STEP I, II, III 2002 Solutions

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  1. Unbounded's Avatar
    141 19-06-2010  00:46
    • TSR Demigod
    Re: STEP I, II, III 2002 Solutions
    (Original post by ziedj)
    Oh.. isn't that unnecessarily complicated when you have already found via sketching that the curve passes through (0,1), so c = 1? Is there a particular reason?
    I suppose it is unnecessary to find 'c' in terms of 'a', since they do tell us the tangent intersects at (0,1). Perhaps he just wanted to briefly examine the general case - before we note that the tangent passes through (0,1).
  2. miml's Avatar
    142 19-06-2010  00:46
    • Overlord in Training
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    Re: STEP I, II, III 2002 Solutions
    (Original post by ziedj)
    Where did that part on the end (not the coefficient of x, everything else) come from?
    For what it's worth I set it out a slightly different way
    Tangent is at x=a (>0) and passes through (0,1)
    y - 1 = \dfrac{-2a}{(1+a^2)^2}x

    and \dfrac{y-1}{x} = \dfrac{\dfrac{1}{1+a^2} - 1}{a} = \dfrac{-a}{1+a^2}

    So we must have

    \dfrac{-2a}{(1+a^2)^2} = \dfrac{-a}{1+a^2}

    Which quickly cancels down to give a^2 = 1
  3. grusini's Avatar
    143 18-07-2010  01:18
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    Re: STEP I, II, III 2002 Solutions
    My solution to question 14 STEP II
    Attached Files
  4. File Type: pdf STEP II.pdf (42.7 KB, 23 views)
    5. Last edited by grusini; 18-07-2010 at 01:52.
  5. grusini's Avatar
    144 19-07-2010  20:30
    • New Member
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    Re: STEP I, II, III 2002 Solutions
    Completing Dadeyemi's work on STEP III Q2, last part
    Attached Files
  6. File Type: pdf STEP III Q2.pdf (32.5 KB, 42 views)
    7. Post rating:
    2
  7. disco1000's Avatar
    145 25-07-2010  15:01
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    Re: STEP I, II, III 2002 Solutions
    (Original post by SimonM)
    STEP III, Question 1

    Spoiler:
    Show


    As a \to \infty, the volume tends to 2\pi
    Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

    Also a little suggestion maybe it would have been easier to use the substitution u = ln x.
  8. SimonM's Avatar
    146 25-07-2010  15:08
    • TSR Idol
    • Posts: 9,193
    Re: STEP I, II, III 2002 Solutions
    (Original post by Diego Granziol)
    Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

    Also a little suggestion maybe it would have been easier to use the substitution u = ln x.
    It's 2 pi, in latex that is 2\pi. Perhaps I forgot the tags?
  9. welshenglish's Avatar
    147 26-07-2010  14:09
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    Re: STEP I, II, III 2002 Solutions
    Solution to 2002 STEP II Q6
    Draw a decent diagram so you can see the lines and angles clearly.
    Let l1, l2 and l3 be the lengths of lines l1, l2 and l3 from a horizontal plane to the point of intersection. Let vertical height of point of intersection above the plane be h. Then: -
    l3 sin /4 = l1 sin /6 = l2 sin = h
    l3 sin = l1 cos = l2
    Therefore
    l1 cos sin = l1 sin /6
    cos sin = 1/2
    l3 sin sin = l3 sin /4
    sin sin = 1/
    (sin sin )^2 + (cos sin )^2 = 3/4
    sin = /2
    = /3
    Same diagram for second part but
    replace with and
    /4, and /6 with 2, /3 and
    l3 sin 2 = l1 sin = l2 sin /3 = h
    l3 sin = l1 cos = l2
    l1 / l3 = tan = sin 2 / sin
    tan = 2 cos
    l1 sin = l1 cos sin /3
    (tan ) ^ 2 + 3 (cos ) ^ 2 = 4
    (tan ) ^ 4 - 3 (tan ) ^ 2 - 1= 0
    (tan ) ^ 2 = (3 + )/2
    (3 - )/2 is not a solution as (tan ) ^ 2
    Last edited by welshenglish; 26-07-2010 at 22:12.
  10. welshenglish's Avatar
    148 26-07-2010  14:50
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    (Original post by SimonM)
    STEP I:
    1: Solution by Unbounded
    2: Solution by Unbounded
    3: Solution by nota bene
    4: Solution by SimonM
    5: Solution by nota bene
    6: Solution by Unbounded
    7: Solution by sonofdot
    8: Solution by Unbounded
    9: Solution by cliverlong
    10: Solution by Dadeyemi
    11:Solution by Unbounded
    12: Solution by Robbie10538
    13: Solution by Unbounded
    14:

