SimonM
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(Updated as far as post #191.) SimonM - 27.05.2009

Solutions for earlier STEP papers are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved (red above); if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

N.B. the "usual" past paper website may have the papers mislabelled; please check the front of the paper, not the filename, to see which paper you're doing. (Mathematics = I, further mathematics A = II, further mathematics B = III.)

STEP I (Mathematics):
1: Solution by justinsh and Dystopia
2: Solution in Advanced Problems in Mathematics (Siklos booklet I)
3: Solution by DFranklin
4: Solution by justinsh
5: Solution by generalebriety
6: Solution by nota bene
7: Solution by brianeverit
8: Solution by justinsh
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15: Solution by brianeverit
16: Solution by brianeverit


STEP II (Further Maths Paper A):
1: Solution by Decota
2: Solution by Rabite and DFranklin
3: Solution by justinsh (alternative by khaixiang)
4: Solution by Rabite
5: Solution by Rabite
6: Solution by Glutamic Acid
7: Solution by Rabite
8: Solution by Rabite
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by Glutamic Acid
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15: Solution by toasted-lion
16: Solution by brianeverit


STEP III (Further Mathematics Paper B):
1: Solution by generalebriety (alternative to (b) by justinsh)
2: Solution by generalebriety
3: Solution by Rabite
4: Solution by mikelbird
5: Solution by Speleo
6: Solution by Rabite
7: Solution by Rabite
8: Solution by squeezebox
9: Solution by mikelbird
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15:
16: Solution by toasted-lion


Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
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khaixiang
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I/2, II/5, II/6 can be found in Siklos' Advanced Problems in Mathematics, I will gladly scan them in here since this booklet is not available for free, but it's 7am here, I need to get some sleep so I will do it tomorrow
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nota bene
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I'm attempting I/3 I'll see if I sort it out...


Okay, I'll post what I have now (probably should scan a sketch tomorrow as well), but it is not finished, and likely contains a few mistakes
Spoiler:
Show

Some observations I made was that if we let the first complex number have a length a the second will have a+ad etc. as long as we are in the first quadrant. This gives z_{n+1}=z_n+ad^{n-1} This forms a G.P. z_{n+1}=\frac{a(1-d^{n-1})}{1-d}=\frac{a(d^{n-1}-1)}{d-1}.
Considering the other quadrants it seems to hold under vector addition, however I'm a bit worried when I have a vertical vector for ad^{n-1}
By looking at this G.P. and how the numbers behave graphically it is evident that d<1 will mean it is a finite shape ending in a point (as z_{n+1}&lt;z_n) and that d>1 will lead to an infinite shape (as z_{n+1}&gt;z_n. This is because we can see that the value of the angle theta is irrelevant as long as it is not zero or a right angle (vector addition will not work there).

Now to the question, expressing the complex number as \frac{z_{n+1}-z_n}{z_n-z_{n-1}}. This leads us to \frac{z_{n+1}-z_n}{z_n-z_{n-1}}=\frac{ad^{n-1}}{ad^{n-2}}=d. Therefore we can conclude that for all z the expression \frac{z_{n+1}-z_n}{z_n-z_{n-1}} will represent d.

If z_1=0 \text{, and, } z_2=1 we have a=1 so it nicely vanishes.
(I am insecure here if I need an explicit formula valid for all z, or just a recurrence relation, but I assume they want an explicit one)

As for the very last part it is easy to verify by a graph, but probably we need an algebraic formula instead, involving theta. I think this works: z_{n+1}=r_ncis(n\theta), where r_n is determined by the G.P. described earlier i.e. r_n=\frac{(1-d^{n-2}}{1-d}.

(As I am going along I realise I shouldn't be doing this straight into latex! Everything comes in a strange order, this formula would probably be good to mention earlier.)

When d=2 \text{, and, } \theta=\frac{\pi}{3} we have z_{n+1}=\frac{2^{n-2}-1}{2-1}(cis(\theta)) and as n gets large so does r_n and as the trig functions are periodic we will have an infinite structure geometrically.


I'm tired of this one, I'll take it up tomorrow again and try to sort it out.
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generalebriety
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Doing III/1.

