STEP I, II, III 1999 solutions

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  1. squeezebox's Avatar
    61 06-04-2008  13:38
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    Re: STEP I, II, III 1999 solutions
    Ok, so heres my working:

    I get the integral down to;
    \frac{3a^{2}}{8} \displaystyle\int^{2\pi}_{0} \sin^{2}(2t) \, dt

    which becomes:

    \frac{3a^{2}}{16} \displaystyle\int^{2\pi}_{0} 1-cos(4t) \, dt

    which is equal to \frac{3a^{2}\pi}{8}?
  2. Dystopia's Avatar
    62 06-04-2008  13:46
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    Re: STEP I, II, III 1999 solutions
    Yeah, that's right. The curve is called an astroid; it has some interesting properties.
  3. squeezebox's Avatar
    63 06-04-2008  13:47
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    Re: STEP I, II, III 1999 solutions
    Cheers , I'll have to check my working more carefully in the future .
  4. nota bene's Avatar
    64 02-05-2008  23:24
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    Re: STEP I, II, III 1999 solutions
    (Original post by Square)
    Sorry to dig up the old thread.

    Looking at the solution for II/2 part i). Why can you not factorise 4n^2-60n+200<0 to 4(n-5)(n-10)<0 and then propose that there are no real roots for (*) when 5 < n < 10.

    I hope I've not done something stupid here :/

    http://www.thestudentroom.co.uk/show...9&postcount=16
    You're right, don't know what's happened with my mental arithmetic on that question. Will try and fix that now.
  5. Square's Avatar
    65 02-05-2008  23:38
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    Re: STEP I, II, III 1999 solutions
    (Original post by nota bene)
    You're right, don't know what's happened with my mental arithmetic on that question. Will try and fix that now.
    Okay! I just saw all those \sqrt{57} and panicked!

    I'm trying to repost with a better typed solution to ii) which gave me the right answer but I still think turned out to be rather weak.
    Last edited by Square; 02-05-2008 at 23:47.
  6. Square's Avatar
    66 02-05-2008  23:45
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    Re: STEP I, II, III 1999 solutions
    Okay II/2

    Spoiler:
    Show


    ii) The discriminant factorises down to:

    (p-r)n^2-sn+q

    =(p-r)(n^2-\frac{sn}{p-r}+\frac{q}{p-r}

    complete the square:

    (p-r)((n-\frac{s}{2(p-r)})^2-\frac{s^2}{4(p-r)^2}+\frac{q}{p-r}) (**)

    Now we are told p<r this implies p-r<0

    Now, for (*) to have no real roots the discriminant (**)<0

    If p-r<0 (**)<0 iff ((n-\frac{s}{2(p-r)})^2-\frac{s^2}{4(p-r)^2}+\frac{q}{p-r})&gt;0

    Here is where I'm not sure, I said that the bracket would ALWAYS be greater than zero if:

    \frac{q}{p-r}&gt;\frac{s^2}{4(p-r)^2}

    (discarding the squared term as it is always positive)

    which gives the desired result.

    However surely the bracket could still be positive if: \frac{q}{p-r}&lt;\frac{s^2}{4(p-r)^2} AND -\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}&lt;(n-\frac{s}{2(p-r)})^2

    If that makes sense at all!



    re-reading this it doesn't make much sense really!
    Last edited by Square; 03-05-2008 at 00:43.
  7. nota bene's Avatar
    67 02-05-2008  23:58
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    Re: STEP I, II, III 1999 solutions
    (Original post by Square)
    However surely the bracket could still be positive if: \frac{q}{p-r}&lt;\frac{s^2}{4(p-r)^2} AND \frac{q}{p-r}-\frac{s^2}{4(p-r)^2}&lt;(n-\frac{s}{2(p-r)})^2
    Firstly it appears to me that your expression after "AND" should be -\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}&lt;(n-\frac{s}{2(p-r)})^2 (unless I'm being stupid)

    And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities. It's well possible that it's possible to prove this "and" cannot happen, or that it can be disregarded for some reason; however I don't see why directly and you would probably need to motivate why/how in a solution.

