# STEP 2018 SolutionsWatch

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10 months ago
#261
THe solution to question 6
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10 months ago
#262
S3 S14

(o)
Spoiler:
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(i)
Spoiler:
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(ii)
Spoiler:
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*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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#263
STEP III, Question 13:

Recall that so that

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

i) We first find :

so that . Then the PGF of is

Moreover, we have as for .

ii) Note that and so we have

Hence, we then have the PGF of as

which is not always , for example, by taking .
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10 months ago
#264
SIII, Q3.

We have So to match with the given ODE, we require
(I)
(II)
(III) .
We need to find to show this is possible. Note that (II) gives and substitution into (III) gives , which after some rearranging becomes , i.e.

Substitution of this into (II) gives and then substitution into (I) gives

(i) Here we want to solve the ODE when Write the ODE in the form , i.e. Integrating up gives for some constant , and so Now if we can integrate this to get for some constants and thus we get the solution of .

Looking at our expressions for we see that this is exactly for some constants regardless of what sign we chose for in the expression for .

Finally, if , i.e. , then we integrate up to get a instead, and so the solution is .

(ii) In this case we have and . So Thus we have

So provided we can integrate to get for some constant . Dividing by gives . So once again, provided we can integrate again to find the general solution to be (after dividing by )

Finally, if then we have and so we get . Noticing that the first term is just the derivative of , we can integrate again to find that the general solution is .

Note: The ODE in (i) is just the homogeneous equation from that in (ii) (with ). So we expect that the solutions only differ by a particular solution, which is what we see.
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10 months ago
#265
SIII Q12.

Spoiler:
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We have numbers drawn from a distribution. Define to be the th smallest number.

(i) Let . Then for to be , we need at least of the numbers to be . We could have exactly being , or being , and so on up to all being . Hence

For exactly of the numbers to be , the rest must be and thus the probability of any given combination of exactly numbers being is (as these are uniform random variables). But there are exactly ways for exactly of the numbers to be and so

and hence the result.

(ii) Simply note

Also

Note that if has pdf , then , and so differentiating (by the fundamental theorem of calculus if you like) we have Hence differentiating the expression from (i) we get

Note that the second term comes from differentiating the term, which is not present when , so the second sum only goes up to . Then using the above result for the binomial coefficients we have

Now re-index the first sum (set if you will) to see

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

which is the pdf we are after.

As this is a pdf, we know that we must have . Thus we see

(iii) We know that in general

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

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.
1
10 months ago
#266
STEP III Q4
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10 months ago
#267
SIII Q9.

(i)

Spoiler:
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We have two particles, one of mass (called P) and the other of mass (called Q) both on a smooth horizontal floor. P has speed traveling towards a wall, whilst Q is between P and the wall, travelling towards P with speed . 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 collision be and similarly for Q. Then looking at the th collision we have from conversation of momentum (CoM, after dividing by ) and Newton’s law of restitution (NLoR):

Eliminate from these to get

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

We now have two equations for . Equating them and rearranging gives the first result.

(ii)

Spoiler:
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We are told that now and , and that the solution of the equation is , for any . So setting gives

Setting gives We can then find in terms of using our previous CoM and NLoR equations in the , i.e. we already found in (i), . Putting in the numbers gives

Solving these equations for gives and .

Hence if , then and . Now note that , and so in the large limit, the will dominate the expression for , and so as its coefficient () is negative, we will have for large .

Thoughts:

Spoiler:
Show

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 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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10 months ago
#268
SIII Q11.

Spoiler:
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We have a particle (of mass say) attached to a string of length . 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 with the upward vertical, when it has speed .

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

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 . The string will become taut again when the coordinates of the particle lie on the circle of radius about 0. Note that the coordinates of the point when the particle starts this projectile motion are .

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

So the string becomes taut again when . Plugging in these expressions for and using the expression for to eliminate , we find that , 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 . Now is just the gradient of its trajectory at this point, which is the gradient of the path, or . So hence as

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

where we have used the fact that . 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 ,

Using the fact that and to eliminate first and then from the expressions for we find

.

Then using that , we can solve the resulting equation for to find

.

Treating this as a quadratic in and using the quadratic formula, we find , since we must have . Thus we are done.

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10 months ago
#269
SIII Q10.

Spoiler:
Show

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 from 0. Then from basic trigonometry we must have

, i.e. .

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

.

So as we get which gives the result.

The disc is now rotated about and then released, so that the angle between OP and the horizontal at a time after is . 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 is the moment of inertia of the disc about the pivot, then its kinetic energy will be 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

where we have used our expression for 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 . Thus the change in GPE is: 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:

.

We are told that and . Using our expression for found originally we have , so that . Plugging these values in (cancelling the and the ) we find

where we have called . For small oscillations, , and so this equation becomes one of SHM with frequency . Thus the period of small oscillations is

as required.

Thoughts:

Spoiler:
Show

First two parts up to finding 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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10 months ago
#270
(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 .
So we have the equation of the tangent to the hyperbola at is and by similarities we reach the tangent at being .

From the perpendicular requirement,

Subtracting the two tangent equations from each other and rearranging:

Multiplying the tangent equation for by and the tangent equation for by , subtracting the two new equations and rearranging:

Now yeah that does look ugly, but you can see that the term is generated from both squares and is cancelled out - so inside the big bracket you're left with:

Note that this can be factorised to:

and so

Apply trig Pythagoras' and this becomes

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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5 months ago
#271
The OP has been edited to include the remaining solutions.
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4 months ago
#272
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)
Spoiler:
Show

Expected winnings is . Let . Graph sketch below by considering things like and as .

The integer which maximises will either be or by that graph. So first we need to find :

.

First note the inequality (*) which is true for all . This could be easily shown in various ways (e.g. graph sketch). We'll need this later.

Note the following:

which is true by (*).

,

which is true by (*).

Thus , which implies either or maximises . But so both options actually maximise .

(ii)

Spoiler:
Show

Probability of getting heads in this scenario is .

So expected winnings is .

When , . By previous part this is maximised when so .

by using approximation given in question.

So .
Last edited by I hate maths; 4 months ago
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2 months ago
#273
(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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1 month ago
#274
In part (iv), when resolving the inequality don't you have to allow , giving u = v = 1 and maximum value 3?
(Original post by DFranklin)
Note that:

Spoiler:
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If you go "the other way" in part (iii) you get and then and you can then use the result from the first part. (And similarly I assume with the "= 3" variant).
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