STEP Maths I, II, III 1988 solutions

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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.

\cos 15 = \dfrac{1}{h} \implies h = \frac{1}{\cos 15}

\cos 15 = \cos(45-30) = \cos 45 \cos 30 + \sin 45 \sin 30

= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}

= \dfrac{\sqrt{3} + 1}{2\sqrt{2}}

\implies h = \dfrac{2\sqrt{2}}{\sqrt{3}+1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} = \dfrac{2\sqrt{6} - 2\sqrt{2}}{2} = \sqrt{6} - \sqrt{2}

triangle.JPG30.0KB

Reply 41

squeezebox

Glutamic Acid

II/3.

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

b = -\dfrac{k^2+1}{k}

and d = 1.

- Since we know c = 1, the quadratic will be of the form

k^2 + bk + 1 = 0

Also

(1-k)^2 + b(1-k) + 1 = 0

will satisfy it. This gives

1-2k+k^2+b-bk+1=0 \implies k^2 + (-2-b)k + (2+b) = 0

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

x^2 - x + 1 = 0

.

- Call the roots of the equation

\alpha, 1 - \alpha, \dfrac{1}{\alpha}

.

Therefore

Unparseable latex formula:

\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\[br](1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\[br]\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0

.

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

\alpha^3 + \dfrac{q}{r}\alpha^2 + \dfrac{p}{r}\alpha + \dfrac{1}{r} = 0

.

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.

Unparseable latex formula:

1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\[br]\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\[br]\implies \alpha^3 + (-3-p)\alpha^2 + (3+3p)\alpha - (2p + 2) = 0

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is

x^3 - \dfrac{3}{2}x^2 - \dfrac{3}{2}x + 1 = 0

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.

Reply 42

Glutamic Acid

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 :smile:

. Thanks for that, I think I've gone about the first part in the wrong way.

Reply 43

squeezebox

STEP III - Question 2

au_{n+2} + bu_{n+1} + cu_{n} = 0

u_{n} = A\alpha^{n} + B\beta^{n}

, where

\alpha

and

\beta

satisfy

ax^{2} + bx +c = 0 (*)

, then

au_{n+2} + bu_{n+1} + cu_{n} = aA\alpha^{n+2} + aB\beta^{n+2} + bA\alpha^{n+1} + bB\beta^{n+1} + cA\alpha^{n} + cB\beta^{n} = A\alpha( a\alpha^{2} + b\alpha + c) + B\beta^{n}(a\beta^{2} + b\beta + c) = 0

. (as alpha and beta satisfy (*).)

So the difference equation is satisfied by

u_{n} = A\alpha^{n} + B\beta^{n}

, for all A,B.

We also need:

u_{0} = A+B

and

u_{1} = A\alpha + B\beta

, and so:

B = \frac{u_{1} - \alpha u_{0}}{ \beta - \alpha}

and

A = \frac{u_{1} - \beta u_{0}}{\alpha - \beta}

.

~~~~~~~~~

Define

u'_{n}

to be a possible solution, so that

u'_{n} \neq u_{n}

. And define also:

w_{n} = u_{n} - u'_{n}

.

Suppose that

w_{k-1} = w_{k}=0

. We shall show that

w_{k+1} =0

We have:

Unparseable latex formula:

u_{k-1} - u_'{k-1} = 0

and

Unparseable latex formula:

u_{k} - u_'{k} = 0

.

Now:

au_{k+1} + bu_{k} + cu_{k} = 0

and

au'_{k+1} + bu'_{k} + cu'_{k} = 0.

\Rightarrow aw_{k+1} + bw_{k} + cw_{k} = 0

\Rightarrow w_{k+1} = 0

, as

w_{k} = w_{k-1} = 0

.

Also, for the determined values of A and B, the fist two terms are

u_{0} and u_{1}

, and so there is no alternative

u'_{0}

and

u'_{1}

. Hence by induction,

u_{n}

are the only solutions.

~~~~~~~~

Put

v_{n} = (n+1)t_{n}

.

Then you get:

(n+1)(n+2)(n+3)[ 8t_{n+2} - 2t_{n+1} - t_{n}] = 0

\Rightarrow 8t_{n+2} - 2t_{n+1} - t_{n} = 0

.

