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Year 13 Maths Help Thread

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Original post by Zacken
Huh? Says who?

Take f(x)=sin⁡xf(x) = \sin x then F(x)=−cos⁡xF(x) = -\cos x so ∫π2πf(x)=F(2π)−F(π)=−1−1=−2<0\int_{\pi}^{2\pi} f(x) = F(2\pi) - F(\pi) = -1 - 1 = -2 < 0. So how is F(b)−F(a)F(b) - F(a) positive?


Well I should have said "if" F(b)>F(a), my bad and then I should have said the vice versa for when F(a)>F(b)


Thanks I didn't consider that
(edited 7 years ago)
Image.jpeg stuck on first part.
Original post by Coolsul98
stuck on first part.


A bit awkward to explain.

We know the k part of the vector because the info tells us the top of the roof is 5m above the base.

You can work out the i part of the vector by considering the lengths of 14m and 6m.

You can work out j part by considering the triangle OCD and the length along the j direction.
Reply 583
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A BMO 2 question some years ago asked:

Find positive integer solutions to a+b=2009\sqrt{a}+\sqrt{b}=\sqrt{2009}.

I happen to be good at detecting stuff to do with square numbers (writing down lists of square numbers was a hobby of mine when I was younger) so I took a shortcut by writing 2009=2025−16=452−42=(45+4)(45−4)=(49)(41)=741\sqrt{2009}=\sqrt{2025-16}=\sqrt{45^2-4^2}=\sqrt{(45+4)(45-4)}=\sqrt{(49)(41)}=7\sqrt{41}.

As 41 is prime we know that 41\sqrt{41} can't be simplified further so we can deduce that both a\sqrt{a} and b\sqrt{b} is in the form m41m\sqrt{41}, n41n\sqrt{41} respectively, m,n∈Nm,n \in \mathbb{N}. Hence m+n=7m+n=7 which simplifies the problem a lot.

But I don't know if this is 'cheating' (I,e. missing what the examiner was testing) or not.
(edited 7 years ago)
Original post by Palette
A BMO 2 question some years ago asked:

Find positive integer solutions to a+b=2009\sqrt{a}+\sqrt{b}=\sqrt{2009}.

I happen to be good at detecting stuff to do with square numbers (writing down lists of square numbers was a hobby of mine when I was younger) so I took a shortcut by writing 2009=2025−16=452−42=(45+4)(45−4)=(49)(41)=741\sqrt{2009}=\sqrt{2025-16}=\sqrt{45^2-4^2}=\sqrt{(45+4)(45-4)}=\sqrt{(49)(41)}=7\sqrt{41}.

As 41 is prime we know that 41\sqrt{41} can't be simplified further so we can deduce that both a\sqrt{a} and b\sqrt{b} is in the form m41m\sqrt{41}, n41n\sqrt{41} respectively, m,n∈Nm,n \in \mathbb{N}. Hence m+n=7m+n=7 which simplifies the problem a lot.

But I don't know if this is 'cheating' (I,e. missing what the examiner was testing) or not.


I wouldn't say so. I think you're expected to know your squares and make use of them wherever appropriate in a BMO so that's a valid method.
Original post by Palette

But I don't know if this is 'cheating' (I,e. missing what the examiner was testing) or not.


It isn't. The only other thing you could do is the proof that a and b have to be of the form root(41)*k, but I suspect that isn't even necessary.
Original post by ValerieKR
It isn't. The only other thing you could do is the proof that a and b have to be of the form root(41)*k, but I suspect that isn't even necessary.


what's the alternative way to do that question?
Original post by metrize
what's the alternative way to do that question?


What I said was in he's hand-wavily said that a and b have to have a factor of 41^(2n+1), which is a true thing to say but some of the marks may require that you prove that (although I suspect they won't).

(As in the proof that ksqrt(a) + msqrt(b) can't equal lsqrt(c) with none of a, b, c being equal.

I can't think of an alternative way to do it off the top of my head. :s-smilie:
(edited 7 years ago)
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Original post by ValerieKR
What part of it are you weak at? Making sense of breaking down the recurrence system? If so 15, 17 and 35 of advanced problems in mathematics are good

http://www.mathshelper.co.uk/110501_Advanced_Problems_in_Mathematics.pdf


A shame that he took down his website as it was a useful bank of STEP and AEA papers.
Original post by Palette
A shame that he took down his website as it was a useful bank of STEP and AEA papers.

Hi I see you posting quite a bit of these types of questions, are you going to study Maths at uni?
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Original post by metrize
Hi I see you posting quite a bit of these types of questions, are you going to study Maths at uni?


I am undergoing mathematical training via solving various questions from different papers so that I can make the transition to maths at university as smooth as possible. Some of the questions are merely a product of my own curiosity.
Original post by Palette
I am undergoing mathematical training via solving various questions from different papers so that I can make the transition to maths at university as smooth as possible. Some of the questions are merely a product of my own curiosity.

Ah cool, what unis you thinking of applying to?
Original post by Palette
Liverpool John Moores, Glyndwr, Southampton Solent, London South Bank and Swansea Metropolitan.

Ok, it's actually Oxford (probably won't get in), Imperial (uncertain), UCL (hopefully yes), Bristol (hopefully yes) and Manchester (ditto).

P.S. I think Glyndwr has a circumflex accent on top of the w.

I see, good luck!
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Original post by metrize
I see, good luck!


Where are you applying to?
Original post by Palette
Where are you applying to?


Cambridge Bristol Bath Imperial Loughborough for engineering
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Original post by metrize
Cambridge Bristol Bath Imperial Loughborough for engineering


Engineering sounds very cut-throat- I believe it's the most competitive course at Cambridge.

Is it mechanical engineering you're interested in?
Original post by Palette
Engineering sounds very cut-throat- I believe it's the most competitive course at Cambridge.

Is it mechanical engineering you're interested in?

Yeah it probably is, I think nat sci is the only one with more applicants.

And yeah mechanical
Why is the derivative of e^3x -> 3e^3x?
Original post by jamestg
Why is the derivative of e^3x -> 3e^3x?


You know that derivative of e^x = e^x
use chain rule with x goes to 3x
d(3x)/dx = 3

d(e^3x)/d(x)
= d(3x)/dx * d(e^3x)/d(3x)
=3*e^3x
=3e^3x
Original post by ValerieKR
You know that derivative of e^x = e^x
use chain rule with x goes to 3x
d(3x)/dx = 3

d(e^3x)/d(x)
= d(3x)/dx * d(e^3x)/d(3x)
=3*e^3x
=3e^3x


Thanks! The bit I put in bold, is that just a principle we have to know then?

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