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STEP Maths I,II,III 1987 Solutions

(edited 12 years ago)

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Reply 1
There isn't one. Some 1987 solutions were posted in the 1991 thread.
Square
Is there a thread for 1987 STEP solutions? I cant find it if there is.

If not we should make one :biggrin:

Go for it... :wink: That is, if you're ready to handle updating the first post...
Reply 3
Well I'd need some contributions from other regulars, especially with II/III.

Got some of the STEP I questions written up though.
Reply 4
I can help keep it updated if you want. Go for it...
Reply 5
I'll just make it this thread then, post your solutions gogo.

Gonna copy the format from the other threads and gonna put it into my opening post, will change the thread name also.
Reply 6
Anyone have a link to the papers?

Edit: Got them.
Reply 7
Go for it, I'll do some questions this night - gonna go out to the cinema and then go to the pub for dinner, but will be back before midnight:wink:

edit: I've PMd the link
Reply 8
Bugger, think there might actually be a mistake in my solution, taken it down for now.
Reply 9
STEP II, Q7.

The trapezium rule approximates the area y=f(x) under the curve by trapezia, all having a base of width h. The side lengths are f(n), f(n+h) for n from the lower limit to h less than the upper limit. The area of each trapezium is therefore h/2(f(n) + f(n+1)), which, when added together, results in the expression given.

1nlnx  dx12(ln1+2ln2++2ln(n1)+lnn)=ln(n!)12lnn\displaystyle \int^{n}_{1} \ln x \; \mathrm{d}x \approx \frac{1}{2}\left(\ln 1 + 2\ln 2 + \cdots + 2\ln(n-1) + \ln n\right) = \ln(n!) - \frac{1}{2}\ln n

1nlnx  dx=[xlnxx]1n=nlnn+1n\displaystyle \int^{n}_{1} \ln x \; \mathrm{d}x = \left[x\ln x - x\right]^{n}_{1} = n\ln n + 1 - n

ln(n!)(n+12)lnn+1n\Rightarrow \ln(n!) \approx (n + \frac{1}{2})\ln n + 1 - n

n!nn+12e1n=g(n)\Rightarrow n! \approx n^{n + \frac{1}{2}} e^{1-n} = g(n)

f(x)=lnx,  f(x)=1x2f(x) = \ln x, \; f''(x) = -\frac{1}{x^{2}}

Therefore ln x is a convex function, so the area under the curve is underestimated by trapezia.

n!<nn+12e1n\Rightarrow n! < n^{n + \frac{1}{2}} e^{1-n}

1nlnx  dx12k(ln1+2lnk+1k+2lnk+2k++2lnn)12klnn\displaystyle \int^{n}_{1} \ln x \; \mathrm{d}x \approx \frac{1}{2k}\left(\ln 1 + 2\ln \frac{k+1}{k} + 2\ln\frac{k+2}{k} + \cdots + 2\ln n\right) - \frac{1}{2k}\ln n

=1k(ln(k+1)+ln(k+2)++lnknk)k(n1)klnk12klnnnlnn+1n= \frac{1}{k}(\ln(k+1) + \ln(k+2) + \cdots + \ln\frac{kn}{k}) - \frac{k(n-1)}{k} \ln k - \frac{1}{2k}\ln n \approx n\ln n + 1 - n

ln(kn)!k!k(n1)lnk+(kn+12)lnn+k(1n)\Rightarrow \ln\frac{(kn)!}{k!} \approx k(n-1)\ln k + (kn+\frac{1}{2})\ln n + k(1-n)

(kn)!k!×kk(n1)×nkn+12×ek(1n)\Rightarrow (kn)! \approx k! \times k^{k(n-1)} \times n^{kn+\frac{1}{2}} \times e^{k(1-n)}

(kn)!k!  nkn+12  (ek)k(1n)\Rightarrow (kn)! \approx k! \; n^{kn+\frac{1}{2}} \; \left(\frac{e}{k}\right)^{k(1-n)}

As required. This is closer to (kn)!(kn)! than g(kn) because the intervals are closer together for the second approximation and so it is more accurate.
Reply 10
Differentiating
f(x)=eaxcosbx[br][br]f(x)=aeaxcosbxbeaxsinbx[br][br]=eax(acosbxbsinbx)\displaystyle f(x)=e^{\text{a}x}\text{cosb}x[br][br]f'(x)=\text{ae}^{\text{a}x}\text{cosb}x-\text{be}^{ax}\text{sinb}x[br][br]= \text{e}^{\text{a}x}(\text{acosb}x-\text{bsinb}x)

