STEP maths I, II, III 1992 solutions

STEP III 1992, Question 1

Unparseable latex formula:

\\f(x)=\ln(1+e^{x})}\\ f'(x)=\frac{e^{x}}{(1+e^{x})}\\ \ln{f'(x)}=x-\ln{(1+e^{x})}\\ \ln{f'(x)}=x-f(x)

Differentiate with respect to x,

\\ \frac{f''(x)}{f'(x)}=1-f'(x)\\ f''(x)=f'(x)-[f'(x)]^{2}

As required.

\\f(0)=\ln{2},\\ f'(0)=0.5\\ f''(0)=f'(0)-[f'(0)]^2\\ f''(0)=0.5-0.25=0.25

Now differentiate with respect to x the equation:

\\f''(x)=f'(x)-[f'(x)]^{2}\\ f'''(x)=f''(x)-2f'(x)f''(x)\\ f'''(0)=0.25-2(0.5)(0.25)=0

Differentiate again:

\\f''''(x)=f'''(x)-2[f''(x)]^{2}-2f'(x)f'''(x)\\ f''''(0)=f'''(0)-2[f''(0)]^{2}-2f'(0)f'''(0)\\ f''''(0)=-1/8

f(x)=f(0)+f'(0)x+\frac{f''(0)}{4}x^2+\frac{f'''(0)}{6}x^3+\frac{f''''(0)}{24}x^4+\dots

f(x)=\ln(2)+\frac{x}{2}+\frac{x^2}{8}-\frac{x^4}{192}+\dots

Now the fun part... lots of binomial expansion, series and latexing.

\\g(x)=\frac{1}{\sinh(x)\cosh(2x)}\\ g(x)=\frac{1}{\sinh(x)(1+2\sinh^{2}(x))}\\ g(x)=(\sinh(x))^{-1}(1+2\sinh^{2}(x))^{-1}

Using binomial expansion,

\\g(x)=(\sinh(x))^{-1}(1-2\sinh^{2}(x)+(2\sinh^{2}(x))^{2}-(2\sinh^{2}(x))^{3}+\dots\\ g(x)=\frac{1}{\sinh(x)}-2\sinh(x)+4\sinh^{3}(x)-8\sinh^{5}(x)+\dots

As you can see, g(x) cannot be expanded as a series of non-negative powers of x since 1/sinh(x) when expanded in series contain the term 1/x. Note that the series expansion of sinh(x) is

x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots

, i.e all odd powers of x, so also g(x) when expanded according to our equation (sinh(x) terms all raised to odd powers) just now contains all odd powers of x. So then xg(x) must contain even powers of x.

Now to expand xg(x) as a series as far as the term in x^4, we first expand g(x) up to x^3 term:

\\g(x)=\frac{1}{\sinh(x)}-2\sinh(x)+4\sinh^{3}(x)-8\sinh^{5}(x)+\dots\\ g(x)=(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots)^{-1}-2(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots)+4(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots)^{3}+\dots\\ g(x)=x^{-1}(1+\frac{x^2}{3!}+\frac{x^4}{5!}+\dots)^{-1}-(2x+\frac{2x^3}{3!}+\dots)+4(x^3+\frac{3x^5}{3!}+...)

\\g(x)=x^{-1}(1-\frac{x^2}{3!}-\frac{x^4}{5!}+(\frac{x^2}{3!}-\frac{x^4}{5!})^{2}-\dots)-(2x+\frac{2x^3}{3!}+\dots)+4(x^3+\frac{3x^5}{3!}+\dots)\\ g(x)=(\frac{1}{x}-\frac{x}{6}+\frac{7x^3}{(360)}+\dots)-(2x+\frac{2x^3}{3!}+\dots)+(4x^3+\dots)\\ g(x)=\frac{1}{x}-\frac{13x}{6}+\frac{1327x^{3}}{(360)}+\dots\\ xg(x)=1-\frac{13x^{2}}{6}+\frac{1327x^{4}}{(360)}+\dots

Do tell me if there's any mistakes.

Reply 101

khaixiang

Do tell me if there's any mistakes.

I haven't looked through your actual maths, but \e isn't a LaTeX tag and hence hasn't appeared in your solution. :smile:

Reply 102

khaixiang

Do tell me if there's any mistakes.

I didn't check the first one, but I got the same answer for the 2nd bit by a very different method, which has to be a good sign!

Reply 103

generalebriety

Doing II/10.

(i) arg(z-1) - arg z = alpha
(ii) arg(z-1) - arg z = alpha - arg(-1)
arg(1-z) - arg z = alpha
(iii) |z-1| = |z|

Stuck on (iv)... anyone?

w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})}

Shuffle around the negative signs...

