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    I've been having a mental battle with Q1 G on the same paper (2011), thinking you have to somehow translate/transform the graph, then find the area underneath using some means. However finally looking at the solutions, I have no idea what they are talking about and how they have got to certain inequalities. Anyone can explain?
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    (Original post by TheFuture001)
    Yeah thanks, I got it! I had ended with a quadratic earlier on, but I discarded it as meaningless since I thought the P's and A's were just x's and y's in the end. I didn't expect to do that.

    @dutchmaths, I'm not sure I really like your method aha. For example, to my eyes the first option after subbing in looks like it could probably be true, just by thinking of how large the number could get.
    Hey guys, Im quite new to this, and i have managed this question, but i think there is potential mistake in the question.
    if we say the rectangle has sides x and y, then the P^3 is
    (2x+2y)^3
    = 8x^3 + 24x^2y + 24xy^2 + 8y^3

    i would think that this is greater than the area xy for all values of x and y, which makes the first inequality P^3 > A seem like the answer
    Ive probably made a mistake somewhere, cause the people who design these papers arent exactly stupid
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    (Original post by MelvyW)
    Hey guys, Im quite new to this, and i have managed this question, but i think there is potential mistake in the question.
    if we say the rectangle has sides x and y, then the P^3 is
    (2x+2y)^3
    = 8x^3 + 24x^2y + 24xy^2 + 8y^3

    i would think that this is greater than the area xy for all values of x and y, which makes the first inequality P^3 > A seem like the answer
    Ive probably made a mistake somewhere, cause the people who design these papers arent exactly stupid
    No, that inequality is not always true.

    If we set x = y = \epsilon then

    P^3 = 64\epsilon^3

    and

    A = \epsilon^2

    So we can get P^3 \leq A provided 64\epsilon^3 \leq \epsilon^2 - that is provided \epsilon \leq \dfrac{1}{64}
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    Does anyone know where the answering space for the questions (that are not multiple choice) will be?

    I usually take up a lot of space for my working out.

    And, what does A(theta) and B(pi/2 * theta) mean on question 4?
    http://www.maths.ox.ac.uk/system/fil...e/0/test07.pdf

    Is A and B a constant, function or what?
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    (Original post by Noble.)
    No, that inequality is not always true.

    If we set x = y = \epsilon then

    P^3 = 64\epsilon^3

    and

    A = \epsilon^2

    So we can get P^3 \leq A provided 64\epsilon^3 \leq \epsilon^2 - that is provided \epsilon \leq \dfrac{1}{64}
    Ahh okay i see. Thanks!

    I very much doubt i could something like that out in exam though -.-
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    (Original post by Sayonara)
    Does anyone know where the answering space for the questions (that are not multiple choice) will be?

    I usually take up a lot of space for my working out.

    And, what does A(theta) and B(pi/2 * theta) mean on question 4?
    http://www.maths.ox.ac.uk/system/fil...e/0/test07.pdf

    Is A and B a constant, function or what?
    From what i understand A(theta) just means the area A. The area is dependent on the value of theta, and hence A(theta)
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    (Original post by MelvyW)
    Ahh okay i see. Thanks!

    I very much doubt i could something like that out in exam though -.-
    Well, it is somewhat intuitive. When you see a potential answer that says "If you cube the perimeter, it'll always be greater than the area" the first thing that should spring to mind is "Well, what if the perimeter is less than 1? Then when we cube it, it's going to be even smaller". This is partially why I, subconsciously, chose \epsilon to use in the notation because \epsilon is typically used to denote a very small value (it doesn't necessarily have to be, but in analysis the \epsilon - \delta arguments are only really interesting for small \epsilon).

    This is why you can instantly discount (b) as a potential answer, as once you apply some thought to it you realise how utterly absurd it is. It's suggesting A^2 &gt; 2P + 1 as a possible answer, yet you can make an area of less than 1 with a perimeter of more than one (i.e. a 3x0.25 box has area <1 yet perimeter 6.5) and clearly a number less than one squared is also less than one, yet the number on the RHS is bounded below by 1.
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    (Original post by MelvyW)
    Hey guys, Im quite new to this, and i have managed this question, but i think there is potential mistake in the question.
    if we say the rectangle has sides x and y, then the P^3 is
    (2x+2y)^3
    = 8x^3 + 24x^2y + 24xy^2 + 8y^3

    i would think that this is greater than the area xy for all values of x and y, which makes the first inequality P^3 > A seem like the answer
    Ive probably made a mistake somewhere, cause the people who design these papers arent exactly stupid
    Hey, I think you misinterpreted my comment as claiming that (a) was the answer, which is perhaps what made you think there was a mistake or something. I meant to say that I prefer pursuing the question as the guy I initially quoted prompted me to do so, since there's no ambiguity as you get the final answer dead on.
    I'd suggest you try this method too, I'll hint you. With the two equations you have, eliminate either x or y and see what happens. And don't be scared off seeing x's/y's and P's and A's in the same equation like I did



    Q1 G in the following paper http://www.mathshelper.co.uk/Oxford%...est%202011.pdf I am having trouble with. The solutions claim -1 <= x^2 - 1 <= 0 . How they came to this conclusion, and what follows, I can not understand. I would appreciate at least a hint in the correct understanding, thanks!
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    (Original post by Noble.)
    Well, it is somewhat intuitive. When you see a potential answer that says "If you cube the perimeter, it'll always be greater than the area" the first thing that should spring to mind is "Well, what if the perimeter is less than 1? Then when we cube it, it's going to be even smaller". This is partially why I, subconsciously, chose \epsilon to use in the notation because \epsilon is typically used to denote a very small value (it doesn't necessarily have to be, but in analysis the \epsilon - \delta arguments are only really interesting for small \epsilon).

