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# STEP I, II, III 2011 Solutions Thread.

1. Since most of the discussion has died down now and the most common thing being posted in those threads are now solutions, I thought I may as well make a thread for it. If you spot any mistakes, feel free to post in this thread and we'll make changes. Cheers!

STEP I Paper here.
STEP II Paper here.
STEP III paper here.

Completed - Nice one, everyone!

STEP I:
1. Solution by Zuzuzu
2. Solution by Zuzuzu
3. Solution by Zuzuzu
4. Solution by Zuzuzu
5. Solution by Farhan.Hanif93
6. Solution by Zuzuzu
7. Solution by Farhan.Hanif93
8. Solution by Zuzuzu
9. Solution by jbeacom600
10. Solution by brianeverit
11. Solution by Farhan.Hanif93
12. Solution by DFranklin
13. Solution by Farhan.Hanif93

STEP II:
1. Solution by mikelbird
2. Solution by matt2k8, Solution by mikelbird
3. Solution by mikelbird
4. Solution by mikelbird
5. Solution by mikelbird
6. Solution by Farhan.Hanif93, Solution by mikelbird
7. Solution by mikelbird
8. Solution by mikelbird
9. Solution by cpdavis
10. Solution by Farhan.Hanif93
11. Solution by DFranklin
12. Solution by Farhan.Hanif93
13. Solution by ben-smith

STEP III:
1. Solution by Farhan.Hanif93
2. Solution by mikelbird
3. Solution by mikelbird
4. Solution by mikelbird
5. Solution by mikelbird
6. Solution by Farhan.Hanif93, Solution by piecewise
7. Solution by Schnecke
8. Solution by mikelbird
9. Solution by jbeacom600
10. Solution by brianeverit
11. Solution by DFranklin
12. Solution by ben-smith
13. Solution by DFranklin

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007 - 2008
2. Doing a solution for 9 STEP II
3. Done solution for 9 will scan it due to diagrams
4. STEP III Q7

(i)

for n = 2,

Check this satisfies the condition given that

therefore true in the basis case n=2

Assume where k is even

Therefore

and

Showing these satisfy , proves that if true for k, also true for k+2, hence true for all even n by induction.

(ii)

where m is even

Of the form required with and

Check this satisfies and you're done

(iii)

For even n,

and

as required

For odd n,

and

as required

Apologies in advance for any mistakes, I rushed this a little bit.
5. Solution for 9, scanned in sorry if the quality is not brilliant
Attached Images

6. OK, let's finish STEP II. Here's Q11 (I hope. This was very fiddly even before I started Latexing it...)

To start, lets ignore the vertical dimension, and try to get a mental picture of where i and j go. It's not that hard. If we do the normal 2D thing of j being the y-axis and i being the x-axis, then OA lies on the Y axis, OC is at roughly 4 o'clock (i.e. x > 0, y < 0) and OB is at roughly 8 o'clock (i.e. x < 0, y < 0). As we progress, we're going to have quite a few possible sign ambiguities (basically we're going to want to work out a unit vector (sin t, cos t) knowing only what tan t is); this should be obvious given the diagram and a little thought.

(i) Let the unit vector in direction PB = ai + bj + ck. If we write h for the size of the horizontal component (so that h^2 = a^2 + b^2), then we have
h^2+c^2 = 1 (since we have a unit vector), and also . So and so .

So in fact we have (so that , since c > 0) and so that .
Now consider a/b (keeping in mind OB is at roughly 8 o'clock). We have , so , so . So , so and since "x < 0" we choose a = -1/3.
Then .

Thus our unit vector is as desired.

(ii) Having done (i), it's fairly clear the plan should be to find unit vectors for PA and PC also. The force corresponding to each string will just be U, V, W times the appropriate unit vector.

So again, write the unit vector in direction PA in the form ai + bj + ck. In this case we know a = 0 (since OA is in the direction of j), and we also know . So , from which , giving a final unit vector .
Finally write the unit vector in direction PC in the form ai + bj + ck.
Again, let h^2 = a^2 + b^2, then and so .
Now we want to find a and b. We have a^2+b^2 = 3/4. Considering the position of C we have a > 0, b < 0 and . So .
So . So . So . Thus our final unit vector is
And of course the unit vector in the direction of W is just -k.

