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STEP Maths I, II, III 1988 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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    (Original post by maltodextrin)
    I'm probably wrong but to show that the integral lies between 0 and 1 would it not be sufficient to say that

    (x^2)(e^-x)/(x + 1) < (x^2)(e^-x)/x = x(e^-x) for 0 < x < infinity

    then integrate x(e^-x), which equals 1. Seems a lot simpler?
    Another way to answer the second part is as follows
    x^2-x-1=(x- \frac{1}{2})^2- \frac{5}{4}&lt;0 \text{ for all positive }x \Rightarrow \dfrac{x^2}{1+x}&lt;1 \Rightarrow \dfrac{x^2 \text{e}^{-x}}{1+x}&lt; \text{e}^{-x}
     \text{so }\displaystyle \int_0^\infty \dfrac{x^2 \text{e}^{-x}}{1+x}dx&lt; \displaystyle \int_0^\infty \text{e}^{-x}dx=[-\text{e}^{-x}]_o^\infty=1
     \text{and obviously } \displaystyle \int_o^\infty \dfrac{x^2 \text{e}^{-x}}{1+x}dx&gt;0 \text{ since curve is above the }x \text{ axis}
     \text{hence }0&lt;  \displaystyle \int_o^\infty \dfrac{x^2 \text{e}^{-x}}{1+x}dx&lt;1
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    (Original post by SimonM)
    ....
    STEP III Q4
    Using the formula for arc length:
    Total distance=2(R-a)+2\displaystyle\int^a_b(\sqrt{  1+(\frac{dz}{dx})^2})dx+b\pi.
    So, if
    z=\frac{1}{2a}(a^2-x^2) \Rightarrow \frac{dz}{dx}=-\frac{x}{a}
    so, total distance,
    T=2(R-a)+2\displaystyle\int^a_b\sqrt{1  +\frac{x^2}{a^2}}dx+b\pi.
    To find the minimum we must differentiate but, before we do that, let us turn
    \displaystyle\int^a_b\sqrt{1+ \frac{x^2}{a^2}}dx into
    F(a)-F(b) where F is the antiderivative operator acting on \sqrt{1+\frac{x^2}{a^2}}.
    So, differentiating w.r.t. b:
    \dot{T}=2\frac{d}{db}[-F(b)]+b\pi
    The derivative and antiderivative cancel giving:
    \dot{T}=-2\sqrt{1+\frac{b^2}{a^2}}+\pi.
    To find stationary point, let the derivative be 0:
    \dot{T}=0 \Rightarrow \sqrt{1+\frac{b^2}{a^2}}=\frac{ \pi}{2} \Rightarrow b=a \sqrt{(\frac{\pi}{2})^2-1}
    (Notice this is the only stationary point in the interval (0,a) which is the range we care about).
    Now, to see whether this really is minimum, we must find the 2nd derivative:
    \dot{T}=2\sqrt{1+\frac{b^2}{a^2}  }+\pi \Rightarrow \ddot{T}=2 \times \frac{1}{2}\sqrt{1+\frac{b^2}{a^  2}}(\frac{2b}{a^2}).
    Since \ddot{T}&gt;0, \forall b&gt;0 \Rightarrow b=a \sqrt{(\frac{\pi}{2})^2-1} gives the minimum the length of road.
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    (Original post by ben-smith)
    STEP III Q4
    Using the formula for arc length:
    Total distance=2(R-a)+2\displaystyle\int^a_b(\sqrt{  1+(\frac{dz}{dx})^2})dx+b\pi.
    So, if
    z=\frac{1}{2a}(a^2-x^2) \Rightarrow \frac{dz}{dx}=-\frac{x}{a}
    so, total distance,
    T=2(R-a)+2\displaystyle\int^a_b\sqrt{1  +\frac{x^2}{a^2}}dx+b\pi.
    To find the minimum we must differentiate but, before we do that, let us turn
    \displaystyle\int^a_b\sqrt{1+ \frac{x^2}{a^2}}dx into
    F(a)-F(b) where F is the antiderivative operator acting on \sqrt{1+\frac{x^2}{a^2}}.
    So, differentiating w.r.t. b:
    \dot{T}=2\frac{d}{db}[-F(b)]+b\pi
    The derivative and antiderivative cancel giving:
    \dot{T}=-2\sqrt{1+\frac{b^2}{a^2}}+\pi.
    To find stationary point, let the derivative be 0:
    \dot{T}=0 \Rightarrow \sqrt{1+\frac{b^2}{a^2}}=\frac{ \pi}{2} \Rightarrow b=a \sqrt{(\frac{\pi}{2})^2-1}
    (Notice this is the only stationary point in the interval (0,a) which is the range we care about).
    Now, to see whether this really is minimum, we must find the 2nd derivative:
    \dot{T}=2\sqrt{1+\frac{b^2}{a^2}  }+\pi \Rightarrow \ddot{T}=2 \times \frac{1}{2}\sqrt{1+\frac{b^2}{a^  2}}(\frac{2b}{a^2}).
    Since \ddot{T}&gt;0, \forall b&gt;0 \Rightarrow b=a \sqrt{(\frac{\pi}{2})^2-1} gives the minimum the length of road.
    I am afraid I disagree with this result...see attached file...
    Attached Files
  1. File Type: pdf Step1988Paper3Question4.pdf (92.0 KB, 72 views)
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    (Original post by brianeverit)
    1988 STEP III number 3.
    I'm afraid I disagree with the solutions published so far. Here is my solution.

