STEP maths I, II, III 1991 solutions

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Reply 80

Rabite

[edit]
I was basically gonna say that but rather less eloquently.
Nice one, Speleo :smile:

Reply 81

Speleo

No need to apologise :tongue:

Reply 82

DFranklin

nota bene

It certainly isn't finished. I think I said somewhere that you should remove it from the OP and leave it free to do for anyone; I might return to it, but I'm a bit fed up with that question :p:

I've forced my way through it - it's a really dull (but fiddly) question.

Start from

Unparseable latex formula:

\displaymath \frac{z_{n+1}-z_n}{z_n-z_{n-1}} = d e^{i\theta}

(justify by looking at the arg and modulus of the numerator/denominator).

So

z_{n+1} -z_n = (de^{i\theta})^{n-1} (z_2-z_1)

z_{n+1} = z_1 + \sum_1^n z_{n+1} - z_n = z_1 + \sum_1^n (de^{i\theta})^{n-1} (z_2-z_1)

. Use

z_1=0,z_2=1

to get:

z_{n+1} = \sum_1^n (de^{i\theta})^{n-1} = \frac{d^ne^{in\theta}-1}{de^{i\theta}-1}

When

d=2, \theta = \pi/3

, we get

z_{n+1} = \frac{2^ne^{in\theta}-1}{2\cos\theta + 2i\sin\theta - 1}

It's tempting to jump into making the denominator real, but realise instead that

2 \cos \theta = 1/2

, so we have

z_{n+1} = \frac{2^ne^{in\theta}-1}{2 i \sin \theta} = \frac{2^ne^{in\theta}-1}{i \sqrt{3}}

\Im(z_{n+1}) = \frac{1 - 2^n \cos n\theta}{\sqrt{3}}

So showing the path crosses the real line infinitely often boils down to showing

1 - 2^n \cos \frac{n\pi}{3}

changes sign infinitely often. When n = 6k+1 (k an integer greater than 0) we have

1-2^{6k+1} \cos \frac{\pi}{3} = 1 - 2^{6k} < 0

. When n = 6k+2 we have

1-2^{6k+2} \cos \frac{2\pi}{3} = 1 + 2^{6k+1} > 0

. So the path crosses the real line between points 6k+1 and 6k+2 for each such k.

Reply 83

generalebriety

Rabite

(On another note, I got an anonymous neg rep in this forum with the reason "Ahh"... bleh. :frown:

)

Most likely it'll be someone searching to neg rep you in particular and doing a search for your last post, which just happened to be in here. Probably won't be anyone from this forum. :s-smilie:

Reply 84

justinsh

I've got an alternative to part 2 of STEP II question 4.

1-e^{i17\theta}=(1-e^{i\theta})(e^{i\theta}+e^{i2\theta}+...+e^{i16 \theta })=0

but

(1-e^{i\theta})

is not 0...

Reply 85

Speleo

There should be a 1 + ... in the second bracket, which would prove the actual result when written as

e^{i0\theta}

Reply 86

Rabite

Most likely it'll be someone searching to neg rep you in particular and doing a search for your last post, which just happened to be in here. Probably won't be anyone from this forum.

I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors. :p:

I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.

mod thing.pdf73.3KB

Reply 87

khaixiang

Rabite

Anyway - I did III Q3, and I'll type it out if I can't find it...

Okay, attached - not sure if it's all right since it seemed quite straight forward.

Hi, nice work again. But I've got a few disagreements, the curve at x=-1 is differentiable. If you evaluate the value of its' derivative at this point using both equations, i.e approaches from both sides and both intervals, the value of derivative is -6 so it's differentiable. (So the curve should be smooth at x=-1, and not of a sharp corner) And I agree that f(x) is continuous at all points but not differentiable at x=1. The quadratic curve in the interval [1,2] is plotted wrongly as well, though the equation is correct. Here's a sketch of f(x) that maybe you can put into your post to make things neater for GE. Cheers. :smile:

3.jpg12.7KB

Reply 88

Rabite

Oh yeah! Whoops! For the minimum point of that section, I had 2x-4=0 which gave me x=½ using the 1=2 formula, and so I thought the turning point was in the middle as opposed to in the outside. :smile:

Thanks for that, I'll edit it into the document.