Edit: seems Speleo's doing this in another thread. The cheek! :p:

III/2:

\displaystyle

\text{Similarity } \Leftrightarrow \\

P_1P_2 : P_1P_3 = Q_1Q_2 : Q_1Q_3 \text{ and } \angle P_2P_1P_3 = \angle \Q_2Q_1Q_3 \\

\Leftrightarrow \biggl\lvert \frac{z_1-z_2}{z_1-z_3} \biggr\rvert = \biggl\lvert \frac{w_1-w_2}{w_1-w_3} \biggr\rvert \text{ and } \arg \bigg(\frac{z_1-z_2}{z_1-z_3}\bigg) = \arg \bigg(\frac{w_1-w_2}{w_1-w_3}\bigg) \\

\Leftrightarrow \frac{z_1-z_2}{z_1-z_3} = \frac{w_1-w_2}{w_1-w_3} \\

\Leftrightarrow (z_1-z_2)(w_1-w_3) = (w_1-w_2)(z_1-z_3) \\

\Leftrightarrow z_1w_2 - z_1w_3 - z_2w_1 + z_2w_3 + z_3w_1 - z_3w_2 = 0 \\

\Leftrightarrow \det \begin{pmatrix} z_1&z_2&z_3 \\ w_1&w_2&w_3 \\ 1&1&1 \end{pmatrix} = 0.

(i)
\displaystyle

\frac{z_1-z_2}{z_1-z_3} = \frac{z^2_1-z^2_2}{z^2_1-z^2_3}\\

\Rightarrow \frac{z^2_1 - z^2_3}{z_1 - z_3} = \frac{z^2_1 - z^2_2}{z_1 - z_2} \\

\Rightarrow z_1 + z_3 = z_1 + z_2 \\

\Rightarrow z_2 = z_3 \text{ which is a contradiction. Hence triangles are not similar}.

(ii)
\displaystyle

\text{Similarity } \Leftrightarrow \frac{z^3_1 - z^3_3}{z_1 - z_3} = \frac{z^3_1 - z^3_2}{z_1 - z_2} \\

\Leftrightarrow z^2_1 + z_1z_3 + z^2_3 = z^2_1 + z_1z_2 + z^2_2 \\

\Leftrightarrow z_1(z_3-z_2) = -(z_2+z_3)(z_3-z_2) \\

\Leftrightarrow z_1+z_2+z_3 = 0 \Leftrightarrow \text{ centroid is origin}.

(iii)
\displaystyle

\text{Equilateral } \Leftrightarrow \triangle P_1P_2P_3 \text{ similar to } P_2P_3P_1 \\

\Leftrightarrow \frac{z_1-z_2}{z_1-z_3} = \frac{z_2-z_3}{z_2-z_1} \\

\Leftrightarrow 2z_1z_2 - z^2_1 - z^2_2 = z^2_3 + z_1z_2 - z_1z_3 - z_2z_3 \\

\text{and result follows}.
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Speleo
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It's yours if you want it, I'm not going to be able to contribute properly for a day or two. I'll finish it if it's still left after that.
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generalebriety
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(Original post by Speleo)
It's yours if you want it, I'm not going to be able to contribute properly for a day or two. I'll finish it if it's still left after that.
Might as well give it a go then.

III/1:

(a)
\displaystyle

\frac{6}{r(r+1)(r+3)} = \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \text{ (partial fractions)} \\

\therefore \sum_1^n \frac{6}{r(r+1)(r+3)} = \sum_1^n \bigg[ \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \bigg] \\

= \bigg[ \frac{2}{1} + \frac{2}{2}+ \frac{2}{3} + \dots + \frac{2}{n} \bigg] - \bigg[ \frac{3}{2} + \frac{3}{3} + \frac{3}{4} + \dots + \frac{3}{n+1} \bigg] + \bigg[ \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \dots + \frac{1}{n+3} \bigg] \\

= \frac{2}{1} + \frac{2}{2} + \frac{2}{3} - \frac{3}{2} - \frac{3}{3} - \frac{3}{n+1} + \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} \\

= \frac{7}{6} - \frac{2}{n+1} + \frac{1}{n+2} + \frac{1}{n+3}.

(b)
\displaystyle \ln (1 + x + x^2 + x^3) = \ln (1 - x^4) - \ln (1 - x) \\

= x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^3}{3} + \frac{x^3}{3} + \dots - x^4 - \frac{x^8}{2} - \frac{x^{12}}{3} - \dots \\

= x + \frac{x^2}{2} + \frac{x^3}{3} - \frac{3x^4}{4} + \dots + \frac{x^{4r-3}}{4r-3} + \frac{x^{4r-2}}{4r-2} + \frac{x^{4r-1}}{4r-1} - \frac{3x^{4r}}{4r} + \dots

(c)
\displaystyle

f(x) = e^{x\ln (x+1)} = e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!} + \dots \\

y = x\ln (x+1) = x\Big[ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots\Big] = x^2 - \frac{x^3}{2} + \frac{x^4}{3} + \dots \\

\therefore e^{x\ln (x+1)} = 1 + \Big[ x^2 - \frac{x^3}{2} + \frac{x^4}{3}\Big] + \frac{[x^2 + \dots]^2}{2!} + \frac{[x^2 + \dots]^3}{3!} + \dots \\

