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STEP Maths I, II, III 1993 Solutions

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Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
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    Nor do I, to be honest. Not entirely sure what I was thinking. It does follow that you need the sign of f between the two turning points to be different if you are to have 3 roots (otherwise there must be another turning point between the 2 turning points). But I think you need to do a little case bashing to prove it the other way.
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    STEP II, Question 16

    Attached is my solution. It's a bit ugly, so improvements are welcome.

    -\delta\nu
    Attached Files
  1. File Type: pdf 1993_2_16.pdf (50.5 KB, 59 views)
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    where are the questions that need to be answered?
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    See the first page...but I dunno how accurate it is...
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    I/14:

    After n(g + r) means distance travelled < 1 so vn(g+r) < 1 so v <1/(n(g+r))
    Before n(g + r) + g means distance travelled > 1 so v[n(g + r) + g] > 1 so v > 1/[n(g+r)+g]

    So this probability is \displaystyle \int_{\beta}^{\gamma} \alpha v^{-2}\exp(- \alpha v^{-1}) \, \text{d}v = \left[ \exp(- \alpha v^{-1}) \right]^{\gamma}_{\beta}

    = \exp(-\alpha n(g + r)) - \exp(-\alpha(n(g+r)+g)), where \beta = (n(g+r) + g)^{-1} and \gamma = (n(g+r))^{-1}.

    P(alpha) is the sum of this from n = 0 to infinity: \displaystyle \sum_{n = 0}^{\infty} \exp(-\alpha n(g + r))(1 - \exp(-\alpha g))
    = \dfrac{1 - \exp(- \alpha g)}{1 - \exp(-\alpha(g + r))}.

    As alpha goes to infinity, exp(- alpha (stuff)) goes to zero so it becomes 1 / 1 = 1.

    As alpha goes to 0, 1 - exp(- \alpha) = 1 - (1 - alpha g + ...) = alpha g; 1 - exp(-alpha(g + r)) = 1 - (1 - \alpha (g + r) + ...) = alpha g / (alpha (g + r)) = g / (g + r).
    I'm late for college and eating my breakfast, so expect some errors. Hmm, this question was set in 1993. I wonder whatever happened to Mr. Toad. He's probably dead now.
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    (Original post by Rabite)
    I think I did Q2 in the Further Maths A.
    But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

    [edit] Here it is anyway.
    \int \cos{mx}\cos{nx} dx = ½ \int \cos (m+n)x + \cos (m-n)x dx
    By the product/sum formulae that no one remembers.
     = ½[\frac{1}{m+n} \sin(m+n)x + \frac{1}{m-n} \sin(m-n)x]_0 ^{2\pi}

    But if m=±n, one of the fractions explodes. So in that case the question is:

     \int \cos^2 {mx} dx

     =½ \int 1+ cos{2mx}dx

     = ½[x+\frac{1}{2m} \sin{2mx}]_0 ^{2\pi}

    =\pi

    If m=n=0, the integral turns to 2\pi.

    As for the second bit.
    Let x = sinh²t

    dx = 2sinhtcosht dt

    I = \int \sqrt{\frac{x+1}{x}}dx

     = \int \sqrt{\frac{\sinh^2 t +1}{\sinh^2 t}}(2\sinh t\cosh t)dt

     = \int 2 \frac{\cosh t}{\sinh t} (\sinh t \ cosh t) dt

     = \int 2 \cosh^2 t dt

     = \int 1+ \cosh 2t dt = t + ½ \sinh 2t Ignoring the +c for now

     = t + \sinh t \cosh t = \sinh ^{-1}(\sqrt{x}) + \sqrt(x(x+x)) +c
    Which you can rewrite using the log form of arsinh.
    I'm writing this even realising that the poster of this solution has not been active on TSR for well over a year.
    I'm very concerned about the substitution. The problem I see is that the function f(x) = \sqrt{1+1/x} is defined for x \leq -1 and x&gt;0. Clearly using the substitution "x = sinh²t" will not consider for the possible negative values of x, or we'd be going into complex numbers (I actually tried the substitution x = \tan^2u at first, and soon realised that f(x) is actually well defined for negative values of x, meaning that my solution is pretty much invalid.