So adding these all up, we must have:

Dotting with i, we get: , and so .
Dotting next with j we get: . Replacing U with we end up with .
Finally dot with k to get: . Replace U and T with the appropriate multiples of V and we find and so .
So finally .
7. STEP I - Q11
Draw yourself the necessary diagram. (I'll attach a copy tomorrow, if I get time)
For this question; let AP=a, BP=b and define the point where the line perpendicular to AB through P meets AB as X.

Let the tensions in the string AP and BP be of magnitude T and take moments about G.

It follows that , as required.

Note that the length of the string is given by .
Using the relation we've just found; the fact that ; and the fact that alpha is acute, it follows that and thus that . It's worth noting that:

By considering triangle APB, and .

Therefore .

Note that, if is the angle of inclination of the bar to the horizontal, then . By considering triangle APX (also right angled), it follows that . Similarly, .

Note that and considering triangle XPG, it follows that , as required.

8. STEP I, Q12:

(i) Can always give change unless the first person has a £2 coin. There are m+1 people, only one of which has a £2 coin. Therefore p = m/(m+1).
(ii) As in (i), we fail if the first person has a £2 coin. This has probability 2/(m+2)
We also fail if the first person has a £1 coin but the next 2 people have 2 pound coins. This has probability .
In all other cases we succeed.
So p(fail) = .
So p(succeed) = 1-p(fail) = as desired.
(iii) So now, the failure cases are:
2, any, ... with p = 3/(m+3)
1,2,2, any, with p = .
1,1,2,2,2, any with p =
1,2,1,2,2, any, with p =
Adding these, p(fail) =
.
Then p(succeed) = 1-p(fail) = as desired.
9. STEP I question 10

]
10. STEP I Q 9

To do this equation, we need only the equation for the trajectory of the particle, in terms of , and the inital velocity, which I choose to be . We let as time.

It is quite easy to derive the equation of motion, which is:

. Now and both lie on the trajectory of the particle. Hence, on rearranging:

(*)

Now, and we may assume that so that if then . So and we may divide as follows:

Rearranging gives:

and so we have that:

Now rearranging (*) gives

.

Now it is straightforward enough to show that the range of the particle is given by :

. We can rewrite as

Hence, the range is

and so on cancelling out like terms and simplifying.
11. (Original post by Farhan.Hanif93)
...
STEP III Q12
let denote the kth derivative of f(t) w.r.t. t.
To find the expected value of Y we have to use the fact that
so, using the chain rule:
as required.
Now, to find the variance:
as required.
For the next part, notice the perfect fit between what the examiners have asked us to prove in the first part and the scenario of the second, the only thing left to see that we haven't been explicitly told is that is a random variable denoting the outcome of the ith toss (1=heads, 0=tails).
Using the first part, since G(t) is the pgf of N and H(t) is the pgf of Xi then the pgf of Y is G(H(f)) so we need to find H(t) and G(t):

To find the expected value:

(You can alternatively do this by using . I have done both methods and I got the same answer for both).
To find the variance, use the first few steps in your derivation of Var(Y) in the first bit of the question to give yourself a shortcut where you can do it in terms of Y's pgf alone:
.
Now to to find P(Y=r). First, consider G(H(t)):

Our most immediate problem is the 't' term which makes it difficult to get the term we want on it's own so let's differentiate r times and then set t=0 to isolate the P(Y=r) term:
which means we only have to divide both sides by r! to get the required result:
12. What do you guys reckon will be the question that the fewest number of people will attempt? After doing STEP III Q12 I have a feeling no-one will do that one. What modules are probability generating functions even on?
13. (Original post by ben-smith)
What do you guys reckon will be the question that the fewest number of people will attempt? After doing STEP III Q12 I have a feeling no-one will do that one. What modules are probability generating functions even on?
I didn't think STEP III Q12 was that bad, but for sure, not many will probably have covered PGFs. Generally no-one touches the "post-S2" probability questions, so I suspect you might be right.

STEP III, Q11 looks horrendous, but I suspect is not that bad (I am assuming that you get the right answer by assuming the vertical velocity = 0 at the point where the strings go taut - if that doesn't give the right answer then the difficulty has suddenly taken a quantum leap).