Khai

Cheers.

Oh no, cheers to you. :wink:

Reply 89

Rabite

Bump.
Anyone done III, 6?
I tried it, but I have no idea if it's right or not.
I got v1+v2=-u for one part...um.
And that l2's image is l1 reflected in the y axis.

Reply 90

generalebriety

I don't think anyone's done it, no.

Bored, might try a question myself.

Reply 91

generalebriety

I/5:

Unparseable latex formula:

\mathbf{a}_1 + \mathbf{a}_2 + \dots + \mathbf{a}_n = \mathbf{0}. \\[br]\therefore \mathbf{a}_1 \cdot [\mathbf{a}_1 + \mathbf{a}_2 + \dots + \mathbf{a}_n] = 0 \\[br]\therefore \alpha^2 + \underbrace{\beta + \beta + \dots + \beta}_{n+1} = 0 \\ [br](n-1)\beta = -\alpha^2.

Unparseable latex formula:

\mathbf{a}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} ; \mathbf{a}_1 \cdot \mathbf{a}_1 = \alpha^2 = 1. \\[br]\text{Let } \mathbf{a}_i = \begin{pmatrix} x_i \\ y_i \end{pmatrix}. \\[br]\mathbf{a}_i \cdot \mathbf{a}_i = |\mathbf{a}_i | = 1 \text{ for all } 0 < i \leq n \Rightarrow \text{ vectors }\mathbf{a}_i \text{ lie on a circle about O} . \\[br]\mathbf{a}_1 \cdot \mathbf{a}_i = \mathbf{a}_1 \cdot \mathbf{a}_j \Rightarrow x_i = x_j \text{ for all } 0 < i \leq j \leq n. \\[br]\therefore n = 2 \text{ or } 3. \\[br]n = 2 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1 \\ 0 \end{pmatrix} \\[br]n = 3 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1/2 \\ \sqrt{3} /2 \end{pmatrix} , \mathbf{a}_3 = \begin{pmatrix} -1/2 \\ -\sqrt{3} /2 \end{pmatrix} \text{ or vice-versa} .

Unparseable latex formula:

\mathbf{a}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \Rightarrow \alpha^2 = 1, x_i = x_j \text{ for all } 1 < i \leq j \leq n. \\[br]n \leq 4. \\[br]n = 2 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} \\[br]n = 3 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1/2 \\ \sqrt{3} /2 \\ 0 \end{pmatrix} , \mathbf{a}_3 = \begin{pmatrix} -1/2 \\ -\sqrt{3} /2 \\ 0 \end{pmatrix} \text{ or vice-versa} .\\[br]n = 4 \Rightarrow \text{ vertices of regular tetrahedron} .\\[br]\beta = -1/3. \\[br]\mathbf{a}_2 \cdot \mathbf{a}_1 = -1/3 \Rightarrow x_2 = -1/3. \\[br]a_2 = \frac{1}{9} + y_2^2 = 1 \Rightarrow y_2 = \pm 2\sqrt{2} /3. \\[br]\text{If } y_2 = 2\sqrt{2} /3 \text{ then, by considering } \mathbf{a}_2 \cdot \mathbf{a}_3 , \text{ it is obvious that } y_3 = -\sqrt{2} /3 \text{; similarly, } y_4 = -\sqrt{2} /3. \\[br]\text{By considering } |\mathbf{a}_3 |, \text{ we see that } z_3 = \pm \sqrt{6} /3. \text{If } z_3 = +\sqrt{6} /3, \text{ then } z_4 = -\sqrt{6} /3. \\[br]\therefore \mathbf{a}_2 = \dfrac{1}{3}\begin{pmatrix} -1 \\ 2\sqrt{2} \\ 0 \end{pmatrix} , \mathbf{a}_3 = \dfrac{1}{3}\begin{pmatrix} -1 \\ -\sqrt{2} \\ \sqrt{6} \end{pmatrix} , \mathbf{a}_4 = \dfrac{1}{3}\begin{pmatrix} -1 \\ -\sqrt{2} \\ -\sqrt{6} \end{pmatrix} . \\[br]\text{Similarly, if } y_2 = -2\sqrt{2} /3 , \text{ then} \\[br]\mathbf{a}_2 = \dfrac{1}{3}\begin{pmatrix} -1 \\ -2\sqrt{2} \\ 0 \end{pmatrix} , \mathbf{a}_3 = \dfrac{1}{3}\begin{pmatrix} -1 \\ \sqrt{2} \\ \sqrt{6} \end{pmatrix} , \mathbf{a}_4 = \dfrac{1}{3}\begin{pmatrix} -1 \\ \sqrt{2} \\ -\sqrt{6} \end{pmatrix} .

