STEP maths I, II, III 1990 solutions Watch

SimonM
Badges: 18
Rep:
?
#1
Report Thread starter 11 years ago
#1
(Updated as far as #109) SimonM - 23.03.2009

Okay, I made the new thread, since everyone seemed to be ruining General's one. :p:
Pull up a chair and join the cool table.

For those who don't come here often, a short introduction:
Solutions for earlier STEPs are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved; if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

You may be wondering why STEP I has hyperbolic functions and differential equations.
The "usual" past paper :ninja: website may have the papers mislabelled; please check the front of the paper, not the filename, to see which paper you're doing.

(Mathematics = I, further mathematics A = II, further mathematics B = III.)



If you don't know of this usual past paper site, just ask like...anyone.
Oh also, if you solved a question using a different method, you can post up your solution too so that people might see other ways of looking at the problem.


One last thing, if you post a partially completed solution, make sure to tell us which bits you did/didn't do. Also, remember to tell us clearly which question it is you're posting. Also, I think there's a bit of STEP 1987 here too...


~~~~~~~~~~~~~~~
STEP I (Mathematics):
1: Solution by ukgea
2: Solution by Dystopia
3: Solution by brianeverit
4: Solution by Dystopia
5: Solutions by Nota and Square
6: Solution by ukgea
7: Solution by SimonM
8: Solution by Dystopia
9: Solution by Dystopia
10: Solution by *bobo*
11: Solution by brianeverit
12: Solution by Dystopia
13: Solution by brianeverit
14: Solution by Nota
15: Solution by SimonM
16: Solution by Nota


STEP II (Further Mathematics Paper A):
1: Solution by Nota
2: Solution by Dystopia
3: Solution by brianeverit
4: Solution by Simba
5: Rant/explanation and solution by ukgea
6: Solution by Simba
7: Solution by SimonM
8: Mostly done by *bobo*, middle bit by DFranklin and last part by Rabite
9: Solution by SimonM
10: Solution by nota bene
11: Solution by *bobo*
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15: Solution by SimonM
16: Solution by brianeverit


STEP III (Further Mathematics Paper B):

1: Solution by Dystopia
2: Solution by Simba
3: Solution by ukgea
4: Solution by SimonM
5: Solution by Speleo
6: Solution by Dystopia
7: Solution by SimonM
8: Dodgy solution and disclaimer by Dystopia
9: Solution by Speleo
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by Dystopia
14: Solution by brianeverit
15: Solution by SimonM
16: Solution by SimonM

~~~~~~~~~~~~~~~

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
0
reply
nota bene
Badges: 13
Rep:
?
#2
Report 11 years ago
#2
I am

Edit: Okay I'll start typing up II/1

Prove that both x^4-2x^3+x^2 and x^2-8x+17 are non-negative for all real x.
Factorise x^4-2x^3+x^2 to (x^2-1)^2, which clearly is positive.
We can verify that x^2-8x+17 is positive for all real x by rewriting it as (x-4)^2+1, which clearly is positive.

The equation x^4-2x^3+x^2-8x+17=0 has clearly no real solutions for x\le0 as we have earlier shown that x^4-2x^3+x^2 is always positive, and -8(-x)+17 is always positive since two negative numbers multiplied is positive...

For the interval x>2 we can consider x^4-2x^3 as we have earlier shown thatx^2-8x+17 always is positive for all real x. Factorise it to get x^3(x-2). For positive x x^3 is always positive, and for numbers over 2(x-2) is always postitive (and when x=2 the whole equation is still positive, as we have the x^2 etc. terms as well.)

Now consider the interval 0<x\le2. Factorise x^4-2x^3+x^2 to (x^2-1)^2, which clearly is positive. For 17-8x where x is less than 2 it is obvious that it will never be negative.


Prove that the equation x^4-x^3+x^2-4x+4=0 has no real roots. First, consider the interval x<1, where we can use the factorisation into 4(1-x)+x^2((x-\frac{1}{2})^2+\frac{3}{4}). Now all squared things are clearly positive, as is the constant, and for the chosen interval 4(1-x) is positive as well.
Consider x=1, here everything stays the same, except the 4(1-x) part that disappears, and the expression as a whole is still positive.
Now consider x>1 - factorise intox^3(x-1) (which is positive as the cube of a positive number is positive, and x-1 for x>1 is positive) and (x-2)^2, which obviously is positive as it is an even power.
Therefore , for all x the equation x^4-x^3+x^2-4x+4=0 has no real solutions (as all y-values are above the x-axis).

