# STEP I, II, III 2000 solutionsWatch

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11 years ago
#41
I believe STEP III is the only paper specifically for the FM syllabus.

As to which is easier, everyone agrees STEP III is hardest by some way. Most people reckon the AEA is also easier than STEP II and STEP I. My feeling is that the difference isn't great, and the biggest factor is probably which exam format you get on with best. There's a fair degree of personal perference: some will find AEA much easier than STEP, others will find STEP easier.
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11 years ago
#42
Thanks Franklin, do you think doing STEP before going to uni would improve/prepare me better for maths at uni? I only did a levels maths, but doing maths in uni this fall..

also, whats the difference between STEP, A level and University maths like?

cheers!
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11 years ago
#43
(Original post by johnheyes91)
Zhen just curious, how do you type maths and transfer it to a pdf? Cheers!
LaTeX. The other thread probably explains better than I could.
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11 years ago
#44
ah, thanks!
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11 years ago
#45
(Original post by DFranklin)
Re: Q8. When it comes to solving the recurrence relation, I think you've approached it the wrong way around. Essentially, you've derived a formula for the solution, but you haven't shown it's the only possible solution.

In fact, to answer the question you don't need to derive it. Consider

All you need is to show , and . Then write ; you have and , so trivially b_n = 0 for all n by induction.
Sorry, but why is ?
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11 years ago
#46
If someone could take a look at my solution to 2000/III Q6 I would be thankful!

Spoiler:
Show

If a^2 root then f(a^2)=0

Substituting in the equations above and a lot of algebra later (which I don't really fancy latexing) you do get f(a^2)=0 which implies a^2 is a root.

For the non-negative root proof, I just wrote a small note saying that since a^2 will always be greater than or equal to zero for any value of a, f(u) must always have a non-negative root.

Therefore 9 a root of f(u) when p=-1 q=-6 r=15

Substitute y=x+2 - this will get rid of the y^3 term and make the quartic similar to that in the first part of the question

Several lines of algebra later:

When you compare to the first part of the question you get p=-1 q=-6 and r=15 which is the same as the second part of the question.

This implies that the cubic f(u) from the previous part holds. Since we showed that f(u) has a root at u=9 for p=-1 q=-6 and r=15 AND that f(u) has a root a^2:

a^2=9
a=3

a=3 implies b=3 and c=5

Therefore x^4-x^2-6x+15=(x^2-3x+3)(x^2+3x+5).

Subsitute back for x=y-2

I hope that makes sense, I've condensed most of the algebra and my actual solution has about 3 and a half sides of A4.
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11 years ago
#47
(Original post by speedy_s)
Sorry, but why is ?
Because and and (for all n).
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11 years ago
#48
Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:

Spoiler:
Show
if , then g(0) <=0. if g(0) = 0, 0 is a non-negative root. If g(u)<0, then as g(u) -> +inf as u->inf, g changes sign for some positive u and so has a non-negative root.

I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.
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11 years ago
#49
(Original post by DFranklin)
Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:

Spoiler:
Show
if , then g(0) <=0. if g(0) = 0, 0 is a non-negative root. If g(u)<0, then as g(u) -> +inf as u->inf, g changes sign for some positive u and so has a non-negative root.

I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.
Aha, I see where you're getting at there.

I do have quite large writing (especially when it comes to doing maths) so maybe slightly less with averaged sized writing.

I can put in some more working if you like but some of it was quite long winded and tedious and I really could not be bothered to LaTeX it.

Maybe you have some better algebra skills than I do (which wouldn't be very hard!)
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10 years ago
#50
I / 7

If , show that iff

Note that f(x) is a quotient and an inequality is involved. While tempting to use something such as the product rule, by using the quotient rule the denominator will always be > 0.