    STEP II:
    1: Solution by sonofdot
    2: Solution by dadeyemi
    3: Solution by sonofdot
    4: Solution by dadeyemi
    5: Solution by dadeyemi
    6:
    7: Solution by Glutamic Acid
    8: Solution by Glutamic Acid
    9: Solution by tommm
    10: Solution by tommm
    11: Solution by tommm
    12:
    13:
    14:

    STEP III:
    1: Solution by SimonM
    2: Solution by Dadeyemi
    3: Solution by Daniel Freedman
    4: Solution by Dadeyemi
    5: Solution by Dadeyemi
    6: Solution by Elongar
    7: Solution by Dadeyemi
    8: Solution by Dadeyemi
    9:
    10:
    11:
    12:
    13:
    14:

    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
    Hi Simon,
    I'm working my way through the gaps in the solutions. I've put in an answer for 2002, STEP II Q6. Was your list up to date? If so, can you include this answer.

    Cheers, Peter
  11. odp04y's Avatar
    149 28-07-2010  15:18
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    Re: STEP I, II, III 2002 Solutions
    Answer to Step III 2002 Q9.

    For the case k>\frac{1}{2}\tan\theta.
    Where \theta is the angle of the plane.

    The area of the fluid = ka^2 and forms a trapezium.
    If you cut the trapezium into two parts, a rectangle and a square angled triangle with angle \theta , the dimensions of the parts are:

    Triangle: Width = a
    Height = a\tan\theta

    Rectangle: width = a
    let height = z

    but ka^2 = za+\frac{a^2 \tan\theta}{2}
    ka = z+\frac{a \tan\theta}{2}
    z = ka-\frac{a \tan\theta}{2}
    z = a(k-\frac{\tan\theta}{2})

    Therefore the centre of mass of the triangle is \frac{1}{3}a from the side of the container and a(k-\frac{\tan\theta}{2})+\frac{a\ta n\theta}{3} from the base. The mass of the triangle is \frac{a^2 \tan\theta}{2}units.

    The centre of mass of the rectangle is \frac{1}{2}a from the side of the container and \frac{a}{2} (k-\frac{\tan\theta}{2}) from the base. Its mass is a^2(k-\frac{\tan\theta}{2})units.

    Using: Moments of whole = Sum moments of parts we can find x and y
    To find x:
    ka^2 x= (\frac{1}{3}a)(\frac{a^2 \tan\theta}{2})+(\frac{1}{2}a)(a ^2(k-\frac{\tan\theta}{2}))

    ka^2 x= \frac{a^3 \tan\theta}{6}+\frac{a^3}{2}(k-\frac{\tan\theta}{2})

    \frac{kx}{a}= \frac{ \tan\theta}{6}+\frac{k}{2}-\frac{\tan\theta}{4}

    \frac{kx}{a}=\frac{k}{2}-\frac{ \tan\theta}{12}

    \frac{x}{a}=\frac{1}{2}-\frac{ \tan\theta}{12k}


    To find y:
    ka^2 y= (\frac{a^2 \tan\theta}{2})(ka-\frac{a \tan\theta}{6})+(ka^2-\frac{a^2 \tan\theta}{2})(\frac{ka}{2}-\frac{a\tan\theta}{4})

    ka^2 y= \frac{ka^3 \tan\theta}{2}-\frac{a^3 \tan^2\theta}{12}+\frac{k^2a^3}{ 2}-\frac{ka^3 \tan\theta}{4} -\frac{ka^3 \tan\theta}{4}+\frac{a^3 \tan^2\theta}{8}