\\w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{1}-z_{4})(z_{3}-z_{2})}\\ arg(w)=arg(\frac{z_{1}-z_{2}}{z_{1}-z_{4}})+arg(\frac{z_{3}-z_{4}}{z_{3}-z_{2}})\\ arg(w)=arg(\frac{z_{1}-z_{2}}{z_{1}-z_{4}})-arg(\frac{z_{3}-z_{2}}{z_{3}-z_{4}})

Now note that since z1, z2, z3 and z4 lies in that order on a circle in the complex plane, the line joining z2 and z4 is a chord and z1 lies above this chord while z3 lies below this chord. We then have, deducing from the result in part(i) and part(ii):

\\arg(\frac{z_{1}-z_{2}}{z_{1}-z_{4}})=\alpha\\ arg(\frac{z_{3}-z_{2}}{z_{3}-z_{4}})=\alpha-\pi\\ arg(w)=arg(\frac{z_{1}-z_{2}}{z_{1}-z_{4}})-arg(\frac{z_{3}-z_{2}}{z_{3}-z_{4}})\\ \therefore arg(w)=\pi\\ \implies \Im{\{w\}}=0

Reply 104

generalebriety

I haven't looked through your actual maths, but \e isn't a LaTeX tag and hence hasn't appeared in your solution. :smile:

oops

, corrected it. Thanks. Any idea why tsr would not allow 360 bracketed by {} to be posted? In my final solution, I have to tex up 360 as part of the fraction you see...

DFranklin

I didn't check the first one, but I got the same answer for the 2nd bit by a very different method, which has to be a good sign!

By the way, couldn't have done the second bit without your hint from some thread back then, thanks again. With that kind of arithmetic involved, this type of question is unlikely to come out again.

Reply 105

\frac{1327x^4}{ 360 }

Edit: ah, there we go. Putting spaces around the 360 seems to have pleased it. TSR's LaTeX is messed up. :s-smilie:

Reply 106

insparato

STEP I Question 8

Just attempting the final and 4th part now.

As 24 hours have passed, I assume you're stuck / fed up.

We wish to solve

\displaystyle \int_0^x f(t)\,dt = \int_x^1 t^2f(t)dt +x - \frac{x^5}{5} + C

. As before, differentiate w.r.t. x:

f(x) = -x^2f(x) + (1-x^4) \implies (1+x^2)f(x) = (1-x^4) = (1-x^2)(1+x^2)

and so

f(x) = 1-x^2

.

Substitute back in:

\displaystyle \int_0^x (1-t^2)\,dt = \int_x^1 t^2-t^4dt +x - \frac{x^5}{5} + C

\displaystyle x-\frac{x^3}{3} = \left[\frac{t^3}{3} - \frac{t^5}{5}\right]_x^1 +x - \frac{x^5}{5} + C

\displaystyle x-\frac{x^3}{3} = \frac{1}{3}-\frac{1}{5} - \frac{x^3}{3} + \frac{x^5}{5} +x - \frac{x^5}{5} + C

Deduce C =

\frac{1}{5}-\frac{1}{3} = -\frac{2}{15}

Reply 107

insparato

Oh! Yes i forgot about that, i was going to have one last attempt but i couldnt see where it was wanting me to go. I dont think i really had a handle on that question to start with.

Seems so simple when you see it done though. Oh well atleast i've learnt something, it was definitely something with integration and function notation i havent seen used before, but then again i think thats the point with STEP.

Reply 108

This thread's almost dead. Give it another 12 hours or so and then I'll create a 1991 thread while working out a way to get my hands on pre-1990 past papers... my teacher might have some, but it's half-term...

Reply 109

nota bene

generalebriety

This thread is not stone dead yet though :wink:

I'm trying to read up on groups and see if I can do those questions...

About pre-90 papers IIRC I saw a page from one on a blog on internet, so they could possibly be available, just that we don't know about it :confused:

Would be nice if you could get them though :smile:

Reply 110

8 1992 STEP III Question (Click here)

scan0009.jpg254.2KB

Reply 111

nota bene

This thread is not stone dead yet though :wink:

Would be nice if you could get them though :smile:

I think Francis Woodhouse, who used to be quite active in this forum has the pre-1990 papers according to what I've seen in his blog... so maybe I will try to email and ask him...

Reply 112

khaixiang

I think Francis Woodhouse, who used to be quite active in this forum has the pre-1990 papers according to what I've seen in his blog... so maybe I will try to email and ask him...

