STEP maths I, II, III 1992 solutions

Original post by ben-smith

Sorry but I am slightly confused as to why you chose P(2k+2). Do you mean P(2k) but with k replaced with k+1?

I mean P(2k+2). Of course, since 2(k+1) = 2k+2, P(2k+2) is P(2k) with k replaced by k + 1.

Reply 182

yeah, sorry. stupid question.

Reply 183

OK, thanks. This is what I have done:

\frac{P(2k+2)}{P(2k)}=\frac{4k^2+10k+6}{4k^2+12k}

Set this to =1 and we get k=3

\frac{P(2k+3)}{P(2k+1)}=\frac{4k^2+14k+12}{4k^2+16k}

Set this to =1 and we get k=6.
P(6)>P(13) so maximum when S=6.

however the question says values of S and I notice that P(8)=P(6), so it would be a max to. why wasn't that a solution when solving for k earlier? :confused:

Reply 184

DFranklin

Well, if P(2k+2)/P(2k) = 1, what does that tell you about whether or not P(2k+2) = P(2k)?

[I confess I expected there to be a distinct maximum, so I didn't allow for equality in my post at #180. However, a certain amount of thinking for your self is expected with STEP...]

Edit: So, having looked at the paper for more than 5 seconds, I see it does ask for the values, so I guess we should have expected more than 1 possible choice for k.

(edited 13 years ago)

Reply 185

Yes, I just realised that I was being an idiot and that I had forgotten to follow through on my (*your*) original proposition. If P(2k)=P(k+2) and k=3 then max occurs at P(6,8) because a value of k where P(2k)=P(k+2) holds is like a stationary point. (very badly expressed)

What do you think of the difficulty of this question? how would you rate it?

Reply 186

DFranklin

From my bare glance at it (basically what you've posted), it feels pretty straightforward. But it's not always easy for me to judge difficulty (I am probably less phased by "unusual concepts" than most, but not actually as good at the manipulation asI should be).

Reply 187

When I first saw the question, I thought I could use markov chains and stochastic matrices (I had just been reading about them) but I couldn't think of a feasible way of doing it, woe is me...

Reply 188

Step1992Paper3Question9.pdf73.5KB

Original post by gyrase

STEP II/5

The group has an identity element e, other elements g_1, g_2 ... and the operation is say *. Then the order of an element g_1 is n such that g_1 * g_1 * g_1 * ... = (g_1)^n = e.

Let A be an element of group S. detA = 1.

Unparseable latex formula:

\displaystyle A=\begin{pmatrix} w&x \\ y&z \end{pmatrix}[br]\displaystyle A^-^1=[br]\frac{1}{wz + xy}\begin{pmatrix} z&-x \\ -y&w \end{pmatrix}

but wz + xy = 1 so

Unparseable latex formula:

\displaystyle A^-^1=\begin{pmatrix} z&-x \\ -y&w \end{pmatrix}

S is associative, has identity element

\displaystyle I=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}

and has unique inverse elements as above and closure;

\displaystyle A_1A_2=\begin{pmatrix} w_1&x_1 \\ y_1&z_1 \end{pmatrix}[br]\displaystyle \begin{pmatrix} w_2&x_2 \\ y_2&z_2 \end{pmatrix} = \displaystyle \begin{pmatrix} w_1w_2+x_1y_2&w_1x_2+x_1z_2 \\ y_1w_2+z_1y_2&y_1x_2+z_1z_2 \end{pmatrix}

A^-1 = A
I = A^2, w^2 + xy = 1
.............wx + xz = 0, x(w+z) = 0
.............yw + yz = 0, y(w+z) = 0
.............yx + z^2 = 1, yx + z^2 = w^2 + xy, z = w

(x-y)(w+z) = 0
z = w
2xw = 2yw = 0 .. w(x-y) = 0, x = y
also wz + xy = 1, w^2 = wz = z^2, solving gives A = I.

:will finish later:

I don't follow the bit in bold, surely zw-xy=1?
Maybe I'm going crazy...

Reply 189

mikelbird

8

Have the solution to 1992 Step 3 Q9 as pdf file but dont know how to place it in the list....will post here...ooops slight correction...see latest posting....

(edited 12 years ago)

Reply 190

Original post by SimonM

(Updated as far as post #159.) SimonM - 23.03.2009
........