    \frac{ky}{a}= \frac{k^2}{2}-\frac{\tan^2\theta}{12}+\frac{\t an^2\theta}{8}

    \frac{y}{a}= \frac{k}{2}+\frac{\tan^2\theta}{ 24k}


    For the case k<\frac{1}{2}\tan\theta.
    The area of the fluid = ka^2 and forms a square angled triangle with angle \theta.

    The area can be found using
    \frac{1}{2} bc \sin\theta=ka^2
    where b and c are the hypotenuse and base of the triangle respectively.
    Also \cos\theta = \frac{c}{b}
    b=\frac{c}{\cos\theta}
    Therefore by substitution:

    \frac{1}{2} c^2 \tan\theta = ka^2

    c^2 = \frac{2ka^2}{\tan\theta}

    The distance x of the centre of mass from the side is \frac{1}{3} c .

    Therefore x=\frac{1}{3} c

    x^2=\frac{1}{9} c^2

    9x^2= c^2

    Therefore: 9x^2 = \frac{2ka^2}{\tan\theta}

    \frac{x^2}{a^2} = \frac{2k}{9\tan\theta}

    \frac{x}{a} = \sqrt{\frac{2k}{9\tan\theta}}


    To find y:
    \sin\theta = \frac{d}{b}
    \tan\theta = \frac{d}{c}
    Where d is the side opposite the angle theta.

    b= \frac{d}{\sin\theta}
    c= \frac{d}{\tan\theta}

    By substitution \frac{d^2}{2\tan\theta\sin\theta } \sin\theta = ka^2

    \frac{d^2}{a^2} = 2k\tan\theta

    The distance y of the centre of mass from the base is \frac{1}{3} d .

    Therefore 9y^2= d^2

    \frac{y}{a} = \sqrt{\frac{2k\tan\theta}{9}}


    For the case when k<\frac{1}{2} and \theta>45
    2k<1 and \tan\theta>1
    Therefore \frac{1}{2} \tan\theta>k the case where the fluid forms a triangle.

    Also \tan \phi=\frac{y}{x}
    Where \phi is the angle between the centre of mass and the base of the container. and x and y are the distances of the centre of mass from the side and base of the container.

    \tan \phi=\frac{y}{x}=\frac{\sqrt{\fr ac{2k\tan\theta}{9}}}{\sqrt{\fra c{2k}{9\tan\theta}}}

    \frac{y^2}{x^2}=(\frac{2k \tan\theta}{9})(\frac {9\tan\theta}{2k})

    \frac{y^2}{x^2}=\tan^2\theta

    \tan \phi=\frac{y}{x}=\tan\theta

    \phi=\theta

    Container will topple when \theta+\phi>90

    In the case when \theta>45
    \theta+\phi>90

    Therefore the container will topple.
    Last edited by odp04y; 28-07-2010 at 15:22.
  12. welshenglish's Avatar
    150 11-08-2010  18:30
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    Re: STEP I, II, III 2002 Solutions
    Well done odp04y,

    You've obviously got a better command of LaTex than me.
  13. theOldBean's Avatar
    151 14-10-2010  19:10
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    Re: STEP I, II, III 2002 Solutions
    Here's an alternative for the last part of Q5

    Let the roots be a,b,c,d. We know that

    abcd=576=2^6 * 3^2
    (a+1)(b+1)(c+1)(d+1)=1+22+172... =3^3 * 7^2
    (a-1)(b-1)(c-1)(d-1)=1-22+172-552+576=7 * 5^2.

    All the roots are negative (Since all the signs in the equation are "+".)

    If the "+1" equation has a factor of 7^2 and the "-1" has a factor 5^2, 6 must be a duplicated root. Also since the "-1" has a factor of 7, 8 must be a root. It is now easy to deduce that the other root is 2.