Likely I have a few typos and reversed inequalities in there, point out if you see any
1
reply
Rabite
Badges: 8
Rep:
?
#3
Report 11 years ago
#3
(Original post by nota)
Go open a thread now I know you want it
Not in public.


Ahh...it's not copying the formatting ._.
I don't want to have to type out all those links again!
0
reply
generalebriety
Badges: 14
Rep:
?
#4
Report 11 years ago
#4
(Original post by Rabite)
Not in public.


Ahh...it's not copying the formatting ._.
I don't want to have to type out all those links again!
Was that a hint?

Spoiler:
Show

:ninja:
[U][B]Links to solutions:[/B][/U]
TSR threads:
[URL="http://www.thestudentroom.co.uk/showthread.php?t=398748"]1991[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=395459"]1992[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=394479"]1993[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=367072"]1994[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=359567"]1995[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=358891"]1996[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=358386"]1997[/URL] - [URL="http://www.thestudentroom.co.uk/showthread.php?t=357961"]1998[/URL]
External sites:
[URL="http://www.priorexams.com/"]1999-2003[/URL]
[URL="http://www.meikleriggs.org.uk/"]2001-2006[/URL] Requires sign-up.
0
reply
Rabite
Badges: 8
Rep:
?
#5
Report 11 years ago
#5
:suith:

I was just moaning without intent actually, as I do (I would fit right into HnR). Thanks!
0
reply
Speleo
Badges: 7
Rep:
?
#6
Report 11 years ago
#6
III/9 Easy as \pi e

tant = sinhu
v = sechu = 1/coshu = 1/sqrt(1+sinh^2u)
= 1/sqrt(1+tan^2t) = 1/sqrt(sec^2t) = 1/sect = cost

dt/du = (dv/du)/(dv/dt) = (-sechutanhu)/(-sint)
= sechu.sqrt(1-sech^2u)/sqrt(1-cos^2t)
= v.sqrt(1-v^2)/sqrt(1-v^2) = v

dv/du = -sechutanhu, -2dv/du = 2sechutanhu
= 2v.sqrt(1-v^2) = 2costsint = 2sin2t
d2v/du2 = d/du(-sechutanhu) = sechutanh^2u - sech^3u
-coshud2v/du2 = sech^2u - tanh^2u = v^2 - (1-v^2)
= cos^2t - sin^2t = cos2t

du/dt.d2v/dt2 = 1/v.(-cost) = -cost/cost = -1
dv/dt.d2u/dt2 = -sint.d/dt(1/v) = -sint.d/dt(sect) = -sint.sect.tant
= -tan^2t
(du/dt)^2 = 1/v^2 = 1/cos^2t = sec^2t = 1 + tan^2t
Sum of these parts = - 1 - tan^2t + 1 + tan^2t = 0
0
reply
Speleo
Badges: 7
Rep:
?
#7
Report 11 years ago
#7
I don't want to type up the following questions at the moment, but I've done them if anyone needs hints:

III/1
Spoiler:
Show
Consider Im[(cost+isint)^9]/Re[(cost+isint)^9] for the first part. Consider the 9 roots of the huge equation you've formed. Consider which ones correspond to which brackets. Look at what each of the co-efficients of the cubic equation is in terms of the roots of the equation for the next part. Try multiplying these co-efficients together in certain combinations and taking away other combinations for the last part.

III/2
Spoiler:
Show
See the end of the 1991 thread for several solutions to the first part. Use a dot product for the second part.

III/5 (now written up, see post #11)
Spoiler:
Show
No fancy formulae needed, just write out the binomial co-efficients explicitly and combine them into one fraction. Next two parts are both by induction and are pretty straightforward.


Rabite, you're missing a link to the 1991 thread
0
reply
insparato
Badges: 15
Rep:
?
#8
Report 11 years ago
#8
Im having a go at II/8 atm
0
reply
nota bene
Badges: 13
Rep:
?
#9
Report 11 years ago
#9
I'll have a look at I/7.

edit: having looked more I think II/10 suits me, but I doubt I'll manage before I go to bed:/
0
reply
insparato
Badges: 15
Rep:
?
#10
Report 11 years ago
#10
Im tired. The integrals in the last two parts of II/8 are screaming at me. But i cant think straight when im tired so ill try crack the last two parts(basically one part to be honest) tomorrow .
0
reply
Speleo
Badges: 7
Rep:
?
#11
Report 11 years ago
#11
III/5

\begin{pmatrix} n \\ r \end{pmatrix} + \begin{pmatrix} n \\ r-1 \end{pmatrix} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!(n-r+1)}{r!(n-r+1)!} + \frac{n!r}{r!(n-r+1)!}
= \frac{n!(n-r+1) + n!r}{r!(n-r+1)!} = \frac{n!(n+1)}{r!(n-r+1)!} = \frac{(n+1)!}{r!(n+1-r)!} = \begin{pmatrix} n+1 \\ r \end{pmatrix}