     |z-i|=2 \Rightarrow |x+(y-1)i|=2 \text{ where }z=x+iy
     \Rightarrow x^2+(y-1)^2=4 \text{ so parametric equations are }x=2\cos\theta, y=2\sin\theta+1
     (i)  w=\dfrac{z+i}{z-i} \Rightarrow z=\left( \dfrac{w+1}{w-1} \right)i \text{ and }|z-i|=2 \Rightarrow \left|\dfrac{w+1}{w-1}i-i\right|=2 \Rightarrow left| \dfrac{2i}{w-1} \right|=2
     \Rightarrow |2i|=2|w-1| \text{ i.e. }|w-1|=1 \text{ so it is a circle, centre }w=1 \text{ radius }1
     (ii) z \text{ real } \Rightarrow \text {Im}\left( \dfrac{(u+1)i-v}{(u-1)i-v} \right)= \text{ Im}\left( \dfrac{((u+1)i-v)((u-1)i-v)}{(u-1)^2+v^2} \right)= -\dfrac{2uv}{(u-1)^2+v^2}=0
     \text{hence, }2uv=0 \text{  i.e.  It is the two axes}
     (iii) z\text{ imaginary }\Rightarrow \text{Re} \left( \dfrac{w+1}{w-=1} \right)i=0 \text{ i.e. } \dfrac {v^2-(u^2-1)}{(u-1)^2+v^2}=0 \Rightarrow v^2-u^2+1=0 \text{  i.e. A hyperbola}
    I think that the three parts to this question are entirely separate and do not include the condition of being on the original circle...that is where we disagree. As regards the other question (4) can you see where we differ on the diffentiation...that is where the theorem comes in....
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    (Original post by mikelbird)
    I think that the three parts to this question are entirely separate and do not include the condition of being on the original circle...that is where we disagree. As regards the other question (4) can you see where we differ on the diffentiation...that is where the theorem comes in....
    When you say Question 4, are you talking about the one I have posted an incorrect solution to?
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    yup!!
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    (Original post by Glutamic Acid)
    I/7.
    

f'(x) = 2ax + b \\

f(1) = a + b + c \\

f(-1) = a - b + c \\

f(0) = c
    Therefore
    f(1)(x+1/2) + f(-1)(x - 1/2) - 2f(0)x = (a+b+c)(x + 1/2) + (a-b+c)(x - 1/2) - 2cx
     = ax + bx + cx + 1/2a + 1/2b + 1/2c + ax - bx + cx - 1/2a + 1/2b - 1/2c - 2cx = 2ax + b

    If |f(x)| <= 1, then -1 <= f(x) <= 1.
    If |f'(x)| <= 4, then -4 <= f(x) <= 4.

    f'(x) = f(1)(x + 1/2) + f(-1)(x - 1/2) - 2f(0)x.

    The largest values of f'(x) will be when x = -1 or 1 to maximize the values of (x + 1/2) and -(x-1/2). The largest value of f(1)(x + 1/2) can be 3/2 when f(1) = 1. The largest value of f(-1)(x - 1/2) can be 1/2 when f(-1) = 1. The largest value of -2f(0)x can be 2 when f(0) = -1. 3/2 + 1/2 + 2 = 4, so |f'(x)| <= is satisfied. When x = -1, the largest value can be (-1)(-1/2) + (-1)(-3/2) - (2)(1)(-1) = 4.

