STEP I, II, III 1999 solutions

Cheers

, I'll have to check my working more carefully in the future :redface:

.

Reply 63

nota bene

Square

Sorry to dig up the old thread.

Looking at the solution for II/2 part i). Why can you not factorise 4n^2-60n+200<0 to 4(n-5)(n-10)<0 and then propose that there are no real roots for (*) when 5 < n < 10.

I hope I've not done something stupid here :/

http://www.thestudentroom.co.uk/showpost.php?p=11643319&postcount=16

You're right, don't know what's happened with my mental arithmetic on that question. Will try and fix that now.

Reply 64

nota bene

You're right, don't know what's happened with my mental arithmetic on that question. Will try and fix that now.

Okay! I just saw all those

\sqrt{57}

and panicked!

I'm trying to repost with a better typed solution to ii) which gave me the right answer but I still think turned out to be rather weak.

Reply 65

Okay II/2

Spoiler

re-reading this it doesn't make much sense really!

Reply 66

nota bene

Square

However surely the bracket could still be positive if:

\frac{q}{p-r}<\frac{s^2}{4(p-r)^2}

AND

\frac{q}{p-r}-\frac{s^2}{4(p-r)^2}<(n-\frac{s}{2(p-r)})^2

Firstly it appears to me that your expression after "AND" should be

-\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}<(n-\frac{s}{2(p-r)})^2

(unless I'm being stupid)

And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities. It's well possible that it's possible to prove this "and" cannot happen, or that it can be disregarded for some reason; however I don't see why directly and you would probably need to motivate why/how in a solution.

I understand what you are doing, but can't see how to finish it solidly. Let's see what comments David has.

Reply 67

nota bene

Firstly it appears to me that your expression after "AND" should be

-\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}<(n-\frac{s}{2(p-r)})^2

(unless I'm being stupid)

And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities. It's well possible that it's possible to prove this "and" cannot happen, or that it can be disregarded for some reason; however I don't see why directly and you would probably need to motivate why/how in a solution.

I understand what you are doing, but can't see how to finish it solidly. Let's see what comments David has.

Sorry think that's a typo on my part.

Reply 68

nota bene

Firstly it appears to me that your expression after "AND" should be

-\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}<(n-\frac{s}{2(p-r)})^2

(unless I'm being stupid)

And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities. It's well possible that it's possible to prove this "and" cannot happen, or that it can be disregarded for some reason; however I don't see why directly and you would probably need to motivate why/how in a solution.

I understand what you are doing, but can't see how to finish it solidly. Let's see what comments David has.

The only thing I can think of is somehow using p<r to get one side positive the other side negative or something and find a contradiction. Alas that would be nice but I can't seem to see anything like that :frown:

.

Reply 69

DFranklin

nota bene

Firstly it appears to me that your expression after "AND" should be

-\frac{q}{p-r}+\frac{s^2}{4(p-r)^2}<(n-\frac{s}{2(p-r)})^2

(unless I'm being stupid)

And yes, I think this "and" case would need examination, as otherwise you haven't covered all possibilities.The question only asks you to show that "p < r and 4q(p-r)>s^2" implies the quadratic has no real roots. It's not an "if and only if" question. So what square has done is fine.

If we ignore the wording of the question, then if n can take any real value, then I'm pretty sure (without detailed calculation) we end up with an "if and only if" result; simply choose

n=\frac{s}{2(p-r)}

. But as n can only take positive integers, the true result would be more complicated. Fortunately, that's not what they're asking for...

Reply 70

nota bene

DFranklin

The question only asks you to show that "p < r and 4q(p-r)>s^2" implies the quadratic has no real roots. It's not an "if and only if" question.
Ah, yes sorry! Of course the result is fine then :smile:

Reply 71

nota bene

Ah, yes sorry! Of course the result is fine then :smile:

Yay, that's great.

I'm glad to hear my solution was correct after all. The question didn't take me an awful long time but it was slightly disheartening to come onto the solution thread and see something totally different :frown:

.

Reply 72

1999 III/1 part ii)

I'm slightly confused by Zhen Lin's solution (well by confused I meant it seems a LOT different to how I did it.)

Spoiler

This almost seems FAR too simple.

David seems to have not been around for the last few days, hope everything is okay!

