STEP Maths I, II, III 1993 Solutions

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Reply 260

Original post

by DFranklin

Looks fine (but I confess I can't be bothered to look the question up and do it, so I'm comparing against the LaTeX generalebriety wrote in the roignal solution). In terms of making things look nice, LaTeX recognizes most trig and hyperbolic terms, so \sinh will give you $\sinh$ which looks nicer than $sinh$ . It doesn't recognize arsinh, so you can either write \text{arsinh} to get $\text{arsinh}$ , or you could go \sinh^{-1} to get $\sinh^{-1}$ , which is my personal preference.

(Which isn't to say you need to do any of this - what you wrote is fine, but if you're aiming for something that looks as nice as possible it makes a difference).

Noted, I'll edit it in now.

Reply 261

solC

Original post

by Zacken

Working all seems fine (the solution is already on an earlier page in the thread); note that \sinh and \cosh are LaTeX commands though.

I thought I might as well rewrite it since the LaTeX in part of the solution was broken :smile:

Reply 262

cxs

one thing you should consider
for it is defined that 0<=arg<=2pi, so when arg(z1)+arg(z2)+arg(z3)<2pi, arg(z1z2z3)>arg(z3)
when arg(z1)+arg(z2)+arg(z3)>2pi, as arg(z2)<pi,arg(z3)<pi, so arg (z1z2z3)<z1, so cannot equal.

Original post

by khaixiang

I've done STEP II (paper A), question 2, 3, 8 and STEP III (paper B), question 1, 2, 4, 9 awhile ago. So I won't do them again and will try other questions. Here's question 9 of STEP II:

Sorry, I know this is very ugly, if it's too much of an eyesore I will remove it for conformity's sake.

This question seems way too easy compared to others, so perhaps I've overlooked something.

Reply 263

cxs

for LEP, as it is rounding up, so i think its variance is 0.5^2/12=1/48?

Original post

by toasted-lion

FMA Question 15

For LEP we add 1500 uniform distributions with $\mu = 0, \ \sigma^2 = \frac{1}{12}$ to give a normal distribution with $\mu = 0, \ \sigma^2 = \frac{1500}{12} = 125$ , ie $E\sim N(0,125)$ .

$P(|E|>15)=2P(E>15)=2P(Z>\frac{15}{\sqrt{125}}) \approx 2P(Z>1.34) \\ \approx 2(1-0.9099) = 0.1802 \approx 0.18$

For VOZ, add 1500 uniform distributions with $\mu = 0.5, \ \sigma^2 = \frac{1}{12}$ to give a normal distribution with $\mu = 750, \ \sigma^2 = 125$ , ie $E\sim N(750,125)$ .

$P(|E|>15)=P(E>15)=P(Z>\frac{15-750}{\sqrt{125}}) \approx P(Z>-65.74) \approx 1$

If LEP makes n additions, $E\sim N(0,\frac{n}{12})$ .

$P(|E|<10)=P(E<10)-P(E<-10)=2P(E<10)-1 \\ =2P(Z<\frac{20\sqrt{3}}{\sqrt{n}}) -1 = 0.9 \implies P(Z<\frac{20\sqrt{3}}{\sqrt{n}})=0.95 \implies \frac{20\sqrt{3}}{\sqrt{n}} = 1.6449 \implies n = 443$

Rounding down at the end for obvious reasons. Now, I think I missed a joke here: what do LEP and VOZ stand for?

Reply 264

casperyc

Original post

by brianeverit

I do not follow this part...

I got this, which cannot use any further initial conditions??? (shouldn't it be V=L ?????)

(edited 6 years ago)

Reply 265

RuoyanC

Original post

by 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...

omg this is hard to caculate by hand! is it allowed to use caculator in 1993 STEPs?

Reply 266

gwenstaicy

Original post

by Rabite

I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~
[edit] Here it is anyway.
$\int \cos{mx}\cos{nx} dx = ½ \int \cos (m+n)x + \cos (m-n)x dx$
By the product/sum formulae that no one remembers.
$= ½[\frac{1}{m+n} \sin(m+n)x + \frac{1}{m-n} \sin(m-n)x]_0 ^{2\pi}$
But if m=±n, one of the fractions explodes. So in that case the question is:
$\int \cos^2 {mx} dx$
$=½ \int 1+ cos{2mx}dx$
$= ½[x+\frac{1}{2m} \sin{2mx}]_0 ^{2\pi}$
$=\pi$
If m=n=0, the integral turns to $2\pi$ .
As for the second bit.
Let x = sinh²t
dx = 2sinhtcosht dt
$I = \int \sqrt{\frac{x+1}{x}}dx$
$= \int \sqrt{\frac{\sinh^2 t +1}{\sinh^2 t}}(2\sinh t\cosh t)dt$
$= \int 2 \frac{\cosh t}{\sinh t} (\sinh t \ cosh t) dt$
$= \int 2 \cosh^2 t dt$
$= \int 1+ \cosh 2t dt = t + ½ \sinh 2t$ Ignoring the +c for now
$= t + \sinh t \cosh t = \sinh ^{-1}(\sqrt{x}) + \sqrt(x(x+x)) +c$
Which you can rewrite using the log form of arsinh.

but what if x is negative?

Reply 267

Fury1755

you made a mistake the original integral should be integral of root(1+ (1/x))

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