    The smallest values of f'(x) will be when x = -1 or 1 likewise. When x = 1, be (-1)(3/2) + (-1)(1/2) - 2(1) with f(0) = 1, f(1) = -1 and f(-1) = -1. This gives -4. When x = -1, it will be (1)(-1/2) + (1)(-3/2) - 2(1) = -4, with f(1) = 1, f(-1) = 1 and f(0) = -1, giving -4. Therefore |f'(x) <= 4 is still satisfied.

    Letting f(0) = -1, this gives c = -1. Letting f(1) = 1, this gives a + b - 1 = 1 hence a + b = 2. Letting f(-1) = 1, this gives a - b = 2. Hence 2a = 4, a = 2 and b = 0.
    f(x) = 2x^2 - 1.

    Sorry for any errors.
    Apologies for digging up ancient thread, but just wanted to check.

    STEP I Q7:

    Surely, the conditions in the question also allow f(x)= -2x^2 +1 for the last part? This should also be true simply by symmetry of the modulus signs and the fact that either f'(1)=4 or f'(-1)=4.

    I could of course be wrong, but someone please clarify.
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    (Original post by SimonM)
    (Updated as far as #68) SimonM - 15.06.2009
    ...
    STEP III Q12
    Let  \mathbf{u_1}/u_1, \mathbf{v_1}/v_1, \mathbf{v_2}/v_2 be the velocities/speeds of ball 1 and 2 with u being before the collision and v after.
    Momentum is conserved in a closed system so:
    \mathbf{u_1}=\mathbf{v_1}+\mathb  f{v_2}
    The coefficient of restitution is < 1 so kinetic energy is not conserved amongst the two balls, i.e:
    \mathbf{u_1}.\mathbf{u_1}&gt; \mathbf{v_1}.\mathbf{v_1}+ \mathbf{v_2}.\mathbf{v_2}
    Combining the two:
    (\mathbf{v_1}+\mathbf{v_2})^2 &gt; \mathbf{v_1}^2+\mathbf{v_2}^2 \Rightarrow 2\mathbf{v_1}.\mathbf{v_2}&gt;0

\Rightarrow 2v_2v_1cos\phi&gt;0 \Rightarrow \phi &lt; \frac{\pi}{2}
    Where \phi is the angle between the two velocities as desired.
    For the next part, let \theta be the angle between \mathbf{u_1}\\and\\\mathbf{v_2}.Separating momentum into two components and and considering restitution in the direction of  \mathbf{v_2} we obtain 3 equations, sufficient to eliminate the three speeds and obtain an equation in terms of our two angles, thus confirming the intuitive idea that the initial velocity of the first ball is irrelevant:
    u_1cos\theta=v_2+v_1sin\phi

u_1sin\theta=v_1sin\phi

eu_1cos\theta=v_2-v_1cos\theta
    By adding the first and last and substituting, we obtain:
    (1+e)u_1cos\theta=2v_2

\Rightarrow (1+e)cot\theta sin\phi v_1=2(eu_1cos\theta+v_1cos\phi)

\Rightarrow (1+e)cot\theta sin\phi =2ecot\thetasin\phi+2cos\phi

\Rightarrow (1-e)cot\theta=2cot\phi

\Rightarrow tan\theta=\frac{1-e}{2}tan\phi

    The angle of deflection is  \phi-\theta which we shall call k. Consider  tank=\frac{tan\phi-tan\theta}{1+tan\phitan\theta}=t  an\theta \frac{(1+e)/(1-e)}{1+(2/(1-e))Tan^2\theta}
    Notice that the maximum of Tank will occur for the same value of theta as the maximum of k so, differentiating both sides w.r.t. theta we get:

    D[tan(k)]=\frac{(1+(2/(1-e))Tan^2\theta)((1+e)sec^2\theta  )-((1+e)tan\theta)(2/(1-e)2tan\theta sec^\theta)}{(1+(2/(1-e))Tan^2\theta)^2}
    Which, if we equate it to zero, gives us:
    (1+\frac{2}{1-e}Tan^2\theta)(1+e)sec^2\theta=(  1+e)tan\theta \frac{2}{1-e}2tan\theta sec^2\theta

