STEP Maths I, II, III 1993 Solutions

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Reply 140

Im in the middle of doing STEP III Question 6.

Reply 141

STEP III, Q15:

Clearly P(N=n) = 0 for n < 4. If n>=4, then if N=n, then either: we had 1 tail and n-2 heads after n-1 tosses and the last toss was a tail, or vice versa.

So $\mathbb{P}(N=n) = (n-1)(p^2q^{n-2}+q^2p^{n-2})$

So $\mathbb{E}(N) = \sum_{n=4}^\infty n(n-1)(p^2q^{n-2}+q^2p^{n-2})$ .

Now we know $\sum_0^\infty a^n = \frac{1}{1-a}$ . Differentiating, we get $\sum_0^\infty na^{n-1} = \frac{1}{(1-a)^2}$ . Differentiating again: $\sum_0^\infty n(n-1)a^{n-2} = \frac{2}{(1-a)^3}$ .

So $\sum_0^\infty n(n-1) (p^2q^{n-2}+q^2p^{n-2}) = 2\left(\frac{p^2}{(1-q)^3}+\frac{q^2}{(1-p)^3}\right) = 2\left(\frac{1}{p}+\frac{1}{q}\right)$

Note also $\frac{1}{p}+\frac{1}{q} = \frac{p+q}{pq} = \frac{1}{pq}$ .

It remains to correct for starting the sum from 0 instead of 4. The n=0 and n=1 terms disappear due to the n(n-1) factor. When n=2 the summand becomes $2(p^2+q^2) = 2((p+q)^2-2pq) = 2(1-2pq)$ . When n=3 the summand is $6(p^2q+q^2p)=6pq(p+q) = 6pq$ .

So $\mathbb{E}(N) = 2\left(\frac{1}{pq} - (1-2pq) -3pq \right) = 2\left(\frac{1}{pq} - 1-pq \right)$ as required.

Reply 142

Swayum

Speleo

EDIT: The Tn(r) question is done (although it is listed above), I haven't sorted that out yet and I don't know which one it is.

Correct me if I'm wrong but I don't think the T_n(r) one is done yet. It's STEP I question 2, I don't think anyone's done the very last bit about what happens when r = 1, r = -1 and r > 1.

Reply 143

justinsh

STEP II, Question 4

Justin

EDIT: this is in fact STEP III question 4

STEP II 1993 question 4.pdf55.1KB

Reply 144

DFranklin

justinsh

STEP II, Question 4You've actually done STEP III, Q4 (or at least Further Mathematics Paper B, which is really the same thing) - look at the first page of the PDF! (The site everyone is getting these papers from has mislabelled the exams for 1993 and earlier).

Someone else has already done this question, but to answer your questions: I'm sure it's fine to quote the series for log x for (i). (The alternative approach would be to sum the series x^{2n} and then integrate both sides). The significance of |x|<1 in (ii) is simply to make sure the GP converges, so you don't have to deal separately with the cases |x|=1, |x|>1 (which would behave differently).

Reply 145

justinsh

DFranklin >> thanks for pointing it out

generalebriety >> I don't understand the following bit (Then S=...) Thank you.

generalebriety

STEP III Q4:

(iii)

$\displaystyle S = \sum_{r=2}^\infty \frac{r2^{r-2}}{3^{r-1}} = \frac{1}{3}\sum_{r=2}^\infty r\bigg( \frac{2}{3} \bigg)^{r-2}.\\[br]\text{Let } u_n = \sum_{r=n}^\infty \bigg( \frac{2}{3} \bigg)^{r-2} = 3\bigg( \frac{2}{3} \bigg)^{n-2}.\\[br]\text{Then } S = \frac{1}{3}(2u_2+u_3+u_4+u_5+\dots). \\[br]$

Reply 146

DFranklin

justinsh

DFranklin >> thanks for pointing it out

generalebriety >> I don't understand the following bit (Then S=...) Thank you.

Firstly, I think there's a typo, the 2 on the last line shouldn't be there.

The idea here is:

$a+2a^2+3a^3+4a^4...$
$=a+a^2+a^3+a^4...$
$+0+a^2+a^3+a^4...$
$+0+0+a^3+a^4...$

(where I've only put in the 0s to make it clearer how things line up).