0
10 years ago
#51
II / 1

Guess a solution to you are provided with and also

The solution is clear -

To prove this is so;

Looking at for N is a prime it is apparent that the only factors of N^2 are

This would imply that if distinct natural numbers (a,b) are sought for the equation for N prime then there is only one pair of solutions.

wip for
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10 years ago
#52
II / 2

Given p(x) we differentiate to find;

The roots of the equation

Inspection reults in So if p[x] denotes the polynomial given then p[-2] = 0. Using the factor and remainder theorem; k = 48
1
10 years ago
#53
II / 4

The second part is alot easier if you just have a hunch.

By sheer observation the complex number shows the fact;
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10 years ago
#54
Attempt at I/11
part 1:
Letting the normal reaction force from the ground be R and the normal reaction force from the wall be T, and the friction from the ground be , and from the wall (see if diagram 1).

I attempted to approach the question with proof by contradiction. Assume the friction is limiting at both the ground and the wall

Resolving verticaly,

and resolving horizontally,

From taking moments about A, the point of contact with the ground:

so we have a contradiction, and so our assumption that the friction is limiting at both the contact points with the ground and the wall is false.

part 2:
Letting the distance of the particle from point A be X metres (see the diagram 2).

Resolving vertically
and resolving horizontally

Taking moments about A, the point of contact between the rod and the ground:

0
10 years ago
#55
II / 8

After subbing (0,6) ;

and case 2

There is a vlaue of k such that as x tends to y tends to

(ii) wip. Got
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10 years ago
#56
III/12:

P(3 or fewer | Holding a winning ticket) = P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket).

3 or fewer and you win: you win and two others win; you win and one other wins; you win and no others win. You winning and other people winning is independent, so P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket) = P(2 others or fewer win).

The others can be modelled as a binomial distribution, X ~ B(2n-1, 1/n)
P(X <= 2) = P(X = 2) + P(X = 1) + P(X = 0)
=

=

As n gets large, So the probability is approximately .

.

The last part is slightly dodgy, plus the arithmetic was rather tedious. I've been light on the algebra at several parts; none of it's particularly difficult.

- Human selections of numbers won't be uniform, and some will be more popular than others. So given that I have a winning ticket, it's likelier that more others will have a winning ticket so this probability will decrease. (Not sure if this is what they want or even correct.)

Consider the other people playing: X ~ B(2N - 1, 1/N) E(X) = "np" = 1/N*(2N-1) = 2 - N-1. So in total, given that you have a winning ticket, the expectation is 3 - N-1.
1
10 years ago
#57
III/13:

If the first die to show a six does so on the rth roll then there'll (r-1) rolls where all dice fail to show a six =

On the rth roll: P(1 or more shows a six) = 1 - P(none shows a six) = 1 - q^n. Overall probability = , as required.

Let G(t) denote the probability generating function. r can take values 1, 2, 3, ... so Geometric series, a = , r = . Sum to infinity = .

When n = 2, p = 1/6, q = 5/6 so E(R) = .

.
Less than or equal to r for one die is 1 - q^r; not n failures. So for n-dice: , and similarly for r - 1: . Overall probability:

r can take values 1, 2, 3, ...

=>

.

= 12 when p = 1/6. (I haven't LaTeXed the derivative of G(t), it's too late and it's just tedious.)
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10 years ago
#58
... many of the questions for 2000 are from 1987 and 1990 papers ...
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10 years ago
#59
STEP I, Question 9, 2000
The value of tan(2theta)
Let the point at which the child fires the toy cannon from be P. Let us consider the situation relative to point P. Let d be the distance the shell travels horizontally. Let T be the time for the shell to land after being fired.

SUVAT vertically for the toy shell:

SUVAT horizontally for the toy shell:

differentiating with respect to :

For maximum range, d is max, so

Proof for theta

As a is small compared to g, will be even smaller, so
Lemma 1
Lemma 1:

Proof:

For small ,

So we need to show that for small y:

which is certainly true for small y.
Therefore we have
Proof for the range

from the original equations in the first part,

2
10 years ago
#60
1
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