By induction. Case when n=1 is clearly just the product rule:
(uv)^{(1)} = u^{(1)}v + uv^{(1)}

Assume true for n=k, then differentiate again to get:
(uv)^{(k+1)} = u^{(k+1)}v + u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k-1)}v^{(2)} + \begin{pmatrix} k \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}
= u^{(k+1)}v + (\begin{pmatrix} k \\ 0 \end{pmatrix}+\begin{pmatrix} k \\ 1 \end{pmatrix}) u^{(k)}v^{(1)} + (\begin{pmatrix} k \\ 1 \end{pmatrix}+ \begin{pmatrix} k \\ 2 \end{pmatrix})u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}
= u^{(k+1)}v + \begin{pmatrix} k+1 \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k+1 \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}
Which is the proposition with n=(k+1), therefore by induction it is true for all natural n.

y=\arcsin x =&gt; \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} =&gt; \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^\frac{3}{2}}
Therefore, the proposition for n=0 reduces to:
\frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} = 0
which is obviously true.
Now assume that the proposition is true for n=k and differentiate:
(1-x^2)y^{(k+3)} - 2xy^{(k+2)} - (2k+1)xy^{(k+2)} - (2k+1)y^{(k+1)} - k^2y^{(k+1)} = 0
(1-x^2)y^{(k+3)} - (2(k+1)+1)xy^{(k+2)} - (k+1)^2y^{(k+1)} = 0
Which is the proposition for n=k+1. Therefore, by induction, the proposition is true for all non-negative integer n.

I hate latex, it took me probably 3 times as long to type this up as to actually do it on paper.
0
reply
Speleo
Badges: 7
Rep:
?
#12
Report 11 years ago
#12
II/16 is awesome by the way if anyone wants to attempt it
0
reply
Rabite
Badges: 8
Rep:
?
#13
Report 11 years ago
#13
Aw. *pat pat* (at Insparato)
I tried III/4 - think I got it, but I'm a bit sleepy for LaTeX...
0
reply
Speleo
Badges: 7
Rep:
?
#14
Report 11 years ago
#14
I've been doing III/6, I'm OK down to the part where it says 'state the area of the region bounded by E' (thanks to the parameterisation method that I picked up after I think Rabite used it), and I was wondering whether there was an easy way to work that out based on the original curve, whose area is obviously pi, and the transformation.
0
reply
DFranklin
Badges: 18
Rep:
?
#15
Report 11 years ago
#15
(Original post by insparato)
Im tired. The integrals in the last two parts of II/8 are screaming at me. But i cant think straight when im tired so ill try crack the last two parts(basically one part to be honest) tomorrow .
I got this out, but boy did I find it tricky:

Spoiler:
Show
We have xy''+x(y^2-1)y'+xy = 0, where x' = y. So \int_0^1 xy''+x(y^2-1)y'+xy dt = 0. Rather than keep it on one line, let's crack the two "simple" integrals.

\int_0^1 xy dt = \int_0^1 x'x dt = [\frac{x^2}{2}]_0^1 = 0

\int_0^1 xy'' dt = [xy']_0^1 - \int_0^1 x'y' dt = -[\frac{y^2}{2}]_0^1 = 0

So we're left with  \int_0^1 x(y^2-1)y' dt = 0. Obviously need to IBP with dv/dt = y', u =  x(y^2-1). So du/dt = y(y^2-1)+2xyy'. We get:

[xy(y^2-1)]_0^1 - \int_0^1 y(y(y^2-1)+2xyy') dt = 0

-\int_0^1 y^4 dt + \int_0^1 y^2 - \int_0^1 2xy^2y' dt = 0 \quad(\S)

Set I = \int_0^1 xy^2y' dt. Then 2I = 2\int_0^1 (xy)(yy') dt. Do IBP with dv/dt = 2yy' = d/dt(y^2):
2I = [xy^3]_0^1 - \int_0^1 y^2(xy'+y^2) dt = -I -\int_0^1 y^4 dt

3I = -\int_0^1 y^4 dt so we end up with

\int_0^1 y^2  dt = \int_0^1 y^4\,dt - \frac{2}{3}\int_0^1y^4\,dt and so we get the required result.

