STEP 2018 Solutions Watch

mikelbird
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THe solution to question 6
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username4188924
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S3 S14

(o)
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 G(t) = \sum_{n=0}^{\infty} P(X=n) t^n\\



G(1)+G(-1) = \sum_{n=0}^{\infty} P(X=n) + \sum_{n=0}^{\infty} P(X=n)(-1)^n\\



= 2\sum_{n=0}^{\infty} P(X=2n) = 2P( \cup_{n=0}^{\infty} (X=2n))\\



X\sim \text{Poisson}(\lambda)



G(t) = \sum_{n=0}^{\infty} \frac{e^(-\lambda) \lambda^n}{n!}t^n = e^{-\lambda}\sum_{n=0}^{\infty} \frac{(\lambda t)^n}{n!} = e^{-\lambda + \lambda t}\\









(i)
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(i) \sum_{n=0}^{\infty} kP(X=2n) = \sum_{n=0}^{\infty} P(Y=n) = 1 \\



 \implies k = \frac{1}{ \frac{1}{2} (G(1)+G(-1))} = \frac{2}{1+e^{-2\lambda}}\\ 



 G_Y(t) = \sum_{n=0}^{\infty} P(Y=n)t^n = k \sum_{n=0}^{\infty} P(X=2n)t^{2n}\\ 

 = k\sum_{n=0}^{\infty} \frac{e^{-\lambda}{(\lambda t)}^{2n}}{2n!} = ke^{-\lambda}\cosh(\lambda t) = \frac{2e^{-\lambda}\cosh(\lambda t)}{1+e^{-2\lambda}} = \frac{\cosh(\lambda t)}{\cosh \lambda}\\ 



E(Y) = \dfrac{dG_Y(t)}{dt} |_{t=0} = \dfrac{\lambda \sinh{\lambda t}}{\cosh \lambda}|_{t=0} = 0 < \lambda \\








(ii)
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(ii) k P( \cup_{n=0}^{\infty} (X=4n)) = P( \cup_{n=0}^{\infty} (Y=4n)) \\ 



 = P( \cup_{n=0}^{\infty} (W=2n)) = \frac{1}{2} (G_W(1)+G_W(-1)) \quad \quad (\text{ letting } 2W = Y)\\



G_W(t) = E(e^{tW}) = E(e^{2tY}) = G_Y(2t) = \frac{\cosh(2\lambda t)}{\cosh \lambda}\\    



\sum_{n=0}^{\infty} cP(X=4n) = \sum_{n=0}^{\infty} P(Z=n) = 1 \\ 



 \implies c = \dfrac{2k}{G_W(1) + G_W(-1)} = \dfrac{2e^{\lambda}}{\cosh 2\lambda}  \\ 



 G_Z(t) = c \sum_{n=0}^{\infty}P(X=4n) t^{4n} = c \sum_{n=0}^{\infty}\frac{e^{-\lambda} (\lambda t)^{4n}}{4!} = c {\frac{e^{-\lambda}(\cosh(\lambda t) + \cos( \lambda t))}{2}} \\  



= \dfrac{(\cosh(\lambda t) + \cos( \lambda t))}{\cosh 2\lambda}}\\

 

E(Z) = \dfrac{dG_Z(t)}{dt} |_{t=0} = 0 <\lambda

*There might be a more slicker way to go about this. Maybe trying (G(i)+G(-i)+G(1)+G(-1))/4, but it has been a while since I've worked with pgf and of course you'll have to think about the validity eg. convergence etc...



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Zacken
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STEP III, Question 13:

Recall that G(t) = \sum_{k\geqslant 0} \mathbb{P}(X = k) t^{k} so that

\displaystyle 

\begin{align*}G(1) + G(-1) & = \sum_{k \geqslant 0} \left(\mathbb{P}(X = k) + (-1)^k \mathbb{P}(X=k)\right) \\ & = 2\sum_{k \geqslant 0, \, k \, \mathrm{even}} \mathbb{P}(X=k) + \sum_{k \geqslant 0, \, k \, \mathrm{odd}} \left(\mathbb{P}(X=k) - \mathbb{P}(X=k) \right) = 2 \sum_{k \geqslant 0} \mathbb{P}(X=2k)\end{align*}

as required. If X is Poisson, with parameter \lambda then we have

\displaystyle

\begin{equation*}G(t) = \sum_{k \geqslant 0} e^{-\lambda} \frac{\lambda^k}{k!} t^k  = e^{-\lambda} \sum_{k \geqslant 0} \frac{(\lambda t)^k}{k!}  = e^{-\lambda} e^{\lambda t} = e^{-\lambda(1-t)}  \end{equation*}

i) We first find k:

