STEP I, II, III 1999 solutions

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  1. SimonM's Avatar
    1 24-03-2008  04:07
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    STEP I, II, III 1999 solutions
    Last edited by SimonM; 05-05-2009 at 21:46.
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  2. Zhen Lin's Avatar
    2 24-03-2008  04:10
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    Re: STEP I, II, III 1999 solutions
    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...
    Last edited by Zhen Lin; 24-03-2008 at 04:17.
  3. generalebriety's Avatar
    3 24-03-2008  04:17
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    Re: STEP I, II, III 1999 solutions
    Err... which paper?

    Edit: ah, you edited. I'll have a look.
    Last edited by generalebriety; 24-03-2008 at 04:19.
  4. generalebriety's Avatar
    4 24-03-2008  04:24
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    Re: STEP I, II, III 1999 solutions
    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...
  5. Zhen Lin's Avatar
    5 24-03-2008  07:16
    • Vengeful, Imperial Overlord of The Student Room
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    Re: STEP I, II, III 1999 solutions
    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.)
    Attached Files
  6. File Type: pdf III-1999-Q01.pdf (33.7 KB, 673 views)
  7. File Type: pdf III-1999-Q03.pdf (38.9 KB, 507 views)
  8. File Type: pdf III-1999-Q04.pdf (23.1 KB, 331 views)
  9. File Type: pdf III-1999-Q05.pdf (25.9 KB, 439 views)
  10. File Type: pdf III-1999-Q07.pdf (28.2 KB, 297 views)
  11. DFranklin's Avatar
    6 24-03-2008  12:05
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    Re: STEP I, II, III 1999 solutions
    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.
  12. squeezebox's Avatar
    7 24-03-2008  12:09
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    Re: STEP I, II, III 1999 solutions
    I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.
    Last edited by squeezebox; 24-03-2008 at 12:11.
  13. DFranklin's Avatar
    8 24-03-2008  12:41
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    Re: STEP I, II, III 1999 solutions
    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.
  14. squeezebox's Avatar
    9 25-03-2008  01:03
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    Re: STEP I, II, III 1999 solutions
    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}}
    Last edited by squeezebox; 06-04-2008 at 00:30.
  15. ukstudent2011's Avatar
    10 25-03-2008  01:29
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    Re: STEP I, II, III 1999 solutions
    (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.
    Last edited by ukstudent2011; 25-03-2008 at 01:31.
  16. squeezebox's Avatar
    11 25-03-2008  01:38
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    Re: STEP I, II, III 1999 solutions
    Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).
  17. Zhen Lin's Avatar
    12 25-03-2008  02:06
    • Vengeful, Imperial Overlord of The Student Room
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    Re: STEP I, II, III 1999 solutions
    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...
    Last edited by Zhen Lin; 25-03-2008 at 02:09.
  18. squeezebox's Avatar
    13 25-03-2008  02:15
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    Re: STEP I, II, III 1999 solutions
    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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  19. nota bene's Avatar
    14 25-03-2008  02:20
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    Re: STEP I, II, III 1999 solutions
    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.
    Last edited by nota bene; 25-03-2008 at 03:00. Reason: solution
  20. ukstudent2011's Avatar
    15 25-03-2008  02:26
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    Re: STEP I, II, III 1999 solutions
    (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.
  21. nota bene's Avatar
    16 25-03-2008  12:48
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    Re: STEP I, II, III 1999 solutions
    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.

    Last edited by nota bene; 02-05-2008 at 23:33.
  22. DFranklin's Avatar
    17 25-03-2008  14:16
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    Re: STEP I, II, III 1999 solutions
    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.
  23. nota bene's Avatar
    18 25-03-2008  14:25
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    Re: STEP I, II, III 1999 solutions
    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
  24. nota bene's Avatar
    19 25-03-2008  19:28
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    Re: STEP I, II, III 1999 solutions
    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)

  25. nota bene's Avatar
    20 26-03-2008  01:26
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    Re: STEP I, II, III 1999 solutions
    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)
    Attached Thumbnails
    Click image for larger version.  Name: 99II7.JPG  Views: 332  Size: 287.2 KB  ID: 50219  
    Last edited by nota bene; 29-03-2008 at 10:59.
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