Reply 73

Zhen Lin

12

Well, the thing is, you are assuming that

ak^{-1}, a, ak

are roots of the equation - this is not a given when you start part ii, I think.

Reply 74

lilman91

4

Step I Question 9

When the hare overtakes, both have travelled 0.5km, but the tortoise has travelled for 0.5hours longer. Using

time=\frac{distance}{speed}

for both:

\frac{0.5}{v}=\frac{0.5}{V}+0.5 \Rightarrow \frac{1}{v}=\frac{1}{V}+1=\frac{V+1}{V}

Therefore,

v=\frac{V}{V+1}

...(A)

When the hare passes the tortoise on the way back, the hare has travelled

X+1.25

whilst the tortoise has travelled

X-1.25

. However, the tortoise has been racing for 1 hour longer.

Using

time=\frac{distance}{speed}

again:

\frac{X+1.25}{V}+1=\frac{X-1.25}{v} \Rightarrow \frac{X+1.25+V}{V}=\frac{X-1.25}{v}

Therefore,

v=\frac{V(X-1.25)}{X+1.25+V}

...(B)

From, (A) and (B)

\frac{V}{V+1}=\frac{V(X-1.25)}{X+1.25+V} \Rightarrow \frac{1}{V+1}=\frac{X-1.25}{X+1.25+V}

This gives:

X+1.25+V=V(X-1.25)+X-1.25 \Rightarrow V(X-2.25)=2.5 \Rightarrow V=\frac{2.5}{X-2.5}

. Multiplying top and bottom by 4 gives

V=\frac{10}{4X-9}

as required.

From A we know that

v=\frac{V}{V+1}

. Substituting expression for V gives:

v=\frac{\frac{10}{4X-9}}{\frac{10+4X-9}{4x-9}} = \frac{10}{4X+1}

.
So,

v=\frac{10}{4X+1}

.

The hare finishes 1.5hrs. This means she has travelled 2X in 1.5hrs.
So,

\frac{2X}{V}=1.5 \Rightarrow 4X=3V

Substituting expression for V,

4X=\frac{30}{4X-9} \Rightarrow 16X^2-36X-30=0 \Rightarrow 8X^2-18x-15=0

.
Solving the quadratic but since

X>0

,

X=\frac{18+\sqrt{804}}{16}=\frac{18+2\sqrt{201}}{16}=\frac{9+\sqrt{201}}{8}

.
So, finally

X=\frac{9+\sqrt{201}}{8}

.

Reply 75

lilman91

4

Step I Question 14

(i)Let Y = Score. The score is a discrete random variable whose probabilities need to be calculated.

P(Y=4)=P(0\leq X<\frac{1}{4})=\int^\frac{1}{4}_0 2x \, dx = \frac{1}{16}

P(Y=3)=P(\frac{1}{4}\leq X<\frac{1}{2})=\int^\frac{1}{2}_\frac{1}{4} 2x \, dx = \frac{3}{16}

P(Y=2)=P(\frac{1}{2}\leq X<\frac{3}{4})=\int\frac{3}{4}_\frac{1}{2} 2x \, dx = \frac{5}{16}

P(Y=1)=P(\frac{3}{4}\leq X<1)=\int^1_\frac{3}{4} 2x \, dx = \frac{7}{16}

So now,

E(Y)=\left(\frac{7}{16}\times 1\right)+\left(\frac{5}{16}\times 2\right)+\left(\frac{3}{16}\times 3\right)+\left(\frac{1}{16}\times 4\right) = \frac{15}{8}

Thus,

E(Y)= \frac{15}{8}

as required.

(ii) The ways of winning are a) 3 b) 2,1 (in any order) c)1,1,1

a)

P(\mathrm{3 on first go})=\frac{3}{16}

b)

P(\mathrm{2,1 in any order})=\frac{5}{16} \times \frac{7}{16}\times 2=\frac{70}{256}

c)

P(\mathrm{1,1,1})=\left(\frac{3}{16}\right)^3= \frac{343}{4096}

So

P(\mathrm{win})=\frac{3}{16}+\frac{70}{256}+\frac{343}{4096}=\frac{2231}{4096}

(that takes a bit of long winded arithmetic)

We want to find

P(\mathrm{3 darts|win})=\frac{P(\mathrm{3darts}\land\mathrm{win})}{P(\mathrm{win})}

But

P(\mathrm{3darts}\land\mathrm{win})=P(\mathrm{1,1,1})=\frac{343}{4096}

So,

P(\mathrm{3 darts|win})= \frac{343}{4096} \div \frac{2231}{4096}=\frac{343}{2231}

as required.