    I hope that someone, who is more knowledgeable, such as DFranklin, will give some attention to this issue.
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    I'm not the greatest person at getting these integration problems out (actually, right now, I suck pretty badly at integration), but there's a reasonable argument that you don't need to worry too much about having sqrt(x) in your working as long as it disappears in the final answer. Which is the case, according to Mathematica.

    Silly question, but does the "whole hog" substitution u=\sqrt{1+1/x} work any better?
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    (Original post by DFranklin)
    I'm not the greatest person at getting these integration problems out (actually, right now, I suck pretty badly at integration), but there's a reasonable argument that you don't need to worry too much about having sqrt(x) in your working as long as it disappears in the final answer. Which is the case, according to Mathematica.

    Silly question, but does the "whole hog" substitution u=\sqrt{1+1/x} work any better?
    It gives out a really nice answer to me, actually!
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    (Original post by insparato)
    STEP III Question 6

    .......

    The next part is bugging me here,

    Determine the positions of P for which Q1 and Q2 coincide.

    I can see how the points Q1 and Q2 could coincide they are on the same locus. However the positions of p ? Am i suppose to find where Q1 and Q2 coincide and use one of the formulas given to find where it transforms onto the Z plane?

    Ive proved a formula i have not yet used, some how i think this has got to come in here.
    (Original post by DFranklin)
    You made very heavy weather of the first bit. |a+ib|^2=a^2+b^2 =(a+ib)(a+ib)^*. So

    |z-k| = r \implies (z-k)(z-k)^* = r^2 \\

\implies (z-k)(z^*-k^*) = r^2 \implies zz^*-zk^*-z^*k+kk^* = r^2

    2nd bit is fine.

    3rd bit: Putting k = i, r=1 we have zz^*-zk^*-z^*k+kk^* = r^2.

    Since w_2=z^*,w_2^* = z we get:w_2^*w_2-w_2^*k^*-w_2k+kk^* = r^2. Set k_2 = k^* and we get w_2^*w_2-w_2^*k_2-w_2k_2^*+k_2^*k_2 = r^2. So w_2 has equation of circle radius 1, center k_2 = k^* = -i.

    (Or your reflection in real axis argument is also fine. I just thought I'd show how you can use the first bit).

    I'm not seeing an easy solution to the last part. I'd probably parameterize z as \cos \theta + (1+\sin \theta)i, find a formula for w and solve for w=z^*. (Possibly just writing z=(x+iy) and taking a similar approach will be easier).
    I did this question just now, and the last part also scared the hell out of me. :lolwut:

    I think what the question is really asking is "for which specific instances of z do Q1 and Q2 coincide?".

    It seems that DFranklin's suggestion does probably give a good hint in the right direction, giving \cos(\theta+\frac{\pi}{4}) = \frac{1}{\sqrt{2}} for certain values of \theta, which imply the required positions of P.
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    we know w_2= -z/(z-i) and we also know w_1=z/(z-1)
    we just let w_1=w_2
    then we can find z=1+i (if my calculation is right lol)

    (Original post by insparato)
    STEP III Question 6

    A complex number z lying on a circle centre K and radius where K represents the complex number k.

