Oxford MAT 2013/2014

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

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 $\epsilon \leq \dfrac{1}{64}$

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/files/private/active/0/test07.pdf

Is A and B a constant, function or what?

Ahh okay i see. Thanks!

I very much doubt i could something like that out in exam though -.-

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

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 $A^2 > 2P + 1$ 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

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!

From what i understand A(theta) just means the area A. The area is dependent on the value of theta, and hence A(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.

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.

What are the limits of the integral -- and so the bounds for x?

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?

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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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.