STEP Maths I, II, III 1993 Solutions

Reply 160

16

DFranklin

What do people think about starting some post 2000 threads? I know there are solutions already available, but I'm sure they are more relevant in terms of current papers.

I think it'd be a better idea to get the pre-2000 threads finished first, but yeah, starting post-2000 threads sounds like a good idea, particularly for alternative solutions. Besides which, I'm sure there'll be a syllabus change some time and all the "old" solutions will be taken down. :p:

But let's get these ones finished first. :smile:

When these threads start becoming inactive, I'll start a 1991 thread.

Reply 161

Kolya

17

DFranklin

What do people think about starting some post 2000 threads? I know there are solutions already available, but I'm sure they are more relevant in terms of current papers.

Sounds good. It would be particularly useful if the solutions to post-2000 were done by users in LaTeX, as the current solutions are sometimes hard to read, and have (at least one) mistake.

Reply 162

khaixiang

5

DFranklin

My advice would be not to let experiences of the "pre-1994" questions affect your confidence. To my mind they are of a very different style, and seem to expect knowledge in different areas, so some of them are absolute brutes, while some (particularly the mechanics) seem surprisingly easy.

What do people think about starting some post 2000 threads? I know there are solutions already available, but I'm sure they are more relevant in terms of current papers.

Yeah, I agree that some of the questions are of different style, but for those sitting the papers exactly 1 month from now, I think the more ancient papers are more important. 4 questions repeated in 2006 STEP from those of 1987-1992, 2 questions from 1987(!), 1 from 1990 and 1 from 1992 (this is the "diamond" questions speleo mentioned.)

Unfortunately, I think no one here has the papers from 1987 to 1989. Or if you do, do share it :biggrin:

and hopefully we can complete all of them before 27th June. And timing couldnt be better right, in case any questions repeat, we can remember them well.

Reply 163

Swayum

18

DFranklin

$T_n(r) = T_{3n}(r^2) + r^3 T_{n-1}(r^6) = \frac{1-r^{6n+1}}{1-r} + \frac{r^3(1-r^{6n})}{1-r^6}$

Sorry, can you explain that? I think I did something similar but I got:

$T_n(r) = 1 + \frac{r^2 - r^{6n+2}}{1-r^2} + \frac{r^3 - r^{6n + 3}}{1-r^6}$

If you want to see my method/why I got that, it's in my post on the second page.

*Edit*

Ur, that's not a very good question on my side. What I mean is, if you try n = 1.

T_n(r) = 1 + r^2 + r^3 + r^4 + r^6

Let r = 2

T_n(r) = 93

When I try n = 1 and r = 2 with your formula, I don't get 93.

Reply 164

generalebriety

16

Done II/4, but I'm ****ed if I'm LaTeXing it. I swear TSR's LaTeX is messed up somehow. Written and scanned. :smile:

q4i.gif289.4KB

Reply 165

DFranklin

18

DeathAwaitsU

Sorry, can you explain that? I think I did something similar but I got:

$T_n(r) = 1 + \frac{r^2 - r^{6n+2}}{1-r^2} + \frac{r^3 - r^{6n + 3}}{1-r^6}$ You're right - I mucked up. I should have put $T_n(r) = \frac{1- r^{6n+2}}{1-r^2} + \frac{r^3(1 - r^{6n})}{1-r^6}$ . I've edited the post, thanks.

Reply 166

DFranklin

18

khaixiang

Hmm... IMHO, I think this part is just about solving $z^{*}=\frac{z}{z-1}$ ?Yes, that's all you have to do: I'm just no good at doing this in my head...

N.B. With earlier threads, I actually did most of the questions, so any comments I made were likely to have a bit more reasoning in them. On these threads, in general, I'm much more shooting from the hip. I'm sure there will be the occasional misfire...

Reply 167

DFranklin

18

I, Q7: In this day and age, I'll leave it to you to draw the graphs on a graphing calculator. But the salient points are: $f(x) = x^3+Ax^2+B, f'(x) = 3x^2+2Ax = x(3x+2A), f''(x)=6x+2A$
So f'(0) = 0 when x = 0 or x= -2A/3, so these are the only local maxima/minima. Looking at the 2nd derivative, f''(0)=2A, f''(-2A/3) = -2A.
So when A<0, 0 is a local maximum, f''(-2A/3) is a local minimum (to the right of x=0).
When A>0, 0 is a local minimum, f''(-2A/3) is a local maximum (to the left of x=0).

Now if f has 3 roots, then f'(x) = 0 at some point between each two adjacent roots (this is Rolle's theorem, but if you don't know it a sketch should make it obvious what's going on). So as there are only 2 places (x=0, x=-2A/3) where f'(x) = 0, there are 3 roots iff the sign of f at the two places is different.

So f has 3 roots iff f(0)f(-2A/3)<0, that is, iff $B(-8A^3/27+4A^3B/9+B) < 0$
Since $B(-8A^3/27+4A^3/9+B) = (27B^2+4A^3B)/27$ the result follows.

N.B.: I don't know why they changed from A,B to a,b halfway through the question, but I haven't bothered to follow suit. If you're wondering about the case where 27B^2+4A^3B = 0, it turns out there's a repeated root, so there are only 2 distinct roots.

Reply 168

generalebriety

16

DFranklin

I, Q7: In this day and age, I'll leave it to you to draw the graphs on a graphing calculator.

