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STEP 2006 Solutions Thread

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I did STEP III q. 5 by setting the origin of your axes to be at alpha (making alpha equal to zero), and then saying that if you translate alpha, beta and gamma by the same amount then the result will still hold, since gamma is just beta multiplied by exp (pi/3*i) and translation doesn't affect angles.

Would this be acceptable?

Reply 21

Dadeyemi

DeanK22

STEP III 8

Spoiler

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows;

[br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

We will now prove property 1

Spoiler

We now prove property 2 is met by this operation

Spoiler

We prove property three is met by the above operation.

Spoiler

We now prove the final property;

Spoiler

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler

As required

I'm not sure if the problem really calls for all this, don't you only need to show that the the condition can be reduced to the statement

\Delta x^n = n x ^{n-1}

and then prove that by induction.

Reply 22

Dadeyemi

STEP III 2006, Question 5 (a somewhat neater method for the first part)

\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0

\displaystyle \Leftrightarrow (\alpha-\beta)^2 + (\gamma-\beta)^2 - (\alpha -\beta)(\gamma-\beta) = 0

\displaystyle \Leftrightarrow (\frac{\alpha-\beta}{\gamma-\beta})^2 - \frac{\alpha -\beta}{\gamma-\beta} +1= 0

a quadratic which is solved to give

\displaystyle \frac{\alpha-\beta}{\gamma-\beta} = \frac{1 \pm i\sqrt 3}{2}

which has modulus 1 which is equivilent to saying

|\alpha-\beta|=|\gamma-\beta|

and argument

\pm \pi/3

, SAS gives that this describes an equilateral triangle.

Reply 23

Simply Bread

STEP II - Q8

Spoiler

This isn't quite finished yet - haven't got the time to Latex it all out now, so I shall come back later :smile:

Reply 24

nuodai

STEP I Question 3

Part (i)

Part (ii)

Part (iii)

Reply 25

Oh I Really Don't Care

STEP II, question 1

(ia) Set U_(n-1) = U_n = 2 to obtain k=20

(b) If we want a period of 2 then U_1 = U_3 = U_3 = ... = U_(2n+1). By definition, if we require a period of 2 then the sequence is not constant. In light of the above we can find an epression linking U_1 and U_3 as follows;

Unparseable latex formula:

U_1 = 2 \; U_2 = k - 18 \; U_3 = k - \frac{26}}{k-18}

which leaves us with a quadratic in k that we can solve. It turns out that k(k-20) = 0 <-> k = 0 or k = 20. We discard k = 20 in light of section (a)

(c) Period 4 means that U_1 = U_5. We trawl through the algebra to find the quartic;

k^4 - 20k^3 - 72k^2 + 1440k = 0

which factors as k(k-20)(k^2 - 72) = 0 so it follows that k = (+,-) 6root(2)

(ii) If U_n => 2 it follows that U_n is greater than or equal to 37 - 18 = 19 (and U_1 = 2 so the result follows).

If there is some limit as n tends to infinty then u_n = u_(n+1) which gives us the quadratic;

u_n^2 - 37u_n + 36 = 0

and from (ii) previously it follows that the limit is 36.

Reply 26

Oh I Really Don't Care

STEP II, question 2

First part

Second part

(edited 5 years ago)

Reply 27

rnd

That 21 appears too many times to be a typo :tongue:

Reply 28

Horizontal 8

A surprisingly similar question to question 19 on the special papers thread :smile:

http://www.thestudentroom.co.uk/showpost.php?p=17975815&postcount=3

Reply 29

Horizontal 8

I think these STEP examiners need abit more creativity.. :rolleyes:

Reply 30

nuodai

And voilà another reason why I shall fail STEP! I'll do it when I start procrastinating from Psychology revision (or in the worst case, after the exam on Monday).