\displaystyle [br]\begin{align*}G(1) + G(-1) & = \sum_{k \geqslant 0} \left(\mathbb{P}(X = k) + (-1)^k \mathbb{P}(X=k)\right) \\ & = 2\sum_{k \geqslant 0, \, k \, \mathrm{even}} \mathbb{P}(X=k) + \sum_{k \geqslant 0, \, k \, \mathrm{odd}} \left(\mathbb{P}(X=k) - \mathbb{P}(X=k) \right) = 2 \sum_{k \geqslant 0} \mathbb{P}(X=2k)\end{align*}

as required. If

X

is Poisson, with parameter

\lambda

then we have

Unparseable latex formula:

\displaystyle[br]\begin{equation*}G(t) = \sum_{k \geqslant 0} e^{-\lambda} \frac{\lambda^k}{k!} t^k = e^{-\lambda} \sum_{k \geqslant 0} \frac{(\lambda t)^k}{k!} = e^{-\lambda} e^{\lambda t} = e^{-\lambda(1-t)} \end{equation*}

i) We first find

k

:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sum_{r \geqslant 0} \mathbb{P}(Y=r) = k \sum_{r \geqslant 0} \mathbb{P}(X=2r) = \frac{k}{2} (G(1) + G(-1)) = \frac{k}{2}(1+ e^{-2\lambda}) = 1\end{equation*}

so that

k = \frac{2e^{\lambda}}{e^{\lambda} + e^{-\lambda}} = \frac{e^{\lambda}}{\cosh \lambda}

. Then the PGF of

Y

is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}H(t) = \frac{1}{\cosh \lambda}\sum_{r \geqslant 0} e^{\lambda}t^{2r} \mathbb{P}(X=2r) = \frac{1}{\cosh \lambda}\sum_{r \geqslant 0 } \frac{(\lambda t)^{2r}}{(2r)!} = \frac{\cosh \lambda t}{\cosh \lambda}\end{equation*}

Moreover, we have

\displaystyle \mathbb{E}(Y) = H'(1) = \frac{\lambda \sinh \lambda}{\cosh \lambda} = \lambda \tanh \lambda < \lambda

as

\tanh \in (0,1)

for

\lambda > 0

.

ii) Note that

G(1) + G(-1) + G(i) + G(-i) = 4\mathbb{P}(X = 0,4,8, 12, \ldots)

and so we have

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sum_{r \geqslant 0} \mathbb{P}(Z=r) = c\sum_{r \geqslant 0} \mathbb{P}(X=4r) = \frac{c}{4}(G(1) + G(-1) + G(i) + G(-i) = \frac{ce^{-\lambda}}{2}\left(\cos \lambda + \cosh \lambda \right) = 1 \end{equation*}

Hence, we then have the PGF of

Z

as

Unparseable latex formula:

\displaystyle[br]\begin{equation*}R(t) = \frac{2e^{\lambda}}{\cos \lambda + \cosh \lambda}\sum_{r \geqslant 0}e^{-\lambda} \frac{(\lambda t)^{4r}}{(4r)!} = \frac{\cos \lambda t + \cosh \lambda t}{\cos \lambda + \cosh \lambda } \Rightarrow \mathbb{E}(Z) = R'(1) = \lambda \frac{\sinh \lambda - \sin \lambda}{\cos \lambda + \cosh \lambda} \end{equation*}

which is not always

< \lambda

, for example, by taking

\lambda = \pi

.

(edited 5 years ago)

Reply 263

jtSketchy

6

SIII, Q3.

We have

x^a(x^b(x^cy)’)’ = x^{a+b+c}y’’ + (b+2c)x^{a+b+c-1}y’ + c(b+c-1)x^{a+b+c-2}y.

So to match with the given ODE, we require
(I)

a+b+c = 2

(II)

b+2c = 1-2p

(III)

c(b+c-1) = p^2-q^2

.
We need to find

a,b,c

to show this is possible. Note that (II) gives

b = 1-2p - 2c

and substitution into (III) gives

c(-c-2p) = p^2-q^2

, which after some rearranging becomes

(c+p)^2 = q^2

, i.e.

c = -p \pm q.

Substitution of this into (II) gives

b = 1\mp 2q

and then substitution into (I) gives

a = 1+p \pm q.

(i) Here we want to solve the ODE when

f\equiv 0.

