STEP III, Question 12:
(i)
(ii)Note that by a geometric distribution argument.
Hence, it follows immediately that the probability generating function for each is:
So that their sum is given by the product of each p.g.f, aka: .
Taking the first derivative gives us and the second derivative is similarly .
We then have and .
Taking the th derivative, we see that:
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STEP Maths II, III 2009 Solutions
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Why bother with a post grad? Are they even worth it? Have your say!  26102016 

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 21
 26042016 15:31
STEP III, Question 12:

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 26042016 15:34
STEP III, Question 3:
(iii)The sketch is easy enough, note that it has a maximum at , asymptotic to the axis as and diverges to negative infinity as gives us our nice little graph:
Since then and hence the numerator of our derivative never vanishes, so the entire derivative never vanishes. This is instrumental in the next sketch.
sketch
Which follows pretty much elementarily from our above results.Last edited by Zacken; 26042016 at 15:45. 
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 26042016 16:38
Doing III Q8 right now .

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 26042016 17:26
STEP III, Question 8
Last edited by IrrationalRoot; 26042016 at 17:37. 
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 26042016 17:37
Done Q7 I can try and type it up now.
II 
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 26042016 18:18
STEP II Question 7
(i)Spoiler:(ii)ShowIf , using product rule in factor first and then using product rule in conjunction with the , we obtain
.
So we could write .
(i)
Let , notice that this integrated is in the exam form of the pervious result with n=15, a=4 and b=4.
Plugging these into the expression given for y at the start of the question yields
.Spoiler:(iii)ShowIf we let .
Noticing that at the moment there is a quartic expression instead of a cubic expression, we factorise and it is clear that (x1) is a factor of the quartic.
Factorising and we see that this means that integral, , becomes
.
Plugging in n=23, a=1 and b=12 we see that again the integral takes the exact form of dy/dx found in the first part of the question.
Therefore we have .Spoiler:ShowIf we let .
We immediately run into the problem that not only is the polynomial expression a quartic, it can also not be factorised (at least not in any useful way to this problem).
Realising that it may be useful to split up the quartic expression into a different form, maybe one with (x2) in it somewhere we find that
we can rewrite the integral as
.
Now both of these are in the exact form found in the first part of the question, the first with n=8, a=2 and b=4 and for the second we have n=7, a=2 and b=4 which gives us
.
Cleaning this up gives
.Last edited by B_9710; 26042016 at 23:58. 
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 27
 26042016 18:25
(Original post by B_9710)
STEP III Question 7
(i)Spoiler:(ii)ShowIf , using product rule in factor first and then using product rule in conjunction with the , we obtain
.
So we could write .
(i)
Let , notice that this integrated is in the exam form of the pervious result with n=15, a=4 and b=4.
Plugging these into the expression given for y at the start of the question yields
.Spoiler:(iii)ShowIf we let .
Noticing that at the moment there is a quartic expression instead of a cubic expression, we factorise and it is clear that (x1) is a factor of the quartic.
Factorising and we see that this means that integral, , becomes
.
Plugging in n=23, a=1 and b=12 we see that again the integral takes the exact form of dy/dx found in the first part of the question.
Therefore we have .Spoiler:ShowIf we let .
We immediately run into the problem that not only is the polynomial expression a quartic, it can also not be factorised (at least not in any useful way to this problem).
Realising that it may be useful to split up the quartic expression into a different form, maybe one with (x2) in it somewhere we find that
we can rewrite the integral as
.
Now both of these are in the exact form found in the first part of the question, the first with n=8, a=2 and b=4 and for the second we have n=7, a=2 and b=4 which gives us
.
Cleaning this up gives
. 
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 28
 26042016 19:10
(Original post by B_9710)
STEP II Question 7
(i)Spoiler:ShowIf , using product rule in factor first and then using product rule in conjunction with the , we obtain
.
So we could write .
(i)
Let , notice that this integrated is in the exam form of the pervious result with n=15, a=4 and b=4.
Plugging these into the expression given for y at the start of the question yields
.
then differentiate implicitly to getPost rating:1 
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 29
 26042016 19:15
(Original post by Zacken)
As an alternative to the above, you can also take logs to get:
then differentiate implicitly to get 
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 26042016 19:16
(Original post by IrrationalRoot)
Probably better to use the fact that for the differentiation at the start. 
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 26042016 19:25
(Original post by B_9710)
I tried that to begin with but it didn't seem to make it any less painless. 
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 26042016 20:49
Is there a thread for STEP I 2009 solutions already? I can't seem to find one.

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 26042016 21:59
(Original post by IrrationalRoot)
Is there a thread for STEP I 2009 solutions already? I can't seem to find one. 
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 26042016 22:11
(Original post by Zacken)
Nopes, it's just that the official examiner solutions are already beautifully laid out and intricately detailed. If there'd demand, we can always just add in a STEP I section after we've completed II and III. 
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 26042016 22:13
(Original post by IrrationalRoot)
Oh yeah the I solutions are lovely. I don't get why they don't do that for II and III, would be so nice. 
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 26042016 22:14
(Original post by Zacken)
Different examiners  the examiner for STEP I is a nice dude, or so I've heard. 
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 26042016 22:15
(Original post by IrrationalRoot)
Oh wow, so the II and III examiners are just lazy lol. 
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 27042016 01:02
STEP III Q10
SolutionConsider motion along the axis of the spring, let the spring have spring constant and suppose is of mass . By the equilibrium condition of the particle at rest, note .
Moreover, by conservation of energy from the time the ball is dropped until time after it first makes contact with the spring:
Differentiating w.r.t. and simplifying then yields
A second order ODE that describes the motion, as desired (alternatively, use Newton's 2nd law if you prefer).
Now consider a solution of the given form, then
Using these results, we obtain the solution , as required.
Note that the particle will first lose contact with the spring when the spring next returns to it's natural length. So seeking s.t. :
Where the line follows from the identity for , which is easily derived by considering a right angled triangle defined s.t. one it's other angles are , with corresponding opposite and adjacent sides of length and respectively.
Note that the former possibility in represents when the particle first makes contact with the spring, leaving us with the latter by the condition. That is, T satisfies:
as was to be shown.

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 39
 14052016 07:22
STEP III Q6
SolutionLet and be the points in the complex plane represented by and the midpoint of respectively. By considering the right angled and observing that lie on the unit circle centred at , note:
As desired.
For ease of LaTeX, translate Greek to English. Also, recall .
Then, taking care with the fact that moduli are nonnegative, by :
Applying :
Rearranging the arguments slightly, that is:
Observing that , the middle two terms cancel, allowing us to again apply :
And finally :
It follows that
As was to be shown.
Defining as the point in the complex plane represented by the complex number with defined similarly, and noting that these points define a cyclic quadrilateral , this result states that:
i.e. the sum of the product of the lengths of each pair of opposing sides in a cyclic quadrilateral is given by the product of the lengths of it's pair of diagonals. See Ptolemy's Theorem.

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 40
 14052016 09:07
STEP III Q7
Solution
(i)
(ii)
Last edited by Farhan.Hanif93; 14052016 at 09:10.Post rating:1
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Updated: June 23, 2016
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