SimonM
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(Updated as far as #111) SimonM - 26.04.2009

Done

STEP I:
1: Solution by SimonM
2: Solution by SimonM
3: Solution in Siklos booklet
4: Solution by brianeverit
5: Solution by Aurel-Aqua
6: Solution by SimonM
7: Solution by Aurel-Aqua
8: Solution by SimonM
9: Solution by lilman91
10: Solution by Aurel-Aqua
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by lilman91 and Aurel-Aqua

STEP II:
1: Solution by brianeverit
2: Solution by nota bene, Solution by Square
3: Solution by nota bene, #28, etc.
4: Solution by brianeverit
5: Solution by I hate maths
6: Solution by nota bene
7: Solution by nota bene
8: Solution by brianeverit
9: Solution by brianeverit
10: Solution by Aurel-Aqua
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by SimonM And Aurel-Aqua
14: Solution by SimonM

STEP III:
1: Solution by Zhen Lin (PDF)
2: Solution by Agrippa (PDF)
3: Solution by Zhen Lin (PDF)
4: Solution by Zhen Lin (PDF)
5: Solution by Zhen Lin (PDF)
6: Solution by squeezebox, #61
7: Solution by Zhen Lin (PDF), Post #52
8: Solution by squeezebox
9: Solution by brianeverit
10: Solution by squeezebox
11: Solution by brianeverit
12: Solution by Aurel-Aqua
13: Solution by toasted-lion
14: Solution by brianeverit

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
Last edited by Sir Cumference; 1 year ago
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Zhen Lin
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I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...
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generalebriety
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Err... which paper?

Edit: ah, you edited. I'll have a look.
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generalebriety
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In STEP questions like that, you can be almost certain that the second part relates to the first part in some way. In this case, put y = e^x - 1 - k tan^-1 x, and find dy/dx...
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Zhen Lin
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Aaah. I see. I will try that now.

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)
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DFranklin
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Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.
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squeezebox
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I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.
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DFranklin
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Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.
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squeezebox
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STEP III - Question 8

i)

 0 \leq x \leq 2\pi \Rightarrow n = 1 .

So the differential equation becomes:

 \frac{d^{2}y}{dx^{2}} + y = 0, whose general solution is:  y(x) = A\cos(x) + B\sin(x) .
Now, since when x=0, y=0 and \frac{dy}{dx} = 1 , the particular solution is y(x) =\sin(x) (*).

ii) Since  2\pi \leq x \leq 4\pi, \Rightarrow n = 2 .

So this time the differential equation is:

\frac{d^{2}y}{dx^{2}} + 4y = 0

general solution: y(x) = C\cos(2x) + D\sin(2x) (**).

Now, since y and \frac{dy}{dx} are continuous at x=2\pi, (**) has the same value as (*) at x = 2\pi and also their derivatives take the same value. From this we can work out the values of C and D, (C=0 and D= \frac{1}{2}). Which gives us: y(x) = \frac{1}{2}\sin(2x).
Using the same method in the previous part, we can deduce that y(x) = \frac{1}{n}\sin(nx).

iii)

 \displaystyle\int^\infty_0 y^{2} \, dx = \displaystyle\int^{2\pi}_{0} y^{2} \, dx + \displaystyle\int^{4\pi}_{2\pi} y^{2} \, dx + \displaystyle\int^{6\pi}_{4\pi} y^{2} \, dx + ...........+ \displaystyle\int^{2r\pi}_{2(r-1)\pi} y^{2} \, dx + ....




=  \displaystyle\int^{2\pi}_{0} \sin^{2}(x) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}\sin^{2}(x) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}\sin^{2}(x) \, dx + .....



=  \displaystyle\int^{2\pi}_{0} (1-\cos(2x)) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}(1-\cos(4x)) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}(1-\cos(6x))\, dx + .....

(using  \cos(2x)  = 1 - 2\sin^{2}(x) )

=  \frac{1}{2}[ 2\pi - 0] + \frac{1}{2^{2}}[\frac{1}{2}( 4\pi - 2\pi )] + \frac{1}{3^{2}}[\frac{1}{2}( 6\pi - 4\pi)] + ....


=  \pi + \frac{\pi}{2^{2}} + \frac{\pi}{3^{2}} + ..... = \pi \displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}
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ukstudent2011
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(Original post by Zhen Lin)
I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...
I would find the stationary points by differentiating, then see where there is a +ve/-ve sign change between stationary points in the original equation, if that makes sense.
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squeezebox
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Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).
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Zhen Lin
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Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether \left( e^x - 1 \right) is increasing/decreasing faster than k \tan^{-1} x. I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...
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squeezebox
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I dunno if what I did was correct, but I considered; g(x) =  \frac{e^{x} - 1}{tan^{-1}(x)}. By playing around with it, I noticed that as x \longrightarrow -\infty, g(x) \longrightarrow \frac{2}{\pi}. If you want me to into anyfurther detail , I will.
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nota bene
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I'm currently struggling my way through II/3, will type up when done.
under construction