STEP III Q15
First part:
if

0<r\leq a)[br]\Rightarrow P(R<r)=\frac{\pi r^2}{4}/a^2=\frac{\pi r^2}{4a^2}

if

r>a

then the area we are concerned with will be the quarter of a circle like the last part but minus the bits where it exceeds the square. Call the point where the circle intersects the square X. Now consider the Right-angled triangle OCX. since OC=a and OX=r then the angle

C\hat{O}X=arccos{\frac{a}{r}}

. So, the area of the sector subtended on that angle is

\frac{1}{2}r^2arccos{\frac{a}{r}}

and the area of the triangle OCX is

\frac{1}{2}rasin(arccos\frac{a}{r})=\frac{1}{2}ra \sqrt{1-(\frac{a}{r})^2}

Subtracting one from the other and then doubling we get the area outside of the i.e:

P(R<r)=(\frac{\pi r^2}{4}-2(\frac{1}{2}r^2arccos{\frac{a}{r}}-\frac{1}{2}ra \sqrt{1-(\frac{a}{r})^2}))/a^2=\frac{\pi r^2}{4a^2}+\frac{1}{a}r\sqrt{1-(\frac{a}{r})^2}-\frac{r^2}{a^2}arccos{\frac{a}{r}}

simplifying we get:

((\frac{r}{a})^2-1)^{\frac{1}{2}} +\frac{\pi r^2}{4a^2}-\frac{r^2}{a^2}arccos(\frac{a}{r})

as required.
Now, to find the median. The median will occur when

P(R<r)=\frac{1}{2}

. Since we don't know whether this will occur for r<a or not we will have to choose one case, solve and then see whether the result tallies with our initial assumption. Seeming as the case r<a is much simpler, let's try that:

\frac{\pi r^2}{4a^2}=\frac{1}{2} \Rightarrow r=a(\sqrt{\frac{\pi}{2}})^{-1}

and since

(\sqrt{\frac{\pi}{2}})^{-1}<1

we can accept this result.
Next part:
for r<a,

\frac{\pi r^2}{4a^2}=\displaystyle\int^r_0(freq.dens)dr[br]\Rightarrow \frac{\pi r^2}{4a^2}=[freq.dens.]^r_0

. Differentiating both sides wrt r:
frequency density

=\frac{\pi}{2a^2}r

Now, for r>a:

\displaystyle\int^r_0(freq.dens)dr=((\frac{r}{a})^2-1)^{\frac{1}{2}} +\frac{\pi r^2}{4a^2}-\frac{r^2}{a^2}arccos(\frac{a}{r})

So, again, we differentiate wrt r to get

freq.dens.=\frac{r}{a^2}(\frac{ \pi}{2}-2arccos(\frac{a}{r}))

*Disclaimer: I have obviously omitted some steps here and I did so because Latex is a **** sometimes. However, it was a bit fiddly so, if you get something different to me then it is fairly likely that I have made a mistake*
Last part:
This event will occur when the goat is inside a quarter circle of radius a, centre O and also within the half of the square made up of a rectangle with OC being one of the longer sides (draw a diagram).
let the mid point of OA be X and the point where the arc and the perpendicular bisector of OA meet be Y.
Since OX=a/2 and OY=a then the angle at O, XOY=pi/3 and therefore the angle COX=pi/6. Now, to find the area, all we have to do is find the area of the triangle OXY and the sector COX and add them together i.e:
Area

=\frac{1}{2}a\frac{a}{2}sin(\frac{\pi}{3})+\frac{1}{2}a^2\frac{\pi}{6}

.
Now, to find the probability, divide by

\frac{\pi a^2}{4}

to get:

\frac{\sqrt{3}}{2 \pi}-\frac{1}{6}

Reply 191

Step1992Paper3Question9.pdf73.5KB

Original post by SimonM

(Updated as far as post #159.) SimonM - 23.03.2009
...