Finding x-ordinates of turning points:

f(x)=0\displaystyle f'(x)=0

eax=0[br][br]no solutions\displaystyle \text{e}^{\text{a}x}=0 [br][br]\text{no solutions}

acosbxbsinbx=0[br][br]a=btanbx[br][br]bx=nπ+arctan(ab)\displaystyle \text{acosb}x-\text{bsinb}x=0[br][br]\text{a}=\text{b}\tan\text{b}x[br][br]\text{b}x=\text{n}\pi+\arctan(\frac{a}{b})

Consider any turning point: x=1b(nπ+arctan(ab))x=\frac{1}{b}(\text{n}\pi+\arctan(\frac{a}{b}))

f(1b(nπ+arctan(ab))=ea(nπb+1barctan(ab))×cos(nπ+arctan(ab)\displaystyle f(\frac{1}{b}(\text{n}\pi+\arctan(\frac{a}{b}))=e^{a(\frac{n\pi}{b}+\frac{1}{b}\arctan(\frac{a}{b}))}\times\cos(n\pi+\arctan(\frac{a}{b})

Expand out using addition formula sin terms go to zero:

eanπb×e1barctan(ab)×cosnπarctanab\displaystyle e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b})}\times\cos n\pi \arctan\frac{a}{b}

Now if n is even the expression is going to be:

eanπb×e1barctan(ab)×arctanab\displaystyle e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b})}\times \arctan\frac{a}{b}

And if n is odd the expression will be:
eanπb×e1barctan(ab)×arctanab\displaystyle -e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b})}\times \arctan\frac{a}{b}

Now consider the next turning point directly after the one shown above ie. (n+1)pi

x=1b((n+1)π+arctan(ab))x=\frac{1}{b}(\text{(n+1)}\pi+\arctan(\frac{a}{b}))

This expression will be of the form:

ea(n+1)πb×e1barctan(ab)×cos((n+1)π)×arctanabe^{\frac{a(n+1)\pi}{b}} \times e^{\frac{1}{b}\arctan(\frac{a}{b})}\times \cos((n+1)\pi)\times \arctan\frac{a}{b}

Which is going to equal:

eaπb×eanπb×e1barctan(ab)×cos((n+1)π)×arctanab\displaystyle e^{\frac{a\pi}{b}} \times e^{\frac{an\pi}{b}}\times e^{\frac{1}{b}\arctan(\frac{a}{b})}\times \cos((n+1)\pi) \times \arctan\frac{a}{b}

So, depending on from the last part if n is even or odd, it is now the opposite here, which means that this next turning point will have the opposite sign from the previous one.

So this is going to equal:

eaπbf(1b(nπ+arctan(ab))\displaystyle -e^{\frac{a\pi}{b}}f(\frac{1}{b}(\text{n}\pi+\arctan(\frac{a}{b}))

Since this is true for n and n+1 then all the turning points must be part of a geometric progression with common ratio: eaπb-e^{\frac{a\pi}{b}} as required.

If anyone can think of a better way of doing the explaining bit at the end they are more than welcome. I've missed out quite abit of the individual working stages but have labelled what I have done so hopefully you can follow it :smile:.

Maybe there is a tidier way to do it with GP formulae I dont know.
Reply 11
STEP I, question 4.

Let S=log2elog4e+log16e...S = \mathrm{log}_2 e - \mathrm{log}_4 e + \mathrm{log}_{16} e...

Take each term to logarithm base 2 (change the base rule)

S=log2e1log2e2+log2e4...\displaystyle S = \frac{\mathrm{log}_2 e}{1} - \frac{\mathrm{log}_2 e}{2} + \frac{\mathrm{log}_2 e}{4}...

Unparseable latex formula:

S = (\mathrm{log}_2 e})(1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32...)



Split the second brackets into two parts.

P = 1 + 1/4 + 1/16...
N = -1/2 - 1/8 - 1/32...

So

Unparseable latex formula:

S = (\mathrm{log}_2 e})(P + N)



P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

P=1/(11/4)=4/3P = 1/(1-1/4) = 4/3

N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

N=(1/2)/(11/4)=2/3N = (-1/2)/(1-1/4) = -2/3

Unparseable latex formula:

S = (\mathrm{log}_2 e})(P + N) = (\mathrm{log}_2 e})(4/3 - 2/3)



Unparseable latex formula:

S = 2/3(\mathrm{log}_2 e})



Change the base to e, so we have the natural logarithm.