\Rightarrow 1+\frac{2}{1-e}Tan^2\theta=\frac{4}{1-e}Tan^2\theta

\Rightarrow tan^2\theta=\frac{1-e}{2}
    So:
    Tan(max[k])=max[Tan(k)]=\sqrt{\frac{1-e}{2}}\frac{(1+e)/(1-e)}{2}=\frac{sin(k_{max})}{\sqrt  {1-sin^2(k_{max})}}



\Rightarrow \frac{(1-sin^2(k_{max}))}{2}=\frac{4sin^2  (k_{max})(1-e)}{(1+e)^2}

\Rightarrow (\frac{4(1-e)}{(1+e)^2}+\frac{1}{2})sin^2(k  _{max})=\frac{1}{2}

\Rightarrow sin^2(k_{max})=(\frac{2(1+e)^2}{  8(1-e)+(1+e)^2})\frac{1}{2}= \frac{(1+e)^2}{(3-e)^2} 

\Rightarrow k_{max}=\sin^{-1}(\frac{1+e}{3-e})
    as required.

    Comment: I found this Q particularly difficult and it has taken me many attempts to get it right. Comments are welcome.
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    (Original post by SimonM)
    (Updated as far as #68) SimonM - 15.06.2009
    ...
    STEP III Q14
    i) If no pressure is exerted on the wire by the bead and (by N3) vice versa, then the wire might as well not be there i.e. the bead is projected under gravity alone and follows the path of the wire. The general equation of motion of a particle freely projected under gravity can be derived by considering the suvat equations in the x and y directions so, horizontally (note that the derivative of the equation of the wire at x=0 is one which means the angle of projection is pi/4:
    x=Vcos(\frac{\pi}{4})t \Rightarrow t=\frac{x}{Vcos\frac{\pi}{4}}
    Vertically:
     y=Vsin(\frac{\pi}{4})t-\frac{g}{2}t^2
    Substituting to eliminate t:
    y=Vsin(\frac{\pi}{4})\frac{x}{Vc  os\frac{\pi}{4}}-\frac{g}{2}(\frac{x}{Vcos\frac{\  pi}{4}})^2=x-\frac{g}{V^2}x^2
    We want this to be the same as the equation of the wire so:
    \frac{g}{V^2}=(4a)^{-1} \Rightarrow V=2\sqrt{ga}

    ii) If  \theta is the angle that the velocity makes with the x axis then:
     Tan\theta=\frac{dy}{dx}= 1-\frac{1}{2a}x \Rightarrow x=2a(1-Tan\theta)

\therefore y=2a(1-Tan\theta)-a(1-Tan\theta)^2=a(1-Tan^2(\theta))

\therefore G.P.E=mga(1-Tan^2(\theta))
    This also allows us to calculate the bead's velocity in terms of \theta. Letting it's position be r:
    \mathbf{r}=(2a(1-Tan\theta))\mathbf{i}+(a(1-Tan^2\theta))\mathbf{j}

\Rightarrow \dot{\mathbf{r}}=(-2a\dot{\theta}sec^2(\theta)) \mathbf{i}+(-2a \dot{\theta}Tan(\theta)sec^2( \theta))\mathbf{j}



\Rightarrow |\mathbf{\dot{r}}|=\sqrt{(-2a\dot{\theta}sec^2(\theta))^2+(-2a \dot{\theta}Tan(\theta)sec^2( \theta))^2}=\sqrt{\frac{4a^2\dot  {\theta}}{cos^6(\theta)}}
    and so, by conservation of energy:
    Initial K.E.= K.E. + G.P.E
    \Rightarrow \frac{1}{2}mV^2=\frac{1}{2}m| \mathbf{\dot{r}}|^2+mga(1-Tan^2(\theta))



\Rightarrow V^2=\frac{4a^2\dot{\theta}}{cos^  6(\theta)}+2ga(1-Tan^2(\theta)) as required.
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    (Original post by SimonM)
    Question 8, STEP III
    ...
    STEP III Q16
    This distribution is hypergeometric and, whilst I quoted it seeming as it is not an uncommon distribution in STEP questions, it's PMF can be derived like this:
    By symmetry, all configurations are equally likely so the probability that we pick a certain number of reds is the number of ways we can do this divided by the total number of configurations. Let this event be I:
    P(I)=\frac{\displaystyle \binom{m}{k}\displaystyle \binom{M-m}{n-k}}{\displaystyle \binom{M}{n}}

    i)Forgetting, for the moment, that there are colours, the probability that any particular ball is chosen at the ith selection is, by symmetry, the same i.e \frac{1}{M} and so the probability that a red ball is drawn at the ith selection is the sum of the probabilities that each particular red ball is drawn on the ith go:
    P(X_i=1)= \Sigma(\frac{1}{M})=\frac{m}{M} as required.

