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:
x
Turn on thread page Beta
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Maths
>< Maths Exams

STEP Maths II, III 2009 Solutions watch

Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 21
 26042016 15:31
STEP III, Question 12:

Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 22
 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. 
IrrationalRoot
 Follow
 19 followers
 17 badges
 Send a private message to IrrationalRoot
Offline17ReputationRep: Follow
 23
 26042016 16:38
Doing III Q8 right now .

IrrationalRoot
 Follow
 19 followers
 17 badges
 Send a private message to IrrationalRoot
Offline17ReputationRep: Follow
 24
 26042016 17:26
STEP III, Question 8
Last edited by IrrationalRoot; 26042016 at 17:37. 
 Follow
 25
 26042016 17:37
Done Q7 I can try and type it up now.
II 
 Follow
 26
 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. 
IrrationalRoot
 Follow
 19 followers
 17 badges
 Send a private message to IrrationalRoot
Offline17ReputationRep: Follow
 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
. 
Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 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 get 
 Follow
 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 
 Follow
 30
 26042016 19:16
(Original post by IrrationalRoot)
Probably better to use the fact that for the differentiation at the start. 
Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 31
 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. 
IrrationalRoot
 Follow
 19 followers
 17 badges
 Send a private message to IrrationalRoot
Offline17ReputationRep: Follow
 32
 26042016 20:49
Is there a thread for STEP I 2009 solutions already? I can't seem to find one.

Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 33
 26042016 21:59
(Original post by IrrationalRoot)
Is there a thread for STEP I 2009 solutions already? I can't seem to find one. 
IrrationalRoot
 Follow
 19 followers
 17 badges
 Send a private message to IrrationalRoot
Offline17ReputationRep: Follow
 34
 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. 
Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 35
 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. 
IrrationalRoot
 Follow
 19 followers
 17 badges
 Send a private message to IrrationalRoot
Offline17ReputationRep: Follow
 36
 26042016 22:14
(Original post by Zacken)
Different examiners  the examiner for STEP I is a nice dude, or so I've heard. 
Zacken
 Follow
 318 followers
 22 badges
 Send a private message to Zacken
 Thread Starter
Offline22ReputationRep: Follow
 37
 26042016 22:15
(Original post by IrrationalRoot)
Oh wow, so the II and III examiners are just lazy lol. 
Farhan.Hanif93
 Follow
 52 followers
 17 badges
 Send a private message to Farhan.Hanif93
Offline17ReputationRep: Follow
 38
 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.

Farhan.Hanif93
 Follow
 52 followers
 17 badges
 Send a private message to Farhan.Hanif93
Offline17ReputationRep: Follow
 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.

Farhan.Hanif93
 Follow
 52 followers
 17 badges
 Send a private message to Farhan.Hanif93
Offline17ReputationRep: Follow
 40
 14052016 09:07
STEP III Q7
Solution
(i)
(ii)
Last edited by Farhan.Hanif93; 14052016 at 09:10.
Reply
Submit reply
Turn on thread page Beta
Related discussions:
 STEP Maths I, II, III 1997 Solutions
 STEP Maths I, II, III 2001 Solutions
 STEP Maths I, II, III 1994 Solutions
 STEP Maths I,II,III 1987 Solutions
 STEP Maths I, II, III 1995 Solutions
 STEP maths I, II, III 1990 solutions
 STEP Maths I, II, III 1993 Solutions
 STEP I, II, III 2000 solutions
 STEP II/III 2014 solutions
 STEP I, II, III 1999 solutions
Related university courses:

Mathematics, Statistics and Financial Economics
Queen Mary University of London

University of Lincoln

University of Portsmouth

University of St Andrews

Environmental Science and Mathematics
University of Stirling

Financial Mathematics with Foundation Year
University of Salford

Durham University

Mathematics and Computer Science with an Industrial Year
University of Birmingham

Mathematics (with foundation year)
University of Hull

University of Warwick
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 rayquaza17
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Labrador99
 EmilySarah00
Updated: April 2, 2017
Share this discussion:
Tweet