= 1 + x^2 - \frac{1}{2}x^3 + \frac{5}{6}x^4 + \dots
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nota bene
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(Original post by generalebriety)
Would appreciate if someone could check (b) for me, I'm not sure it's right...
I'm working it through now; firstly, where did you get the 1's from? surely they shouldn't be there? (or am I being stupid? anyway those doesn't affect the result)

I think the (4r-1)x^(4r) seems wrong I'm trying to work out what it should be...

edit: having looked at the first four terms where both series get involved I'm inclined to believe it shall plainly be \frac{3x^{4r}}{4r}
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generalebriety
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(Original post by nota bene)
I'm working it through now; firstly, where did you get the 1's from? surely they shouldn't be there? (or am I being stupid? anyway those doesn't affect the result)

I think the (4r-1)x^(4r) seems wrong I'm trying to work out what it should be...

edit: having looked at the first four terms where both series get involved I'm inclined to believe it shall plainly be \frac{3x^{4r}}{4r}
Oh - of course the 1s shouldn't be there, sorry. I'm crap at the Maclaurin series.

I think you're right about the other bit too, but I shall mostly take it on trust... need to do some French literature work, I shouldn't be doing maths.

Cheers.
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nota bene
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(Original post by generalebriety)
I think you're right about the other bit too, but I shall mostly take it on trust... need to do some French literature work, I shouldn't be doing maths.
Hehe I recognise doing maths instead of what we are supposed to do :ninja:.

About the question, I looked at more terms and as the difference between the two series always is a multiple of 4 I'm quite 100% sure I am right about that 3 now.

In fact I think all questions on power series, maclaurin and taylor seem very accessible and fairly easy on these old papers (what the hell I've been able to do them!).
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generalebriety
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(Original post by nota bene)
Hehe I recognise doing maths instead of what we are supposed to do :ninja:.

About the question, I looked at more terms and as the difference between the two series always is a multiple of 4 I'm quite 100% sure I am right about that 3 now.

In fact I think all questions on power series, maclaurin and taylor seem very accessible and fairly easy on these old papers (what the hell I've been able to do them!).
Yeah, but, I haven't. :hmmm: :p:
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(Original post by nota bene)
I'm working it through now; firstly, where did you get the 1's from? surely they shouldn't be there? (or am I being stupid? anyway those doesn't affect the result)

I think the (4r-1)x^(4r) seems wrong I'm trying to work out what it should be...

edit: having looked at the first four terms where both series get involved I'm inclined to believe it shall plainly be \frac{3x^{4r}}{4r}
The 2nd line of your answer to (b) looks crazy; did you just forget to put in all the denominators in the log expansions? (i.e. a LaTeX error rather than a maths one).
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nota bene
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The 2nd line of your answer to (b) looks crazy; did you just forget to put in all the denominators in the log expansions? (i.e. a LaTeX error rather than a maths one).
Did you mean to quote me there?

I assume you didn't, and yes there should be denominators in general's expression I think someone was lazy latexing:p:.
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STEP I question 1

First part:

\sin^2\theta+sin^2\phi+sin^2\psi  +2\sin\theta\sin \phi \sin\psi\\
=\sin\theta[\cos(\phi+\psi)+2\sin\phi\sin\ps  i]+\sin^2\phi+\sin^2\psi\\
=\sin\theta[\cos\phi\cos\psi+\sin\phi\sin\ps  i]+\sin^2\phi+\sin^2\psi\\
=\cos(\phi+\psi)\cos(\phi-\psi)+\sin^2\phi+\sin^2\psi\\
=\cos^2\phi\cos^2\psi-\sin^2\phi\sin^2\psi+\sin^2\phi+  \sin^2\psi\\
=(1-\sin^2\phi)(1-\sin^2\psi)-\sin^2\phi\sin^2\psi+\sin^2\phi+  \sin^2\psi\\
=1

Second part:

Letting \theta=\phi=\pi/5 and \psi=\pi/10, using the first part we have:

2\sin^2\frac{\pi}{5}+\sin^2\frac  {\pi}{10}+2\sin^2 \frac{\pi}{5}\sin\frac{\pi}{10}-1=0\\
\Rightarrow8\sin^2\frac{\pi}{10}  \cos^2\frac{\pi}{10}+\sin^2\frac  {\pi}{10}+8\sin^3\frac{\pi}{10}\  cos^2\frac{\pi}{10}-1=0\\
\Rightarrow8\sin^2\frac{\pi}{10}-8\sin^4\frac{\pi}{10}+\sin^2\fra  c{\pi}{10}+8\sin^3\frac{\pi}{10}-8\sin^5\frac{\pi}{10}-1=0\\
\Rightarrow(8\sin^3\frac{\pi}{10  }+\sin^2\frac{\pi}{10}-1)+\sin^2\frac{\pi}{10}(1-8\sin^2\frac{\pi}{10}-8\sin^3\frac{\pi}{10})=0
\Rightarrow(8\sin^3\frac{\pi}{10  }+8\sin^2\frac{\pi}{10}-1)\cos^2\frac{\pi}{10}=0\\
\Rightarrow8\sin^3\frac{\pi}{10}  +8\sin^2\frac{\pi}{10}-1=0
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generalebriety
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(Original post by DFranklin)
The 2nd line of your answer to (b) looks crazy; did you just forget to put in all the denominators in the log expansions? (i.e. a LaTeX error rather than a maths one).
I think it was more a fatigue error than either - I certainly seem to have got the rest of it wrong from there on.
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generalebriety
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Did you mean to quote me there?

I assume you didn't, and yes there should be denominators in general's expression I think someone was lazy latexing:p:.
No, I just can't do the Maclaurin series. At all. I always make stupid mistakes and I know none of them off the top of my head.
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Typing up I/6

i)
g(x)=\frac{2x^2+3}{x^4+4} gives g'(x)=\frac{-4x^5-12x^3+16x}{(x^4+4)^2}
Since the denominator is non-negative, for the derivative to =0 the numerator must be zero. Factorizing renders 4x(4-3x^2-x^4) x=0 is a trivial solution, putting u=x^2 we can solve the quadratic that is left, giving u_1=-4 \text{, and, } u_2=1 we can reject -4 as that gives imaginary solutions for x and we are left with x^2=1 i.e. x=\pm1. This agrees with the reasoning in the question.

"Hence the stationary values are given by x=0, g(x)=3/4 and x=\pm1, g(x)=1" True.

"Since 3/4<1, there is a maxima at x=\pm1 and a minimum at x=0"
The statement happens to be true, as we shall notice, but the argument is invalid as we cannot know if any of the points is a point of inflection. Therefore we test the points by looking at the value of the second derivative.
g''(x)=\displaystyle\frac{(x^4+4  )^2(-20x^4-36x^2+16)-(-4x^5-12x^3+16x)(4x^3)}{(x^4+4)^4}
At x=0 g''(x)>0 which indicates a minimum, and at both x=1 and x=-1 g''(x)<0 which indicates a maximum.

"Thus we must have \frac{3}{4} \le g(x) \le1for all x"
This is not true as we know nothing of what happens when x&lt;-1 \text{, or,} x&gt;1. We can easily prove the statement wrong by e.g. considering x=5 which gives g(x)=0.08...

ii) Their first statement that\displaystyle\int(1-x)^{-3}dx=-3(1-x)^{-4} is false as they are differentiating (missing out a sign change as well) instead of integrating. It shall be: \displaystyle\int(1-x)^{-3}dx=\frac{1}{2}(1-x)^{-2}
Evaluating the integral \displaystyle\int_{-1}^3(1-x)^{-3} means we have [\frac{1}{2(1-x)^2}]_{-1}^3
\frac{1}{8}-\frac{1}{8}=0 as proposed (but with faulty argument as they differentiated...)
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nota bene
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(Original post by generalebriety)
No, I just can't do the Maclaurin series. At all. I always make stupid mistakes and I know none of them off the top of my head.
Well, what is the problem? I got the same answer as you (when including the fractions half-way:p:).

I'm probably missing something obvious, must be the lack of sleep...
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(Original post by nota bene)
Well, what is the problem? I got the same answer as you (when including the fractions half-way:p:).

I'm probably missing something obvious, must be the lack of sleep...
Um, because I copied your answer without thinking when you corrected me. Because I hate the Maclaurin series.

There's no problem. I just do stupid things like ln(1+x) = 1 when x=0. Or I differentiate wrongly (especially with complicated functions). Or my arithmetic goes haywire. And I can't remember any of them off the top of my head.
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(Original post by generalebriety)
Um, because I copied your answer without thinking when you corrected me. Because I hate the Maclaurin series.

There's no problem. I just do stupid things like ln(1+x) = 1 when x=0. Or I differentiate wrongly (especially with complicated functions). Or my arithmetic goes haywire. And I can't remember any of them off the top of my head.
Ah, okay, I didn't actually read your post after you edited it with my comment, I just assumed it was okay:p:

ugh, I'm bored of latexing, but I'll finish that question I'm doing now and then try to solve some more...
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(Original post by nota bene)
Did you mean to quote me there?

I assume you didn't, and yes there should be denominators in general's expression I think someone was lazy latexing:p:.
Sorry. I did actually intend to quote you (just because it looked easier than splitting GE's LaTeX post), but I also meant to make it clear I was referring to GE, and I forgot!

Sorry again...
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