\displaystyle 

\begin{equation*}\sum_{r \geqslant 0} \mathbb{P}(Y=r) = k \sum_{r \geqslant 0} \mathbb{P}(X=2r) = \frac{k}{2} (G(1) + G(-1)) = \frac{k}{2}(1+ e^{-2\lambda}) = 1\end{equation*}

so that k = \frac{2e^{\lambda}}{e^{\lambda} + e^{-\lambda}} = \frac{e^{\lambda}}{\cosh \lambda}. Then the PGF of Y is

\displaystyle

\begin{equation*}H(t) = \frac{1}{\cosh \lambda}\sum_{r \geqslant 0} e^{\lambda}t^{2r} \mathbb{P}(X=2r) = \frac{1}{\cosh \lambda}\sum_{r \geqslant 0 } \frac{(\lambda t)^{2r}}{(2r)!} = \frac{\cosh \lambda t}{\cosh \lambda}\end{equation*}

Moreover, we have \displaystyle \mathbb{E}(Y) = H'(1) = \frac{\lambda \sinh \lambda}{\cosh \lambda} = \lambda \tanh \lambda < \lambda as \tanh \in (0,1) for \lambda > 0.

ii) Note that G(1) + G(-1) + G(i) + G(-i) = 4\mathbb{P}(X = 0,4,8, 12, \ldots) and so we have

\displaystyle

\begin{equation*}\sum_{r \geqslant 0} \mathbb{P}(Z=r) = c\sum_{r \geqslant 0} \mathbb{P}(X=4r) = \frac{c}{4}(G(1) + G(-1) + G(i) + G(-i) = \frac{ce^{-\lambda}}{2}\left(\cos \lambda + \cosh \lambda \right) = 1 \end{equation*}

Hence, we then have the PGF of Z as

\displaystyle

\begin{equation*}R(t) = \frac{2e^{\lambda}}{\cos \lambda + \cosh \lambda}\sum_{r \geqslant 0}e^{-\lambda} \frac{(\lambda t)^{4r}}{(4r)!} = \frac{\cos \lambda t + \cosh \lambda t}{\cos \lambda + \cosh \lambda } \Rightarrow \mathbb{E}(Z) = R'(1) = \lambda \frac{\sinh \lambda - \sin \lambda}{\cos \lambda + \cosh \lambda} \end{equation*}

which is not always < \lambda, for example, by taking \lambda = \pi.
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jtSketchy
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SIII, Q3.

We have  x^a(x^b(x^cy)’)’ = x^{a+b+c}y’’ + (b+2c)x^{a+b+c-1}y’ + c(b+c-1)x^{a+b+c-2}y. So to match with the given ODE, we require
(I)  a+b+c = 2
(II) b+2c = 1-2p
(III) c(b+c-1) = p^2-q^2.
We need to find a,b,c to show this is possible. Note that (II) gives b = 1-2p - 2c and substitution into (III) gives c(-c-2p) = p^2-q^2 , which after some rearranging becomes  (c+p)^2 = q^2 , i.e.  c = -p \pm q.

Substitution of this into (II) gives  b = 1\mp 2q and then substitution into (I) gives  a = 1+p \pm q.

(i) Here we want to solve the ODE when f\equiv 0. Write the ODE in the form  x^a(x^b(x^cy)’)’ = 0 , i.e. (x^b(x^cy)’)’ = 0. Integrating up gives  x^b(x^c y)’ = A for some constant  A , and so  (x^c y)’ = Ax^{-b}. Now if  b\neq 1 we can integrate this to get  x^cy = \tilde{A} x^{1-b} + B for some constants \tilde{A},B and thus we get the solution of  y = \tilde{A} x^{1-b-c} + Bx^{-c} .

Looking at our expressions for b,c we see that this is exactly  y= A x^{p+q} + Bx^{p-q} for some constants  A,B regardless of what sign we chose for q in the expression for c.

Finally, if  b = 1 , i.e.  q = 0 , then we integrate up to get a \ln(x) instead, and so the solution is y = Ax^p \ln(x) + Bx^p .