    ii) In a similar way, we can see that the probability of one particular ball being chosen on one particular selection and similarly for another ball is \frac{1}{M(M-1)} but, if we take into account colour, we must times this by 2 as we need to consider the situation in which these two particular balls are swapped as we need not distinguish between any two red balls. Now that we have the probability of one particular pair of i and j values, we need to multiply this by the total number of pairs of reds:
    P(X_i=1 \bigcap X_j=1)= 2\frac{1}{M(M-1)}\displaystyle \binom{m}{2}= 2\frac{1}{M(M-1)}\frac{m!}{(m-2)!2}=\frac{m(m-1)}{M(M-1)} as required.

    To find the mean:
    E(x)=E(\Sigma_{i=1}^nX_i)=\Sigma  _{i=1}^n(E(X_i))=\Sigma_{i=1}^n(  m/M)=n\frac{m}{M}

    To find the variance, we must first find E(X^2):
    E(X^2)=E((\Sigma_{i=1}^nX_i)( \Sigma_{j=1}^nX_j))=E(\Sigma_{i=  1}^n(\Sigma_{i=1}^nX_iX_j))

=\Sigma_{i=1}^n(\Sigma_{i=1}^nE(  X_iX_j))
    Now we can use part ii to evaluate the sum except we must be careful to take into account that it is only valid for  i \not= j. When i=j, this result becomes \frac{m}{M} because the other automatically happens if one of the two happens. This second case happens n times out of the total of n^2 times so:
    E(X^2)=(n^2-n)\frac{m(m-1)}{M(M-1)}+n\frac{m}{M}

\therefore Var(X)=(n^2-n)\frac{m(m-1)}{M(M-1)}+n\frac{m}{M}-(\frac{nm}{M})^2
    Simplifying:
    Var(X)=n\frac{m}{M}(\frac{(n-1)(m-1)}{M-1}+1-\frac{nm}{M}=n\frac{m}{M}(\frac{  (n-1)(m-1)M+M(M-1)-nM(M-1)}{M(M-1)})=n\frac{m}{M}(\frac{M^2-nM-mM+nm}{M(M-1)})=n\frac{m}{M}\frac{(M-n)(M-m)}{M(M-1)}
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    (Original post by SimonM)
    ....
    STEP III Q13
    Throughout, if I don't put a vector in bold it is because I am referring to it's modulus. Also, when I refer to coordinates I am doing so relative to axes with origin the goal line and in the plane of the balls motion. +ve x is away from the goal and +ve y is away from the ground.
    Let's find the resultant force on the ball. The force due to gravity will be m\mathbf{g} and the force due to the wind is mk times the velocity of the ball relative to the wind velocity i.e.  \mathbf{F_{wind}}=-mk(\mathbf{v}-\mathbf{w}) \Rightarrow \mathbf{F}=m\mathbf{g}-mk(\mathbf{v}-\mathbf{w})\Rightarrow \frac{d\mathbf{v}}{dt}=\mathbf{g  }-k(\mathbf{v}-\mathbf{w}) \Rightarrow \frac{d\mathbf{v}}{dt}+\mathbf{v  }=\mathbf{g}+k\mathbf{w} as required.
    Now, this equation can separated into it's horizontal and vertical components as \mathbf{w}&\mathbf{g} are perpendicular. Horizontally:
    \frac{dv_x}{dt}+kv_x=-kw \Rightarrow \frac{d}{dt}[e^{kt}v_x]=-kwe^{kt}

\Rightarrow e^{kt}v_x=-we^{kt}+C
    Now we can consider the initial conditions r(0)=(0,0) and v(0)=(v_0cos\alpha,v_0sin\alpha):
    v_x=-w+(w+v_0cos\alpha)e^{-kt} \Rightarrow x=-wt-(\frac{w+v_0cos\alpha}{k})e^{-kt}+C
    Using the initial conditions in the same way:
    x=-wt-(\frac{w+v_0cos\alpha}{k})e^{-kt}+\frac{w+v_0cos\alpha}{k}=-wt+(\frac{w+v_0cos\alpha}{k})(1-e^{-kt}). Now for the vertical component:
    \frac{dv_y}{dt}+kv_y=-g \Rightarrow \frac{d}{dt}[v_ye^{kt}]=-ge^{kt} \Rightarrow v_ye^{kt}=\frac{-ge^{kt}}{k}+C