...which would not be allowed in a STEP exam. :wink:

Reply 169

DFranklin

18

generalebriety

...which would not be allowed in a STEP exam. :wink:

Sure. What I meant was, I could find some paper, draw some sketches, scan them in, reduce the file size, etc. Or I could assume you all have access to graphing calculators so don't need me to bother. Guess which approach I went for...

Reply 170

generalebriety

16

Indeed.

There's plenty of freeware graphing software around anyway. Besides, if you can't sketch graphs, you don't get offers from Cambridge. Half of your interview is graph sketching.

II/4 done above (post #165).

Reply 171

Speleo

7

I can edit again :biggrin:

Tell me if I've missed anything.

generalebriety

Indeed. There's plenty of freeware graphing software around anyway. Besides, if you can't sketch graphs, you don't get offers from Cambridge. Half of your interview is graph sketching.

I only had one graph question :tongue:

Reply 172

Rabite

8

I didn't have any. But then again, my offer probably came from quotas.

Anyway. Er.
I could have sworn I saw STEP III, Q2 floating around...where is it? Need to check my own solution.... :p:

Reply 173

generalebriety

16

Speleo

I can edit again :biggrin:

Tell me if I've missed anything.

I only had one graph question :tongue:

I feel violated. I had two, and they took up half the interview time. :p:

Reply 174

khaixiang

5

Rabite

I didn't have any. But then again, my offer probably came from quotas.

Anyway. Er.
I could have sworn I saw STEP III, Q2 floating around...where is it? Need to check my own solution.... :p:

Post #156, just a page before. I don't know about you, but the second bit of this question was very difficult for me! I couldn't have done that integral myself without knowing what substitution to make beforehand.

generalebriety

I feel violated. I had two, and they took up half the interview time.

I had actual 2006 STEP III papers as interview test, and had just come to know about STEP papers vaguely before that. So I guess they didn't look much into that test, since I did so badly. :redface:

Reply 175

nota bene

15

Rabite

I could have sworn I saw STEP III, Q2 floating around...where is it? Need to check my own solution.... :p:

#156

edit: Here you wait an hour for an answer and two come simultaneously...

Reply 176

DFranklin

18

khaixiang

Post #156, just a page before. I don't know about you, but the second bit of this question was very difficult for me! I couldn't have done that integral myself without knowing what substitution to make beforehand.Just had a look at post #156: Am I right in thinking justinsh has mistakenly written $\sin \theta + \cos \theta$ instead of $\sin \theta \cos \theta$ in a couple of places near the beginning of where he calculates the area? (I'm sure it's a typo, given he assumes the square of the term is $\sin^2\theta \cos^2\theta$ later on).

Reply 177

Rabite

8

Hmm, I used x=tanT, but it didn't work for some reason.
My "r" seems to be different from yours too.

$r = \frac{3 \cos\theta \sin\theta}{\cos^3 \theta + \sin^3 \theta}$
Is what I get (so do you ignoring the typo)
Then divide everything by cos³:
$r = \frac{3 \tan \theta \sec^2 \theta}{1+ \tan^3 \theta}$
So squaring r gives
$r^2 = 9\frac{\tan^2 \theta \sec^4 \theta}{(1+\tan^3 \theta)^2}$
Which turns to x²(1+x²)/(1+x³)².
x²/(1+x³)² is pretty obious but I haven't been able to do much about the x^4/(1+x³)²
Except for make an algebraic error and evaluate it as a standard form ^.^
But ignoring that...hm.

Though I did this at like three in the morning, so I've probably just done something dumb...

(Also, did anyone ever bother finishing the z and z* question? The brute force z=x+iy method gave two answers, but whether they're right or not is another question. 0+0i and 1-i?)

Reply 178

khaixiang

5

Rabite

Hmm, I used x=tanT, but it didn't work for some reason.
My "r" seems to be different from yours too.

$r = \frac{3 \cos\theta \sin\theta}{\cos^3 \theta + \sin^3 \theta}$
Is what I get (so do you ignoring the typo)
Then divide everything by cos³:
$r = \frac{3 \tan \theta \sec^2 \theta}{1+ \tan^3 \theta}$
So squaring r gives
$r^2 = 9\frac{\tan^2 \theta \sec^4 \theta}{(1+\tan^3 \theta)^2}$
Which turns to x²(1+x²)/(1+x³)².
x²/(1+x³)² is pretty obious but I haven't been able to do much about the x^4/(1+x³)²
Except for make an algebraic error and evaluate it as a standard form ^.^
But ignoring that...hm.

Though I did this at like three in the morning, so I've probably just done something dumb...

Your r is correct, but there's a careless slip in your second formula, it should be sec(theta), no squared.

Rabite

(Also, did anyone ever bother finishing the z and z* question? The brute force z=x+iy method gave two answers, but whether they're right or not is another question. 0+0i and 1-i?)

I've posted about this a page behind (last post). There's no need to put z=x+iy, I got 0 and 1+i, but I honestly can't remember whether it's 1-i or 1+i for the second answer.

Reply 179

Rabite

8

Oh whoops, you're right.
:p:

Sorry 'bout that ~

I think I'll commit suicide if I miss the offer because of something like that.

STEP Maths I, II, III 1993 Solutions

Related discussions

Articles for you

How The Student Room is moderated