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# STEP Maths I, II, III 1995 Solutions

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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1. Re: STEP Maths I, II, III 1995 Solutions
(Original post by Rabite)
Nu. I was gonna give Step III Q2, but it's already up there. Integration is like the only thing I can do, since I lack the creativity to do any others.

You skipped over the other parts though:
For the first part,x=asin²t works (you have to use double angle formulae a bit).

Here's the last bit for completion's sake.

I don't know if I have the right number of 'n's, so could someone check it please?

 Oh, Insparato's editing in the rest. My bad.
You cant use the reduction formula because you havent proved it yet. I think?
2. Re: STEP Maths I, II, III 1995 Solutions
It was proven in the previous part, surely?
(I didn't bother writing that out since you already had it down)
Last edited by Rabite; 06-03-2007 at 20:09.
3. Re: STEP Maths I, II, III 1995 Solutions
First part was show that I_0 is pia^2/8 then it asked you to show that .... is the reduction formula.
Last edited by insparato; 06-03-2007 at 20:32.
4. Re: STEP Maths I, II, III 1995 Solutions
Well in any case, that's the bit that you had already written up, and thus the bit that I left out
Last edited by Rabite; 06-03-2007 at 21:44.
5. Re: STEP Maths I, II, III 1995 Solutions
I know it was my fault because id already done the first part the other day i just did the rest at school without realising i hadnt actually posted it .
Last edited by insparato; 06-03-2007 at 21:15.
6. Re: STEP Maths I, II, III 1995 Solutions
Anyone doing STEP III, question 6 and 8? I've been trying to do them for some time now, but could only reach halfway. And I am not confident with my answer.
7. Re: STEP Maths I, II, III 1995 Solutions
(Original post by khaixiang)
Anyone doing STEP III, question 6 and 8? I've been trying to do them for some time now, but could only reach halfway. And I am not confident with my answer.
So where have you got, and what is the problem?
8. Re: STEP Maths I, II, III 1995 Solutions
For question 6,

We've learnt transformation from z-plane to w-plane in FP3. This case is different right? Since they want us to sketch the locus of z and 1/z in the same argand diagram. I got a horizontal line locus for 1/z and a single point as a locus for the last part, which sounds silly.

Question 8 is rather beyond me, I could find the equation of normal and the perpendicular distance but not the rest.
9. Re: STEP Maths I, II, III 1995 Solutions
(Original post by khaixiang)
For question 6,

We've learnt transformation from z-plane to w-plane in FP3. This case is different right? Since they want us to sketch the locus of z and 1/z in the same argand diagram. I got a horizontal line locus for 1/z and a single point as a locus for the last part, which sounds silly.
The locus for 1/z sounds OK. The other locus doesn't look right. Try starting from 1/w = 1+it (t real).

Spoiler:
Re(1/w) = 1, so write 1/w = 1+it. Then w = (1-it)/(1+t^2). Write 2w = x+iy, so x = 2/(1+y^2), y = -2t/(1+t^2). Then (x-1)^2 +y^2 = ((t^2-1)^2 +4t^2)/(1+t^2)^2 = 1. So w is a circle, radius 1/2, center (1/2, 0). Strictly, we should remove the point w=0 since 1/w is undefined there.

Question 8 is rather beyond me, I could find the equation of normal and the perpendicular distance but not the rest.
So you've really hardly started. You might want to start by just thinking about the conditions on the position of the sphere if C1 is to lie on the sphere. You should find that there's a line that the sphere center has to lie on.
10. Re: STEP Maths I, II, III 1995 Solutions
Thanks DFranklin, I managed to do question 6. (And Re(1/z)=-1, not 1) The first part and the second part are not related and I was trying too hard to use the result from the first part. I've learnt a lesson now. I'll type out the solution later.

I think I'll leave question 8 for someone else, I was doing an old vector question from Siklos and the vectors manipulations and techniques required are much more demanding not to mention that there're a few things that I've not learnt before (vector equation of a sphere). And sorry to others if I've turned this solution thread into something else.
Last edited by khaixiang; 07-03-2007 at 19:20.
11. Re: STEP Maths I, II, III 1995 Solutions
(Original post by khaixiang)
Thanks DFranklin, I managed to do question 6. (And Re(1/z)=-1, not 1)
Oops, that's what comes from reading the question off one window and then bringing the forum window to the front in order to type an answer...!

I think I'll leave question 8 for someone else, I was doing an old vector question from Siklos and the vectors manipulations and techniques required are much more demanding not to mention that there're a few things that I've not learnt before (vector equation of a sphere).
Fair enough. The vector equation of a sphere, center a, radius r is just |x-a| = r. I think you're expected to know this, but I also think it (plus the standard equations for planes, lines) is the only shape you're expected to know the equations for.

