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STEP Maths I,II,III 1987 Solutions

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Just started looking at some of these and not really up to standard. Can anyone just explain on STEP I question 2 the first part, why the angle is pi/n ?
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thestudent
Just started looking at some of these and not really up to standard. Can anyone just explain on STEP I question 2 the first part, why the angle is pi/n ?


As DFranklin said, you take a right angled triangle, hypotenuse R, small side r consider the angle of this triangle theta such that sin(theta)=r/R.

Now if you think of it like this, there are 2 'thetas' for every smaller circle and there are n smaller circles, so n=2pi/2theta so theta=pi/n.

It's kinda hard to explain without a diagram.
thanks for the help and thanks for the thread, a great help
Here is my attempt at Paper 1 number 6. Does anyone agree with it?
Dystopia
STEP II, Q5.

i) f(x)=αxf(x) = \alpha x

[yf(y)]n=yn(1α)n[y - f(y)]^{n} = y^{n}(1-\alpha)^{n}

dn1dyn1[yf(y)]n=n!y(1α)n\Rightarrow \frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}} [y-f(y)]^{n} = n! y (1-\alpha)^{n}

n=01n!n!y(1α)n\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}n!y(1 - \alpha)^{n}

This is an infinite geometric series with a=y(1α),  r=(1α)a=y(1-\alpha), \; r=(1-\alpha). Since 0<α<20 < \alpha < 2, it converges to yαy\frac{y}{\alpha} - y.

f1(y)=y+yαy=yα\Rightarrow f^{-1}(y) = y + \frac{y}{\alpha} - y = \frac{y}{\alpha}

As expected.

ii) Let y=f(x)=xx34y = f(x) = x - \frac{x^{3}}{4}

[yf(y)]n=y3n22n[y - f(y)]^{n} = \frac{y^{3n}}{2^{2n}}

dn1dyn1y3n=(3n)!(2n+1)!y2n+1\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}} y^{3n} = \frac{(3n)!}{(2n+1)!}y^{2n+1}

f1(y)=y+n=1(3n)!y2n+1n!(2n+1)!22n\displaystyle f^{-1}(y) = y + \sum_{n=1}^{\infty} \frac{(3n)!y^{2n+1}}{n!(2n+1)!2^{2n}}

But y=12y=\frac{1}{2}, so

x=12+n=1(3n)!n!(2n+1)!24n+1\displaystyle x = \frac{1}{2} + \sum_{n=1}^{\infty}\frac{(3n)!}{n!(2n+1)!2^{4n+1}}

Note that (3n)!n!(2n+1)!24n+1=12\frac{(3n)!}{n!(2n+1)!2^{4n+1}}=\frac{1}{2} when n=0.

x=n=0(3n)!n!(2n+1)!24n+1\displaystyle x = \sum_{n=0}^{\infty}\frac{(3n)!}{n!(2n+1)!2^{4n+1}}

iii) Let y=f(x)=xeλxy = f(x) = x - e^{\lambda x}

[yf(y)]n=eλny[y - f(y)]^{n} = e^{\lambda n y}

dn1dyn1[yf(y)]n=(λn)n1eλny\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}} [y-f(y)]^{n} = (\lambda n)^{n-1} e^{\lambda n y}

f1(y)=y+n=1(λn)n1n!eλny\displaystyle f^{-1}(y) = y + \sum_{n=1}^{\infty} \frac{(\lambda n)^{n-1}}{n!} e^{\lambda n y}

But y = 0, so

x=n=1(λn)n1n!\displaystyle x = \sum_{n=1}^{\infty} \frac{(\lambda n)^{n-1}}{n!}

Edit: Fixed. Didn't realise it was actually posting last night... :redface:

Been wondering what was wrong with this. Finally realised that in first part you have lower limit of sum as n=0 when it should be n=1
Here are my attempts at numbers 9-12 on 1987 STEP Paper I
I would be grateful for any confirmation or reports of errors.

Some errors have been discopvered. For revised solutions
see post no.237 for question 10
Post 238 for question 11 and post 239 for question 13
(edited 12 years ago)
STEP I numbers 13-16
Again if anyone could check them I would be most grateful.
I think you need to update the links to the papers on the front page, as when I click on them I get sent to some random advertising webiste and am bombarded by pop-ups for 5 minutes.
brianeverit
Been wondering what was wrong with this. Finally realised that in first part you have lower limit of sum as n=0 when it should be n=1

Thank you for pointing that out; is it correct now?

Are you also in Y13 btw?
The links for the papers don't work.
Still not working...
DaveSimpson
Still not working...

That site no longer "works", it hasn't for several months. Shame, it was really good.:frown:
Link

That should work.
lol it's no good, just gives me Page cannot be displayed
Those are the answers not the questions.
DaveSimpson
Those are the answers not the questions.

The purpose of this thead is to provide answers to the 1987 paper questions :s-smilie: .
Yeah but what's the point of answers if no one has the questions?

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