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# STEP Maths I, II, III 1988 solutions

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1. Re: STEP Maths I, II, III 1988 solutions
Shaky try at I/3.

OP, PQ and QO will form a triangle. The triangle with the longest shortest side will have its vertices touching the edges, as otherwise the triangle can always be stretched to increase the shortest side.

Using the sine rule, the shortest side is opposite the sine of the shortest angle etc. and the sine of the shortest side has the least value etc. Therefore, we want to maximize the smallest value of A+B+C = 180, which will be when A=B=C=60 degrees. Ie, an equilateral triangle.

Therefore, the triangle will look something like the sketch I've made.

.

.
Attached Thumbnails

2. Re: STEP Maths I, II, III 1988 solutions
(Original post by Glutamic Acid)
II/3.

- Let the roots be k, 1/k. The sum = (k^2+1)/k and the product = 1. Therefore and d = 1.

- Since we know c = 1, the quadratic will be of the form
Also will satisfy it. This gives

Comparing coefficients, -2-b = b ==> b = -1 This also satisfies 2 - 1 =1. So the quadratic is .

- Call the roots of the equation .

Therefore
.

Multiplying through last equation by alpha^3 and dividing by r, you get:
.

Comparing coefficients, 1/r = 1 so r = 1.
q/r = p p/r = q so p = q.

Replacing q by p and expanding the second equation we get.

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is .
I'm not trying to pick holes in your solution, but you may want to consider the case (in the first part), where the equation has roots -1 and 1. Which would make b zero.
3. Re: STEP Maths I, II, III 1988 solutions
(Original post by squeezebox)
I'm not trying to pick holes in your solution, but you may want to consider the case (in the first part), where the equation has roots -1 and 1. Which would make b zero.
No problem . Thanks for that, I think I've gone about the first part in the wrong way.
4. Re: STEP Maths I, II, III 1988 solutions
STEP III - Question 2

If , where and satisfy , then . (as alpha and beta satisfy (*).)

So the difference equation is satisfied by , for all A,B.

We also need:

and , and so:

and .

~~~~~~~~~

Define to be a possible solution, so that . And define also: .

Suppose that . We shall show that

We have: and .

Now: and

, as .

Also, for the determined values of A and B, the fist two terms are , and so there is no alternative and . Hence by induction, are the only solutions.

~~~~~~~~

Put .

Then you get:

.

Let the roots of this difference equation be p and q. So from the first part, p and q must satisfy: , whose roots are: .

So let p be 0.5, and so q = -0.25. Then . Where C = and D = .

So .

Hence: .
Last edited by squeezebox; 19-04-2008 at 01:31.
5. Re: STEP Maths I, II, III 1988 solutions
(Original post by Glutamic Acid)
No problem . Thanks for that, I think I've gone about the first part in the wrong way.
I supose a small improvement would be letting the roots be a and b, and then say: if b = k, then for the condition to be satisfied k^{-1} is also a root. So then either b = or a = k^{-1}.
6. Re: STEP Maths I, II, III 1988 solutions
I/15

First of all this question looks more or less trivial, so I'm possibly missing something

In Fridge football each team scores two points for a goal and one point for a foul committed by the opposing team. In each game, for each team, the probability that the team scores n goals is for and zero otherwise, while the number of fouls committed against it will with equal probability be one of the numbers from 0 to 9 inclusive. The number of goals and fouls for each team are mutually independent. What is the probability that in some particular game a particular team gains more than half its points from fouls?
P(0 goals)=1/9 => For more than half points any foul is allowed 1-P(no foul)=9/10
P(1 goal)=2/9 => For more than half points: P(3 fouls or more)=7/10
P(2)=3/9=1/3 => For more than half points: P(5 fouls or more)=5/10=1/2
P(3)=2/9 => For more than half points: p(7 fouls or more)=3/10
P(4)=1/9 => For more than half points: P(9 fouls)=1/10

So probability that a team in a game scores more than half its points from fouls committed against it is:

In response to criticisms that the game is boring and violent, the ruling body increases the number of penalty points awarded for a foul, in the hope that this will cause large numbers of fouls to be less probable. During the season following the rule change, 150 games are played and on 12 occasions (out of 300) a team committed 9 fouls. Is this good evidence for a change in the probability distribution of the number of fouls? Justify your answer.
which is larger than 1/10, so there has been a change in the probability for 9 fouls, which suggests the strategy has worked, although not a very drastic decrease. However, only knowing data for 9 fouls creates a bit of insecurity as for the accuracy of our findings.
7. Re: STEP Maths I, II, III 1988 solutions
Here are some of my solutions for paper I
Numbers 4 6 and 8 attached.
Attached Files
8. 1988PAPER I.4.pdf (26.7 KB, 156 views)
9. 1988PAPER I.6.pdf (32.7 KB, 119 views)
10. 1988PAPER I.8.pdf (23.9 KB, 110 views)
11. Re: STEP Maths I, II, III 1988 solutions
Here are 10, 11,12,13,14, and 16 on Paper I
Attached Files
12. 1988PAPER I.10.pdf (42.8 KB, 149 views)
13. 1988PAPER I.11-14.pdf (73.3 KB, 112 views)
14. 1988PAPER I.16.pdf (36.4 KB, 62 views)
15. Re: STEP Maths I, II, III 1988 solutions
Does anyone have this paper anywhere? (If so could they send it to me )
16. Re: STEP Maths I, II, III 1988 solutions
(Original post by SimonM)
Does anyone have this paper anywhere? (If so could they send it to me )

