STEP Maths I, II, III 1993 Solutions

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Reply 100

Speleo

III/9

EDIT: I am going to give up on the challenge for a while, the rest of the questions look really tedious :frown:

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Reply 101

generalebriety

Speleo

Because it's eeeeeeeeeeeeasy :p:

Nein, because it was the first STEP question I ever did. :biggrin:

Reply 102

TheDuck

Just out of interest....if I wanted to try a few of these (since I'm doing Maths at uni and might be able to do a few one would hope!), where can I find the paper itself!.

Reply 103

Speleo

Check your private messages.

Reply 104

DFranklin

Speleo

Is the bus question finished?

I'm pretty sure you're not supposed to assume the bus arrives at 8.05 for the last part I'm afraid. Either that or it's a question answerable in under 60 seconds, and I'm not sure I've ever seen one of those on a STEP paper before. (I have seen one on a CCE: it was a "by induction or otherwise", and there was a 1 line "otherwise"! Cue lots of embarrassment for the examiner).

Ditto the powers of r one.

Think this is done, but the pieces (from different) posts need pasting together.

Reply 105

DFranklin

generalebriety

Trying STEP III Q7, no promises though...

Edit: abandoned for a while, might come back to it. Until I say otherwise I'm not gonna be doing it though, so anyone else is free to steal it off me. :p:

Having given other people a hard time, thought I'd confess that I tried this and failed miserably. After about an hour I could see how you get a quartic, but not the cubic they're looking for.

Reply 106

Rabite

Q5 in StepII (Maths A)
Not sure if it's right though.

r+dr.jpg

If $\tan \alpha = \frac{Opposite}{Adjacent}$ then
We can see that the opposite is (r+dr)sin{dT} (T is theta) and the adjacent is
((r+dr)cos{dT} -r).
In considering the quotient as dT and dr approach zero, we can use the small angle approximations, sin{dT} --> dT and cos{dT} -->1-dT²/2 = ~1 for a sufficiently small dT.
So
$\tan \alpha = \frac{(r+\delta r)\delta \theta}{r+ \delta r - r}$

$= (r+\delta r) \frac {\delta \theta}{\delta r}$

Ignoring the delta{r} in the bracket... (because r would be so much larger) as delta r and delta theta tend to zero...

$\tan \alpha = r \frac{d\theta}{dr}$
And the result follows.

That formula gives the angle that a tangent cuts a radius at. Using it with
$r=e^{\theta \cot \alpha}$
(Alpha is now 'theirs', though it turns out to be the same alpha anyway)
$\frac{dr}{d\theta} = \cot \alpha e^{\theta \cot \alpha}$

$r \frac{d \theta}{dr} = \frac{r}{\frac{dr}{d\theta}} = tan \alpha$

And so the angle is given by
$\tan^{-1} (tan \alpha) = \alpha$

The spiral looks like any other spiral really. The radius grows exponentally of course, but I can't draw that.

$\frac{ds}{d\theta} = \sqrt (r^2+(\frac{dr}{d\theta})^2)$ (Given)

$r^2 = e^{2\theta \cot \alpha}$ and

$(\frac{dr}{d\theta})^2 = e^{2\theta \cot \alpha} \cot^2 \alpha$
So (Choosing the limits T=0 at r=1 and T=2pi after one complete turn)
$s = \int _0 ^{2\pi} \sqrt( e^{2\theta \cot \alpha} (1+\cot^2 \alpha))d\theta$

$= \int _0 ^{2\pi} e^{\theta \cot \alpha} \sqrt (1+\cot^2 \alpha)d\theta$

$= \tan \alpha \sqrt (1+\cot^2 \alpha) [e^{\theta \cot \alpha}]_0 ^{2\pi}$

The bit at the front can be tidied up to sec{alpha} I think, but it doesn't really matter does it?

For the last bit is the same, but the limits are -{infinity} (bottom) and 0 (top). So that ends up as
$\tan \alpha \sqrt (1+\cot^2 \alpha) = \sqrt(tan^2 \alpha +1) = \sec \alpha$ I think?