\Rightarrow v_y=-\frac{g}{k}+(\frac{g}{k}+v_0sin \alpha)e^{-kt}



\Rightarrow y=-\frac{g}{k}t-(\frac{g}{k^2}+\frac{v_0sin \alpha}{k})e^{-kt}+\frac{g}{k^2}+\frac{v_0sin\a  lpha}{k}



=(-gt)/(k)+ [({g}/{k^2}+(v_0sin \alpha)/(k)](1-e^{-kt})

    For the next part, note that if the ball returns to the origin for a second time, an own goal will be scored. So, consider the case x=y=0:
    if y=0 then  1-e^{-kt}=\frac{g}{k} (\frac{g}{k^2}+\frac{v_0sin \alpha}{k})^{-1}t \Rightarrow -wt+ (\frac{w+v_0cos\alpha}{k}) \frac{g}{k} (\frac{g}{k^2}+\frac{v_0sin \alpha}{k})^{-1}t=0.
    We already know about the t=0 solution,we are interested in the second root:
    w=(\frac{w+v_0cos\alpha}{k}) \frac{g}{k} (\frac{g}{k^2}+\frac{v_0sin \alpha}{k})^{-1}= \frac{(w+v_0cos \alpha)g}{g+kv_0 sin\alpha}



\Rightarrow gw+wkv_0sin \alpha=gw+ v_0gcos \alpha \Rightarrow Tan\alpha =\frac{g}{kw}
    As required.
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    (Original post by SimonM)
    ....
    Also, don't forget my solutions to questions 14 and 16
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    Here is my answer for Question 11, which curiously doesn't seem to have been done yet (?) for Paper 3 (i.e. 1988 Paper FM B Q11).

    Just for novelty I've posted my answer as a youtube video, although warning it is 40+ minutes long and the tone will be a bit too didactic (i.e. boring) for most of the kind of people viewing a thread like this (It's primarily intended for students at my school).



    Edit: For anyone just double checking their answer, I got (i)  \mu = \frac{\sqrt{3}}{3} and (ii)  \mu = \frac{3\sqrt{3}}{5}
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    (Original post by waxwing)
    Here is my answer for Question 11, which curiously doesn't seem to have been done yet (?) for Paper 3 (i.e. 1988 Paper FM B Q11).

    Just for novelty I've posted my answer as a youtube video, although warning it is 40+ minutes long and the tone will be a bit too didactic (i.e. boring) for most of the kind of people viewing a thread like this (It's primarily intended for students at my school).



    Edit: For anyone just double checking their answer, I got (i)  \mu = \frac{\sqrt{3}}{3} and (ii)  \mu = \frac{3\sqrt{3}}{5}
    I get the same as you....see file...
    Attached Files
  2. File Type: pdf Step1988Paper3Question11.pdf (70.1 KB, 44 views)
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    (Original post by SimonM)
    STEP II, Question 2

    Spoiler:
    Show
    2a-3y = \dfrac{(z-x)^2}{y} and 2a-3z = \dfrac{(x-y)^2}{z}

    The first two equations give us

    2ay - 3y^2 = (z-x)^2 and
    2az - 3z^2 = (x-y)^2

    Subtracting yields

    2a(y-z)-3(y^2-z^2)= (z-x)^2 - (x-y)^2

    Which is

    (y-z)(2a - 3(y+z)) = (z-y)(z+y-2x)

    Since y \not = z

    2a - 3y - 3z = -z-y + 2x \Rightarrow \boxed{x+y+z = a}

    2a - 3y- 3x = 2z-x-y

    So 2a(y-x) -3(y-x)(y+x) = (y-z)(2z-x-y)

    So 2a(y-x)-3(y^2-x^2) = (z-x)^2 - (y-z)^2

    Subtracting from 2ay - 3y^2 = (z-x)^2 gives

    2ax - 3x^2 = (y-z)^2 \Rightarrow \boxed{2a - 3x = \frac{(y-z)^2}{x}}

    No, take x = \frac{2}{3} a, y = z = \frac{1}{6} a we can easily verify this satisfies all equations.
    x = y = z = \frac{2}{3} a is also a valid solution
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    (Original post by kabbers)
    III/1:

    Sketch y = \frac{x^2 e^{-x}}{x+1}

    I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

    Differentiating (quotient rule is your friend), we get

    \frac{dy}{dx} = e^{-x}\frac{x(2-x^2)}{(x+1)^2}

    So we have turning points at x = 0, and x = +\sqrt{2}, -\sqrt{2}


    Differentiating again, we get

    \frac{d^2 y}{dx^2} = e^{-x}\frac{(2-3x^2)(x+1)^2 - 2(x+1)(2x-x^3) - (2x - x^3)(x+1)^2}{(x+1)^4}

    Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.