I'm not totally sure how you're supposed to answer this question, incidentally. I can see an "intuitive" approach that involves picking some arbitrary vectors out of a hat, and a "pure" approach that relies purely on vector algebra. But the "pure" approach is harder, and I think it tells you far less about what is actually going on.

And sorry to others if I've turned this solution thread into something else.
Actually, I think your questions are an important part of the thread. People don't generally learn much by reading complete solutions someone else has done; particularly on here where it is easy to skim over the details. I think everyone learns more from looking at when problems arise.
12. Re: STEP Maths I, II, III 1995 Solutions
STEP III, Q7 (just for completeness, since according to dvs this is probably no longer on syllabus. This is probably a little sketchier than you would be expected, but I don't expect anyone will read this, so...):

(i) Is a group. if |a|=|b| = 1, then |ab|=|a||b| = 1. So the set is closed under multiplication. Similarly, |1/a||a| = 1 and so |1/a| = 1, so the inverse of a is in the set. Finally 1 is an identity.

(ii) Not a group. 0 x 0 = 0 x 1, so 0 can't have an inverse.

(iii) Is a group. M(a)M(b)=M(c) where c = a+b for a+b < 2pi, a+b-2pi otherwise.

(iv) Is a group. 1 is the identity. 3 x 3 = 5 x 5 = 7 x 7 = 1, so every element is it's own inverse. 3 x 5 = 7, 3 x 7 = 5, 5 x 7 = 3, so set is closed under multiplication.

(v) Not a group. The zero matrix has no inverse (same as ii).

(vi) Is a group. 1 is the identity. 2x3=1, so 2,3 are inverses for each other. 4x4=1, so 4 is it's own inverse. 2x2 = 4, 2x4=3,3x4=2 so set is closed under multiplication.

For the isomorphisms, (i) and (iii) are infinite, (iv) and (vi) have order 4. Isomorphism preserves cardinality, so the only possible isomorphisms are between (i) and (iii) and (iv) and (vi).

(i),(iii) are isomorphic with f defined by f(M(t)) = (cos t + i sin t) being an explicit isomorphism.

(iv),(vi) are not isomorphic since (iv) has no element of order 4, whereas in (vi) 2 and 3 are both of order 4.
13. Re: STEP Maths I, II, III 1995 Solutions
Step II, Q5:

WLOG, pool has radius 1. Consider case (b).

Distance swum = .

Distance run = . So time taken is .

So for all , and so T has no internal minima. (i.e. any points with must be maxima).

Note that our formula for T is actually valid for , which is case (a), and for which is case(c). So we can justifiably say T attains it's minimum value at one of the end points of , that is, either case (a) or case(c).

So case (b), never happens. Case (a) takes time 2, case (c) takes time . So she should use case (a) for , case (c) for , and it makes no difference when .

Edit: very straightforward, as long as you don't actually try to solve for and evaluate T there, which looks very nasty.
Last edited by DFranklin; 08-03-2007 at 12:35.
14. Re: STEP Maths I, II, III 1995 Solutions
Question 6, STEP III, 1995.
(Thanks to DFranklin for his guidance on this question)

So loci of points in an argand diagram which represents z and 1/z are a circle of radius 1 centre at (0,1) and a horizontal line y=-1/2.

Locus of C in argand diagram is a circle with centre (-1/2,0) and radius 1/2.
Last edited by khaixiang; 08-03-2007 at 16:45.
15. Re: STEP Maths I, II, III 1995 Solutions
STEP III Q1 - where in the booklet? I went through it and I must have missed it.
16. Re: STEP Maths I, II, III 1995 Solutions
(Original post by Rabite)
STEP III Q1 - where in the booklet? I went through it and I must have missed it.
http://www.thestudentroom.co.uk/show...23&postcount=8
17. Re: STEP Maths I, II, III 1995 Solutions
Do you think we should start 1994 now? We're only missing two, we can put a note about them in the next thread.
18. Re: STEP Maths I, II, III 1995 Solutions
Also, if someone has time, it would be good to TeX all other solutions and make pdf's. I think we are lacking a few stat/mechanics on most though... would be good if someone managed to do those.

But yes, I think it will be OK to start the 94 now=)
19. Re: STEP Maths I, II, III 1995 Solutions
(Original post by nota bene)
Also, if someone has time, it would be good to TeX all other solutions and make pdf's. I think we are lacking a few stat/mechanics on most though... would be good if someone managed to do those.

But yes, I think it will be OK to start the 94 now=)
I would do this but i really havent got the time now. TeXing them all up would take time, time i dont have .

But yes should be okay to start 94.
20. Re: STEP Maths I, II, III 1995 Solutions
I've been trying to work through the stats from papers I and II (in the later years) when I have time but I haven't done S3 yet which is needed for paper III.
Last edited by datr; 09-03-2007 at 22:07.

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