    So the roots are -2, -6 (twice) and -8.

    By the fundamental theorem of algebra, this is the only solution set.
  14. AnonyMatt's Avatar
    152 10-11-2010  02:35
    • Vengeful, Imperial Overlord of The Student Room
    • Location: Wales
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    Re: STEP I, II, III 2002 Solutions
    STEP III Q14

    Spoiler:
    Show


    Var[X + Y] = E[(X+Y)^2] - E^2[X+Y] = E[X^2 + 2XY + Y^2] - E^2[X+Y]
    = E[X^2] + 2E[XY] + E[Y^2] - (E[X] + E[Y])^2
    = E[X^2] + 2E[XY] + E[Y^2] - E[X]^2 - 2E[X]E[Y] - E[Y]^2
    = E[X^2] - E[X]^2 + E[Y^2] - E[Y]^2 + 2(E[XY] - E[X]E[Y])
    = Var[X] + Var[Y] + 2Cov[X,Y]
    as required.

    i) Var[W + L + D] = E[(W + L + D)^2] - E^2[W + L + D]

    E[W+L+D] = m => E^2[W+L+D] = m^2
    E[(W+L+D)^2] = E[W^2 + L^2 + D^2 + 2(WL+WD+LD)]
    = 3(m/3)^2 + 2.3(m/3)^2 = m^2/3 + 2m^2/3 = m^2

    => Var[W+L+D] = m^2 - m^2 = 0

    ii) Consider W:
    E[W] = m/3
    Var[W] = E[W^2] - E^2[W] =

    E[W^2] = 1*P(W=1) + 4*(P(W=2)) + ... + n^2(P(W=n)) for n = 1, 2, ..., m
    = 1*(m/3)(1-m/3)^m-1 + 4*(m/3)(m/3)(1-m/3)^m-2 + ... + n^2(m/3)^n(1-m/3)^m-n + ... + m^2(m/3)^m
    =



    Very much unfinished, and I don't think I've done part i) correctly...
  15. Asboob's Avatar
    153 19-04-2011  14:31
    • Junior Member
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    Re: STEP I, II, III 2002 Solutions
    OK, I know this is an old thread, but I would be forever grateful if someone could clarify a tiny bit of the solution for II/7:


    (Original post by Glutamic Acid)
    II/7:

    So 1/3 + \alpha(\lambda - 1/3) = 1 + \beta(\lambda - 1)
    2/3 + \alpha(\lambda - 2/3) = 0
    2/3 - \alpha 2/3 = \lambda \beta

    The second equation gives \alpha = \dfrac{2}{2 - 3 \lambda}. Substituting into the third gives \beta = \dfrac{2}{2 - 3 \lambda}. Substituting these into the first:
    1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1); multiplying out gives 2 \lambda = 0 \Rightarrow \lambda = 0, so no non-zero solutions.
    I have the same three equations. From the second equation, I get the same value of \alpha (\dfrac{2}{2 - 3 \lambda}). However, when I substitute it into the third equation, I get \beta = \dfrac{2}{3 \lambda - 2}, the negative of what Glutamic Acid got (which is the answer that works, putting \lambda = 0 into 1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1) gives a false equation). I'm probably just being an idiot, but I'm tearing my hair out over this, can someone put me out of my misery?
  16. ghostwalker's Avatar
    154 19-04-2011  16:40
    • Outcast of Imrryr
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Asboob)
    I have the same three equations. From the second equation, I get the same value of \alpha (\dfrac{2}{2 - 3 \lambda}). However, when I substitute it into the third equation, I get \beta = \dfrac{2}{3 \lambda - 2}, the negative of what Glutamic Acid got (which is the answer that works, putting \lambda = 0 into 1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1) gives a false equation). I'm probably just being an idiot, but I'm tearing my hair out over this, can someone put me out of my misery?
    I suspect it's just a typo on his part, and he meant to put \beta = \dfrac{-2}{2-3 \lambda } since he's used that value in the following line (having multiplied top and bottom by -1).
  17. brianeverit's Avatar
    155 21-06-2011  15:49
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    Re: STEP I, II, III 2002 Solutions
    2002 STEP I question 14