    Noticing the denominator of y, we find that there is an asymptote at x = -1.

    And because of the exponential properties of e^-x, y tends to zero as x \to +\infty

    y tends to -\infty as x \to -\infty since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.


    So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14



    Prove \displaystyle 0 &lt; \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx &lt; 1

    First note that \frac{x^2}{x+1} = \frac{x^2+x}{x+1} - \frac{x}{x+1} = x - \frac{x}{x+1} = x - (\frac{x+1}{x+1} - \frac{1}{x+1}) = x - 1 + \frac{1}{x+1}

    Hence we may split the integral into \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

    Consider \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx

    = \displaystyle\int^{\infty}_0 xe^{-x} dx - \int^{\infty}_0 e^{-x} dx

    = \displaystyle([-xe^{-x}]^{\infty}_0 + \int^{\infty}_0 e^{-x} dx - \int^{\infty}_0 e^{-x} dx

    = 0


    Now consider \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

    I posit the inequality \frac{1}{x+1}e^{-x} &lt; e^{-x} for x > 0

    \frac{1}{x+1} &lt; 1
    1 &lt; (x+1)
    0 &lt; x

    So our inequality holds.

    So:

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx &lt; \int^{\infty}_0 e^{-x} dx

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx &lt; [-e^{-x}]^{\infty}_0

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx &lt; 1

    Thus \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx &lt; 1

    And therefore, \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx &lt; 1


    Now notice that the graph of y = \frac{x^2 e^{-x}}{1+x} is greater than 0 for x > 0, and hence so is the infinite integral.

    So, \displaystyle 0 &lt; \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx &lt; 1

    please point out any mistakes

    Is this correct?

    \displaystyle 0 &lt; \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx &lt; \int^{\infty}_0 \frac{x^2 e^{-x}}{x} dx = \int^{\infty}_0 x e^{-x}dx = 1
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    (Original post by klausw)
    x = y = z = \frac{2}{3} a is also a valid solution
    I don't think it is... \frac{2}{3} a + \frac{2}{3} a + \frac{2}{3} a \neq a
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    (Original post by kabbers)
    relatively straightforward q, pointing out any mistakes will be appreciated as always

    Ok first note that at the end of day n, the firm will owe everything it owed the previous day, the interest on this plus everything it has borrowed that day.

    This may be expressed as \displaystyle S_{n+1} = (1+k)S_n + c

    By evaluating the first few terms of this sequence, it becomes easy to 'guess' and then prove a total value for Sn:

    \displaystyle S_0 = 0

    \displaystyle S_1 = c

    \displaystyle S_2 = (1+k)c + c = c((1+k) + 1)

    \displaystyle S_3 = (1+k)((1+k)c + c) + c = c((1+k)^2 + (1+k) + 1)


    ok my guess is that \displaystyle S_n = c \sum^{n-1}_{r=0} (1+k)^r = c \frac{(1+k)^n - 1}{(1+k) - 1} = c \frac{(1+k)^n - 1}{k} (by geometric series)

    this fits the questions, but we have to prove it (induction!)


    Basis case: \displaystyle S_0 =  c \frac{(1+k)^0 - 1}{k} = 0, therefore the basis case works.

    Inductive step: \displaystyle S_{n+1} = (1+k)S_n + c = (1+k)c \frac{(1+k)^n - 1}{k} + c = c((1+k)\frac{(1+k)^n - 1}{k} + 1)

    \displaystyle  = c(\frac{(1+k)^{n+1} - (k+1)}{k} + 1) = c(\frac{(1+k)^{n+1} - (k+1) + k}{k}) = c\frac{(1+k)^{n+1} - 1}{k})

    This is in the same form as the above, so it works.


    for part 2, firstly we can assert that \displaystyle S_T = c\frac{(1+k)^{T} - 1}{k}) (this is the initial amount owed)

    the firm must pay interest on what it owes then pay back what it has earned.