    This is why you can instantly discount (b) as a potential answer, as once you apply some thought to it you realise how utterly absurd it is. It's suggesting A^2 &gt; 2P + 1 as a possible answer, yet you can make an area of less than 1 with a perimeter of more than one (i.e. a 3x0.25 box has area <1 yet perimeter 6.5) and clearly a number less than one squared is also less than one, yet the number on the RHS is bounded below by 1.
    Youve sat this exam so could you explain the format of the answer book please.

    I realise you are expected to do all workings for the multiple choice in the space underneath each part. So how are you expected to answer the 15 mark questions.

    Is there a question paper bookl and and answer book. are you given say x pages for all of the longer questions. or is it y pages for each of the 15 mark questions. and do they give a certain amount of space for each part of a question dependednt on how much space they think it should take up, or a set amount. Or is there 1 big space for the whole of the q

    cheers
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    (Original post by IceKidd)
    Youve sat this exam so could you explain the format of the answer book please.

    I realise you are expected to do all workings for the multiple choice in the space underneath each part. So how are you expected to answer the 15 mark questions.

    Is there a question paper bookl and and answer book. are you given say x pages for all of the longer questions. or is it y pages for each of the 15 mark questions. and do they give a certain amount of space for each part of a question dependednt on how much space they think it should take up, or a set amount. Or is there 1 big space for the whole of the q

    cheers
    You're just given a massive answer booklet with your answers to the multiple choice on the front. I was worried about not having enough room to write as I usually use a shed load of paper, but they provide more than enough.
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    (Original post by TheFuture001)
    The solutions claim -1 <= x^2 - 1 <= 0 . How they came to this conclusion, and what follows, I can not understand. I would appreciate at least a hint in the correct understanding, thanks!
    What are the limits of the integral -- and so the bounds for x?
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    Hi, does anyone have any tips on how to approach questions, because sometimes I just look at questions and don't even know where to start! Any help would be much appreciated! Thanks XD!
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    (Original post by MelvyW)
    From what i understand A(theta) just means the area A. The area is dependent on the value of theta, and hence A(theta)
    Thanks for the reply but I find your explanation very unclear.

    You said that A and B would be dependent on theta, but I don't understand how it is used in this context:
    A(theta) = B(0.5pi - theta)

    I find that this notation is very unclear and most likely there was quite a few people who sat this MAT paper losing marks due to this.
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    (Original post by Sayonara)
    I find that this notation is very unclear and most likely there was quite a few people who sat this MAT paper losing marks due to this.
    There is nothing more unclear about writing A(theta) than there is f(x) or y(t).
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    (Original post by Sayonara)
    Thanks for the reply but I find your explanation very unclear.

    You said that A and B would be dependent on theta, but I don't understand how it is used in this context:
    A(theta) = B(0.5pi - theta)

    I find that this notation is very unclear and most likely there was quite a few people who sat this MAT paper losing marks due to this.
    Think of it as just a function, as much as it may seem like you can't find an equation to give the area for those sections, you actually can. I fell into the same trap.
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    (Original post by RichE)
    What are the limits of the integral -- and so the bounds for x?
    Yes, but when you square the bounds and take off 1 you don't end with the same limits.

    I think I understand now, the graph shows for x=-1, f(x)=0, so if you square f(x), 0, and take off 1 the limit is -1, yes? Using the same logic for x=0 and x=1, the upper limit becomes 0. Is this the correct way to think about this?
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    (Original post by TheFuture001)

    I think I understand now, the graph shows for x=-1, f(x)=0, so if you square f(x), 0, and take off 1 the limit is -1, yes? Using the same logic for x=0 and x=1, the upper limit becomes 0. Is this the correct way to think about this?
    At no point should you be squaring f(x). All you need to understand is what the function f(x^2-1) equals.

    x is a number between -1 and 1.

    So x^2 is a number between 0 and 1

    and x^2 - 1 is a number between -1 and 0.

    We now need to work out what f(x^2-1) is. The function f is made up of different "rules" depending on the input. From the graph we can see that f is the rule "add 1" for inputs between -1 and 0.

    So f(x^2-1) = (x^2-1) + 1 = x^2.

    So it is x^2 you need to integrate between -1 and 1.
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    (Original post by fizzy pops)
    Hi, does anyone have any tips on how to approach questions, because sometimes I just look at questions and don't even know where to start! Any help would be much appreciated! Thanks XD!
    I've always found it helpful to write down everything you know - all the given information and what you can infer from it. Also, with MAT questions, reading the full question (especially trying to use previous results or given hints) is a good idea.

    Haha might not have been that helpful, just my two cents. Could you give an example of one of those questions?
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    (Original post by RichE)
    .
    (Original post by Noble.)
    .
    Hi, at most cambridge colleges they have a small 1 hour paper on the day of your interview, and your solutions are discussed during your interview and new problems are approached.

    Since Oxford has the MAT already is this the same. Is there also a small test for any of the colleges so that some of your interview involves discussing your answers to some of these problems. Or is there no test/problem sheet for any college?
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    (Original post by RichE)
    At no point should you be squaring f(x). All you need to understand is what the function f(x^2-1) equals.

    x is a number between -1 and 1.

    So x^2 is a number between 0 and 1

    and x^2 - 1 is a number between -1 and 0.

    We now need to work out what f(x^2-1) is. The function f is made up of different "rules" depending on the input. From the graph we can see that f is the rule "add 1" for inputs between -1 and 0.

    So f(x^2-1) = (x^2-1) + 1 = x^2.

    So it is x^2 you need to integrate between -1 and 1.
    Cheers, I see the required thought process now!
 
 
 
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