Write the ODE in the form

x^a(x^b(x^cy)’)’ = 0

, i.e.

(x^b(x^cy)’)’ = 0.

Integrating up gives

x^b(x^c y)’ = A

for some constant

A

, and so

(x^c y)’ = Ax^{-b}.

Now if

b\neq 1

we can integrate this to get

x^cy = \tilde{A} x^{1-b} + B

for some constants

\tilde{A},B

and thus we get the solution of

y = \tilde{A} x^{1-b-c} + Bx^{-c}

.

Looking at our expressions for

b,c

we see that this is exactly

y= A x^{p+q} + Bx^{p-q}

for some constants

A,B

regardless of what sign we chose for

q

in the expression for

c

.

Finally, if

b = 1

, i.e.

q = 0

, then we integrate up to get a

\ln(x)

instead, and so the solution is

y = Ax^p \ln(x) + Bx^p

.

(ii) In this case we have

f(x) = x^n

and

q =0

. So

a = 1+p,\ b = 1,\ c = -p.

Thus we have

(x(x^{-p}y)’)’ = x^{n-p-1}.

So provided

n\neq p

we can integrate to get

x(x^{-p}y)’ = \frac{x^{n-p}}{n-p} + B

for some constant

B

. Dividing by

x

gives

(x^{-p}y)’ = \frac{x^{n-p-1}}{n-p} + Bx^{-1}

. So once again, provided

n \neq p

we can integrate again to find the general solution to be (after dividing by

x^{-p}

)

y = \frac{x^{n}}{(n-p)^2} + Bx^p\ln(x) + Cx^p.

Finally, if

n = p

then we have

(x(x^{-p}y)’)’ = x^{-1}

and so we get

(x^{-p}y)’ = \frac{\ln(x)}{x} + Bx^{-1}

. Noticing that the first term is just the derivative of

\frac{1}{2}(\ln(x))^2

, we can integrate again to find that the general solution is

y = \frac{1}{2}(\ln(x))^2 + Bx^p\ln(x) + Cx^p

.

Note: The ODE in (i) is just the homogeneous equation from that in (ii) (with

q =0

). So we expect that the solutions only differ by a particular solution, which is what we see.

Reply 264

jtSketchy

6

SIII Q12.

Spoiler

I always find it amazing how simple some of these probability questions can be once you write down the probabilities. They essentially become an easier version of a ‘Pure’ question. I would definitely recommend more people to look at them.

Reply 265

Garjun

1

STEP III Q4 :tongue:

STEP III 2018 Q4.pdf80.0KB

Reply 266

jtSketchy

6

SIII Q9.

(i)

Spoiler

(ii)

Spoiler

Thoughts:

Spoiler

Reply 267

jtSketchy

6

SIII Q11.

Spoiler

(edited 5 years ago)

Reply 268

jtSketchy

6

SIII Q10.

Spoiler

Thoughts:

Spoiler

(edited 5 years ago)

Reply 269

username1357565

15

Original post by carpetguy

I'll start assuming the first result was reached and the midpoint of the two meeting points was shown to be at

P

.
So we have the equation of the tangent to the hyperbola at

P

is

bx - ay\sin\theta = ab\cos\theta

and by similarities we reach the tangent at

Q

being

bx - ay\sin\phi = ab\cos\phi

.

From the perpendicular requirement,

\frac{b}{a\sin\theta} \times \frac{b}{a\sin\phi} = -1

\therefore b^2 = -a^2\sin\theta\sin\phi

Subtracting the two tangent equations from each other and rearranging:

y = \frac{b\,(\cos\phi \,-\, \cos\theta)}{\sin\theta \,-\, \sin\phi} = \frac{\sqrt{-a^2\sin\theta\sin\phi}\,(\cos \phi \,-\, \cos\theta)}{\sin\theta \,-\, \sin\phi}[br]

Multiplying the tangent equation for

P

by

\sin\phi

and the tangent equation for

Q

by

\sin\theta

, subtracting the two new equations and rearranging:

x = \frac{a\,(\cos\phi\sin\theta \,-\, \cos\theta\sin\phi)}{\sin\theta \,-\, \sin\phi}

\therefore x^2 + y^2 = \frac{a^2}{(\sin\theta \,-\, \sin\phi)^2}\,\{(\cos\phi \sin \theta \,-\, \cos\theta\sin\phi)^2 \,- \, \sin\theta\sin\phi\,(\cos\phi \,-\, \cos\theta)^2 \}