II/3

S_n(x)=e^{x^3}\frac{\mathrm{d}^n  }{\mathrm{d}x^n}(e^{-x^3})
Then let's compute some derivatives

\frac{\mathrm{d}}{\mathrm{d}x}(e  ^{-x^3})=-3x^2e^{-x^3} so S_1(x)=-3x^2 and we can see that for all S_n we will get e^(x^3) and e^(-x^3) cancelling out.
\frac{\mathrm{d}^2}{\mathrm{d}x^  2}(e^{-x^3})=\frac{\mathrm{d}}{\mathrm{  d}x}(-3x^2e^{-x^3})=9x^4e^{-x^3}-6xe^{-x^3}=(9x^4-6x)e^{-x^3} So S_2(x)=9x^4-6x
And similarily \frac{\mathrm{d}^3}{\mathrm{d}x^  3}(e^{-x^3})=\frac{\mathrm{d}}{\mathrm{  d}x}((9x^4-6x)e^{-x^3})=e^{-x^3}(-27x^6+54x^3-6)
So S_3(x)=-27x^6+54x^3-6

To see the pattern I computed S_4-S_6 as well, won't type up the calculations but they are as follows:
S_4(x)=81x^8-324x^5+180x^2 \newline S_5(x)=-243x^{10}+1620x^7-2160x^4+360x \newline S_6(x)=729x^{12}-7290x^9+17820x^6-9720x^3+360

From this we can form the following closed form S_n(x)=\displaystyle\sum_{k=0}^{  n-\lceil\frac{n}{3}\rceil}(-1)^ka_kx^{2n-3k} where \lceil b\rceil denotes ceiling[b] i.e. the nearest integer above, or equal to b.
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ukstudent2011
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(Original post by Zhen Lin)
Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether \left( e^x - 1 \right) is increasing/decreasing faster than k \tan^{-1} x. I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...
You know the roots must be between turning points, before the first turning point, or after the last one, so if there is a sign change between two turning points there must be one root between the turning points. It is quite easy to work out if there is another root before or after the last turning point.
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nota bene
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II/2
solution

nx^2+2x\sqrt{pn^2+q}+rn+s=0 p\not=r, p>0\ and\ n\in \Bbb{Z}^{+}

i) Where p=3 q=50 r=2 and s=15 find the set of values for n where the equation above has no real roots.

For non-real roots we need 4(pn^2+q)-4n(rn+s)<0 \Leftrightarrow pn^2+q<n^2r+ns \Leftrightarrow n^2(p-r)-ns+q<0
Inserting the given values of p, q, r and s:
n^2-15n+50<0 \Leftrightarrow (n-\frac{15}{2})^2-\frac{25}{4}<0\Rightarrow \pm (n-\frac{15}{2})<\frac{5}{2}
Solving for the positive case we have n<10 and the negative case -n<\frac{1}{2}(5-15) \Leftrightarrow n>5
I.e. the equation lacks real roots when n>5 and n<10

ii) Prove that if p<r and 4q(p-r)&gt;s^2 then the above equation has no real roots fo any n.

Recall from i) that for non-real roots n^2(p-r)-ns+q&lt;0
Solving for n gives n= \frac{s^2\pm\sqrt{s^2-4q(p-r)}}{2(p-r)} and \sqrt{s^2-4q(p-r)} is obviously complex when 4q(p-r)&gt;s^2 and thus there are no real solutions for any n.


iii) Find when the above equation has real roots when n=1, (p-r)=1 and q=\frac{s^2}{8}
For real roots (p-r)n^2-ns+q\geq0, and with the above values this transfers to finding where 1-s+\frac{s^2}{8}\geq0 \Leftrightarrow (\frac{s}{2\sqrt{2}}-\sqrt{2})^2\geq1 \Rightarrow \pm(\frac{s}{2\sqrt{2}}-\sqrt{2})\geq1
Solving the positive case gives s\geq2\sqrt{2}(1+\sqrt{2})=4+2\s  qrt{2} as desired and the negative case gives -s\geq2\sqrt{2}(1-\sqrt{2})=2\sqrt{2}-4\Leftrightarrow s\geq4-2\sqrt{2} as desired.

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DFranklin
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For (ii), I don't see why you say that s is positive. Given the conditions they've given, it's pretty obvious you're supposed to use "B^2-4AC<0" on n^2(p-r)-ns+q.
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nota bene
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I'll fix that - I thought I'd take a shortcut but turns out I didn't check carefully as obviously \pm(\sqrt{4q(p-r)})&gt;s
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nota bene
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II/6
solution
y=\frac{ax+b}{cx+d} then \frac{dy}{dx}=\frac{a(cx+d)-c(ax+b)}{(cx+d)^2}=\frac{ad-bc}{(cx+d)^2}

Now, this can be used to find integrals such as \displaystyle\int_0^1 \frac{1}{(x+3)^2}\ln(\frac{x+1}{  x+3})dx Substitute u=\frac{x+1}{x+3} and then we have \frac{du}{dx}=\frac{2}{(x+3)^2} \Leftrightarrow dx=\frac{1}{2}(x+3)^2du
This transfers our integral to
\frac{1}{2}\displaystyle\int_{\f  rac{1}{3}}^{\frac{1}{2}}\ln(u)du  =\frac{1}{2}[u(\ln(u)-1)]_{\frac{1}{3}}^{\frac{1}{2}}= \newline\frac{1}{4}\ln(\frac{1}{  2})-\frac{1}{4}-\frac{1}{6}\ln(\frac{1}{3})+\fra  c{1}{6}=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln  (3)-\frac{1}{12}
as desired.