STEP II, Q11 not only looks horrendous, but I found it to be pretty long and tricky too. I suspect it will be one of the least answered questions on STEP II.
14. (Original post by DFranklin)
STEP III, Q11 looks horrendous, but I suspect is not that bad (I am assuming that you get the right answer by assuming the vertical velocity = 0 at the point where the strings go taut - if that doesn't give the right answer then the difficulty has suddenly taken a quantum leap).
This question is pretty tough. The fact that they've given so many 'show that's gives a good reflection of it's trickiness. I've given it a little thought so far and I'm already stuck on the magnitude of the couple, even though it feels like I shouldn't be. Probably has something to do with the fact that I don't know much of the stuff on M4 and thus know very little about couples. I should really be learning rotational dynamics properly first...
15. (Original post by DFranklin)
I didn't think STEP III Q12 was that bad, but for sure, not many will probably have covered PGFs. Generally no-one touches the "post-S2" probability questions, so I suspect you might be right.

STEP III, Q11 looks horrendous, but I suspect is not that bad (I am assuming that you get the right answer by assuming the vertical velocity = 0 at the point where the strings go taut - if that doesn't give the right answer then the difficulty has suddenly taken a quantum leap).

STEP II, Q11 not only looks horrendous, but I found it to be pretty long and tricky too. I suspect it will be one of the least answered questions on STEP II.
STEP III Q12 is definitely not necessarily 'hard' by STEP standards, I mean, I have only done S1 and had never heard of a pgf before today and I managed to get it out so it can't be that bad.
On the other hand, I've been looking and I can't find generator functions anywhere on the whole edexcel syllabus which is a bit tragic for those on edexcel who made the effort to do up to S4. Did you do generator functions at a-level?
OMG, I hadn't noticed STEP III Q11. That is a stomach wrenchingly horrible question. The amount of reading you have to do before you could even get into the question was already a bad sign.
16. OK, so let's have a go at STEP III, Q11...

Consider the point P. It's hanging from the point (a, 0, 0). After rotation, P has position , where h is the height of the disk.
So, what's the horizontal displacement? It's . So its size is . But of course the horizontal displacement is also , hence the first result.

Now let bT be the tension in the string. Then the horizontal component of the force from the string is just . We want the size of the tangential component of this, which is going to be .

So each string provides a turning moment .
Suppose we have n strings. Resolving vertically, .
But .

So .

So the n strings provide a total couple of as desired.

At this point, the disc is below the ceiling. When the strings go taut, the disc is b below the ceiling.

So the loss in GPE is

This must equal the rotational KE.

So , where I is the moment of inertia of the disk.

That is,

So as desired.

This took about 26 minutes, including LaTeX. I'd say that puts it in the "not too bad" category for STEP III

Edit: on a little thought, I'm not 100% convinced about the method for calculating the "tangential component" I've used; in terms of the picture I had in my mind, I think there are 2 compensatory sign errors (one for each x-component). I'd be very surprised to lost more than 1 mark for it though - it would be fine for a different mental picture. On the other hand, if you draw an actual diagram, it's obvious - it's just that it's hard to draw diagrams on here.
17. (Original post by ben-smith)
STEP III Q12 is definitely not necessarily 'hard' by STEP standards, I mean, I have only done S1 and had never heard of a pgf before today and I managed to get it out so it can't be that bad.
On the other hand, I've been looking and I can't find generator functions anywhere on the whole edexcel syllabus which is a bit tragic for those on edexcel who made the effort to do up to S4. Did you do generator functions at a-level?
I believe pgfs and mgfs used to be in the S5 module that no longer exists.

When I did Further Maths, the applied was very mechanics heavy. (From memory, only 2 of 12 questions on the applied paper would be probability based, although it might have been 3). I also did "Maths with Stats", and I believe they may have been mentioned there, although I'm not sure if that was our teacher going beyond the syllabus. It must have been either taught or in a textbook I had, however, as I recall doing CCE questions involving pgfs. (And we didn't have t'internet then).
18. Solutions to Q2,Q3,Q4
Attached Images
19. Step2011Paper3Question2.pdf (69.2 KB, 669 views)
20. Step2011Paper3Question3.pdf (66.4 KB, 489 views)
21. Step2011Paper3Question4.pdf (72.6 KB, 434 views)
22. Solution to Paper3 Question 8
Attached Images
23. Step2011Paper3Question8.pdf (63.4 KB, 418 views)
24. Solution to Paper 3 Question 5
Attached Images
25. Step2011Paper3Question5.pdf (61.1 KB, 403 views)

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