Reply 107

Rabite

Frank

Having given other people a hard time, thought I'd confess that I tried this and failed miserably. After about an hour I could see how you get a quartic, but not the cubic they're looking for.

Is that the one about t=sinh²y or something? I did that one after vastly overcomplicating it...

Reply 108

DFranklin

Rabite

Is that the one about t=sinh²y or something? I did that one after vastly overcomplicating it...

Yep. Couldn't get a handle on it at all.

Reply 109

Rabite

Start with the good old
cosh²y=1+sinh²y (=1+t) [1]

We're given
sinh(2x)=coshy [2] and
sinh(2y)=2coshx [3]
Sub the first one in:
([2] into [1])
sinh²(2x)=1+t
4sinh²xcosh²x = 1+t *

[3] can be written as
sinhycoshy=coshx, so
cosh²x=sinh²ycosh²y=t(1+t)
cosh²x-1=sinh²x = (t+t²-1).
Sub those results into *:

4t(1+t)(t²+t-1)=1+t
(edit: t=/=-1 because t>0)
4t(t²+t-1)=1 as required.

The rest of the question is quite easy.

Reply 110

DFranklin

Rabite

Nice job. I obviously don't know my hyperbolic identities as well as I should!

Reply 111

Speleo

Damnit Rabite I just did that one :tongue:

Uploading it anyway since you didn't do the later parts...

III/7

not entirely sure I did the very last part correctly...

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Reply 112

nota bene

Okay I've looked at the matrix questions 6 and 10 (on STEP II) and can do parts of them, but I get stuck (finding suitable a on 10, and for iv) on 6 I have no idea where to start, the statement looks true to me at a glance (is a quadratic with matrices), I've not found and obvious counterexamples and don't know where to start a proof.).

I think I'll give up on this paper now unless I suddenly realise something smart :p:

Reply 113

DFranklin

nota bene

I don't think Q6, iv is true. Consider what happens if you square

(0 a)
(0 0)

As for Q10, substitute $P{\bf x_1} + {\bf a}$ for x into $x^TAx + b^Tx$ . Keep things as matrices until you get as far as you can, then substitute in actual components for x_1,y_1,a_x,a_y (where a_x,a_y are the components of a) and you should be able to find a_x,a_y so that you end up with ${\bf x_1}^TPAP{\bf x_1} + k$ ). (i.e. you can make the b^Tx bit disappear).

It's quite tricky and easy to make mistakes though, and LaTeXing it will be horrific...

Edit: Based on the two I've seen, STEP questions involving diagonalisation of matrices tend to be rather on the long side.

Reply 114

nota bene

DFranklin

I don't think Q6, iv is true. Consider what happens if you square

(0 a)
(0 0)

Okay yeah, that's surely a counterexample... I'll type up my solutions later I think it is finished (although I'm not quite convinced of my interpretation of iii), where the statement is false if I set detX=0 and let detA be non-zero)

DFranklin

As for Q10, substitute $P{\bf x_1} + {\bf a}$ for x into $x^TAx + b^Tx$ . Keep things as matrices until you get as far as you can, then substitute in actual components for x_1,y_1,a_x,a_y (where a_x,a_y are the components of a) and you should be able to find a_x,a_y so that you end up with ${\bf x_1}^TPAP{\bf x_1} + k$ ). (i.e. you can make the b^Tx bit disappear).

It's quite tricky and easy to make mistakes though, and LaTeXing it will be horrific...

Yeah, I started with this on paper and got stuck repeatedly, I know what to do but it is a grind of algebra it seems... I'll give it a try though :smile:

. Thanks for the hints!