    I understand what you are doing, but can't see how to finish it solidly. Let's see what comments David has.
  8. Square's Avatar
    68 03-05-2008  00:38
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    Re: STEP I, II, III 1999 solutions
    (Original post by nota bene)
    Firstly it appears to me that your expression after "AND" should be -\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}&lt;(n-\frac{s}{2(p-r)})^2 (unless I'm being stupid)

    And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities. It's well possible that it's possible to prove this "and" cannot happen, or that it can be disregarded for some reason; however I don't see why directly and you would probably need to motivate why/how in a solution.

    I understand what you are doing, but can't see how to finish it solidly. Let's see what comments David has.
    Sorry think that's a typo on my part.
  9. Square's Avatar
    69 03-05-2008  00:45
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    Re: STEP I, II, III 1999 solutions
    (Original post by nota bene)
    Firstly it appears to me that your expression after "AND" should be -\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}&lt;(n-\frac{s}{2(p-r)})^2 (unless I'm being stupid)

    And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities. It's well possible that it's possible to prove this "and" cannot happen, or that it can be disregarded for some reason; however I don't see why directly and you would probably need to motivate why/how in a solution.

    I understand what you are doing, but can't see how to finish it solidly. Let's see what comments David has.
    The only thing I can think of is somehow using p<r to get one side positive the other side negative or something and find a contradiction. Alas that would be nice but I can't seem to see anything like that .
  10. DFranklin's Avatar
    70 03-05-2008  12:12
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    Re: STEP I, II, III 1999 solutions
    (Original post by nota bene)
    Firstly it appears to me that your expression after "AND" should be -\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}&lt;(n-\frac{s}{2(p-r)})^2 (unless I'm being stupid)

    And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities.
    The question only asks you to show that "p < r and 4q(p-r)>s^2" implies the quadratic has no real roots. It's not an "if and only if" question. So what square has done is fine.

    If we ignore the wording of the question, then if n can take any real value, then I'm pretty sure (without detailed calculation) we end up with an "if and only if" result; simply choose n=\frac{s}{2(p-r)}. But as n can only take positive integers, the true result would be more complicated. Fortunately, that's not what they're asking for...
  11. nota bene's Avatar
    71 03-05-2008  12:40
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    Re: STEP I, II, III 1999 solutions
    (Original post by DFranklin)
    The question only asks you to show that "p < r and 4q(p-r)>s^2" implies the quadratic has no real roots. It's not an "if and only if" question.
    Ah, yes sorry! Of course the result is fine then
  12. Square's Avatar
    72 03-05-2008  13:47
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    Re: STEP I, II, III 1999 solutions
    (Original post by nota bene)
    Ah, yes sorry! Of course the result is fine then
    Yay, that's great.

    I'm glad to hear my solution was correct after all. The question didn't take me an awful long time but it was slightly disheartening to come onto the solution thread and see something totally different .
  13. Square's Avatar
    73 21-05-2008  22:45
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    Re: STEP I, II, III 1999 solutions
    1999 III/1 part ii)

    I'm slightly confused by Zhen Lin's solution (well by confused I meant it seems a LOT different to how I did it.)

    Spoiler:
    Show

    ii) r=\frac{p^3}{q^3}

    from part i) r=a^3 => a=p/q. Since a is a root of the cubic above therefore p/q must also be a root.

    The other two roots are ak and a/k, the product of these is a^2, a^2=(p/q)^2 as required.

    The three roots are: ak, a, a/k. These are obviously in geometric progression with common ratio 1/k.



    This almost seems FAR too simple.

    David seems to have not been around for the last few days, hope everything is okay!
    Last edited by Square; 21-05-2008 at 22:49.
  14. Zhen Lin's Avatar
    74 22-05-2008  01:57
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    Re: STEP I, II, III 1999 solutions
    Well, the thing is, you are assuming that ak^{-1}, a, ak are roots of the equation - this is not a given when you start part ii, I think.
  15. lilman91's Avatar
    75 24-07-2008  21:50
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    Re: STEP I, II, III 1999 solutions
    Step I Question 9

    When the hare overtakes, both have travelled 0.5km, but the tortoise has travelled for 0.5hours longer. Using time=\frac{distance}{speed} for both:

    \frac{0.5}{v}=\frac{0.5}{V}+0.5 \Rightarrow \frac{1}{v}=\frac{1}{V}+1=\frac{ V+1}{V}

    Therefore, v=\frac{V}{V+1}...(A)

    When the hare passes the tortoise on the way back, the hare has travelled X+1.25 whilst the tortoise has travelled X-1.25. However, the tortoise has been racing for 1 hour longer.