Next integral: \displaystyle\int_0^1 \frac{1}{(x+3)^2}\ln(\frac{x^2+3  x+2}{(x+3)^2}dx= \displaystyle\int_0^1 \frac{1}{(x+3)^2}\ln(\frac{(x+1)  (x+2)}{(x+3)^2}dx=  \displaystyle\int_0^1 \frac{1}{(x+3)^2}(\ln(\frac{x+1}  {x+3})+\ln(\frac{x+2}{x+3}))dx =\displaystyle\int_0^1 \frac{1}{(x+3)^2}\ln(\frac{x+1}{  x+3})+\displaystyle \int_0^1 \frac{1}{(x+3)^2}\ln(\frac{x+2}{  x+3})
Let u=\frac{x+2}{x+3} thus dx=(x+3)^2du which gives
\displaystle \int_{\frac{2}{3}}^{\frac{3}{4}}  \ln(u)du=[u(\ln(u)-1)]_{\frac{2}{3}}^{\frac{3}{4}}
And similarily let v=\frac{x+1}{x+3} thus dx=\frac{1}{2}(x+3)^2dv which gives
\frac{1}{2}\displaystyle\int_{\f  rac{1}{3}}^{\frac{1}{2}} \ln(v)dv=\frac{1}{2}[v(\ln(v)-1)]_{\frac{1}{3}}^{\frac{1}{2}}
Evaluating these two integrals, and adding together gives
\frac{3}{4}\ln(3)-\frac{3}{2}\ln(2)-\frac{2}{3}\ln(2)+\frac{2}{3}\ln  (3)-\frac{1}{4}\ln(2)+\frac{1}{6}\ln  (3)=\newline \frac{19}{12}\ln(3)-\frac{29}{12}\ln(2)-\frac{1}{6}

Now to the last integral:
\displaystyle\int_0^1\frac{1}{(x  +3)^2}\ln(\frac{x+1}{x+2})=\disp  laystyle\int_0^1\frac{1}{(x+3)^2  }\ln(\frac{x+1}{x+2}\frac{x+3}{x  +3})=\newline \displaystyle\int_0^1\frac{1}{(x  +3)^2}(\ln(\frac{x+1}{x+3})-\ln(\frac{x+2}{x+3}))
which both have been evaluated in the part above.
-\frac{1}{4}\ln(2)-\frac{1}{4}+\frac{1}{6}\ln(3)+\f  rac{1}{6} - (\frac{3}{4}\ln(3)-\frac{3}{2}\ln(2)-\frac{3}{4}-\frac{2}{3}\ln(2)+\frac{2}{3}\ln  (3)+\frac{2}{3})= \newline (\frac{1}{6}-\frac{3}{4}-\frac{2}{3})\ln(3)+(\frac{3}{2}+  \frac{2}{3}-\frac{1}{4})\ln(2)=\frac{23}{12}  \ln(2)-\frac{5}{4}\ln(3)

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nota bene
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II/7

This question seems too easy, so I've probably made a mistake:

y=\frac{x}{\sqrt{x^2-2x+a}} Then \frac{dy}{dx}= - \frac{x\frac{x-1}{\sqrt{x^2-2x+a}}-\sqrt{x^2-2x+a}}{x^2-2x+a}= - \frac{x(x-1)-(x^2-2x+a)}{(x^2-2x+a)^{\frac{3}{2}}}= - \frac{x-a}{(x^2-2x+a)^{\frac{3}{2}}} = \frac{a-x}{(x^2-2x+a)^{\frac{3}{2}}}
Stationary points can occur when dy/dx=0 i.e. the numerator needs to be zero <=> 0=a-x <=>x=a Hence there is at most one stationary point. The derivative is undefined if x^2-2x+a=0 i.e. x=\frac{2\pm\sqrt{4-4a}}{2} and for a>1 this has no (real) solutions and thus the function has more than zero stationary points. I.e. there is one stationary point, as desired.

i) a=2 f(0)=0 f(2)=\sqrt{2} so stationary point at (2,\sqrt{2}) f'(1.9)&lt;0and f'(2.1)&gt;0 and thus the point is a maximum.
Also note \displaystyle\lim_{x\to\pm\infty  }[f(x)]=\pm1

ii) a=1 f(0)=0 From above we have that there are no stationary points because the denominator of the function (and it's derivative) goes to 0 as a goes to 1 i.e. vertical asymptote at x=1 Same as in i) note that \displaystyle\lim_{x\to\pm\infty  }[f(x)]=\pm1


Sketches see attached. (Very rough, would put much more effort in during a real exam)
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