EDIT: Okay here we go with STEP II Question 6

i) Statment: If AB=0 then BA=0
Consider $A=\begin{pmatrix} 1&1 \\ 1&1 \end{pmatrix}$ and $B=\begin{pmatrix} -2&-3 \\ 2&3 \end{pmatrix}$
Then $AB=\begin{pmatrix} 1&1 \\ 1&1 \end{pmatrix}\begin{pmatrix} -2&-3 \\ 2&3 \end{pmatrix}=\begin{pmatrix} 0&0 \\ 0&0 \end{pmatrix}$
and $BA=\begin{pmatrix} -2&-3 \\ 2&3 \end{pmatrix}\begin{pmatrix} 1&1 \\ 1&1 \end{pmatrix}=\begin{pmatrix} -5&-5 \\ 5&5 \end{pmatrix}$

$\begin{pmatrix} 0&0 \\ 0&0 \end{pmatrix}\not= \begin{pmatrix} -5&-5 \\ 5&5 \end{pmatrix}$
Therefore proven false by counterexample.

ii) Statement: $(A-B)(A+B)=A^2-B^2$ consider the same matrices A and B as in i) then for the LHS:
$(A-B)(A+B)=\begin{pmatrix} 3&4 \\ -1&-2 \end{pmatrix}\begin{pmatrix} -1&-2 \\ 3&4 \end{pmatrix}=\begin{pmatrix} 9&6 \\ -5&-6 \end{pmatrix}$
For the RHS:
$A^2-B^2=\begin{pmatrix} 2&2 \\ 2&2 \end{pmatrix} - \begin{pmatrix} -2&-3 \\ 2&3 \end{pmatrix}=\begin{pmatrix} 4&5 \\ 0&-1 \end{pmatrix}$

$\begin{pmatrix} 9&6 \\ -5&-6 \end{pmatrix}\not= \begin{pmatrix} 4&5 \\ 0&-1 \end{pmatrix}$
Therefore proven false by counterexample.

iii) Statement: AX=0 has a non-zero solution for X iff detA=0
Consider the matrix $A=\begin{pmatrix} 1&0 \\ 3&1 \end{pmatrix}$ which doesn't have a zero determinant.
Chose a matrix X with a zero determinant, e.g. $X=\begin{pmatrix} 2&1 \\ 4&2 \end{pmatrix}$
Then $AX=\begin{pmatrix} 1&0 \\ 3&1 \end{pmatrix}\begin{pmatrix} 2&1 \\ 4&2 \end{pmatrix}=\begin{pmatrix} 2&1 \\ 10&5 \end{pmatrix}$
This matrix has solutions non-zero solutions (2x=y). Therefore it is not necessary for the matrix A to have a determinant 0 for AX=0 to have non-zero solutions and the statement is proven false by counterexample.

(somewhat a disclaimer here, I think (without having investigated it) that the statement is true provided X doesn't have a determinant of 0, but as this is not a criterion in the statement it is proven false by my example.)

iv) Credit to David for this one! (hope I've done it okay, it's getting late and I've not checked my working)
Statement: For any matrix A and B there are at most two matrices such that $X^2+AX+B=0$ .

Consider $X=\begin{pmatrix}0&a \\0&0 \end{pmatrix}$
Then $X^2=\begin{pmatrix}0&0 \\0&0 \end{pmatrix}$
Consider $A=\begin{pmatrix}1&1 \\1&1 \end{pmatrix}$ multiplied by X will be:
$AX=\begin{pmatrix}1&1 \\1&1 \end{pmatrix}\begin{pmatrix}0&a \\0&0 \end{pmatrix}=\begin{pmatrix}0&a \\0&a \end{pmatrix}$

We can see that we can chose a arbitrarily and end up with more solutions than two. Therefore proven by counterexample.

edit: I'm not quite sure about iv) maybe it is better to say
Consider $A=\begin{pmatrix}0&0 \\1&0 \end{pmatrix}$ and then X^2 and AX vanishes leaving B=0 so that there are infinitely many X for which it holds?

Reply 115

TheDuck

Superb...must have spent ages typing up the answer to this question only to realise that not only was the paper I was looking at not in fact STEP III but also someone else had answered the bloody question!
:P

Reply 116

generalebriety

II/10 done, will type up tomorrow though. Too tired. :redface:

Edit: II/10:

$\displaystyle P=\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix}, A=\begin{pmatrix} -1&8 \\ 8&11 \end{pmatrix}.\\[br]PAP = \begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix}\begin{pmatrix} -1&8 \\ 8&11 \end{pmatrix}\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix} = \begin{pmatrix} 15&30 \\ -10&5 \end{pmatrix}\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix} \\[br]= 5\begin{pmatrix} 3&6 \\ -2&1 \end{pmatrix}\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix} = 5\begin{pmatrix} 15&0 \\ 0&-5 \end{pmatrix}\\ \\[br][br]\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} , \mathbf{x}_1 = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} .\\[br]\mathbf{x} = P \mathbf{x}_1 + \mathbf{a} \\[br]\mathbf{x} = \begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix}\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \\[br]\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_1+2y_1+a_1 \\ 2x_1-y_1+a_2 \end{pmatrix} .\\ \\[br][br]\mathbf{x}^T A\mathbf{x} + \mathbf{b}^T \mathbf{x} - 11 = 0 \\[br]\begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} -1&8 \\ 8&11 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 18&6 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} - 11 = 0 \\[br]\begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} -x+8y \\ 8x+11y \end{pmatrix} + 18x+6y-11 = 0 \\[br]-x^2 + 8xy + 8xy + 11y^2 + 18x + 6y - 11 = 0 \\[br]-x^2 + 11y^2 + 16xy + 18x + 6y - 11 = 0. \\[br]\therefore -(x_1+2y_1+a_1)^2 + 11(2x_1-y_1+a_2)^2[br]+ 16(x_1+2y_1+a_1)(2x_1-y_1+a_2) + 18(x_1+2y_1+a_1) + 6(2x_1-y_1+a_2) - 11 = 0. \\[br]\therefore x_1^2 (-1+44+32) + y_1^2 (-4+11-32) + x_1y_1 (-4-44+48) + x_1 (-2a_1 + 44a_2 + 16a_2 + 32a_1 + 18 + 12) + y_1 (-4a_1 - 22a_2 + 32a_2 - 16a_1 + 36 - 6) + (-a_1^2 + 11a_2^2 + 16a_1a_2 + 18a_1 + 6a_2 - 11) = 0 \\[br]\therefore 75x_1^2 - 25y_1^2 + (30a_1+60a_2+30)x_1 + (-20a_1+10a_2+30)y_1 + (-a_1^2 + 11a_2^2 + 16a_1a_2 + 18a_1 + 6a_2 - 11) = 0. \\[br]\text{Solving simultaneously: } 30a_1+60a_2+30=0, \; -20a_1+10a_2+30 = 0 \\[br]\therefore a_2 = -1, a_1 = 1. \\[br]75x_1^2 - 25y_1^2 - 5 = 0 \\[br]3x_1^2 - y_1^2 = \tfrac{1}{5}.$

Reply 117

generalebriety

nota bene

Oops, didn't realise you were doing 10. Sorry. :redface:

I'm not entirely sure my answer's right anyway, but my method is right, I know that. Will type it up tomorrow, anyway. :smile:

Reply 118

nota bene

generalebriety

Oops, didn't realise you were doing 10. Sorry. :redface:

I'm not entirely sure my answer's right anyway, but my method is right, I know that. Will type it up tomorrow, anyway. :smile:

Well, I've still not finished the question, so feel free to type it up! I'll see if we have come to the same conclusions to the parts we've both done :smile:

I'll be attempting a few probability questions tomorrow, but it is likely I'll need help with parts, unless someone else feels up to do them (doesn't seem to be many that get done!).

Reply 119

insparato

Ive had a go at STEP II Question 6 myself. First two parts are easy.

The third and fourth are trickier,

Jonah

Statement: AX=0 has a non-zero solution for X if and only if detA=0.

Are you sure this is false?

The detX isnt really important i think.

This is the way i see it.

If DetA = 0, this means it has no inverse. If A had an inverse you could do this.

$AX = 0$

$A^{-1}(AX) = A^{-1}(0)$

$(A^{-1}A)X = 0$

$IX = 0$

$X = 0$

Therefore if DetA doesnt equal zero, X has a zero solution.

So i think DetA has to be zero for X to have a non zero solution. I dont know how important it is to consider what the actual matrix A is or what the actual matrix of X is, as its only asking about X being non zero.

On a side note
Jonah, Billy jesus! get some sleep, how can you expect to function adequately without sleep?

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