    Using time=\frac{distance}{speed} again:
    \frac{X+1.25}{V}+1=\frac{X-1.25}{v} \Rightarrow \frac{X+1.25+V}{V}=\frac{X-1.25}{v}

    Therefore, v=\frac{V(X-1.25)}{X+1.25+V}...(B)

    From, (A) and (B)

    \frac{V}{V+1}=\frac{V(X-1.25)}{X+1.25+V} \Rightarrow \frac{1}{V+1}=\frac{X-1.25}{X+1.25+V}
    This gives: X+1.25+V=V(X-1.25)+X-1.25 \Rightarrow V(X-2.25)=2.5 \Rightarrow V=\frac{2.5}{X-2.5}. Multiplying top and bottom by 4 gives V=\frac{10}{4X-9} as required.

    From A we know that v=\frac{V}{V+1}. Substituting expression for V gives:

    v=\frac{\frac{10}{4X-9}}{\frac{10+4X-9}{4x-9}} = \frac{10}{4X+1}.
    So, v=\frac{10}{4X+1}.

    The hare finishes 1.5hrs. This means she has travelled 2X in 1.5hrs.
    So, \frac{2X}{V}=1.5 \Rightarrow 4X=3V
    Substituting expression for V,
    4X=\frac{30}{4X-9} \Rightarrow 16X^2-36X-30=0 \Rightarrow 8X^2-18x-15=0.
    Solving the quadratic but since X&gt;0, X=\frac{18+\sqrt{804}}{16}=\frac {18+2\sqrt{201}}{16}=\frac{9+\sq rt{201}}{8}.
    So, finally X=\frac{9+\sqrt{201}}{8}.
    Last edited by lilman91; 24-07-2008 at 22:03.
  16. lilman91's Avatar
    76 24-07-2008  22:51
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    Re: STEP I, II, III 1999 solutions
    Step I Question 14

    (i)Let Y = Score. The score is a discrete random variable whose probabilities need to be calculated.

    P(Y=4)=P(0\leq X&lt;\frac{1}{4})=\int^\frac{1}{4}_ 0 2x \, dx = \frac{1}{16}
    P(Y=3)=P(\frac{1}{4}\leq X&lt;\frac{1}{2})=\int^\frac{1}{2}_ \frac{1}{4} 2x \, dx = \frac{3}{16}
    P(Y=2)=P(\frac{1}{2}\leq X&lt;\frac{3}{4})=\int\frac{3}{4}_\ frac{1}{2} 2x \, dx = \frac{5}{16}
    P(Y=1)=P(\frac{3}{4}\leq X&lt;1)=\int^1_\frac{3}{4} 2x \, dx = \frac{7}{16}

    So now, E(Y)=\left(\frac{7}{16}\times 1\right)+\left(\frac{5}{16}\time s 2\right)+\left(\frac{3}{16}\time s 3\right)+\left(\frac{1}{16}\time s 4\right) = \frac{15}{8}
    Thus, E(Y)= \frac{15}{8} as required.

    (ii) The ways of winning are a) 3 b) 2,1 (in any order) c)1,1,1

    a) P(\mathrm{3 on first go})=\frac{3}{16}
    b) P(\mathrm{2,1 in any order})=\frac{5}{16} \times \frac{7}{16}\times 2=\frac{70}{256}
    c) P(\mathrm{1,1,1})=\left(\frac{3} {16}\right)^3= \frac{343}{4096}

    So P(\mathrm{win})=\frac{3}{16}+\fr ac{70}{256}+\frac{343}{4096}=\fr ac{2231}{4096} (that takes a bit of long winded arithmetic)

    We want to find P(\mathrm{3 darts|win})=\frac{P(\mathrm{3dar ts}\land\mathrm{win})}{P(\mathrm {win})}

    But P(\mathrm{3darts}\land\mathrm{wi n})=P(\mathrm{1,1,1})=\frac{343} {4096}

    So, P(\mathrm{3 darts|win})= \frac{343}{4096} \div \frac{2231}{4096}=\frac{343}{223 1} as required.
  17. Aurel-Aqua's Avatar
    77 02-01-2009  15:59
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    Re: STEP I, II, III 1999 solutions
    STEP III Question 12