     |z - k | = r^2

     |x + iy - (a+bi)| = r^2

     |x - a + i(y-b)| = r^2

     (x - a)^2 + (y - b)^2 = r^2
     x^2 - 2ax + a^2 + y^2 - 2yb + b^2 = r^2

    Consider zz*

     (x + iy)(x - iy) = x^2 + y^2

    kz*

     (a + bi)(x - iy) = ax -aiy + bix + by

    k*z

     (a - bi)(x + iy) = ax + aiy - bix + by

    kk*

    -k*z - kk* = ax + aiy - bix - by - ax - aiy + bix - by = -2ax - 2by

     (a - bi)(a + bi) = a^2 + b^2

     x^2 - 2ax + a^2 + y^2 - 2yb + b^2 = r^2

     (x^2 + y^2) -2ax - 2by  + (a^2 + b^2) = r^2

     zz^* - k^*z - kk^* + kk^* - r^2 = 0

    The locus of P is which represents the complex number z

    | z - i | = 1

     w_1 = \frac{z}{z-1}

     w_1z - w_1 = z

     z = \frac{w_1}{w_1 - 1}

     | z - i | = | \frac{w_1}{w_1 - 1} - i |

     1 =  \frac{|w_1 - i(w_1 - 1)|}{|w_1-1|}

     |w_1 - 1| = |w_1 - iw_1 + i |

    let w_1 = x + iy

     |x + iy - 1| = x + iy -i(x + iy) + i|

     \sqrt{(x-1)^2 + y^2} = |x + y + iy - ix + i |

     \sqrt{(x-1)^2 + y^2} = \sqrt{(x+y)^2 + (y - x + 1)^2}

     (x-1)^2 + y^2 = (x+y)^2 + (y - x + 1)^2

     x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2xy + y^2 - xy + y - xy + x^2 - x + y - x + 1

     x^2 - 2x + 1 + y^2 = 2x^2 + 2y^2 + 2y - 2x + 1

     x^2 + y^2 + 2y = 0

     x^2 + (y+1)^2 - 1 = 0

     x^2 + (y+1)^2 = 1

    The locus of W_1 is a circle with centre (0,-i) and radius 1 on the argand diagram.

    Thus the locus L is

    |w_1 + i | = 1

    Now this is where it gets tricky.

    w_2 = z*

    This formula i proved is screaming at me.

     zz* - k*z - kk* + kk* - r^2 = 0

    We know it has to be some circle and you can guess as its going to be of radius one if its the conjugate of z.

    Im tempted to just say if locus of z is |z - i | = 1

    therefore the locus of the conjugate of z is |z* - i | = 1

    this incidentally happens to come out as the locus of L.

    Just had another thought. What is the conjugate of a complex number?, as far as im aware its a reflection in the x axis of the original complex number on the argand diagram. So if the locus of Z is a circle, if you take all the specific points you could make its conjugates by reflecting in the x axis.

    So if the locus of Z is a circle centered at (0,i) and has a radius of 1 it touches(0,0) a reflection of this is simply a circle centered (0,-i) with radius 1.

    This happens to be x^2 + (y + 1)^2 = 1

    Which is the same locus as L.

    The next part is bugging me here,

    Determine the positions of P for which Q1 and Q2 coincide.

    I can see how the points Q1 and Q2 could coincide they are on the same locus. However the positions of p ? Am i suppose to find where Q1 and Q2 coincide and use one of the formulas given to find where it transforms onto the Z plane?

    Ive proved a formula i have not yet used, some how i think this has got to come in here.
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    (Original post by squidfuji)
    we know w_2= -z/(z-i) and we also know w_1=z/(z-1)
    we just let w_1=w_2
    then we can find z=1+i (if my calculation is right lol)
    I had a crack at this again, god it is so weird to have done that question 2 years ago nearly. Anyways without looking at what you told me, as i said 2 years ago, the equation i proved in the first part was screaming at me and indeed for good reason, why I did not click no idea because with it, the 3rd part came out fine.

    I got w_2 = -iz/(z - i) btw.

    Then the last part just fell out really. I might write it up later on, unless you want to do it ?
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    oh mistyped w_2! yeh w_2 is -iz/(z-i)!!


    (Original post by insparato)
    I had a crack at this again, god it is so weird to have done that question 2 years ago nearly. Anyways without looking at what you told me, as i said 2 years ago, the equation i proved in the first part was screaming at me and indeed for good reason, why I did not click no idea because with it, the 3rd part came out fine.

    I got w_2 = -iz/(z - i) btw.

    Then the last part just fell out really. I might write it up later on, unless you want to do it ?
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    STEP III, Question 5

    Spoiler:
    Show
    This question is essentially a discussion of Quaternions, and the Cayley-Dickson construction

    Under this interpretation we can simply state the answers, rather than plugging thought the algebra already available on the net.

    The elements are (1, -1, i, -i, j, -j, k, -k ) with orders (1,2,4,4,4,4,4,4) respectively

    The subgroup is isomorphic
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    FMA Question 15

    For LEP we add 1500 uniform distributions with  \mu = 0, \ \sigma^2 = \frac{1}{12} to give a normal distribution with  \mu = 0, \ \sigma^2 = \frac{1500}{12} = 125 , ie  E\sim N(0,125) .