STEP III Q16
First part:
Given a particular order of m+n trials, the probability of getting m successes and n failures is

p^m(1-p)^n

However, we need to account for all the possible orderings of them which means we have to multiply this probability by the number of ways this event occur, that is:
P(exactly m successes and n failures)

=p^m(1-p)^n\begin{pmatrix} m+n \\ m\end{pmatrix}

Second part:
P(fail to spot exactly n|spot exactly m)=

\frac{P(failto spotn\cap spotm)}{P(spot m)}

P(fail to spot n and you spot m)

=\frac{e^{-\lambda}\lambda^{m+n}}{(m+n)!}p^m(1-p)^n\frac{(m+n)!}{m!n!}

P(spot m)

= \frac{e^{-\lambda}\lambda^m}{m!}p^m \begin{pmatrix} m \\ m \end{pmatrix} + \frac{e^{-\lambda}\lambda^{m+1}}{(m+1)!}(1-p)p^m \begin{pmatrix} m+1 \\ m \end{pmatrix} + \frac{e^{-\lambda}\lambda^{m+2}}{(m+2)!}(1-p)^2p^m \begin{pmatrix} m+1 \\ m \end{pmatrix}...

Or more concisely:
P(spot m)

=\sum^{\infty}_{r=0} \frac{e^{-\lambda}\lambda^{m+r}}{(m+r)!}(1-p)^rp^m \begin{pmatrix} m+r \\ m \end{pmatrix}=\frac{e^{-\lambda}p^m\lambda^m}{m!} \displaystyle\sum^{\infty}_{r=0} \frac{\lambda^r(1-p)^r}{r!}

\therefore

P(fail to spot exactly n|spot exactly m)

=\frac{\lambda^n(1-p)^n}{n!}(\displaystyle\sum^{\infty}_{r=0} \frac{\lambda^r(1-p)^r}{r!})^{-1}

.
Now, note that

\displaystyle\sum^{\infty}_{r=0} \frac{\lambda^r(1-p)^r}{r!}

is the maclaurin series expansion of

e^{\lambda (1-p)}

which means that:
P(fail to spot exactly n|spot exactly m)

=\frac{\lambda^n(1-p)^n}{n!}e^{-(1-p)\lambda}

as required.

Reply 192

mikelbird

8

Original post by mikelbird

this is the correct version...

Reply 193

STEP II question 12
Let tension in string be T then since extension is

Unparseable latex formula:

2\lambdar\sin\theta=\frac{d}{2V\cos\theta}

Taking moments about the centre of the wheel

C_1

for forces acting on

C_1

mgr=Tr\cos\theta

Hence,

Unparseable latex formula:

mg=\frac{2\lambdar\sin\theta\cos\theta}{d}\Rightarrow\sin2\theta=\frac{gd}{r\lambda}

for equilibrium
(i) If there is no equilibrium position then there is no solution for

\sin2\theta\Rightarrow\lambda< \frac{gd}{r}

(ii) There will be just one equilibrium position if

\sin2\theta= \frac{gd}{r\lambda}

has only one solution
i.e.

2\theta=90^{\circ}\Rightarrow \lambda=\frac{gd}{r}

(iii) If

0<\sin2\theta<1

then there are two solutions so two equilibrium positions for

\lambda>\frac{gd}{r}

(iv) We cannot have more than two equilibrium positions.

Reply 194

STEP III number 13

MI of disc and particle is

\frac{1}{2}4b^2m+4b^2m=6b^2m

In initial position, extension of the string is

Unparseable latex formula:

\{pi}b

So elastic energy stored in the string is

\frac{kmg\pi b}{2}

P reaches C if this is greater than the gain in PE of the system if P reaches C
i.e. if

\frac{kmg \pi b}{2}>2bmg \Rightarrow k > \frac {4bmg}{mg \pi b}= \frac {4}{ \pi}

so since we are given that

k> \frac {4}{\pi}

it follows that P does reach C.
The system will have KE at this point of

\frac{kmg\pi b}{2}-2bmg \Rightarrow6b^2m \omega^2=\frac{kmg \pi b}{2}-2bmg

I.e.

