STEP I, II, III 2011 Solutions Thread.

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Farhan.Hanif93
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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
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Slowbro93
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Doing a solution for 9 STEP II
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Slowbro93
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Done solution for 9 will scan it due to diagrams
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Schnecke
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STEP III Q7

(i)

for n = 2,

 T_2 = (\sqrt{a+1} + \sqrt{a})^2 = (a+1) + 2\sqrt{a}\sqrt{a+1} + a = (2a+1) + 2\sqrt{a}\sqrt{a+1}



\Rightarrow A_2 = 2a+1, B_2 = 2

Check this satisfies the condition given that

 A_n^2 = a(a+1)B_n^2 + 1





LHS = 4a^2+4a+1, RHS = a(a+1)(4)+1 = 4a^2+4a+1 = LHS


therefore true in the basis case n=2


Assume  T_k = A_k + B_k\sqrt{(a)(a+1)} where k is even


 T_{k+2} = (A_k + B_k\sqrt{(a)(a+1)})((2a+1) + 2\sqrt{a}\sqrt{a+1})



 = (2a+1)A_k + (2a+1)\sqrt{(a)(a+1)}B_k + 2\sqrt{a}\sqrt{a+1}A_k + 2a(a+1)B_k 



= [(2a+1)A_k + 2a(a+1)B_k] + [(2a+1)B_k + 2A_k]\sqrt{(a)(a+1)}

Therefore  A_{k+2} = (2a+1)A_k + 2a(a+1)B_k

and

 B_{k+2} = (2a+1)B_k + 2A_k

Showing these satisfy  A_n^2 = a(a+1)B_n^2 + 1 , proves that if true for k, also true for k+2, hence true for all even n by induction.


(ii)

 T_m = A_m + B_m\sqrt{(a)(a+1)} where m is even

 \Rightarrow T_{m+1} = (A_m + B_m\sqrt{(a)(a+1)})(\sqrt{a+1} + \sqrt{a}) = [A_m + aB_m]\sqrt{a+1} + [A_m + (a+1)B_m]\sqrt{a}

Of the form required with C_{m+1} = A_m + aB_m and  D_{m+1} = A_m + (a+1)B_m

Check this satisfies  (a+1)C_n^2 = aD_n^2 +1 and you're done



(iii)

For even n,  T_n = A_n + B_n\sqrt{(a)(a+1)}

and

 A_n = \sqrt{a(a+1)B_n^2 + 1}

 \Rightarrow T_n = \sqrt{a(a+1)B_n^2 + 1} + \sqrt{B_n^2(a)(a+1)} as required

For odd n,  T_n = C_n\sqrt{a+1} + D_n\sqrt{a}

and

 C_n = \sqrt{\frac{aD_n^2+1}{a+1}}





\Rightarrow T_n = \sqrt{aD_n^2+1} + \sqrt{aD_n^2}
as required



Apologies in advance for any mistakes, I rushed this a little bit.
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Slowbro93
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Solution for 9, scanned in sorry if the quality is not brilliant
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DFranklin
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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 \dfrac{h}{c} = \tan(90-\theta) = \dfrac{1}{\tan \theta} = \dfrac{1}{\sqrt{2}}. So h = \dfrac{c}{\sqrt{2}} and so h^2+c^2 = \dfrac{1}{2}c^2 + c^2 = \dfrac{3}{2}c^2.

So in fact we have c^2 = 2/3 (so that c=\dfrac{\sqrt{2}}{\sqrt{3}}, since c > 0) and h^2=1/3 so that a^2+b^2 = 1/3.
Now consider a/b (keeping in mind OB is at roughly 8 o'clock). We have \dfrac{b}{a} = \tan \theta = \sqrt{2}, so b = a \sqrt{2}, so a^2+b^2 = 3a^2. So 3a^2 = 1/3, so a=\pm \dfrac{1}{3} and since "x < 0" we choose a = -1/3.
Then b = a \sqrt{2} = \dfrac{-\sqrt{2}}{3}.