    \text{P(I get money at first attempt) is }\dfrac{1}{n}, \text{P(at second attempt) is }\dfrac{n-1}{n} \times \dfrac{1}{n-1} \times \text{e}^{-\lambda /r}
    \text{P(at third attempt} is }\dfrac{n-1}{n}\times\dfrac{n-2}{n-1}\times\dfrac{1}{n-2}\times \text{e}^{-2\lambda/r} \text{ etc}
    \text{hence, P(I get my money) is}\dfrac{1}{n}+\dfrac{1}{n} \displaystyle \sum_{k=1}^n \text{e}^{-k\lambda/r}=\dfrac{1}{n} \displaystyle \sum_{k=1}^n\text{e}^{-\lambda(k-1)/r}
    \text{so Probability that police arrive before I get my mone is }1-\dfrac{1}{n} \displaystyle \sum_{k=1}^n\text{e}^{-\lambda(k-1)r}=1-\dfrac{1}{n} \left( \dfrac{1-\text{e}^{-n\lambda/r}}{1-\text{e}^{-\lambda/r}}\right)
    \text{If I punch in numbers at random, the probability of getting the correct one or an incorrect one }
    \text {remain constant at } \dfrac{1}{n} \text{ and } \dfrac{n-1}{n} \text{ respectively, hence Prob(I get my money) is now }
    \dfrac{1}{n}+ \dfrac{n-1}{n}\times\dfrac{1}{n}\text{e}^ {-\lambda/r}+ \left(\dfrac{n-1}{n}\right)^2\times \dfrac{1}{n}\text{e}^{-2\lambda/r} + \dots
    = \dfrac{1}{n} \left(1+ \displaystyle \sum_{k=1}^{\infty} \left( \dfrac{n-1}{n} \right)^k \text{e}^{-k \lambda/r} \right)= \dfrac{1}{n} \left( \dfrac{1}{1- \frac{n-1}{n} \text{e}^{- \lambda/r}} \right)=1- \dfrac{1}{n-(n-1)\text{e}^{-\lambda/r}}
  18. brianeverit's Avatar
    156 21-06-2011  16:46
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    Re: STEP I, II, III 2002 Solutions
    2002 STEP II question 12

    \text{Consider a single coin thrown }M\text{ times, then number of heads }X\text{ say is a binomial variable}Bin(M,p)
    \text{We use the normal approximation }X \approx(1-p))
    \text{so }P(X<m)=\Phi\left(\dfrac{m-\frac{1}{2}-Mp}{\sqrt{Mp(1-p)}}\right)=\Phi\left(\dfrac{2m-1-2MP}{2\sqrt{Mp(1-p)}}\right)= \dfrac{h}{L} h\text{ defined as in question}
    \text{Now consider }L\text{ such trials. The number of heads }Y \text{ say will have a }Bin\left(L,\dfrac{h}{L}\right) \text{ distribution}
    \text{it is reasonable to assume that }\dfrac{h}{L} \text{ is small, so we use a Poisspon approximation }Po(h)
    \text{if }h \text{ is not too small we may approximate this with }N(h.h)
    \text{so we now have }P(Y>l)=1- \Phi \left( \dfrac{1-0.5-h}{\sqrt{h}}\right)=1-\Phi\left(\dfrac{2l-1-2h}{2\sqrth}\right)= \Phi \left( \dfrac{2h-2l-1}{2\sqrt{h}}\right)=q
    \text{If each such set of trials is performed }K\text{ times then the number of successful outcomes is }B(K,q)
    \text{hence, P(}k\text{ successful outcomes })=\displaystyle\binom{K}{k}q^k( 1-q)^K-k \text{ as required}
    \text{With }K=7,k=2,L=500,l=4,M=1)),m=48 \text{ and }p=0.6
    M\text{ is large and }p\text{ is close to }0.5 \text{ so first normal approximation is acceptable}
    h=500\Phi\left(\dfrac{96-1-120}{2\sqrt24}\right)=500\Phi \left(- \dfrac{25}{2\sqrt24}\right)=2.7
    \dfrac{h}{L}=0.0054 \text{ so Poisson approximation }Po(h) \text{ is acceptable but the normal approximation to this is not}
    \text{so I would not expect theresult to be very accurate}
  19. brianeverit's Avatar
    157 21-06-2011  19:29
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    Re: STEP I, II, III 2002 Solutions
    2002 STEP question 13