17. Re: STEP Maths I, II, III 1988 solutions
(Original post by Horizontal 8)
Done
18. Re: STEP Maths I, II, III 1988 solutions
STEP II, Question 2

Spoiler:
Show
and

The first two equations give us

and

Subtracting yields

Which is

Since

So

So

Subtracting from gives

No, take we can easily verify this satisfies all equations.
Last edited by SimonM; 23-03-2009 at 22:12.
19. Re: STEP Maths I, II, III 1988 solutions
STEP II, Question 6

Spoiler:
Show
(i)

Therefore as y is increasing

(ii)

(by (i))

Therefore

(iii)

(by (ii))

Consider

Differentiating,

Since if there are turning points that expression is zero

(by (iii))

And we're done
20. Re: STEP Maths I, II, III 1988 solutions
STEP II, Question 9

Spoiler:
Show
Let . Given we have

We know Lagrange's Theorem states that the order of a subgroup divides the order of a group. If, for the sake of contradiction, neither of is then

However, since

For our example, consider the Klein four-group, an abelian group, order 4, where ever element has order two. That is

Consider the subgroups .

Why aren't group theory questions in STEP any more :'(
21. Re: STEP Maths I, II, III 1988 solutions
STEP III, Question 9

Spoiler:
Show

Therefore

But . Which contradicts n's minimality. Therefore

(ii) There is a trivial way of doing this using induction and associativity. However, we can say something stronger, which proves this result. That is (conjugation) is an Inner automorphism. (This isn't hard to verify) and for any homomorphism.

(iii) as required.

Therefore . If the order of h is less than 31, then it divides 31. However, 31 is prime. Therefore it is the order of h.
22. Re: STEP Maths I, II, III 1988 solutions
(Original post by Glutamic Acid)
II/3.

- Both k and 1/k will satisfy the equation.
.
So 1/c = c hence c^2 = 1 and c = +-1.
And b/c = b so b = cb. When c = 1, b can be any real value. When c = -1, b = 0.

- Since we know c = 1, the quadratic will be of the form , as k^2 - 1 = 0 does not satisfy the conditions (because -1 is a root, but 1--1 = 2 isn't a root).
Also will satisfy it. This gives

Comparing coefficients, -2-b = b ==> b = -1 This also satisfies 2 - 1 =1. So the quadratic is .

- Call the roots of the equation .

Therefore
.

Multiplying through last equation by alpha^3 and dividing by r, you get:
.

Comparing coefficients, 1/r = 1 so r = 1.
q/r = p p/r = q so p = q.

Replacing q by p and expanding the second equation we get.

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is .
I'm probably missing something really obvious here but I'm confused as to how c can equal -1 in the first part. I understand how you've come to that conclusion but if c = -1 and b = 0 then the equation is x^2 - 1 = 0 with roots 1 and -1. But -1 isn't 1/1 so doesn't that mean the conditions aren't satisfied?
23. Re: STEP Maths I, II, III 1988 solutions
(Original post by maltodextrin)
I'm probably missing something really obvious here but I'm confused as to how c can equal -1 in the first part. I understand how you've come to that conclusion but if c = -1 and b = 0 then the equation is x^2 - 1 = 0 with roots 1 and -1. But -1 isn't 1/1 so doesn't that mean the conditions aren't satisfied?
That's true, and on reflection my method of doing it was... odd. It's clear the product of the roots = c = k*k^(-1) = 1. I must have been on crack. Thanks.
24. Re: STEP Maths I, II, III 1988 solutions
(Original post by Glutamic Acid)
That's true, and on reflection my method of doing it was... odd. It's clear the product of the roots = c = k*k^(-1) = 1. I must have been on crack. Thanks.
Ahh thanks that's a relief, I thought i must be wrong. To be fair the method you used seems to be the only way of solving the second part of the question so I can see why you'd use it
25. Re: STEP Maths I, II, III 1988 solutions
(Original post by nota bene)
STEP I Q1

For max/min set h'(x)=0 i.e.
To find the nature of the stationary point, consider e.g. h'(e-0.2)=positive and h'(e+0.2)=negative hence a (global) maximum. For sketching purposes might be worth noting that and . Graph see attached (Mathematica because of lack of scanner).

Solving is equivalent of solving (last step valid as neither m nor n can be 0).
m=n is the trivial solution.
Consider the line f=c (i.e. a constant). For there are two solutions. There's one solution when or . And for there are no solutions.

Now if then ().

Thus and .

(What am I missing here, this looks trivial for a STEP question...)

----

STEP I Q2

()

Let

Now WLOG

Has solutions cos(x)=0 and sin(x)=1/2
i.e.

where
By quadratic formula the solutions are i.e. u=-1, 2
has no real solutions and has the root (which we have already found).

where
Solutions to this are and as sin(x)=u will have no real solutions.

, so a (local)min
, so a (global) min
, so a (global) max
, so a (local) max

could someone explain the reasoning behind the last bit to question 2 (the nature of the points)

i also dont get the last partof question 1, dont we just see from the graph the only possible values that one of m or n must be is 2 (ignoring m=n) because that is the only integer between e and 1, and we just use trial and error for the other value???

thanks
26. Re: STEP Maths I, II, III 1988 solutions
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