Reply 147

DFranklin

DeathAwaitsU

Correct me if I'm wrong but I don't think the T_n(r) one is done yet. It's STEP I question 2, I don't think anyone's done the very last bit about what happens when r = 1, r = -1 and r > 1.

I think it has, but there are bits and pieces scattered all other. Here's a (hopefully correct!) full answer:

Write $G_n(x) = \sum_0^n x^n = \frac{1-x^{n+1}}{1-x}$ .

$1+r^2+r^4+...+r^{2n} = G_n(r^2) = \frac{1-r^{2n+2}}{1-r^2}$

$S_n(r) = G_{3n}(r)-G_n(r^3) = \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3}$ .

If |r|<1, then as $n\to\infty; r^{3n+1}, r^{3n+3} \to 0$ , so $S_n(r) \to \frac{1}{1-r} - \frac{1}{1-r^3}$ .

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

If r > 1, all terms in T_n are positive and tend to infinity. So $T_n(r) \to \infty$ .
If r = 1, T_n = 4n+1 and tends to infinity.
If r=-1, T_n = 2n+1 and tends to infinity.

Reply 148

justinsh

Thank you DFranklin for the explanation.

Considering part (i) of question 2 STEP I (I hope it has not already been done) a trick that works is to put m=a-b and n=a+b (a and b uniquely defined integers).

It then all simplifies out. (The specials cases arise when a or b=0)

Reply 149

DFranklin

justinsh

Again, that's not STEP I. Look at the front.

In case of any doubt:

MATHEMATICS = STEP I
FURTHER MATHEMATICS PAPER A = STEP II
FURTHER MATHEMATICS PAPER B = STEP III

Reply 150

justinsh

Ok, so if I got it right:

When the usual site gives STEP I it is in fact STEP II, when STEP II it is in fact STEP III and when STEP III it is in fact STEP I.

Reply 151

insparato

STEP III Question 6

A complex number z lying on a circle centre K and radius where K represents the complex number k.

$|z - k | = r^2$

$|x + iy - (a+bi)| = r^2$

$|x - a + i(y-b)| = r^2$

$(x - a)^2 + (y - b)^2 = r^2$
$x^2 - 2ax + a^2 + y^2 - 2yb + b^2 = r^2$

Consider zz*

$(x + iy)(x - iy) = x^2 + y^2$

kz*

$(a + bi)(x - iy) = ax -aiy + bix + by$

k*z

$(a - bi)(x + iy) = ax + aiy - bix + by$

kk*

-k*z - kk* = ax + aiy - bix - by - ax - aiy + bix - by = -2ax - 2by

$(a - bi)(a + bi) = a^2 + b^2$

$x^2 - 2ax + a^2 + y^2 - 2yb + b^2 = r^2$

$(x^2 + y^2) -2ax - 2by + (a^2 + b^2) = r^2$

$zz^* - k^*z - kk^* + kk^* - r^2 = 0$

The locus of P is which represents the complex number z

| z - i | = 1

$w_1 = \frac{z}{z-1}$

$w_1z - w_1 = z$

$z = \frac{w_1}{w_1 - 1}$

$| z - i | = | \frac{w_1}{w_1 - 1} - i |$

$1 = \frac{|w_1 - i(w_1 - 1)|}{|w_1-1|}$

$|w_1 - 1| = |w_1 - iw_1 + i |$

let w_1 = x + iy

$|x + iy - 1| = x + iy -i(x + iy) + i|$

$\sqrt{(x-1)^2 + y^2} = |x + y + iy - ix + i |$

$\sqrt{(x-1)^2 + y^2} = \sqrt{(x+y)^2 + (y - x + 1)^2}$

$(x-1)^2 + y^2 = (x+y)^2 + (y - x + 1)^2$

$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2xy + y^2 - xy + y - xy + x^2 - x + y - x + 1$

$x^2 - 2x + 1 + y^2 = 2x^2 + 2y^2 + 2y - 2x + 1$

$x^2 + y^2 + 2y = 0$

$x^2 + (y+1)^2 - 1 = 0$

$x^2 + (y+1)^2 = 1$

The locus of W_1 is a circle with centre (0,-i) and radius 1 on the argand diagram.

Thus the locus L is

|w_1 + i | = 1

Now this is where it gets tricky.

w_2 = z*

This formula i proved is screaming at me.