    \text{G}(y)=\dfrac{\text{F}(y)}{ 2-\text{F}(y)} \text{ so G}(a)=\dfrac{\text{F}(a)}{2-\text{F}(a)}=\dfrac{0}{2}=0 \text{ and G}(b)=\dfrac{\text{F}(b)}{}2-\text{F}(b)}=\dfrac{1}{1}=1
    \text{G}'(y)= \dfrac{(2-\text{F}(y))\text{F}'(y)-\text{F}(y)(-\text{F}'(y))}{(2-\text{F}(y))^2}= \dfrac{2}{(2-\text{F}(y))^2}>0
    0\leq \text{F}(y) \leq1 \Rightarrow 1\leq(2-\text{F}(y))^2 \leq4 \Rightarrow\dfrac{1}{2} \leq \dfrac{2}{(2-\text{F}(y))^2 }\leq2
    \text{E}(Y)=\displaystyle \int_0^1 y\text{G}'(y) dy=\displaystyle \int_0^1 \dfrac{2y\text{F}'(y)}{(2-\text{F}(y))^2}dy \Rightarrow \displaystyle \int_0^1 \dfrac{1}{2}y\text{F}'(y)dy \leq \text{E}(y) \leq \displaystyle \int_0^1 2y\text{F}'(y)dy \text{ from previous result}
    \text{This may also be written as }\displaystyle \int \dfrac{1}{2}x\terxt{F}'(x)dx \leq\text{E}(Y) \leq \dfisplaystyle \int_0^1 2x\text{F}'(x)dx \text{ i.e. }\dfrac{1}{2} \text{E}(X) \leq\text{E}(Y) \leq 2\text{E}(X)
    \text{E}(Y^2)= \displaystyle \int_0^1y^2\text{G}'(y)dy= \displaystyle \int_0^1 \dfrac{2y^2\text{F}'(y)}{(2-\text{F}(y))^2} \leq \displaystyle \int_0^1 2y^2\text{F}'(y)dy=2 \displaystyle \int_0^1 x^2 \text{F}'(x)dx=2 \text{E}(X^2)
    \text{i.e. E}(Y^2)=2\text{Var}(X)+2[\text{E}(X)]^2
    \text{hence, Var}(Y)=2\text{Var}(X)+2[\text{E}(X)]^2-[\text{E}(Y)]^2 \text { but [E}(Y)]^2 \geq \dfrac{1}{4}[\text{E}(X)]^2 \text{ by earlier result}
    \text{so Var}(Y) \leq 2\text{Var}(X)+2[\text{E}(X)]^2- \dfrac{1}{4}[\text{E}(X)]^2=2 \text{Var}(X)+ \dfrac{7}{4}[\text{E}(X)]^2 \text{ as required}
  20. brianeverit's Avatar
    158 23-06-2011  10:53
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    Re: STEP I, II, III 2002 Solutions
    2002 STEP II question 14

    \text{Area of region }R_n \text{ is } \pi \left[n^2-(n-1)^2 \right]= \pi(2n-1)
    \text{Probability that observer selects region }R_n \text{ is } k\pi(2n-1) \text{ where } k\pi N^2=1 \Rightarrow k=\dfrac{1}{\pi N^2}
    \text{so probability that he selects }R_n \text{ is } \dfrac{2n}{N^2}