(ii) In this case we have  f(x) = x^n and q =0 . So a = 1+p,\ b = 1,\ c = -p. Thus we have (x(x^{-p}y)’)’ = x^{n-p-1}.

So provided n\neq p we can integrate to get x(x^{-p}y)’ = \frac{x^{n-p}}{n-p} + B for some constant B . Dividing by  x gives (x^{-p}y)’ = \frac{x^{n-p-1}}{n-p} + Bx^{-1} . So once again, provided  n \neq p we can integrate again to find the general solution to be (after dividing by  x^{-p})

 y = \frac{x^{n}}{(n-p)^2} + Bx^p\ln(x) + Cx^p.

Finally, if  n = p then we have  (x(x^{-p}y)’)’ = x^{-1} and so we get  (x^{-p}y)’ = \frac{\ln(x)}{x} + Bx^{-1}. Noticing that the first term is just the derivative of \frac{1}{2}(\ln(x))^2 , we can integrate again to find that the general solution is  y = \frac{1}{2}(\ln(x))^2 + Bx^p\ln(x) + Cx^p.

Note: The ODE in (i) is just the homogeneous equation from that in (ii) (with  q =0 ). So we expect that the solutions only differ by a particular solution, which is what we see.
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jtSketchy
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SIII Q12.

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We have n numbers drawn from a U[0,1] distribution. Define Y_k to be the kth smallest number.

(i) Let 0\leq y\leq 1. Then for Y_k to be \leq y, we need at least k of the numbers to be \leq y. We could have exactly k being \leq y, or k+1 being \leq y, and so on up to all n being \leq y. Hence

P(Y_k\leq y) = \sum^n_{m=k} P(\text{Exactly }m \text{ of the numbers are }\leq y).

For exactly m of the numbers to be \leq y, the rest must be >y and thus the probability of any given combination of exactly m numbers being  \leq y is y^m(1-y)^{n-m} (as these are uniform random variables). But there are exactly {{n}\choose{m}} ways for exactly m of the numbers to be  \leq y and so

P(\text{Exactly } m \text{ of the numbers are }\leq y) = {{n}\choose{m}}y^m(1-y)^{n-m}

and hence the result.

(ii) Simply note n {{n-1}\choose{m-1}} = n\cdot\frac{(n-1)!}{(m-1)!(n-m)!} = \frac{n!}{(m-1)!(n-m)!} = m\cdot\frac{n!}{m!(n-m)!} = m{{n}\choose{m}}.

Also (n-m){{n}\choose{m}} = (n-m)\cdot\frac{n!}{m!(n-m)!} = \frac{n!}{m!(n-m-1)!} = n\cdot \frac{(n-1)!}{m!((n-1)-m)!} = n{{n-1}\choose{m}}.

Note that if Y_k has pdf f, then P(Y_k\leq y) = \int^y_0 f(x)\ \mathrm{d}x, and so differentiating (by the fundamental theorem of calculus if you like) we have f(y) = \frac{\mathrm{d}}{\mathrm{d}y} P(Y_k\leq y). Hence differentiating the expression from (i) we get

 f(y) = \sum^n_{m=k} m{{n}\choose{m}}y^{m-1}(1-y)^{n-m} - \sum^{n-1}_{m=k} (n-m){{n}\choose{m}}y^m(1-y)^{n-m-1}

Note that the second term comes from differentiating the  (1-y)^{n-m} term, which is not present when m = n, so the second sum only goes up to  n-1. Then using the above result for the binomial coefficients we have

f(y) = n\sum^n_{m=k}{{n-1}\choose{m-1}}y^{m-1}(1-y)^{n-m} - n\sum^{n-1}_{m=k}{{n-1}\choose{m}}y^m(1-y)^{n-m-1}

Now re-index the first sum (set  l = m-1 if you will) to see

 f(y) = n\sum^{n-1}_{m=k-1} {{n-1}\choose{m}} y^m(1-y)^{n-m-1} - n\sum^{n-1}_{m=k} {{n-1}\choose{m}}y^m(1-y)^{n-m-1}

Note that both sums are the same, except the first one contains the extra term in  m = k-1. So after cancelling we see

 f(y) = n{{n-1}\choose{k-1}}y^{k-1}(1-y)^{n-k}

which is the pdf we are after.

As this is a pdf, we know that we must have  \int^1_0 f(y) \mathrm{d} y = 1 . Thus we see

 \int^1_0 y^{k-1}(1-y)^{n-k}\ \mathrm{d}y = \frac{1}{n{{n-1}\choose{k-1}}} = \frac{(k-1)!(n-k)!}{n!}.