Thus our unit vector is \dfrac{-1}{3}{\bf i} - \dfrac{\sqrt{2}}{3}{\bf j} + \dfrac{\sqrt{2}}{\sqrt{3}}{\bf k} 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 b = c\tan 30^o = c /\sqrt{3}. So 1 = b^2+c^2 = 4c^2/3, from which b = \dfrac{1}{2}, c = \dfrac{\sqrt{3}}{2}, giving a final unit vector \dfrac{1}{2}{\bf j} + \dfrac{\sqrt{3}}{2}{\bf k}.
Finally write the unit vector in direction PC in the form ai + bj + ck.
Again, let h^2 = a^2 + b^2, then h = c \tan 60^o = c \sqrt{3} and so c = \dfrac{1}{2}, h = \dfrac{\sqrt{3}}{2}.
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 \dfrac{b}{a} = -\tan \phi. So b = -a \tan \phi = \dfrac{-a\sqrt{2}}{4}=\dfrac{-a}{\sqrt{8}}.
So b^2+a^2 = \dfrac{a^2}{8} + a^2 = \dfrac{9}{8}a^2. So a^2 = \dfrac{3}{4}\dfrac{8}{9} = \dfrac{2}{3}. So a = \dfrac{\sqrt{2}}{\sqrt{3}}, b = -\dfrac{\sqrt{2}}{\sqrt{3}\sqrt{8}}=\dfrac{-1}{2\sqrt{3}}. Thus our final unit vector is \dfrac{\sqrt{2}}{\sqrt{3}}{\bf i} + \dfrac{-1}{2\sqrt{3}}{\bf j} + \dfrac{1}{2} {\bf k}
And of course the unit vector in the direction of W is just -k.

So adding these all up, we must have:

T(\dfrac{1}{2}{\bf j} + \dfrac{\sqrt{3}}{2}{\bf k}) +  U (\dfrac{-1}{3}{\bf i} - \dfrac{\sqrt{2}}{3}{\bf j} + \dfrac{\sqrt{2}}{\sqrt{3}}{\bf k}) + V(\dfrac{\sqrt{2}}{\sqrt{3}}{\bf i} + \dfrac{-1}{2\sqrt{3}}{\bf j} + \dfrac{1}{2} {\bf k}) - W{\bf k} = 0

Dotting with i, we get: \dfrac{-U}{3} + \dfrac{V\sqrt{2}}{\sqrt{3}} = 0, and so U = \dfrac{V3\sqrt{2}}{\sqrt{3}} = V\sqrt{6}.
Dotting next with j we get: \dfrac{T}{2} - \dfrac{\sqrt{2}}{3}U - \dfrac{1}{2\sqrt{3}}V = 0. Replacing U with V \sqrt{6} we end up with T = V \sqrt{3}.
Finally dot with k to get: \dfrac{T\sqrt{3}}{2} + \dfrac{U \sqrt{2}}{\sqrt{3}}+\dfrac{V}{2} = W. Replace U and T with the appropriate multiples of V and we find \dfrac{3V}{2} + \dfrac{V \sqrt{2}\sqrt{6}}{\sqrt{3}} + \dfrac{V}{2} = W and so \left(\dfrac{3}{2} + 2 + \dfrac{1}{2}\right)V = W \implies V = \dfrac{W}{4}.
So finally T = \dfrac{\sqrt{3} W}{4}, U =\dfrac{W\sqrt{6}}{4} = \dfrac{W\sqrt{3}}{2\sqrt{2}}.
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Farhan.Hanif93
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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 3dT\sin \alpha - 4dT\sin \beta =0 \implies \boxed{3\sin \alpha = 4\sin \beta}, as required.

Note that the length of the string is given by \l = a+b.
Using the relation we've just found; the fact that \cos \beta = \dfrac{4}{5}; and the fact that alpha is acute, it follows that \sin \alpha = \cos \beta and thus that \angle APB = \dfrac{\pi}{2}. It's worth noting that:
\sin \beta = \dfrac{3}{5}
\sin \alpha = \dfrac{4}{5}
\cos \alpha = \dfrac{3}{5}

By considering triangle APB, \sin \beta = \dfrac{a}{7d} \implies a = \dfrac{21d}{5} and \sin \alpha = \dfrac{b}{7d} \implies b = \dfrac{28d}{5}.