    \text{overall average number of accidents is thus }\displaystyle \sum_{n=1}^N \left( \dfrac{2n-1}{N^2} \right) \left( 2- \dfrac {n}{N} \right)= \displaystyle \sum_{n=1}^N \left( \dfrac {2n-1}{N^2} \right) \left( \dfrac{2N-n}{N} \right)
    = \dfrac{1}{N^3} \displaystyle \sum_{n=1}^N (4nN-2N-2n^2+n)=\dfrac{4}{N^2} \displaystyle \sum_{n=1}^Nn-\dfrac{2N^2}{N^3}-\dfrac{2}{N^3} \displaystyle \sum_{n=1}^Nn^2+\dfrac{1}{N^3} \displaystyle \sum_{n=1}^Nn
    = \dfrac{4}{N^2} \dfrac{N(N+1)}{2}- \dfrac{2}{N}- \dfrac{2}{N^3} \dfrac{N(N+1)(2N+1)}{6}+ \dfrac{1}{N^3} \dfrac{N(N+1)}{2}= \dfrac{2(N+1)}{N} - \dfrac{2}{N} - \dfrac{(N+1)(2N+1)}{3N^2}+ \dfrac{N+1}{2N^2}
    =2- \dfrac{(N+1)(2N+1)}{3N^2}+ \dfrac{N+1}{2N^2}=2- \dfrac{4N^2+3N-1}{6N^2}=2- \dfrac{(N+1)(4N-1)}{6N^2}=2-\dfrac{1}{6} \left(1+\dfrac{1}{N} \right) \left(4-\dfrac{1}{N} \right)
    Pr (X=k)= \displaystyle \sum_{n=1}^N \dfrac{2n-1}{N^2} \times \dfrac{ \text{e}^{-2+n/N}(2-n/N)^k}{k!}= \dfrac{ \text{e}^{-2}N^{-k-2}}{k!} \displaystyle \sum_{n=1}^N(2n-1)(1N-n)^k \text{e}^{n/N}
    \text{with }N=3, Pr(X=2)= \dfrac{\text{e}^{-2}}{2.3^4} \displaystyle \sum_{n=1}^3(2n-1)(6-n)^2\text{e}^{n/3}= \dfrac{1}{162\text{e}^2}\left[25\text{e}^{1/3}+48\text{e}^{2/3}+45\text{e} \right]}
    Pr(R_2 \text{ selected and }X=2)= \dfrac{3}{N^2} \times \dfrac{\text{e}^{-4/3} (4/3)^2}{2!} c \text{ hence}
    Pr(R_2 \text{ selected }|X=2)= \dfrac{\frac{1}{3} \times \frac{16}{18}\text{e}^{-4/3}}{\frac{1}{162\text{e}^2}[25\text{e}^{1/3}+48\text{e}^{2/3}+45\text{e}]}= \dfrac{48}{48+45 \text{e}^{1/3}+25 \text{e}^{-1/3}}
  21. brianeverit's Avatar
    159 23-06-2011  12:02
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP I, II, III 2002 Solutions
    2002 STEP III question 9