3b \omega^2=\frac{1}{2}(kg \pi-4g)\Rightarrow \omega^2= \frac{g(k\pi-4)}{6}

Resolving radially for particle at Q, if

R

is the reaction of the disc on Q we have

mg-R=2m\omega^2b

\Rightarrow R=m(g-2b\omega^2)=m\left(g-\frac{g(k\pi-4)}{3}\right)=\frac{mg}{3}(7-k\pi)

Reaction on axis consistsn of ca force supporting the weight of the disc plus the force required to cause particle to have circular motion, i.e.

m+\frac{mg}{3}(7-k\pi)=\frac{mg}{3}(10-k\pi)

as required

(edited 12 years ago)

Reply 195

STEP II number 5(second part)

\text{A}^{-1}= \text{A iff } w=z \text { and } x=y=0

So all elements of the form

\begin{pmatrix} x & 0\\ 0 & x \end {pmatrix}

\begin{pmatrix} a & b\\c & d \end{pmatrix}

has order 2 if

a^2+bc=bc+d^2 \text{ and }ab+bd=ac+cd \text{ with also }ad-bc=1

\text{i.e. }b=c=0 \text{ and }a^2=d^2=1 \text{ with } ad=1 \text{ i.e. } \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \text{ or }\begin{pmatrix} -1 & 0\\0 & -1 \end{pmatrix}

\text{ or if }b,c \not=0 \text{ then we must have }a+d=0,bc=1 \text{ and }bc=-1 \text{ (contradiction) }

\text{So we only have the two elements given above}

.

\text{If A }=\begin{pmatrix} w & x \\ y & z \end{pmatrix} \text{ has order 3 then A}^{-1}=\text{A}^2

Unparseable latex formula:

\text{ so }\dfrac{1}{wz-xy}\begin{pmatrix}z & -x\\-y & w\end{pmatrix}}=\begin{pmatrix}w^2+xy & wx+xz \\wy+yz & xy+z \end{pmatrix}

\Rightarrow\dfrac{z}{wz-xy}=w^2+xy\text{ ; }\dfrac{-y}{wz-xy}=wx+xz\text{ ;}\dfrac{-y}{wz-xy}=wy+yz \text{ and }\dfrac{w}{wz-xy}=xy+y^2

-x=wx+xz \text{ and }-y=wy+yz\Rightarrow x=y=0 \text{ or }-1=w+z

\text{but }x=y=0\Rightarrow z=w^2 \Rightarrow w=z=1 \text{ and A is the identity matrix}

\text{Hence we must have }1+w+z=0

\text{One such element is}\begin{pmatrix}\omega & 0\\0 & \omega^2 \end{pmatrix} \text{ where }\omega \text{ is a complex cube root of unity}

Reply 196

STEP I number 6
I'm a beginner at using latex so I hope there are not too many mistakes.

\text{Let P be the point }(\theta,\text{f}(\theta)) \text{ and Q }((\theta+\delta\theta),(\text{f}(\theta+\delta\theta))\text{ in diagram below}

\text{AB}=\delta\theta \text{ and the gradient at P is f}^/(\theta) \text{ so RS}=\text{f}^/(\theta)\delta\theta

Unparseable latex formula:

\text{If }\theta\text{ is small then SQ }\approx RS}

\Rightarrow\text{f}(\theta + \delta \theta))\approx\text{f}(\theta)+\text{f}^/(\theta)\delta\theta

\text{With reference to the diagram on the question paper let D be the}[br]\text{ foot of the perpendicular from C to AB then if BD}=x

\text{we have }h=x\tan(\theta+\phi)=(a+x)\tan \theta

\text{f}(\theta)=\tan \theta \Rightarrow\text{f}^/(\theta)=\sec^2\theta=(1+\tan^2 \theta)

\text{if }h\text{ is very large then }\phi \text{ will be small so by the above result}\tan (\theta+ \phi)\approx\tan (\theta+\phi\sec^2 \theta)

\text{so }h=x(\tan \theta+\phi\sec^2 \theta)=a\tan \theta+x\tan \theta

\text{i.e. }x=\dfrac{a\tan \theta}{\phi\sec^2 \theta}

\phi \ \mathrm{\ very\ small\ } \rightarrow h \approx x\tan\theta

\Rightarrow h=\dfrac{a \tan^2 \theta}{\phi\sec^2 \theta}=\dfrac{asin^2 \theta}{\phi}\text{ as required}

\text{AC}=\dfrac{h}{\sin \theta}=\dfrac{a\sin \theta}{\phi}

\dfrac{d}{d\phi}\text{(AC)}=-\dfrac{a\sin \theta}{\phi^2}\Rightarrow\text{approx error in AC is }=\dfrac{a\sin \theta}{\phi^2}\delta\phi

\text{the error when we double }a\text{ will thus be }-\dfrac{2a\sin \theta}{(2\phi)^2}\delta\phi

\text{since doubling }a\text{ will double }\phi\text{ since }h\text{ is constant}

\text{hence, ratio of errors is }\dfrac{1}{2}\text{ so approximate error is halved by doubling }a

(edited 10 years ago)

1992.I.6Picture.jpg6.5KB

Reply 197

SimonM

OP

Original post by brianeverit

STEP I number 6
I'm a beginner at using latex so I hope there are not too many mistakes.