     P(|E|&gt;15)=2P(E&gt;15)=2P(Z&gt;\frac{15  }{\sqrt{125}}) \approx 2P(Z&gt;1.34) \\ \approx 2(1-0.9099) = 0.1802 \approx 0.18

    For VOZ, add 1500 uniform distributions with  \mu = 0.5, \ \sigma^2 = \frac{1}{12} to give a normal distribution with  \mu = 750, \ \sigma^2 = 125 , ie  E\sim N(750,125) .

     P(|E|&gt;15)=P(E&gt;15)=P(Z&gt;\frac{15-750}{\sqrt{125}}) \approx P(Z&gt;-65.74) \approx 1

    If LEP makes n additions,  E\sim N(0,\frac{n}{12}) .

     P(|E|&lt;10)=P(E&lt;10)-P(E&lt;-10)=2P(E&lt;10)-1 \\ =2P(Z&lt;\frac{20\sqrt{3}}{\sqrt{n}  }) -1 = 0.9 \implies P(Z&lt;\frac{20\sqrt{3}}{\sqrt{n}})  =0.95 \implies \frac{20\sqrt{3}}{\sqrt{n}} = 1.6449 \implies n = 443

    Rounding down at the end for obvious reasons. Now, I think I missed a joke here: what do LEP and VOZ stand for?
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    (Original post by toasted-lion)
    Rounding down at the end for obvious reasons. Now, I think I missed a joke here: what do LEP and VOZ stand for?
    They are both simple Caesar encryptions of "IBM".

    I J K L M N O P Q R S T U V
    B C D E F G H I J K L M N O
    M N O P Q R S T U V W X Y Z


    So I'm guessing that's the joke. (There's a long running rumour that the name of the computer HAL from "2001: A Space Odyssey" was a reference to IBM; the names here are simply taking the joke further).
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    (Original post by deltinu)
    STEP II, Question 16

    Attached is my solution. It's a bit ugly, so improvements are welcome.

    -\delta\nu
    You went a lot further than I did, here's mine:

    a)  \left(\frac{3}{4}\right)^{n-1}\left(\frac{1}{4}\right)

    b)  \frac{e^{-20}20^m}{m!}

    c) We require  \displaystyle\sum_{n\geq1} \left(\frac{3}{4}\right)^{n-1}\left(\frac{1}{4}\right)P\left  (X&gt;25 | X \sim Po(4n)\right)

    For 1\leq n\leq3, P(X&gt;25) \approx 0 so neglect these terms.

    For 4\leq n\leq13, use a normal approximation to the Poisson distribution and calculate each term manually.

    For 14\leq n, P(X&gt;25) \approx 1 so sum the rest of the terms as a geometric progression with  a = \frac{3^{13}}{4^{14}} and  r = \frac{3}{4} .
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    (Original post by DFranklin)
    They are both simple Caesar encryptions of "IBM".

    I J K L M N O P Q R S T U V
    B C D E F G H I J K L M N O
    M N O P Q R S T U V W X Y Z


    So I'm guessing that's the joke. (There's a long running rumour that the name of the computer HAL from "2001: A Space Odyssey" was a reference to IBM; the names here are simply taking the joke further).
    Ah, I wouldn't have guessed that. Thanks!

    I do love the STEP probability questions.
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    At this level, I presume we can use  \displaystyle\sum_{r\geq 0} x^r = \frac{1}{1-x} without proof? What about  \displaystyle\sum_{r\geq 0} rx^r = \frac{x}{(1-x)^2} ?
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    I'd say the 1st is quotable, the 2nd isn't. Although you'd probably get away with quoting the 2nd as well (maybe lose 1 mark).
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    (Original post by justinsh)
    =9\int_1^2\frac{dv}{3v^2}\\
    =3(\frac{-1}{2}+1)=\frac{3}{2}\\
    I think there is a mistake in that line, the integral becomes 9 \int_1^2\frac{1}{3v}dv and the answer is thus 3\ln2?


    edit: III/Q2 part ii)

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