    \text{The problem is equivalent to finding the centre of mass of the solid on the righrt of the diagram}
    \text{with respect to axes OA and OB where the volume is }k^3 \text{, i.e. }a^2b+ \dfrac{1}{2}a^3 \tan \tehta= ka^3 \Rightarrow b= \dfrac{a(2k-\tan \theta)}{2}
    \text{providing }2k> \tan \theta \text{ i.e. }k> \dfrac{1}{2} \tan \theta
    \text{splitting into a rectangular block and a triangular prism we have distance of c of m from OA given by }
    ka^3 \bar{x}=a^2b \times \dfrac{a}{2}+\ \dfrac{1}{2}a^3 \tan \thewta \times \left(b+\dfrac{a \tan \theta}{3} \right)= \dfrac{1}{2}a^3b+ \dfrac{1}{6}a^4 \tan \theta= \dfrac{1}{4}a^4(2k- \tan \theta)+ \dfrac{1}{6}a^4 \tan \theta
    \Rightarrow \dfrac{\bar{x}}{a}=\dfrac{1}{2}- \dfrac{ \tan \theta}{12k}\text{ and similarly distance from OB is given by}
    ka^3 \bar{y}=a^2b \times \dfrac{1}{2}b+ \dfrac{1}{2}a^3 \tan \theta \times \left(b+ \dfrac{a \tan \theta}{3} \right)= \dfrac{1}{2}a^2b^2+ \dfrac{1}{2}a^3b \tan \theta+ \dfrac{a^4 \tan^2 \theta}{6}
    \text { so } \drac{\bar{y}}{a}= \dfrac{k}{2}+ \dfrac{\tan \theta}{24k} \text{ as required}
    \text{If }k< \dfrac{1}{2} \tan \theta \text{ then the volume of fluid is a triangular prism, (see second diagram)}
    \text{Let base now be a rectangle }a \times b \text{ then } \dfrac{1}{2}ab^2 \tan \theta=ka^3 \text{ i.e. }b^2= \dfrac{2ka^2}{\tan \theta}
    \text{We now have } \dfrac{ \bar{x}}{a}= \dfrac{b}{3a}= \dfrac{\sqrt{2k}}{3\sqrt{ \tan \theta}} \text{ and } \dfrac{ \bar{y}}{a}= \dfrac{b \tan \theta}{3}a= \dfrac{\sqrt{2k\tan\theta}}{3}
    \text{In this latter case container will topple if } \dfrac{ \bar{x}}{y}< \tan \theta \Rightarrow \tan \theta > \dfrac{ \sqrt{2k}}{3 \sqrt{ \tan \theta}}\times \dfrac{3}{ \sqrt{2k\tan \theta}}= \dfrac{1}{ \tan \theta}
    \text{i.e. if }\tan^2 \theta<1 \Rightarrow \theta>45 ^\circ
    Attached Thumbnails
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  22. brianeverit's Avatar
    160 23-06-2011  14:04
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP I, II, III 2002 Solutions
    2002 STERP III question 10

    \text{By conservation of energy, if angular velocity is } \dot{\theta} \text{ and the second particle}
    \text{does not slip then we have, loss of PE }=mga\sin \theta-mga(1- \cos \theta)
    \text{Gain of KE is }ma^2 \dot{\theta}^2 \text{ hence } a \dot{\theta}^2=g(\sin \theta+ \cos \theta-1) \Rightarrow \dot{\theta}^2= \dfrac{g}{a}(\sin \theta+\cos \theta-1)
    \text{Differentiating w.r.t. time we have }2 \dot{\theta} \ddot{\theta}=\dfrac{g}{a}( \cos \theta- \sin \theta) \dot{\theta} \text{ i.e. } \ddot{\theta}= \dfrac{g}{2a} (\cos \theta- \sin \theta)
    \text{For second particle not to slip we must have }F-mg \sin \theta=ma \ddot{\theta}= \dfrac{mg(\cos \thetra- \sin \theta)}{2}
    \text{also }R-mg \cos \theta=m \dot{\theta}^2a \Rightarrow R=mg(\sin \theta+2 \cos \theta-1) \text{ and } F= \dfrac{mg(\cos \theta+ \sin\theta)}{2}
    \text{ so when } \theta=60^\circ,R=mg \dfrac{\sqrt3}{2} \text{ and }F=\dfrac{1}{2}mg \left( \dfrac{1+\sqrt3}{2} \right)
    \text{Particle is about to slip if }F> \mu R \Rightarrow \dfrac{1+\sqrt3}{2}>\mu \sqrt3 \Rightarrow \mu< \dfrac{1+\sqrt3}{2\sqrt3}= \dfrac{3+\sqrt3}{6}
    \text{so since this is the given value of }\mu, \text{ it follows that particle slips when cylinder has rotated through }60^\circ
    Attached Thumbnails
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