Therefore \boxed{l = \dfrac{49d}{5}}.

Note that, if \theta is the angle of inclination of the bar to the horizontal, then \angle XPG = \theta. By considering triangle APX (also right angled), it follows that \sin \alpha = \dfrac{PX}{21d/5} \implies PX = \dfrac{84d}{25}. Similarly, \cos \alpha = \dfrac{AX}{a} \implies AX = \dfrac{63d}{25}.

Note that XG = 3d - \dfrac{63d}{25} = \dfrac{12d}{25} and considering triangle XPG, it follows that \tan \theta = \dfrac{XG}{XP} = \dfrac{1}{7} \implies \boxed{\theta = \arctan \dfrac{1}{7}}, as required.

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DFranklin
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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 \dfrac{m}{m+2}\dfrac{2}{m+1} \dfrac{1}{m} = \dfrac{2}{(m+1)(m+2)}.
In all other cases we succeed.
So p(fail) = \dfrac{2}{m+2}\left(1 + \dfrac{1}{m+1}\right) = \dfrac{2}{m+2}\dfrac{m+2}{m+1} = \dfrac{2}{m+1}.
So p(succeed) = 1-p(fail) = \dfrac{m-1}{m+1} as desired.
(iii) So now, the failure cases are:
2, any, ... with p = 3/(m+3)
1,2,2, any, with p = \dfrac{m}{m+3}\dfrac{3}{m+2} \dfrac{2}{m+1} = \dfrac{6}{m+3} \dfrac{m}{(m+2)(m+1)}.
1,1,2,2,2, any with p = \dfrac{m}{m+3}\dfrac{m-1}{m+2}\dfrac{3}{m+1}\dfrac{2}{m}\dfrac{1}{m-1}=\dfrac{6}{m+3}\dfrac{1}{(m+2)(m+1)}
1,2,1,2,2, any, with p = \dfrac{m}{m+3}\dfrac{3}{m+2} \dfrac{m-1}{m+1}\dfrac{2}{m}\dfrac{1}{m-1}=\dfrac{6}{m+3}\dfrac{1}{(m+2)(m+1)}
Adding these, p(fail) = \dfrac{3}{m+3}\left(1 + \dfrac{2m}{(m+2)(m+1)} + \dfrac{2}{(m+2)(m+1)}+\dfrac{2}{(m+2)(m+1)}\right)
=\dfrac{3}{m+3}\left(1 + \dfrac{2(m+2)}{(m+2)(m+1)}\right) = \dfrac{3}{m+3}\left(1+\dfrac{2}{m+1}\right) = \dfrac{3}{m+3}\dfrac{m+3}{m+1} = \dfrac{3}{m+1}.
Then p(succeed) = 1-p(fail) = \dfrac{m-2}{m+1} as desired.
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brianeverit
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STEP I question 10
 \text{Let collision occur at a height }h \text{ above the ground}
 \text{Taking upwards as positive}
 \text{Let velocities of A and B at collision be }u_A \text{ and }u_B \text{ then, since both have same displacement }
 \text{from starting position }u_A^2=2g(h-x) \text{ and }u_B^2=2g(h-x) \implies u_A=u_B=u \text{ say}
 \text{If velocities after collision are }v_A \text{ and }v_B \text{ then by conservation of momentum}
 Mu-mu=Mv_A+mv_B \text{ and since collision is perfectly elastic}
v_B-v_A=2u \implies v_B=2u+v_A
 \text {so }(M-m)u=Mv_A+2mu+mv_A \impliesa (M+m)v_A=(M-3m)u \implies v_A= \dfrac{(M-3m)u}{M+m}&gt;0 \text{ since }M&gt;3m
v_B=2u+ \dfrac{(M-3m)u}{M+m} \text { which is clearly also positive so both particles move upwards }
 \text {after the collision }
\text{considering motion of B before collision }h-x= \dfrac{u^2}{2g}
 \text{considering motion of B after collision, if maximum height attained is }y \text{ above point of collision}
y= \dfrac{v_B^2}{2g}= \dfrac{(3M-m)^2u^2}{2(M+m)^2g} \text { so maximum height above the plane is} \dfrac{(3M-m)^2u^2}{2(M+m)^2g}+h- \dfrac{u^2}{g}
=h+ \dfrac{[(3M+m)^2-(M+m)^2]u^2}{2g(M+m)^2} =h+ \dfrac{(8M^2-8Mn)u^2}{2g(M+m)^2}=h+ \dfrac{4M(M-m)u^2}{g(M+m)^2} \text { as required}]
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STEP I Q 9