(iii) We know that in general E(Y_k) = \int^1_0 y f(y)\ \mathrm{d}y = n{{n-1}\choose{k-1}}\int^1_0y^k(1-y)^{n-k}\ \mathrm{d}y.

Note that this integral is the same as the one just found, except changing  k \to k+1 and n\to n+1. So hence we have from the end of (ii),

E(Y_k) = n{{n-1}\choose{k-1}}\cdot\frac{k!(n-k)!}{(n+1)!} = \frac{k}{n+1}.



I always find it amazing how simple some of these probability questions can be once you write down the probabilities. They essentially become an easier version of a ‘Pure’ question. I would definitely recommend more people to look at them.
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Garjun
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STEP III Q4
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jtSketchy
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SIII Q9.

(i)

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We have two particles, one of mass m (called P) and the other of mass km (called Q) both on a smooth horizontal floor. P has speed u_0 traveling towards a wall, whilst Q is between P and the wall, travelling towards P with speed v_0. In the subsequent motion Q collides between P and the wall. Since the wall has coefficient of restitution 1, the speed it has after colliding with P will be the speed it has when it collides with P again.

So let the speed of P after the n collision be u_n and similarly v_n for Q. Then looking at the nth collision we have from conversation of momentum (CoM, after dividing by m) and Newton’s law of restitution (NLoR):

 u_{n-1}-kv_{n-1} = u_n + kv_n

 (v_{n-1}+u_{n-1})e = v_n - u_n

Eliminate v_{n-1} from these to get k(1+e)v_n = u_n(k-e) + e(1+k)u_{n-1}.

This is good, but we do not know how to eliminate v_n to find an equation just including the u_n. To do this, note that if we look at the (n+1)th collision, this will involve v_n and v_{n+1}, and we can eliminate v_{n+1}. So replacing n with n+1 in our previous CoM and NLoR equations, and now eliminating v_{n+1} we find

 k(1+e)v_n = (1-ke)u_n - (1+k)u_{n+1}.

We now have two equations for  k(1+e)v_n. Equating them and rearranging gives the first result.



(ii)

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We are told that now  e = 1/2 and  k = 1/34 , and that the solution of the equation is u_n = A\left(\frac{7}{10}\right)^n + B\left(\frac{5}{7}\right)^n, for any n\geq 0. So setting n=0 gives

u_0 = A+B.

Setting n = 1 gives u_1 = \frac{7A}{10} + \frac{5B}{7}. We can then find u_1 in terms of  u_0, v_0 using our previous CoM and NLoR equations in the n=1, i.e. we already found in (i),  k(1+e)v_0 = u_0(1-ke) - (1+k)u_1. Putting in the numbers gives

\frac{7A}{10} + \frac{5B}{7} = \frac{67}{70}u_0 - \frac{3}{70}v_0.

Solving these equations for A,B gives A = 3v_0 - 17u_0 and B = 18u_0 - 3v_0.

Hence if 0<6u_0<v_0, then B<0 and A > u_0 >0. Now note that \frac{7}{10}<\frac{5}{7}, and so in the large n limit, the \left(\frac{5}{7}\right)^n will dominate the expression for u_n , and so as its coefficient (B) is negative, we will have u_n<0 for large n.



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Once again, I feel this question demonstrates well the gems which can be found in the mechanics and probability sections of STEP. The only sort of ‘trick’ is to realise that you can eliminate all the v_n by considering two consecutive collisions. Once you have realised that the rest of the question is simply just solving simulataneous equations.
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jtSketchy
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SIII Q11.

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We have a particle (of mass m say) attached to a string of length b. The other end is fixed at the origin, and the particle initially hangs vertically below. It then receives a horizontal impulse, and moves in a circular arc. The string becomes slack when it makes an angle \alpha with the upward vertical, when it has speed V.

When the string becomes slack the tension is 0. Hence resolving forces radially, we have (as it is circular motion)

mg\cos(\alpha) = \frac{mV^2}{b}

which leads to V = \sqrt{bg\cos(\alpha)}.

For the next part we want to show that the string becomes taut again. When it becomes slack, the particle starts moving as a projectile with constant downward acceleration g. The string will become taut again when the coordinates of the particle lie on the circle of radius b about 0. Note that the coordinates of the point when the particle starts this projectile motion are (b\sin(\alpha),b\cos(\alpha)).