Yeah, good idea. :smile:

I emailed OCR to ask if they would let me buy any, but I doubt they will.

Reply 113

gyrase

STEP II/5

The group has an identity element e, other elements g_1, g_2 ... and the operation is say *. Then the order of an element g_1 is n such that g_1 * g_1 * g_1 * ... = (g_1)^n = e.

Let A be an element of group S. detA = 1.

Unparseable latex formula:

\displaystyle A=\begin{pmatrix} w&x \\ y&z \end{pmatrix}[br]\displaystyle A^-^1=[br]\frac{1}{wz + xy}\begin{pmatrix} z&-x \\ -y&w \end{pmatrix}

but wz + xy = 1 so

Unparseable latex formula:

\displaystyle A^-^1=\begin{pmatrix} z&-x \\ -y&w \end{pmatrix}

S is associative, has identity element

\displaystyle I=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}

and has unique inverse elements as above and closure;

\displaystyle A_1A_2=\begin{pmatrix} w_1&x_1 \\ y_1&z_1 \end{pmatrix}[br]\displaystyle \begin{pmatrix} w_2&x_2 \\ y_2&z_2 \end{pmatrix} = \displaystyle \begin{pmatrix} w_1w_2+x_1y_2&w_1x_2+x_1z_2 \\ y_1w_2+z_1y_2&y_1x_2+z_1z_2 \end{pmatrix}

A^-1 = A
I = A^2, w^2 + xy = 1
.............wx + xz = 0, x(w+z) = 0
.............yw + yz = 0, y(w+z) = 0
.............yx + z^2 = 1, yx + z^2 = w^2 + xy, z = w

(x-y)(w+z) = 0
z = w
2xw = 2yw = 0 .. w(x-y) = 0, x = y
also wz + xy = 1, w^2 = wz = z^2, solving gives A = I.

:will finish later:

Reply 114

I/16. Doesn't warrant LaTeXing, so I won't.

P(only open route is ABCD) = P(AB open) * P(BC open) * P(CD open) * P(AC closed) * P(BD closed)
= (1-p)*(1-p)*(1-p)*p*p = (1-p)³p².

Possible routes from A to D:
1. AB + BC + CD
2. AB + BD
3. AC + CD
4. AC + CB + BD

Only routes 2 and 4 matter in this problem.
P(only route is ABD) = (1-p)²p³
P(only route is ACBD) = (1-p)³p²
AB better choice

\Longleftrightarrow

(1-p)²p³ < (1-p)³p²

\Longleftrightarrow

p < 1/2.
CB better choice

\Longleftrightarrow

(1-p)²p³ > (1-p)³p²

\Longleftrightarrow

p > 1/2.
Equally advantageous

\Longleftrightarrow

p = 1/2.

It is easy to show that AD is only blocked:
if AB, AC are blocked (probability 1/100)
or AB, BC, CD are blocked (probability 1/1000)
or AC, CB, BD are blocked (probability 1/1000)
or BD, CD are blocked (probability 1/100).
Therefore P(AD is blocked) = 0.022 - they go ahead with it.

Reply 115

generalebriety

Yeah, good idea. :smile:

I emailed OCR to ask if they would let me buy any, but I doubt they will.

OCR did not even reply me the last few times I emailed them, so don't think this will work out. I've yet to receive a reply from Francis Woodhouse, he might be too busy preparing for his tripos exam to check his gmail. Anyone knows how else we can get the pre-'90 papers?

Also, maybe someone else can have a look at III/10, not sure if I am doing it the way it's intended to.

Reply 116

khaixiang

Hmm. It's not important yet, anyway - we still have 1991 and 1990 to do. :smile:

Reply 117

khaixiang

Also, maybe someone else can have a look at III/10, not sure if I am doing it the way it's intended to.

For the lengths you should get 2 very similar integrals; do a

\phi = 2\theta

substitution on the first one to get the forms even closer, and then use the symmetry of cos,sin to account for the different limits.

For the areas, both integrals should be straightforward to evaluate (or even quote, in the case of the ellipse).

Does that help, or do you want more detail?

Reply 118

DFranklin

For the lengths you should get 2 very similar integrals; do a

\phi = 2\theta

Thanks, this is exactly what I did, I wasn't so confident on using integration though, the line between "using integration" and "evaluating the integral" isn't so clear to me. And I am sure you remember one STEP question where a geometrical approach is intended instead of integration, I thought this is one of them. But there're no easy ways to know isn't it, except trying for half and hour and coming to the conclusion that a geometrical approach is not possible.

Reply 119