    \displaystyle S_{T+m+1} = (1+k)S_{T+m} - d

    Again, let's try a few values:

    \displaystyle S_{T+1} = (1+k)S_T - d

    \displaystyle S_{T+2} = (1+k)((1+k)S_T - d) - d = (1+k)^2 S_T - d(1+k) - d

    \displaystyle S_{T+3} = (1+k)((1+k)((1+k)S_T - d) - d) - d = (1+k)^3 S_T - d(1+k)^2 - d(1+k) - d

    hypothesis: \displaystyle S_{T+m} = (1+k)^m S_T - d\sum^{m-1}_{r=0}(1+k)^r

    \displaystyle  = (1+k)^m S_T - d\frac{(1+k)^m - 1}{k} = (1+k)^m(S_T - \frac{d}{k}) + \frac{d}{k} (again by geometric)


    we need to prove this, again by induction.

    basis case: when m = 0, \displaystyle S_T = (1+k)^0 (S_T - \frac{d}{k}) + \frac{d}{k} = S_T basis works

    inductive step: \displaystyle S_{T+m+1} = (1+k)S_{T+m} - d = (1+k)((1+k)^m(S_T - \frac{d}{k}) + \frac{d}{k}) - d

    \displaystyle  = (1+k)^{m+1}(S_T - \frac{d}{k}) + (1+k)\frac{d}{k} - d

    \displaystyle  = (1+k)^{m+1}(S_T - \frac{d}{k}) + \frac{d}{k} + d - d

    \displaystyle  = (1+k)^{m+1}(S_T - \frac{d}{k}) + \frac{d}{k}

    Hence the sum works. However the question did not ask for us to express S_T, so we must substitute in (see our above def. for S_T)


    \displaystyle S_{T+m} = (1+k)^m(c\frac{(1+k)^{T} - 1}{k} - \frac{d}{k}) + \frac{d}{k}

    \displaystyle S_{T+m} = (1+k)^m(c\frac{(1+k)^{T} - 1 - d}{k}) + \frac{d}{k}

    \displaystyle S_{T+m} = \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}



    for part 3, note that if the sequence is decreasing, the firm will pay off its debt. so the conditions are:

    \displaystyle S_{T+m+1} &lt; S_{T+m}

    \displaystyle \frac{c}{k} (k+1)^{T+m+1} - \frac{c+d}{k}(k+1)^{m+1} + \frac{d}{k} &lt; \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}

    Letting x = k+1


    \displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} + \frac{d}{k} &lt; \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m + \frac{d}{k}

    \displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} &lt; \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m

    taking everything on to one side


    \displaystyle x^{T+m}(\frac{c}{k}x - \frac{c}{k}) - x^m(\frac{c+d}{k}x-\frac{c+d}{k}) &lt; 0

    \displaystyle (x-1)(\frac{c}{k}x^{T+m} - \frac{c+d}{k}x^m) &lt; 0

    \displaystyle x^m(x-1)(\frac{c}{k}x^{T} - \frac{c+d}{k}) &lt; 0

    k+1 > 0, so both x^m and (x-1) are greater than zero. Therefore, for the inequality to be satisfied,

    \displaystyle \frac{c}{k}x^{T} - \frac{c+d}{k} &lt; 0

    \displaystyle cx^{T} - (c+d) &lt; 0

    \displaystyle x^{T} - 1 &lt; \frac{d}{c}

    \displaystyle (1+k)^{T} - 1 &lt; \frac{d}{c} as required.

    If being decreasing is sufficient for the sequence to become negative finally? Maybe there is a positive limit there for it to reach as it decreases. So far I have no idea how to prove that there is no such a limit. Hope that you could have a check on this if you have time. Thx!
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    (Original post by hassi94)
    I don't think it is... \frac{2}{3} a + \frac{2}{3} a + \frac{2}{3} a \neq a
    I think in the question they wrote “ determine whether this last equation holds only if y does not equal x” and this equation should be 2a-3x=(y-z)^2/x, so I think it doesn't matter what is x+y+z here. What do you think of this?

    btw, how do you find the value of x, y, z here? I did it by let y=z in the last equation so we get x and then substitute back to the first equations. Then get a quadratic equation of y. Two roots, 2/3a and 1/6a
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    [/I]
    (Original post by mikelbird)
    I get the same as you....see file...
    Haven't you ignored the weight of the ladder itself?

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