So using SUVAT, the coordinates of the particle at a later time t are (since the particle will be moving in the negative x direction and positive y direction initially):

x = b\sin(\alpha) - Vt\cos(\alpha)

y = b\cos(\alpha) + Vt\sin(\alpha) - \frac{1}{2}gt^2.

So the string becomes taut again when x^2+y^2 = b^2. Plugging in these expressions for x,y and using the expression for V to eliminate \cos(\alpha), we find that gT = 4V\sin(\alpha), as required.

Just before when the string becomes taut again, we need to find the angle the particle trajectory makes with the horizontal, which we will call \beta. Now \tan(\beta) is just the gradient of its trajectory at this point, which is the gradient of the path, or \frac{\mathrm{d} y}{\mathrm{d} x}. So hence as

\frac{\mathrm{d} x}{\mathrm{d} t} = -V\cos(\alpha)
\frac{\mathrm{d} y}{\mathrm{d} t} = V\sin(\alpha) - gt

(either from differentiating our previous expressions or just SUVAT again) we therefore have:

\tan(\beta) = \frac{\mathrm{d} y}{\mathrm{d} t}/\frac{\mathrm{d} x}{\mathrm{d} t} = \frac{V\sin(\alpha) - gT}{-V\cos(\alpha)} = \frac{V\sin(\alpha)-4V\sin(\alpha)}{-V\cos(\alpha)} = 3\tan(\alpha)

where we have used the fact that gT = 4V\sin(\alpha). This gives this result.

Finally, we are told that when the string becomes taut the momentum of the particle in the direction of the string is 0. Hence for the particle to be instantaneously at rest at this point, we need the motion to be entirely radial (this is an if and only if as well). This would mean that the angle the particle trajectory makes is actually just that made of a radial line from the origin, i.e. from basic trigonometry, we mean at  t=T,

\tan(\beta) = \frac{y}{x}.

Using the fact that T = \frac{4V\sin(\alpha)}{g} and V^2 = bg\cos(\alpha) to eliminate first T and then V from the expressions for x,y we find

x = b\sin(\alpha) - 4b\sin(\alpha)\cos^2(\alpha)
y = b\cos(\alpha)-4b\cos(\alpha)\sin^2(\alpha).

Then using that \frac{y}{x} = \tan(\beta) = 3\tan(\alpha), we can solve the resulting equation for \sin(\alpha) to find

8\sin^4(\alpha) - 4\sin^2(\alpha) - 1 = 0.

Treating this as a quadratic in \sin^2(\alpha) and using the quadratic formula, we find \sin^2(\alpha) = \frac{1+\sqrt{3}}{4}, since we must have \sin^2(\alpha)\geq 0. Thus we are done.

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jtSketchy
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SIII Q10.

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Draw a nice diagram of the scenario, with 0 and P horizontal. To be in equilibrium the centre of mass must lie directly underneath the pivot point. The centre of mass is a distance \frac{ma}{M+m} from 0. Then from basic trigonometry we must have

a\sin(\beta) = \frac{ma}{M+m} , i.e. \sin(\beta) = \frac{m}{M+m}.

To find the distance AP we can use the cosine rule to get

AP^2 = a^2+a^2 - 2a^2\cos(\pi/2-\beta) = 2a^2(1-\cos(\pi/2-\beta)).

So as \cos(\pi/2-\beta) = \sin(\beta) = \frac{m}{M+m} we get AP^2 = \frac{2a^2M}{m+M} which gives the result.

The disc is now rotated about L and then released, so that the angle between OP and the horizontal at a time t after is \theta. The expression given clearly looks like an energy, which we know is constant in the motion. The energies we must consider are the kinetic energies of the disc and particle as well as their GPE.

Indeed, if I is the moment of inertia of the disc about the pivot, then its kinetic energy will be \frac{1}{2}I\dot{\theta}^2 from standard theory. Next we look at the particle. In the motion, we can treat P as moving with circular motion about the pivot A. Thus as it is a distance AP from A (so this is the radius of the circular motion), its kinetic energy is

\frac{1}{2}m(AP\dot{\theta})^2 = \frac{1}{2}m(AP)^2\dot{\theta}^2 = \frac{1}{2}m(2a^2(1-\sin(\beta)))\dot{\theta}^2 = (1-\sin(\beta))ma^2\dot{\theta}^2

where we have used our expression for AP^2 found earlier from the cosine rule. This is the second term in the expression we are after. Finally we must find the GPE. To do this, simply treat the system as a point particle located at the centre of mass. Then the change in GPE is simply that of a pendulum, where the length of the pendulum is the distance from A to the centre of mass, which is a\cos(\beta). Thus the change in GPE is: (M+m)ga\cos(\beta)(1-\cos(\theta)) which again us one of the terms we were after.

