you 8x(x^2-6)^3. You want what differentiates to give you just x(x^2-6)^3, so you want 1/8(x^2-6)^4 to get that. It's kinda hard to explain so https://www.youtube.com/watch?v=Fag7U3NtUmw might be a better bet if that doesn't make sense
    Ah thank you so much! Yes that makes sense, the video is really useful too he goes through a lot of examples so thank you
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    (Original post by ilovecake123)
    lol i wouldn't know where to start if you gave me a substution then i could do it
    There is no substitution. No parts.
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    (Original post by FluorescentM)
    Ah thank you so much! Yes that makes sense, the video is really useful too he goes through a lot of examples so thank you
    No problemo
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    (Original post by humzaakram1)
    One to one just means one value of x maps onto one value of y, and it depends on the function for example y=x^2 can have a domain for all real values but we know that f(x)>or equal to 0 as f(x) can't be negative, so just look at what they give and try a few numbers with your calculator both positive and negative and you will see its range.
    Thank you so much, I managed to get the awnser so I think I've got it now
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    Any ideas for 5B june 2014??
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    (Original post by AlexM10)
    Any ideas for 5B june 2014??
    I think it's because the inverse function's range is the domain of the original function, so you know that the range of f^-1(x) is f^-1(x)>=3, and if you had the minus one then you'd get values less than 3 so it has to be the plus one?
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    (Original post by Ano123)
    Attachment 550055
    Try this integral.
    My attempt, no idea if correct.

    EDIT: +C of course
    Attached Images
     
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    (Original post by MahuduElec)
    My attempt, no idea if correct.

    EDIT: +C of course
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    hey guys, does anyone know if we have to know double angle formulae for C3 or do they only come up in C4???

    Also, does anyone know if this is the last year of the maths spec (like it was in chemistry)? Because today's chemistry paper was quite horrible and apparently it's because it was the last one from this syllabus.. I wonder if it will be the same for maths?
 
 
 
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