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MAT Prep Thread - 2nd November 2016 Watch

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    (Original post by KloppOClock)
    what is this
    Attachment 585580
    Ah, I know which question you're talking about - it's the multiple choice question with 3 different integrals, yes?

    Just so you know, for that question, you don't need to evaluate the integrals at all. You technically can, but it's not on the syllabus and it'll probably take longer.

    Some more help if you need it:
    Spoiler:
    Show
    Think of it as 1/(cos3x). Is there a point where the graph's undefined between 0 and pi/8? If the graph starts at a certain value, can it ever "cross over" this undefined value?
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    (Original post by lewman99)
    Ah, I know which question you're talking about - it's the multiple choice question with 3 different integrals, yes?

    Just so you know, for that question, you don't need to evaluate the integrals at all. You technically can, but it's not on the syllabus and it'll probably take longer.

    Some more help if you need it:
    Spoiler:
    Show
    Think of it as 1/(cos3x). Is there a point where the graph's undefined between 0 and pi/8? If the graph starts at a certain value, can it ever "cross over" this undefined value?
    i just dont get why the dx is above the cos3x
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    Can anyone help explain G of the multiple choice questions from 2011?
    Thank you
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    (Original post by RichE)
    Yes, the bold is right.

    Well the next ones are hyperbolae, can you sketch those?

    And if you think about it you only need those hyperbolae to answer the rest of the second part.
    Hey, I still really can't follow the question feel so stupid. Im guessing that the max for x^2+y^2+4xy is 3? (Probably wrong) since its = (x+y)^2+2xy and (x+y)^2 is dominant term and we know the max for x+y so we just sub it in to get 3? I don't really know how to do the next 2 parts though, would you please guide me in the right direction?
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    (Original post by Moogle679)
    Can anyone help explain G of the multiple choice questions from 2011?
    Thank you
    think of it as a compound function
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    (Original post by danielhx)
    Hey, I still really can't follow the question feel so stupid. Im guessing that the max for x^2+y^2+4xy is 3? (Probably wrong) since its = (x+y)^2+2xy and (x+y)^2 is dominant term and we know the max for x+y so we just sub it in to get 3? I don't really know how to do the next 2 parts though, would you please guide me in the right direction?
    You're overcomplicating things. When is x^2+y^2 largest, when is xy largest, is there somewhere where both these things happen?
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    (Original post by Moogle679)
    Can anyone help explain G of the multiple choice questions from 2011?
    Thank you
    What is the function f for inputs between -1 and 0? Note x^2-1 is in that range.
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    dont get this
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    Anyone else find the specimen paper B much easier that the other papers??
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    (Original post by KloppOClock)
    think of it as a compound function
    (Original post by RichE)
    What is the function f for inputs between -1 and 0? Note x^2-1 is in that range.
    Name:  IMG_1261.jpg
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    This is what I did but I'm pretty sure it's wrong, nothing is clicking with me for this question for some reason
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    (Original post by Moogle679)
    Name:  IMG_1261.jpg
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    This is what I did but I'm pretty sure it's wrong, nothing is clicking with me for this question for some reason
    looks right
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    Ahhhh I'm so ***king nervous But not studying 👻🙀
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    (Original post by KloppOClock)
    x
    dont get this
    Spoiler:
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    the function  \displaystyle f^{i}gf^{j}gf^{k} = 4x + 4k + 2j + i , you worked this out in iii)

    so if you want to find the different ways to get  \displaystyle 4x + 4m then comparing constant terms of your large function (the fgfgf thing) and the required function

    we get  \displaystyle 4x + 4k + 2j + i = 4x + 4m so therefore  \displaystyle 4k + 2j + i = 4m

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    (Original post by 17lina)
    Ahhhh I'm so ***king nervous But not studying 👻🙀
    only 23 days left already, time is going so fast :eek:
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    (Original post by DylanJ42)
    Spoiler:
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    the function  \displaystyle f^{i}gf^{j}gf^{k} = 4x + 4k + 2j + i , you worked this out in iii)

    so if you want to find the different ways to get  \displaystyle 4x + 4m then comparing constant terms of your large function (the fgfgf thing) and the required function

    we get  \displaystyle 4x + 4k + 2j + i = 4x + 4m so therefore  \displaystyle 4k + 2j + i = 4m
    but surely thats just calculating the number of possibilities for when you are doing the function
    \displaystyle f^{i}gf^{j}gf^{k}

    what about if you did something like \displaystyle g^{2}f^{m}

    EDIT: nvm i get it now, j and k would be zero in that case, I understand now
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    (Original post by KloppOClock)
    i just dont get why the dx is above the cos3x
    \int_0^{\pi/8} \! \frac{\mathrm{d}x}{\mathrm {cos} 3x} can just be treated as \int_0^{\pi/8} \! \frac{1}{\mathrm {cos} 3x} \, \mathrm{d}x, as it is basically treating \mathrm{d}x as a scalar multiple, thus being on the numerator:

    \frac{1}{\mathrm {cos} 3x} \, \times \mathrm{d}x = \frac{\mathrm{d}x}{\mathrm {cos} 3x}
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    (Original post by some-student)
    \int_0^{\pi/8} \! \frac{\mathrm{d}x}{\mathrm {cos} 3x} can just be treated as \int_0^{\pi/8} \! \frac{1}{\mathrm {cos} 3x} \, \mathrm{d}x, as it is basically treating \mathrm{d}x as a scalar multiple, thus being on the numerator:

    \frac{1}{\mathrm {cos} 3x} \, \times \mathrm{d}x = \frac{\mathrm{d}x}{\mathrm {cos} 3x}
    Okay, I didn't know you could treat \displaystyle dx like that, thanks. Are there any other ways you can use \displaystyle dx where its not\displaystyle f'(x) [x] dx or \displaystyle f'(x) [dx/x] ?
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    Also, is anyone here going to that MAT support event thing tomorrow at manchester uni?
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    (Original post by KloppOClock)
    Also, is anyone here going to that MAT support event thing tomorrow at manchester uni?
    No, but could you do me a favour and let me know what they say? Thanks!
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    Can someone explain MAT15 Q3 Part VI?

    I have absolutely no idea where they got anything in their solution from.
 
 
 
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