Adding together these three terms gives the expression we were after, which we know must be constant as the energy is conserved.

Differentiate this expression (which must be 0 since it is constant) to get:

I\dot{\theta}\ddot{\theta} + 2(1-\sin(\beta))ma^2\dot{\theta} \ddot{\theta} + (m+M)ga\cos(\beta)\dot{\theta} \sin(\theta) = 0.

We are told that m = \frac{3}{2}M and I = \frac{3}{2}Ma^2 . Using our expression for \sin(\beta) found originally we have \sin(\beta) = \frac{3}{5}, so that \cos(\beta) = \frac{4}{5}. Plugging these values in (cancelling the m,M and the \dot{\theta}) we find

\ddot{\theta} + \omega^2\sin(\theta) = 0

where we have called \omega^2 = \frac{20g}{27a}. For small oscillations, \sin(\theta)\approx \theta, and so this equation becomes one of SHM with frequency \omega. Thus the period of small oscillations is

 \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{27a}{20g}} = 3\pi\sqrt{\frac{3a}{5g}}

as required.



Thoughts:

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First two parts up to finding AP are quite simple. The hardest part is finding the expression for the energy, and realising how to deal with the latter two terms. Once you have the energy expression the rest is quite straightforward.
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carpetguy
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(Original post by carpetguy)
I'll start assuming the first result was reached and the midpoint of the two meeting points was shown to be at P.
So we have the equation of the tangent to the hyperbola at P is bx - ay\sin\theta = ab\cos\theta and by similarities we reach the tangent at Q being bx - ay\sin\phi = ab\cos\phi.

From the perpendicular requirement, \frac{b}{a\sin\theta} \times \frac{b}{a\sin\phi} = -1
\therefore b^2 = -a^2\sin\theta\sin\phi

Subtracting the two tangent equations from each other and rearranging:
y = \frac{b\,(\cos\phi \,-\, \cos\theta)}{\sin\theta \,-\, \sin\phi} = \frac{\sqrt{-a^2\sin\theta\sin\phi}\,(\cos \phi \,-\, \cos\theta)}{\sin\theta \,-\, \sin\phi}

Multiplying the tangent equation for P by \sin\phi and the tangent equation for Q by \sin\theta, subtracting the two new equations and rearranging:
x = \frac{a\,(\cos\phi\sin\theta \,-\, \cos\theta\sin\phi)}{\sin\theta \,-\, \sin\phi}

\therefore x^2 + y^2 = \frac{a^2}{(\sin\theta \,-\, \sin\phi)^2}\,\{(\cos\phi \sin \theta \,-\, \cos\theta\sin\phi)^2 \,- \, \sin\theta\sin\phi\,(\cos\phi \,-\, \cos\theta)^2 \}

Now yeah that does look ugly, but you can see that the -2\cos\phi\cos\theta\sin\phi\sin \theta term is generated from both squares and is cancelled out - so inside the big bracket you're left with:
\cos^2\phi\sin^2\theta + \cos^2\theta\sin^2\phi - \sin\theta\sin\phi\cos^2\phi - \sin\theta\sin\phi\cos^2\theta

Note that this can be factorised to:
(\sin\theta \,-\, \sin\phi)(\cos^2\phi\sin\theta \,-\, \cos^2\theta\sin\phi)

and so x^2 + y^2 = \frac{a^2}{(\sin\theta \,-\, \sin\phi)}\,(\cos^2\phi\sin \theta \,-\, \cos^2\theta\sin\phi)

Apply trig Pythagoras' and this becomes \frac{a^2}{(\sin\theta \,-\, \sin\phi)}\,(\sin\theta \,-\, \sin\phi \,+\, \sin^2\theta\sin \phi \,-\, \sin^2\phi\sin \theta)

\therefore x^2 + y^2 = \frac{a^2}{(\sin\theta \,-\, \sin\phi)}\,(\sin\theta \,-\, \sin\phi)(1 + \sin\theta\sin\phi)

= a^2 + a^2\sin\theta\sin\phi = a^2 - b^2

Phew! You could probably introduce some shorthand or something if you wanted to save on the writing but that's pretty much a slog.
Bumping as a full solution.