To do this equation, we need only the equation for the trajectory of the particle, in terms of  t = \tan (\theta) ,  x and the inital velocity, which I choose to be  u ms^{-2} . We let  t = \tan ( \theta) as time.

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

 y = x t - \dfrac{gx^{2}}{2u^2}(1+t^2) . Now  (d_1, d_2) and  (d_2, d_1) both lie on the trajectory of the particle. Hence, on rearranging:

 d_1 t -d_2 = \dfrac{gd_1^{2}}{2u^2}(1+t^2)
 d_2 t -d_1 = \dfrac{gd_2^{2}}{2u^2}(1+t^2) (*)

Now,  (1+t^2) \geq 1 and we may assume that d_1 \neq d_2 so that if  d_1 =0 then  d_2 = 0 . So  \dfrac{gd_1^{2}}{2u^2}(1+t^2)  \neq 0 and we may divide as follows:

 \dfrac{d_1 t -d_2}{d_2 t -d_1} = \dfrac{\frac{gd_1^{2}}{2u^2}(1+t^2)}{\frac{gd_2^{2}}{2u^2}(1+t^2)} = \dfrac{d_1^2}{d_2^2}

Rearranging gives:

  (d_1d_2^2-d_1^2d_2)t = d_1d_2(d_2-d_1)t=d_2^3-d_1^3=(d_2-d_1)(d_2^2+d_1d_2+d_1^2) and so we have that:

 \tan (\theta) = \dfrac{d_1^2+d_1d_2+d_2^2}{d_1d_2}

Now rearranging (*) gives

 \dfrac{u^2}{g} = \dfrac{d_2^2(1+t^2)}{2(d_2t-d_1)} .

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

 x_{max} = \dfrac{ u^2 \sin (2 \theta)}{g} . We can rewrite  \sin (2 \theta) as

 \sin (2 \theta) = 2 \sin \theta \cos \theta = \dfrac{ 2 \sin \theta}{\cos \theta \sec ^2 \theta} = \dfrac{ 2t} {1+t^2}

Hence, the range is

 x_{max} = \dfrac{ 2tu^2}{g(1+t^2)} = \dfrac{2t}{1+t^2} . \dfrac{ d_2^2(1+t^2)}{2(d_2t-d_1)}

and so  x_{max} =  \dfrac{ d_2^2 + d_1d_2 +d_1^2}{d_2 + d_1} on cancelling out like terms and simplifying.
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Dirac Spinor
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(Original post by Farhan.Hanif93)
...
STEP III Q12
let D^{(k)}[f(t)] 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 E(Y)=D'[G(H(t))]|_{t=1}
so, using the chain rule:
E(Y)=D'[G(H(t))]|_{t=1}=G'(H(t))H'(t)|_{t=1}=G'(H(1))H'(1)=G'(1)H'(1)=E(N)E(X_i) as required.
Now, to find the variance:
Var(Y)=E(Y^2)-E(Y)^2=E(Y^2)-E(Y)+E(Y)-E(Y)^2=E(Y(Y-1))+E(Y)-E(Y)^2

=D''[G(H(t))]|_{t=1}+D'[G(H(t))]|_{t=1}-(D'[G(H(t))]|_{t=1})^2]