    \displaystyle P(X=j,Y=k) = e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k! }

    (i) Find \displaystyle P(X+Y=n) for each n&gt;0:
    This means that we're looking at combinations of (n,0), (n-1,1), (n-2,2), \ldots (2,n-2), (1,n-1), (0,n) of (j,k). Therefore,
    \displaystyle P(X+Y=n) = e^{-1}n\lambda^n\left(\frac{1}{n!0!} +\frac{1}{(n-1)!1!}+\cdots+\frac{1}{1!(n-1)!}+\frac{1}{0!n!}\right), which is equivalent to:

    \displaystyle P(X+Y=n) = e^{-1}n\lambda^n\sum_{r=0}^{n}\frac{ 1}{(n-r)!r!} = e^{-1}n\lambda^n\sum_{r=0}^{n} \frac{1}{n!}\frac{n!}{(n-r)!r!}.

    We can now see that this produces a binomial series, and can thus simplify down to:

    \displaystyle P(X+Y=n) = e^{-1}n\lambda^n\sum_{r=0}^{n}\frac{ 1}{(n-r)!r!} = \frac{e^{-1}n\lambda^n}{n!}\sum_{r=0}^{n}\ frac{n!}{(n-r)!r!} = \frac{e^{-1}n\lambda^n2^n}{n!}=

    \therefore \displaystyle P(X+Y=n) =\frac{e^-1(2\lambda)^n}{(n-1)!}.

    (ii) Show that \displaystyle 2\lambda e^{2\lambda-1} = 1:

    For a discrete probability density function, we know that the cumulative sum of probabilities will equal 1. Thus, \displaystyle \sum_{n=1}^{\infty} P(X+Y=n) = 1. This implies that:

    \displaystyle 1 = \sum_{n=1}^{\infty} \frac{e^-1(2\lambda)^n}{(n-1)!} = 2\lambda e^{-1}\sum_{n=1}^{\infty} \frac{(2\lambda)^{n-1}}{(n-1)!}.

    We notice that \displaystyle e^x = \sum_{r=0}^{\infty}\frac{x^r}{r! }, and so \displaystyle 2\lambda e^{-1}\sum_{n=1}^{\infty} \frac{(2\lambda)^{n-1}}{(n-1)!} = 2\lambda e^{-1}e^{2\lambda} = 2\lambda e^{2\lambda - 1}.

    (iii) Show that \displaystyle 2x e^{2x-1} is an increasing function of \displaystyle x for \displaystyle x &gt; 0 and deduce that the equation in has at most one solution and hence determine \lambda:

    Let \displaystyle f(x) = 2xe^{2x-1}. \displaystyle f'(x) = 4xe^{2x-1} + 2e^{2x-1} = (4x+2)e^{2x-1}, so \displaystyle f'(x) &gt; 0 for \displaystyle x&gt;0, and so there are no turning points and thus there will only be one root of \displaystyle x in this range.

    Finding the value of f(\lambda) = 1 simply takes some observation: \displaystyle f(\frac{1}{2}) = 2\cdot\frac{1}{2}e^{2\cdot\frac{ 1}{2}-1} = 1 = 1, and so \displaystyle\lambda = \frac{1}{2}.

    (iv) Calculate the expectation E\left[2^{X+Y}\right]:

    \displaystyle E\left[2^{X+Y}\right] = \sum_{n=1}^{\infty} 2^n P(X+Y=n) = 4\lambda e^{-1}\sum_{n=1}^{\infty} \frac{(4\lambda)^{n-1}}{(n-1)!} =
    \displaystyle =4\lambda e^{-1}e^{4\lambda} = 2e.

    My first attempt, I hope someone verifies this .
  18. SimonM's Avatar
    78 23-03-2009  22:10
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    Re: STEP I, II, III 1999 solutions
    STEP I, Question 1

    Spoiler:
    Show
    Using the principle of inclusion-exclusion, there will be

    10^6- (\frac{10^6}{2}+\frac{10^6}{5} - \frac{10^6}{10}) = 10^5 ( 10 +1-5-2) = \boxed{4 \times 10^5}

    positive integers less than one million.