The last part is much much more straightforward I think. Start from \int_0^1 y'^2 dt - \int_0^1 y^2 dt, use IBP on the LH integral, combine, use the diff equation. You end up wanting to show \int y^3y' dt = 0, which follows after another IBP.
0
reply
Rabite
Badges: 8
Rep:
?
#16
Report 11 years ago
#16
I knew there was something funny about that!
It's an ellipse, I'm quite sure. I think you're meant to work out that

"An ellipse centered at the origin can be viewed as the image of the unit circle under a linear map associated with a symmetric matrix A = PDPt, D being a diagonal matrix with the eigenvalues of A, both of which are real positive, along the main diagonal, and P being a real unitary matrix having as columns the eigenvectors of A."
(From Wikipedia)

And the enclosed by an ellipse is pi*ab where a and b are the 'radii' along the y axis and x axis.
0
reply
insparato
Badges: 15
Rep:
?
#17
Report 11 years ago
#17
I got the first two parts out. For the 3rd

Im stuck with

 2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)

The thing is for  \int 2[x(t)]^2y(t)y'(t)

could you just do

 u = blah

 du = blah

 dv = x(t)y(t)

 v = \int x(t)y(t) = 0

Therefore  \int [x(t)]^2y(t)y'(t) = blah(0) - \int blah(0) = 0 [/latex]

Hmm i havent looked at the spoiler because this very frustrating. Even if i get it out, i doesn't make me a better mathematician for being able to do it. There seems to be loads of ways you can do by parts with these integrals all leading on to different integrals. It just seems thats its luck you pick the right combination of u = .. and dv/dx = ... and maybe a bit of experience in these matters.
0
reply
DFranklin
Badges: 18
Rep:
?
#18
Report 11 years ago
#18
(Original post by Speleo)
I've been doing III/6, I'm OK down to the part where it says 'state the area of the region bounded by E' (thanks to the parameterisation method that I picked up after I think Rabite used it), and I was wondering whether there was an easy way to work that out based on the original curve, whose area is obviously pi, and the transformation.
You just need to multiply by the determinant =\frac{4}{25}(9 \times 6 - (-2) \times (-2)) = 8. (If the determinant had turned out to be -ve you would need to take the modulus too).
0
reply
Rabite
Badges: 8
Rep:
?
#19
Report 11 years ago
#19
Oh. When does this work?
Is that just something that textbooks happened to forget to tell us? I hardly know what the determinant is... (Except for how to calculate it)

Though Matrices have disappeared off the syllabus, it's nice to know anyway.
0
reply
DFranklin
Badges: 18
Rep:
?
#20
Report 11 years ago
#20
(Original post by insparato)
I got the first two parts out. For the 3rd

Im stuck with

 2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)
Just to confirm: you're trying to show the \int y^2 = \frac{1}{3}\int y^4 relation, yes?

Edit: At first I didn't even really bother looking at what you'd done because to be honest, I thought it was a dead end. But I just looked a little harder, and thought, actually, maybe that might work. But it's a bit hard to tell unless you fill in the details a bit. The problem I'm expecting is that you don't know v(t) = 0, you only know \int_0^1 v dt = 0. So I think v is going to end up in some of your integrals, which looks like a showstopper. But I might be wrong here...

Hmm i havent looked at the spoiler because this very frustrating. Even if i get it out, i doesnt make me a better mathematician for being able to do it. There seems to be loads of ways you can do by parts with these integrals all leading on to different integrals. It just seems thats just luck you pick the right combination of u = .. and dv/dx = ... and maybe abit of experience in these matters.
For what it's worth, the main thing that was obvious to me was that we haven't got a clue what to do with \int x(t) dt, so whenever you have to choose what to IBP, the choice should be aimed at getting rid of the x(t)'s. Also, I ended up with terms like yy'^2y'', and I found it sped things up a little to split this as (yy')(y'y'') as both those terms are things we know we can integrate.

If it makes you feel better, there's no way I'd have got this question out under exam conditions. (Well, not unless I was following my advice of "better to get one complete question out than 6 with tiny mistakes" ).
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Arts University Bournemouth
    Art and Design Foundation Diploma Further education
    Sat, 25 May '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Wed, 29 May '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Thu, 30 May '19

How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (489)
31.05%
The paper was reasonable (607)
38.54%
Not feeling great about that exam... (258)
16.38%
It was TERRIBLE (221)
14.03%

Watched Threads

View All