Now yeah that does look ugly, but you can see that the

-2\cos\phi\cos\theta\sin\phi\sin \theta

term is generated from both squares and is cancelled out - so inside the big bracket you're left with:

\cos^2\phi\sin^2\theta + \cos^2\theta\sin^2\phi - \sin\theta\sin\phi\cos^2\phi - \sin\theta\sin\phi\cos^2\theta

Note that this can be factorised to:

(\sin\theta \,-\, \sin\phi)(\cos^2\phi\sin\theta \,-\, \cos^2\theta\sin\phi)

and so

x^2 + y^2 = \frac{a^2}{(\sin\theta \,-\, \sin\phi)}\,(\cos^2\phi\sin \theta \,-\, \cos^2\theta\sin\phi)

Apply trig Pythagoras' and this becomes

\frac{a^2}{(\sin\theta \,-\, \sin\phi)}\,(\sin\theta \,-\, \sin\phi \,+\, \sin^2\theta\sin \phi \,-\, \sin^2\phi\sin \theta)

\therefore x^2 + y^2 = \frac{a^2}{(\sin\theta \,-\, \sin\phi)}\,(\sin\theta \,-\, \sin\phi)(1 + \sin\theta\sin\phi)

= a^2 + a^2\sin\theta\sin\phi = a^2 - b^2

Phew! You could probably introduce some shorthand or something if you wanted to save on the writing but that's pretty much a slog.

Bumping as a full solution.

Oh wait @Garjun's is full as well. I didn't notice the other pages.
Oh well.

Step III Q4 btw. The original post had some more info in it. I also put together what I thought the question was since the paper wasn't uploaded yet.

(edited 5 years ago)

Reply 270

Notnek

21

The OP has been edited to include the remaining solutions.

Reply 271

username3555092

14

STEP 2 Question 12 alternate solution. I'm a little bit rusty so hopefully no mistakes but let me know if there are. I think this is different enough to a solution that was given earlier.

(i)

Spoiler

(ii)

Spoiler

(edited 5 years ago)

Reply 272

JudeH2001

4

Original post by Zacken

Yes.

The fact that you sat it early and didn't learnt all the relevant material is squarely on you, universities won't care - if you sat the exam, they (or at least Cambridge) presume that you were ready for it and treat you as such. I don't know how Oxford works, but it would certainly detract from a Cambridge application.

Haha funny thing is I'm only now noticing your the same person as from the forum about the STEP day! You responded to another one of my messages over there! And btw I DID get a 2 in the paper last year and I DID apply to Cambridge for maths at Kings and I got an offer!

Reply 273

VerySuspicious

3

In part (iv), when resolving the inequality don't you have to allow

\beta = 1

, giving u = v = 1 and maximum value 3?

Original post by DFranklin

Note that:

Spoiler

Reply 274

oscar.garcia.v

2

Quick question, do you know if every Imperial offer requires STEP or only some?

Reply 275

Rohan77642

14

Original post by oscar.garcia.v

Quick question, do you know if every Imperial offer requires STEP or only some?

some

Reply 276

Aksjdhakzjdj

1

no need to expand out brackets you have a^2x = x(b-x)^2. You subtract x(b-x)^2 from both sides giving a^2x - x(b-x)^2 = 0 then factorise x to give x(a^2 - (b-x)^2) = 0. You are then left with either x = 0 or a^2 - (b-x)^2 =0 and thus a = b-x or -a= b-x and this x = b a or x = b-a.

Reply 277

lillylove123

7

Original post by username3555092

Step 2 question 2 (if there are mistakes let me know)

I might do the sketch for the stem of the question later if I'm feeling it. Edit: sketch is here, and thank you @A02 for making a suggestion which reduces workload in part i.

Sketch:

Spoiler

Commentary on the sketch:

Spoiler

For this question how do people know which values of t,u and v to pick

Reply 278

DFranklin

18

Original post by lillylove123

For this question how do people know which values of t,u and v to pick

Which part of the question are you talking about? u and v are only mentioned in part (i) of the question, and you need to prove the result for all possible values of them, so you can't pick them.

[I suspect you are talking about variables used in some specific solution/workings that you're looking at, but if so, you need to provide those workings].

STEP 2018 Solutions

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