Reply 76

Aurel-Aqua

5

STEP III Question 12

\displaystyle P(X=j,Y=k) = e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!}

(i) Find

\displaystyle P(X+Y=n)

for each

n>0

:
This means that we're looking at combinations of

(n,0), (n-1,1), (n-2,2), \ldots (2,n-2), (1,n-1), (0,n)

of

(j,k)

. Therefore,

\displaystyle P(X+Y=n) = e^{-1}n\lambda^n\left(\frac{1}{n!0!}+\frac{1}{(n-1)!1!}+\cdots+\frac{1}{1!(n-1)!}+\frac{1}{0!n!}\right)

, which is equivalent to:

\displaystyle P(X+Y=n) = e^{-1}n\lambda^n\sum_{r=0}^{n}\frac{1}{(n-r)!r!} = e^{-1}n\lambda^n\sum_{r=0}^{n} \frac{1}{n!}\frac{n!}{(n-r)!r!}

.

We can now see that this produces a binomial series, and can thus simplify down to:

\displaystyle P(X+Y=n) = e^{-1}n\lambda^n\sum_{r=0}^{n}\frac{1}{(n-r)!r!} = \frac{e^{-1}n\lambda^n}{n!}\sum_{r=0}^{n}\frac{n!}{(n-r)!r!} = \frac{e^{-1}n\lambda^n2^n}{n!}=

\therefore \displaystyle P(X+Y=n) =\frac{e^-1(2\lambda)^n}{(n-1)!}

.

(ii) Show that

\displaystyle 2\lambda e^{2\lambda-1} = 1

:

For a discrete probability density function, we know that the cumulative sum of probabilities will equal 1. Thus,

\displaystyle \sum_{n=1}^{\infty} P(X+Y=n) = 1

. This implies that:

\displaystyle 1 = \sum_{n=1}^{\infty} \frac{e^-1(2\lambda)^n}{(n-1)!} = 2\lambda e^{-1}\sum_{n=1}^{\infty} \frac{(2\lambda)^{n-1}}{(n-1)!}

.

We notice that

\displaystyle e^x = \sum_{r=0}^{\infty}\frac{x^r}{r!}

, and so

\displaystyle 2\lambda e^{-1}\sum_{n=1}^{\infty} \frac{(2\lambda)^{n-1}}{(n-1)!} = 2\lambda e^{-1}e^{2\lambda} = 2\lambda e^{2\lambda - 1}

.

(iii) Show that

\displaystyle 2x e^{2x-1}

is an increasing function of

\displaystyle x

for

\displaystyle x > 0

and deduce that the equation in has at most one solution and hence determine

\lambda

:

Let

\displaystyle f(x) = 2xe^{2x-1}

.

\displaystyle f'(x) = 4xe^{2x-1} + 2e^{2x-1} = (4x+2)e^{2x-1}

, so

\displaystyle f'(x) > 0

for

\displaystyle x>0

, and so there are no turning points and thus there will only be one root of

\displaystyle x

in this range.

Finding the value of

f(\lambda) = 1

simply takes some observation:

\displaystyle f(\frac{1}{2}) = 2\cdot\frac{1}{2}e^{2\cdot\frac{1}{2}-1} = 1 = 1

, and so

\displaystyle\lambda = \frac{1}{2}

.

(iv) Calculate the expectation

E\left[2^{X+Y}\right]

:

\displaystyle E\left[2^{X+Y}\right] = \sum_{n=1}^{\infty} 2^n P(X+Y=n) = 4\lambda e^{-1}\sum_{n=1}^{\infty} \frac{(4\lambda)^{n-1}}{(n-1)!} =

\displaystyle =4\lambda e^{-1}e^{4\lambda} = 2e

.

My first attempt, I hope someone verifies this :biggrin:

.

Reply 77

SimonM

OP

STEP I, Question 1

Spoiler

Reply 78

SimonM

OP

STEP I, Question 6

Spoiler

Reply 79

SimonM

OP