    By symmetry, their average value will be \boxed{5 \times 10^5}

    Using PIE again

    4179 - (\frac{4179}{3} + \frac{4179}{7} - \frac{4179}{21}) = \boxed{2388}

    To find the sum of these values, we have

    \displaystyle \frac{4179(4180)}{2} - \left ( 3 \times \frac{1393(1394)}{2} + 7 \times \frac{597(598)}{2} - 21 \times \frac{199(200)}{2} \right ) = 4179/2
  19. SimonM's Avatar
    79 23-03-2009  22:24
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    Re: STEP I, II, III 1999 solutions
    STEP I, Question 6

    Spoiler:
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    Let y = bx + a, \x \in [-10,10]

    If b=0, y is constant, so \max y = \min y = a
    If b>0, y is strictly increasing so \min y = a-10b, \max y = a+10b
    If b<0, y is strictly decreasing so \min y = a+10b, \max y = a-10b

    Let y = cx^2 + bx+a
    If c=0, treat it as above

    \frac{dy}{dx} = 2cx+b

    This is the case where b&gt;0 from earlier.

    If b-20c \ge 0 then it is strictly increasing and \min y = 100c-10b+a, \max y = 100c+10b+a

    If b+20c&lt;0 then it is strictly decreasing \min y = 100c+10b+a, \max y = 100c-10b+a

    Otherwise 2cx+b changes sign in our interval. This means that

    \min y = a - \frac{b^2}{4c}

    The maximum will occur on an end point. Therefore it depends solely on the sign of b. Therefore if b>0

    \max y = 100c+10b+c otherwise 100c-10b+c
  20. SimonM's Avatar
    80 23-03-2009  22:51
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    Re: STEP I, II, III 1999 solutions
    STEP I, Question 8

    Spoiler:
    Show
    \displaystyle \int_0^1 \frac{t}{n(n-t)} \, dt = \frac{1}{n} \int_0^1 \frac{t}{n-t} \, dt = \frac{1}{n} \int_0^1 \left ( -1 +\frac{n}{n-t} \right ) \, dt =
    \displaystyle \frac{1}{n} \left [ -t - n \ln ( n-t) \right ]_0^1 = \frac{1}{n} \left ( n \ln \left ( \frac{n}{n-1}\right) - 1 \right) = \boxed{\ln \left ( \frac{n}{n-1}\right) - \frac{1}{n}}

    However

    \displaystyle 0 \leq \frac{t}{n(n-t)} = \frac{1}{n-t} - \frac{1}{n} \leq \frac{1}{n-1} - \frac{1}{n}

    Therefore

    \displaystyle 0 \leq \int_0^1 \frac{t}{n(n-t)} \, dt \leq \frac{1}{n-1} - \frac{1}{n}
    or
    \displaystyle \boxed{ 0 \leq \ln \left ( \frac{n}{n-1}\right) - \frac{1}{n} \leq \frac{1}{n-1} - \frac{1}{n}} (as required)

    \displaystyle \sum_{n=2}^N 0 \leq \sum_{n=2}^N \left ( \ln \left ( \frac{n}{n-1}\right) - \frac{1}{n} \right ) \leq \sum_{n=2}^N \frac{1}{n-1} - \frac{1}{n}
    \displaystyle 0 \leq \sum_{n=2}^N \left ( \ln n -\ln (n-1) \right ) - \sum_{n=2}^N \frac{1}{n} \leq 1 - \frac{1}{n} \leq 1
    \displaystyle \boxed{0 \leq \ln N - \sum_{n=2}^N \frac{1}{n} \leq 1}

    \displaystyle \sum_{n=1}^{N} \frac{1}{n} \leq 1 + \sum_{n=2}^{10^{30}} \frac{1}{n} \leq 1+\ln 10^{30} \leq 1+\ln 2^{100} = 1+100 \ln 2

    Crudely approximating e &gt; 1+1 +\frac{1}{2} = \frac{5}{2}

    Therefore \ln 2 &lt; 1

    So \boxed{ \sum_{n=1}^{N} \frac{1}{n} &lt; 101}
    Last edited by SimonM; 23-03-2009 at 22:55.
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