Oh wait Garjun's is full as well. I didn't notice the other pages.
Oh well.

Step III Q4 btw. The original post had some more info in it. I also put together what I thought the question was since the paper wasn't uploaded yet.
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Notnek
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The OP has been edited to include the remaining solutions.
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I hate maths
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STEP 2 Question 12 alternate solution. I'm a little bit rusty so hopefully no mistakes but let me know if there are. I think this is different enough to a solution that was given earlier.

(i)
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Expected winnings is hp^h. Let f(h)=hp^h. Graph sketch below by considering things like f(0) and f(h) \rightarrow 0 as h \rightarrow \infty .



The integer h which maximises f(h) will either be \lfloor h_s \rfloor or \lceil h_s \rceil by that graph. So first we need to find h_s:

\displaystyle f'(h)=p^h+hp^h \ln p \Rightarrow 0=p^{h_s}+h_sp^{h_s}\ln p \Rightarrow 0=p^{h_s}(1+h_s \ln p) \Rightarrow h_s = \frac{-1}{\ln p}=1/ \ln \Big( \frac{N+1}{N} \Big).

First note the inequality \ln(1+x) < x (*) which is true for all x \neq 0. This could be easily shown in various ways (e.g. graph sketch). We'll need this later.

Note the following:

\displaystyle N < 1/ \ln \Big( \frac{N+1}{N} \Big) \iff \frac{1}{N}> \ln\Big( 1+\frac{1}{N} \Big) which is true by (*).

Additionally,

\displaystyle N+1> 1/ \Big( \ln \frac{N+1}{N} \Big) \iff \frac{1}{N+1} < \ln \Big( \frac{N+1}{N} \Big) \iff \frac{-1}{N+1}> \ln \Big(1-\frac{1}{N+1}\Big),

which is true by (*).

Thus \displaystyle N < -\frac{1}{\ln p} < N+1, which implies either h=N or h=N+1 maximises f(h). But N(\frac{N}{N+1})^N=(N+1)(\frac{N  }{N+1})^{N+1} \Rightarrow f(N)=f(N+1) so both options actually maximise f(h).


(ii)

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Probability of getting h heads in this scenario is p^h+p^h(1-p) {h \choose 1} + p^h(1-p)^2 {h \choose 2}+...+p^h(1-p)^h {h \choose h} = p^h \Big[ 1+1/(N+1) {h \choose 1} + 1/(N+1)^2 {h \choose 2} +...+ 1/(N+1)^h {h \choose h} \Big] = p^h(1+1/(N+1))^h = \frac{N^h(N+2)^h}{(N+1)^{2h}}.

So expected winnings is E=\frac{hN^h(N+2)^h}{(N+1)^{2h}}.

When N=2, \displaystyle E=h(\frac{8}{9})^h. By previous part this is maximised when h=8 so \displaystyle E=8(\frac{8}{9})^8.

\log_{3}E=\log_3 8(\frac{8}{9})^8=3\log_3 2+24\log_3 2-8\log_3 9=1.01 by using approximation given in question.

So E \approx 3.
Last edited by I hate maths; 4 months ago
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JudeH2001
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(Original post by Zacken)
Yes.



The fact that you sat it early and didn't learnt all the relevant material is squarely on you, universities won't care - if you sat the exam, they (or at least Cambridge) presume that you were ready for it and treat you as such. I don't know how Oxford works, but it would certainly detract from a Cambridge application.
Haha funny thing is I'm only now noticing your the same person as from the forum about the STEP day! You responded to another one of my messages over there! And btw I DID get a 2 in the paper last year and I DID apply to Cambridge for maths at Kings and I got an offer!
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VerySuspicious
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In part (iv), when resolving the inequality don't you have to allow \beta = 1, giving u = v = 1 and maximum value 3?
(Original post by DFranklin)
Note that:

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If you go "the other way" in part (iii) you get \alpha = 1 + \dfrac{1}{\beta} and then \dfrac{1}{u}+\dfrac{1}{v}+uv = -\dfrac{\alpha}{\beta} + \beta = \beta - \dfrac{1}{\beta} - \dfrac{1}{\beta^2} and you can then use the result from the first part. (And similarly I assume with the "= 3" variant).
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