=[G''(H(t))H'(t)^2+H''(t)G'(H(t))]|_{t=1}+E(N)E(X_i)-(E(n)E(X_i))^2

=E(N(N-1))E(X_i)^2+E(X_i(X_i-1))E(N)+E(N)E(X_i)-(E(n)E(X_i))^2

=E(N^2)E(X_i)^2-E(X_i)^2E(N)+E(X_i^2)E(N)-(E(X_i)E(N))^2

=Var(N)E(X_i)^2+E(N)Var(X_i) 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 X_i 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):
G=E(t^N)=\displaystyle\sum^{\infty}_{n=1}P(N=n)t^n=\displaystyle\sum^{\infty}_{n=1}(\frac{1}{2})^nt^n=\frac{t/2}{1-t/2}=\frac{t}{2-t}

H(t)=E(t^{X_i})= \displaystyle\sum^1_{x=0}P(X_i=x)t^x=\frac{1}{2}\times 1+\frac{1}{2}\times t=\frac{1}{2}(1+t)

\therefore pgf_Y=G(H(t))=\frac{1+t}{2(2-0.5(1+t))}=\frac{1+t}{3-t}
To find the expected value:
E(Y)=D'[\frac{1+t}{3-t}]|_{t=1}=\frac{4}{(3-t)^2}|_{t=1}=1
(You can alternatively do this by using E(Y)=E(N)E(X_i). 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:
Var(Y)=D''[\frac{1+t}{3-t}]|_{t=1}+D'[\frac{1+t}{3-t}]|_{t=1}-(D'[\frac{1+t}{3-t}]|_{t=1})^2=[4(-2)(3-t)^{-3}(-1)]|_{t=1}+1-1=1.
Now to to find P(Y=r). First, consider G(H(t)):
G(H(t))=P(Y=1)t+P(Y=2)t^2+...+P(Y=r)t^r
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:
D^{(r)}[G(H(0))]=r!P(Y=r) which means we only have to divide both sides by r! to get the required result:
P(Y=r)=\frac{F^{(r)}[G(H(0))]}{r!}=\frac{4r!(3)^{-(r+1)}}{r!}=4(3)^{-(r+1)}
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Dirac Spinor
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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?
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DFranklin
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(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.
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Farhan.Hanif93
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(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...
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Dirac Spinor
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(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.
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DFranklin
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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 (a \cos \theta, a \sin \theta, h), where h is the height of the disk.
So, what's the horizontal displacement? It's (a, 0) - (a\cos \theta, a \sin \theta) = a(1-\cos \theta, \sin \theta). So its size is a \sqrt{(1-\cos \theta)^2 + \sin^2 \theta} = a \sqrt{1 - 2 \cos \theta + \cos^2 \theta + \sin^2\theta}= a \sqrt{2(1-cos\theta)} = a \sqrt{4 \sin^2 \dfrac{\theta}{2}} = 2a \sin \dfrac{\theta}{2}. But of course the horizontal displacement is also b\sin \phi, 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 Ta(1-cos \theta, \sin \theta). We want the size of the tangential component of this, which is going to be Ta(1-cos \theta, \sin \theta)\cdot (a \sin \theta, a \cos\theta) = Ta^2 \sin \theta.

So each string provides a turning moment Ta^2 \sin \theta.
Suppose we have n strings. Resolving vertically, nbT \cos \phi = mg.
But b \cos \phi = \sqrt{b^2 - b^2 \sin^2 \phi} = \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}}.

So T = \dfrac{mg}{n \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}}}.

So the n strings provide a total couple of nTa^2 \sin \theta = \dfrac{mga^2 \sin \theta}{\sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}}} as desired.

At this point, the disc is  b \cos \phi = \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}} below the ceiling. When the strings go taut, the disc is b below the ceiling.

So the loss in GPE is mg(b - \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}})

This must equal the rotational KE.

So \dfrac{1}{2} \omega^2 I = mg(b - \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}}), where I is the moment of inertia of the disk.

That is, \dfrac{1}{2}\omega^2 \dfrac{ma^2}{2} = mg(b - \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}})

So \dfrac{a^2 \omega^2}{4g} = b - \sqrt{b^2 - 4a^2\sin^2\dfrac{\theta}{2}} 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.
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DFranklin
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(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).
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mikelbird
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Solutions to Q2,Q3,Q4
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mikelbird
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Solution to Paper3 Question 8
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mikelbird
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Solution to Paper 3 Question 5
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