$zz* - k*z - kk* + kk* - r^2 = 0$

We know it has to be some circle and you can guess as its going to be of radius one if its the conjugate of z.

Im tempted to just say if locus of z is |z - i | = 1

therefore the locus of the conjugate of z is |z* - i | = 1

this incidentally happens to come out as the locus of L.

Just had another thought. What is the conjugate of a complex number?, as far as im aware its a reflection in the x axis of the original complex number on the argand diagram. So if the locus of Z is a circle, if you take all the specific points you could make its conjugates by reflecting in the x axis.

So if the locus of Z is a circle centered at (0,i) and has a radius of 1 it touches(0,0) a reflection of this is simply a circle centered (0,-i) with radius 1.

This happens to be x^2 + (y + 1)^2 = 1

Which is the same locus as L.

The next part is bugging me here,

Determine the positions of P for which Q1 and Q2 coincide.

I can see how the points Q1 and Q2 could coincide they are on the same locus. However the positions of p ? Am i suppose to find where Q1 and Q2 coincide and use one of the formulas given to find where it transforms onto the Z plane?

Ive proved a formula i have not yet used, some how i think this has got to come in here.

Reply 152

DFranklin

You made very heavy weather of the first bit. $|a+ib|^2=a^2+b^2 =(a+ib)(a+ib)^*$ . So

$|z-k| = r \implies (z-k)(z-k)^* = r^2 \\[br]\implies (z-k)(z^*-k^*) = r^2 \implies zz^*-zk^*-z^*k+kk^* = r^2$

2nd bit is fine.

3rd bit: Putting k = i, r=1 we have $zz^*-zk^*-z^*k+kk^* = r^2$ .

Since $w_2=z^*,w_2^* = z$ we get: $w_2^*w_2-w_2^*k^*-w_2k+kk^* = r^2$ . Set $k_2 = k^*$ and we get $w_2^*w_2-w_2^*k_2-w_2k_2^*+k_2^*k_2 = r^2$ . So w_2 has equation of circle radius 1, center $k_2 = k^* = -i$ .

(Or your reflection in real axis argument is also fine. I just thought I'd show how you can use the first bit).

I'm not seeing an easy solution to the last part. I'd probably parameterize z as $\cos \theta + (1+\sin \theta)i$ , find a formula for w and solve for $w=z^*$ . (Possibly just writing z=(x+iy) and taking a similar approach will be easier).

Reply 153

insparato

1st part: Nice definitely made more work there.
2nd part: Yay
3rd part: I actually tried using the formula last night, but i didnt jumped to the w_2 = z* part. I did something horrible with the algebra getting z* on its own. But thats nice.
4th part: Ill have a go later on.

Always seems like an uphill struggle.

Reply 154

DFranklin

insparato

Always seems like an uphill struggle.

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.

Reply 155

justinsh

STEP III Question 2

First part: Proving that y'' is never zero

$x^3+y^3=3xy\\$ (*)
$\Rightarrow3x^2+3y^2y'=3y+3xy'\\$ (differentiating (*))
$\Rightarrow x^2+y^2y'=y+xy'\\$
$\Rightarrow y'=\frac{y-x^2}{y^2-x}\\$
$\Rightarrow y''=\frac{(y'-2x)(y^2-x)-(2yy'-1)(y-x^2)}{(y^2-x)^2}\\$

We don't care about the denominator, so let us study the numerator:

$(y'-2x)(y^2-x)-(2yy'-1)(y-x^2)\\$
$=y'(2x^2y-x-y^2)+x^2+y-2xy^2\\$

Replacing $y'$ and putting all on a common denominator we get, for the numerator, the following expression:

$6x^2y^2-3xy-2xy(x^3+y^3)\\$
$=3xy(2xy-1-2xy)$ (using (*))
$=-3xy\\$

This is never zero since the origin not an inflection point and the inflection point is the only point where x=0 or y=0. Indeed, if x=0 then y=0 and vice versa.