You don't need to put everything in latex tags. You can leave the text outside. Looks good otherwise :smile:

Keep up the good work.

Reply 198

Original post by SimonM

...

STEP II Q15
P(n)=number of ways of falling on those numbers times

(\frac{1}{2})^n

let a +ve step=i and a -ve step=j.
So:

P(n)=P(\Sigma i-\Sigma j =2,3,4,5)

where

\Sigma i+\Sigma j =n[br]\therefore \Sigma j = n- \Sigma i \Rightarrow \Sigma i-\Sigma j = 2\Sigma i -n

let n=5:

P(5)=P(2\Sigma i -5 =3,5)

We can omit 2 and 4 because even minus odd is odd.
So

\Sigma i=5,4

no. of ways of getting 5is =1
no. of ways of getting 4js =5 choose 4=5
so, number of ways of falling on 2,3,4,5=6.
Therefore

P(5)=6/32=3/16

as required

P(6)=P(2 \Sigma i -6=2,4)

occurs when

\Sigma i=5,4

no. of ways of getting 5is =6 choose 5 =6
no. of ways of getting 4js =6 choose 4=15
so, number of ways of falling on 2,3,4,5=21.
therefore

P(6)=21 \frac{1}{2^6}=\frac{21}{64}

as required

P(2k)=P(2 \Sigma i -2k)=2,4

so

\Sigma i=k+1,k+2

no. of ways of getting (k+1)is =2k choose k+1
no. of ways of getting (k+2)js =2k choose k+2
so, number of ways of falling on 2,3,4,5=

\begin{pmatrix} 2k \\ k+1\end{pmatrix} + \begin{pmatrix} 2k \\ k+2\end{pmatrix}=\frac{2k!}{(k-1)!(k+1)!} + \frac{2k!}{(k-2)!(k+2)!} =\frac{(2k+1)!}{(k+2)!(k-1)!}=\begin{pmatrix} 2k+1 \\ k-1\end{pmatrix}

so

P(2k)=\begin{pmatrix} 2k+1 \\ k-1\end{pmatrix}(\frac{1}{2})^{2k}

as required.

P(2k+1)=P(2 \Sigma i -(2k+1))=3,5

so

\Sigma i=k+2,k+3

no. of ways of getting (k+2)is =2k+1 choose k+2
no. of ways of getting (k+3)js =2k choose k+2
Adding these two together we get:

\frac{(2k+1)!}{(k-1)!(k+2)!} + \frac{(2k+1)!}{(k-2)!(k+3)!}=\frac{(2k+2)!}{(k-1)!(k+3)!}=\begin{pmatrix} 2k+2 \\ k+3\end{pmatrix}

So,

P(2k+1)=\begin{pmatrix} 2k+2 \\ k+3\end{pmatrix} (\frac{1}{2})^{2k+1}

.
Now, it seems reasonable to assume that, as s gets bigger, P(s) gets smaller and so that maximums will occur at "stationary points" where either

\frac{P(2k+2)}{P(2k)}=1

or

\frac{P(2k+3)}{P(2k+1)}=1

Solving the first case:

\frac{P(2k+2)}{P(2k)}=\frac{4k^2+10k+6}{4k^2+12k}=1\Rightarrow 6=2k \Rightarrow k=3 \Rightarrow S=6,8

Similarly for the second case:

\frac{P(2k+3)}{P(2k+1)}=\frac{4k^2+14k+12}{4k^2+16k}=1 \Rightarrow 12=2k \Rightarrow k=6 \Rightarrow S=13,15

Notice that P(6)>P(13) and so the max values will occur at

S=6,8

and, if I remember correctly, this is confirmed by a spreadsheet I made.

(edited 12 years ago)

Reply 199