Unparseable latex formula:

\displaystyle S = 2/3(\mathrm{log}_2 e}) = 2/3(\frac{\mathrm{log}_e e}{\mathrm{log}_e 2}) = 2/3(\frac{1}{\mathrm{log}_e 2})


S=23ln2\displaystyle S = \frac{2}{3\mathrm{ln} 2}

Using the power rule

S=23ln2=2ln8\displaystyle S = \frac{2}{3\mathrm{ln} 2} = \frac{2}{\mathrm{ln} 8}

S=21212ln8\displaystyle S = \frac{2*\frac{1}{2}}{\frac{1}{2}\mathrm{ln} 8} (multiplying top and bottom by 1/2)

S=112ln8=1ln8\displaystyle S = \frac{1}{\frac{1}{2}\mathrm{ln} 8} = \frac{1}{\mathrm{ln} \sqrt 8}

S=1ln22\displaystyle S = \frac{1}{\mathrm{ln} 2\sqrt 2}

WWWWW.
Square
If anyone can think of a better way of doing the explaining bit at the end they are more than welcome.Write c=arctan(ab),xn=1b(nπ+c)c = \arctan(\frac{a}{b}), x_n = \frac{1}{b}(n\pi + c) (so xnx_n is the nth turning point).

Then f(xn)=ea(nπ+c)/bcos(nπ+c)f(x_n) = e^{a(n\pi+c)/b}\cos(n\pi + c), so

f(xn+1f(xn=ea((n+1)π+c)/bcos((n+1)π+c)ea(nπ+c)/bcos(nπ+c)=eaπ/bcos((n+1)π+c)cos(nπ+c)\displaystyle \frac{f(x_{n+1}}{f(x_n} = \frac{ e^{a((n+1)\pi+c)/b}\cos((n+1)\pi + c)}{ e^{a(n\pi+c)/b}\cos(n\pi + c)} = e^{a\pi/b} \frac{\cos((n+1)\pi + c)}{\cos(n\pi+c)}

But cos(y+π)=cos(y)cos(π)sin(π)sin(y)=cos(y)\cos(y+\pi) = \cos(y)\cos(\pi)-\sin(\pi)\sin(y) = -\cos(y), so cos((n+1)π+c)cos(nπ+c)=1\frac{\cos((n+1)\pi + c)}{\cos(n\pi+c)} = -1.

So f(xn+1)f(xn)=eaπ/b\frac{f(x_{n+1})}{f(x_n)} = -e^{a\pi/b} and so the values of f at the turning points form a GP with ratio eaπ/b -e^{a\pi/b} as required.
Reply 13
STEP I Q3

Our differential equation, referred to from here on as (*) is:

x3dvdx+x2v=1+x2v2(1+x2)v\displaystyle x^3\frac{dv}{dx}+x^2v=\frac{1+x^2v^2}{(1+x^2)v}

Let y=xv    v=yx\displaystyle y=xv \implies v=\frac{y}{x}

    dvdx=xdydxyx2\displaystyle \implies \frac{dv}{dx}=\frac{x\frac{dy}{dx}-y}{x^2}

Hence in (*):

x2dydx=x(1+y2)y(1+x2)\displaystyle x^2\frac{dy}{dx}=\frac{x(1+y^2)}{y(1+x^2)}

After cancelling an xx and rearranging, (*) reduces to:

y1+y2dydx=1x(1+x2)\displaystyle \frac{y}{1+y^2}\frac{dy}{dx}=\frac{1}{x(1+x^2)}

Hence, after dealing with partial fractions (which should be routine)

y1+y2dy=1xx1+x2dx\displaystyle \int{\frac{y}{1+y^2}}\,dy=\int{\frac{1}{x}-\frac{x}{1+x^2}}\,dx

Thus 12ln(1+y2)=lnx12ln(1+x2)+c\displaystyle \frac12 \ln{(1+y^2)}=\ln{|x|}-\frac12 \ln{(1+x^2)}+c

Now x=1,v=1    y=1    c=ln2x=1,v=1 \implies y=1 \implies c=\ln{2}

ln(1+y2)=ln(4x21+x2)\displaystyle \therefore \ln{(1+y^2)}=\ln{\left(\frac{4x^2}{1+x^2}\right)}

    1+x2v2=4x21+x2\displaystyle \implies 1+x^2v^2=\frac{4x^2}{1+x^2}

    x2v2=3x211+x2\displaystyle \implies x^2v^2=\frac{3x^2-1}{1+x^2}

    v2=31+x21x2(1+x2)\displaystyle \implies v^2=\frac{3}{1+x^2}-\frac{1}{x^2(1+x^2)}

Hence as x,v0x \to \infty, v \to 0

This completes the question, although I would like someone to have a quick scan for errors as I did this quite roughly.
Reply 14
III/7

arctant=0t11+x2dx\displaystyle \arctan t = \int_0^t \frac 1{1 + x^2}dx

=0t1+(1)nx2n+2+(1)n+1x2n+21+x2dx\displaystyle = \int_0^t \frac{1 + (-1)^nx^{2n+2} + (-1)^{n+1}x^{2n+2}}{1+x^2}dx