Second part: Area of the first quadrant using the formula $A=\frac{1}{2}\int_{\theta_{min}}^{\theta_{max}}r^2d\theta$

First, we convert in polar coordinates and use (*) to find r in terms of theta:

$r^3(\cos^3\theta+\sin^3 \theta)=3(r\cos\theta)(r\sin\theta)\\$
$\Rightarrow r=3\frac{\cos\theta \sin\theta}{\cos^3\theta+\sin^3 \theta}\\$

Then, we apply the formula:

$A=\frac{1}{2}\int_0^{\pi/2}\frac{9\cos^2\theta\sin^2\theta}{(\cos^3\theta+ \sin^3 \theta)^2}d\theta\\$
$=9\int_0^{\pi/4}\frac{\cos^2\theta\sin^2\theta}{(\cos^3\theta+ \sin^3 \theta)^2}d\theta\\$ (x=y is a symetry axis because of the symetric roles played by x and y in (*))
$=9\int_0^{\pi/4}\frac{\tan^2\theta\sec^2\theta}{(1+\tan^3\theta)^2}d\theta\\$
$=9\int_0^1\frac{u^2du}{1+u^3}\\$
$=9\int_1^2\frac{dv}{3v^2}\\$
$=3(\frac{-1}{2}+1)=\frac{3}{2}\\$

The successive changes of variable were $u=\tan\theta, v=1+u^3$ .

NOTES:

In the first part, I did not consider the case where the denominator $y^2-x$ equals zero. In this case, using (*), we find that $y^6+y^3=3y^3$ which means that either $y=0$ (this is the origin, no problem) or $y=2^{1/3}$ (which is more of a problem because y' is undefined).

In the second part, r is well-defined on the considered interval because $\cos^3\theta+\sin^3 \theta=(\cos\theta+\sin\theta)(1-\cos\theta\sin\theta)=\sqrt{2}\sin(\theta+\pi/4)(1-\frac{\sin2\theta}{2})$ which is never zero.

In the second part, I don't know if my symetry axis argument is correct and/or rigorous.

Reply 156

Rabite

Anyone done STEP I Q7? I looked and couldn't find it, but I may have missed it since it's a piddly little thing.
It starts with f(x)=x³+Ax²+B and is about determining how many roots it has.

Reply 157

generalebriety

DFranklin

Nope, the 2 should be there. Consider:
$\displaystyle 3S = \sum_{r=2}^\infty r\bigg( \frac{2}{3} \bigg)^{r-2} \\[br]= 2\bigg( \frac{2}{3} \bigg)^0 + 3\bigg( \frac{2}{3} \bigg)^1 + 4\bigg( \frac{2}{3} \bigg)^2 + 5\bigg( \frac{2}{3} \bigg)^3 + \dots \\[br]= 2\bigg[\bigg( \frac{2}{3} \bigg)^0 + \bigg( \frac{2}{3} \bigg)^1 + \bigg( \frac{2}{3} \bigg)^2 + \bigg( \frac{2}{3} \bigg)^3 + \dots\bigg] + \bigg[\bigg( \frac{2}{3} \bigg)^1 + \bigg( \frac{2}{3} \bigg)^2 + \bigg( \frac{2}{3} \bigg)^3 + \dots\bigg] \\ + \bigg[\bigg( \frac{2}{3} \bigg)^2 + \bigg( \frac{2}{3} \bigg)^3 + \dots\bigg] + \bigg[\bigg( \frac{2}{3} \bigg)^3 + \dots\bigg] + \dots \\[br]= 2u_0 + u_1 + u_2 + u_3 + \dots$

Does that make it any clearer? I know it wasn't clear to begin with... :p:

Reply 158

DFranklin

Rabite

Anyone done STEP I Q7? I looked and couldn't find it, but I may have missed it since it's a piddly little thing.
It starts with f(x)=x³+Ax²+B and is about determining how many roots it has.

Not been done I think. It's pretty easy.

Reply 159

khaixiang

DFranklin

I'm not seeing an easy solution to the last part. I'd probably parameterize z as $\cos \theta + (1+\sin \theta)i$ , find a formula for w and solve for $w=z^*$ . (Possibly just writing z=(x+iy) and taking a similar approach will be easier).

Hmm... IMHO, I think this part is just about solving $z^{*}=\frac{z}{z-1}$ ?

You have from the first part $z^{*}=\frac{-iz}{z-i}$ , if I am not mistaken, everything should factor out nicely giving z=0 and z=1+i, without having substitute z=x+iy. You can also find the locus of z* using $z^{*}=\frac{-iz}{z-i}$ .

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