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

8x^{2} -2x -1 = 0

, whose roots are:

\frac{1}{2} and \frac{-1}{4}

.

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

t_{n} = C(\frac{1}{2})^{n} + D(\frac{-1}{4})^{n}

. Where C =

\frac{t_{1} - q t_{0}}{ p -q}

and D =

\frac{t_{1} - p t_{0}}{ q-p}

.

So

t_{n} = \frac{4}{3}(\frac{1}{2})^{n} -\frac{4}{3}(\frac{-1}{4})^{n}

.

Hence:

v_{n} = (n+1)[\frac{4}{3}(\frac{1}{2})^{n} -\frac{4}{3}(\frac{-1}{4})^{n}]

Reply 44

squeezebox

Glutamic Acid

No problem :smile:

. 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 =

\pm 1

or a = k^{-1}. :smile:

Reply 45

nota bene

I/15

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

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

\frac{3-|2-n|}{9}

for

0\le n \le 4

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:

\frac{1}{9}\frac{9}{10}+\frac{2}{9}\frac{7}{10}+ \frac{1}{3}\frac{1}{2}+\frac{2}{9}\frac{3}{10}+ \frac{1}{9}\frac{1}{10}= \frac{45}{90}=\frac{1}{2}

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.

\frac{12}{300}=\frac{1}{25}

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.

Reply 46

brianeverit

Here are some of my solutions for paper I
Numbers 4 6 and 8 attached.

1988PAPER I.4.pdf27.3KB

1988PAPER I.6.pdf33.5KB

1988PAPER I.8.pdf24.5KB

Reply 47

brianeverit

Here are 10, 11,12,13,14, and 16 on Paper I

1988PAPER I.10.pdf43.9KB

1988PAPER I.11-14.pdf75.1KB

1988PAPER I.16.pdf37.2KB

Reply 48

SimonM

Does anyone have this paper anywhere? (If so could they send it to me :smile:

)

Reply 49

Horizontal 8

SimonM

Does anyone have this paper anywhere? (If so could they send it to me :smile:

)

Check your PM inbox :smile:

Reply 50

SimonM

8 Horizontal

Check your PM inbox :smile:

Done

Reply 51

SimonM

STEP II, Question 2

Spoiler

Reply 52

SimonM

STEP II, Question 6

Spoiler

Reply 53

SimonM

STEP II, Question 9

Spoiler

Why aren't group theory questions in STEP any more :'(

Reply 54

SimonM

STEP III, Question 9

Spoiler

Reply 55

maltodextrin

Glutamic Acid

II/3.

- Both k and 1/k will satisfy the equation.

Unparseable latex formula:

x^2 + bk + c = 0 \\[br]\dfrac{1}{k^2} + \dfrac{b}{k} + c = 0 \implies 1 + bk + ck^2 = 0 \implies k^2 + \dfrac{b}{c}k + \dfrac{1}{c} = 0

.
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

k^2 + bk + 1 = 0

, as k^2 - 1 = 0 does not satisfy the conditions (because -1 is a root, but 1--1 = 2 isn't a root).
Also

(1-k)^2 + b(1-k) + 1 = 0

will satisfy it. This gives

1-2k+k^2+b-bk+1=0 \implies k^2 + (-2-b)k + (2+b) = 0

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

x^2 - x + 1 = 0

.

- Call the roots of the equation

\alpha, 1 - \alpha, \dfrac{1}{\alpha}

.

Therefore

Unparseable latex formula:

\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\[br](1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\[br]\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0

.

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

\alpha^3 + \dfrac{q}{r}\alpha^2 + \dfrac{p}{r}\alpha + \dfrac{1}{r} = 0

.

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.

Unparseable latex formula:

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is

x^3 - \dfrac{3}{2}x^2 - \dfrac{3}{2}x + 1 = 0

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?

Reply 56

Glutamic Acid

maltodextrin

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. :smile:

Reply 57

maltodextrin

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. :smile:

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 :smile:

Reply 58

cheena

nota bene

STEP I Q1

h(x)=\frac{\log x}{x}

h'(x)=-\frac{\log x}{x^2}+\frac{1}{x^2}=\frac{1}{x^2}(1-\log x)

For max/min set h'(x)=0 i.e.