=0t1+(1)nx2n+21+x2dx+(1)n+10tx2n+21+x2dx\displaystyle = \int_0^t \frac{1 + (-1)^nx^{2n+2}}{1+x^2} dx + (-1)^{n+1}\int _0^t\frac{x^{2n+2}}{1+x^2}dx

(using formula for sum of geometric progression with common ratio -x^2)

=0t(1x2+x4+(1)nx2n)dx+(1)n+10tx2n+21+x2dx\displaystyle = \int_0^t \left(1 - x^2 + x^4 - \cdots + (-1)^nx^{2n}\right) dx + (-1)^{n+1}\int_0^t \frac{x^{2n+2}}{1+x^2}dx

=tt33+t55+(1)nt2n+12n+1+(1)n+10tx2n+21+x2dx\displaystyle = t - \frac{t^3}{3} + \frac{t^5}{5} - \cdots + \frac{(-1)^n t^{2n+1}}{2n+1} + (-1)^{n+1}\int_0^t \frac{x^{2n+2}}{1+x^2}dx

as required.

Rearranging this, we get

arctantr=0n(1)rt2r+12r+1=0tx2n+21+x2dx\displaystyle \left| \arctan t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1}\right| = \int_0^t \frac{x^{2n+2}}{1+x^2}dx

But for 0t10 \leq t \leq 1

0tx2n+21+x2dx0tx2n+22dx=t2n+32(2n+3)\displaystyle \int_0^t \frac{x^{2n+2}}{1+x^2}dx \geq \int_0^t \frac{x^{2n+2}}{2}dx = \frac{t^{2n+3}}{2(2n+3)}

and

0tx2n+21+x2dx0tx2n+21dx=t2n+32n+3\displaystyle \int_0^t \frac{x^{2n+2}}{1+x^2}dx \leq \int_0^t \frac{x^{2n+2}}{1}dx = \frac{t^{2n+3}}{2n+3}.

The last three statements together prove

t2n+32(2n+3)arctantr=0n(1)rt2r+12r+1t2n+32n+3\displaystyle \frac{t^{2n+3}}{2(2n+3)} \leq \left| \arctan t - \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1}\right| \leq \frac{t^{2n+3}}{2n+3} (*)

as required.

Let us set t=1t=1 in (*). Using the right half, we get

π4r=0n(1)r2n+112n+3\displaystyle \left|\frac{\pi}{4} - \sum_{r=0}^n \frac{(-1)^r}{2n+1} \right| \leq \frac{1}{2n+3}

Since the RHS of this inequality clearly approches zero as n grows towards positive infinity, we have by squeezing that

r=0n(1)r2n+1\displaystyle \sum_{r=0}^n \frac{(-1)^r}{2n+1}

approaches π/4\pi / 4 as nn \rightarrow \infty, and the next required result immediately follows.

Let the error in the approximation of π\pi be EE. The error is given by

E=π4r=0n(1)r2r+1\displaystyle E = \left|\pi - 4\sum_{r=0}^n \frac{(-1)^r}{2r+1}\right|,

which is exactly four times the middle term in (*) for t=1t=1. Thus from the left part of (*) it follows

E42(2n+3)\displaystyle E \geq \frac{4}{2(2n+3)}

For n98n \leq 98 we have

42(2n+3)4398>102\displaystyle \frac{4}{2(2n+3)} \geq \frac{4}{398} > 10^{-2}

Thus we have

E>102\displaystyle E > 10^-2

as required.
Reply 15
On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?
DeathAwaitsU
On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?

Depends. Are you talking about 1987? No. :wink: If you're talking about 2004-present, then maybe. But you'll improve.
Reply 17
generalebriety
Depends. Are you talking about 1987? No. :wink: If you're talking about 2004-present, then maybe. But you'll improve.


what is good if you have only done c1 and c2?
eponymous
what is good if you have only done c1 and c2?

If you've only done C1 and C2, I'd be surprised if you could do more than the odd STEP I question at all. The 1992 STEP II paper had quite a few nice questions, I seem to remember.
Reply 19
generalebriety
The 1992 STEP II paper had quite a few nice questions, I seem to remember.


thanks, i'll go look at them now :smile:

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