Reply 92

Rabite

Generalebriety

I don't think anyone's done it, no.

The youth these days. Tsk tsk.

Hey wow, I'm not the only one up.

Doing III/7, rather not-elegantly...

Reply 93

nota bene

Okay, will be nice to see Rabite :smile:

I got stuck on one of the parts in a) on that one:/

I'm looking at I/9 right now, but I'm a bit tired so I doubt I'll work it out :tongue:

Reply 94

Rabite

Awh ~
Oh thanks for the rep btw :p:

Also, here's that question. My confidence in its accuracy was, of course, a sharply decreasing function of time.

Rabite, for the limit of

\frac{\ln(x)}{x}

David did that on the last page, it cancels down to

\frac{k}{\sqrt{x}}

, which we know goes to zero. Alternatively (as seemingly both me and khaixiang did) use L'Hôpital once and the result follows immediately (as we get lim of 1/x).

The part I missed was the part with 2I and log rules on a :p:

Reply 96

Rabite

Yes, L'Hopital's rule did scream out "Use me", and I would have done so in the real exam; but sometimes I like to be obedient and use the hints given :p:

I don't know. Sometimes I feel like it's ruining the problem if I just apply something that's obviously not intended :redface:

Yeah, the lnx /x limit was quite straight forward; but the xlnx limit confuses me; I couldn't really think of a way of getting round the x>1 condition in the hint.

Reply 97

DFranklin

Rabite

Yeah, the lnx /x limit was quite straight forward; but the xlnx limit confuses me; I couldn't really think of a way of getting round the x>1 condition in the hint.

Seems what you did is fine. Basically, given

\ln x < 2 \sqrt{x} (x\ge 1)

, by putting y = 1/x it's immediate that

\ln 1/x < 2 /\sqrt{x} (x \le 1)

, so

-x \ln x <2\sqrt{x} (x\le 1)

, so you get

0 \le -x \ln x \le 2\sqrt{x}

, and then letting

x\to 0+

gives the desired limit.

Reply 98

Rabite

Yay, okay.
Could someone look/has anyone looked at the summation in III, 8?
I have absolutely no idea how to proceed.
First I considered induction. But reaching the {m+1} stage would involve multiplying by a lot of weird things for the cosAcosB identity, not to mention using some identity with the nCrs (there's one like {n+1}Cr = nCr+nC{r-1} isn't there?).
Then I thought it might be some funny expansion of, say, (1+cos2T)^m, but I couldn't see where that was going, since coses are missing and the powers are off.
Then I tried considering the summation backwards and seeing if anything funny happened by virtue of the symmetry of nCrs, but no such luck.
But that aside, I can't find any deductive route, since I have no idea how to handle the nCr things...

Anyway. Help please.

Reply 99

DFranklin