0=\frac{1}{x^2}(1-\log x) \Rightarrow x=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

\displaystyle\lim_{x\to0^+}h(x)=-\infty

and

Unparseable latex formula:

\displastyle\lim_{x\to\infty}h(x)=0

. Graph see attached (Mathematica because of lack of scanner :p:

).

Solving

n^m=m^n

is equivalent of solving

\log n^m=\log m^n\Leftrightarrow mlog n=n\log m \Leftrightarrow \frac{\log n}{n}=\frac{\log m}{m}

(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

0< c< \frac{1}{e}

there are two solutions. There's one solution when

c=\frac{1}{e}

c\le0

. And for

c>\frac{1}{e}

there are no solutions.

Now if

m\not=n

then

\frac{n}{m}=k

(

k\not=1

\frac{\log n}{n}=\frac{\log \frac{n}{k}}{\frac{n}{k}}

\log n=\frac{nk(\log n-\log k)}{n}=k(\log n - \log k)

\log n=\frac{k}{k-1}\log k

Thus

n=k^{\frac{k}{k-1}}

and

m=k^{\frac{ 1}{k-1}}

.

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

----

STEP I Q2

g(x)=f(x)+\frac{1}{f(x)}

(

f(x)\not=0

)

g'(x)=f'(x)-\frac{f'(x)}{(f(x))^2}=f'(x)(1-\frac{1}{(f(x))^2})

g''(x)=f''(x)(1-\frac{1}{(f(x))^2})+f'(x)(\frac{2f'(x)}{(f(x))^3})=f''(x)(1-\frac{1}{(f(x))^2})+\frac{2(f'(x))^2}{(f(x))^3}

Let

f(x)=4+\cos(2x)+2\sin(x)

f'(x)=-2\sin(2x)+2\cos(x)=2\cos(x)(-2\sin(x)+1)

Now WLOG

g'(x)=0\Leftrightarrow f'(x)=0 \text{or} (f(x))^2=1

f'(x)=0 \Leftrightarrow 2\cos(x)(1-2\sin(x))=0

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

x=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}

f(x)=1\Leftrightarrow 1=4+1-2\sin^2(x)+2\sin(x) \Leftrightarrow 0=2(2-u^2+u)

where

u=\sin(x)

By quadratic formula the solutions are

u=\frac{1\pm\sqrt{1+4\times2}}{2}

i.e. u=-1, 2

\sin(x)=2

has no real solutions and

\sin(x)=-1

has the root

x=\frac{3\pi}{2}

(which we have already found).

f(x)=-1\Leftrightarrow -1=4+1-2\sin^2(x)+2\sin(x)\Leftrightarrow 0=2(3-u^2+u)

where

u=\sin(x)

Solutions to this are

u=\frac{1\pm\sqrt{1+4\times3}}{2}

and as

\frac{1\pm\sqrt{1+4\times3}}{2}>1

sin(x)=u will have no real solutions.

f'(\frac{\pi}{2}-\frac{\pi}{16})<0

f'(\frac{\pi}{2}+\frac{\pi}{16})>0

so a (local)min

f(\frac{\pi}{2})=4-1+2=5 \Rightarrow g(\frac{\pi}{2})=5+\frac{1}{5}

f'(\frac{3\pi}{2}-\frac{\pi}{16})<0

f'(\frac{3\pi}{2}+\frac{\pi}{16})>0

so a (global) min

f(\frac{3\pi}{2})=4-1-2=1 \Rightarrow g(\frac{3\pi}{2})=2

f'(\frac{\pi}{6}+\frac{\pi}{18})<0

f'(\frac{\pi}{6}-\frac{\pi}{18})>0

so a (global) max

f(\frac{\pi}{6})=4+\frac{\sqrt{2}}{2}+1=5+\frac{\sqrt{2}}{2} \Rightarrow g(\frac{\pi}{6})=5+\frac{\sqrt{2}}{2}+\frac{1}{5+ \frac{\sqrt{2}}{2}}

f'(\frac{5\pi}{6}+\frac{\pi}{18})<0

f'(\frac{5\pi}{6}-\frac{\pi}{18})>0

so a (local) max

f(\frac{5\pi}{6})=4+\frac{1}{2}+1=\frac{11}{2} \Rightarrow g(\frac{5\pi}{6})=\frac{11}{2}+\frac{2}{11}

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 :cool: