# STEP Maths II, III 2009 Solutions

(edited 7 years ago)

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STEP II, Question 1

First part

Second part

Third part

(edited 7 years ago)
STEP II, Question 4

First part

Second part

Third part

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STEP II, Q5:

(i)

(ii)

(iii)

(edited 7 years ago)
STEP III, Question 1

p and q

st and s+t

p+q=0

(edited 7 years ago)
Original post by Zacken
(Updated as far as #5)

STEP II:
1: Solution by Zacken
2:
3:
4: Solution by Zacken
5: Solution by Zacken
6:
7:
8:
9:
10:
11:
12:
13:

STEP III:

2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:

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I'll take ii q11 for now. Will probably focus on the applied seeing as no-one else seems to like it :-P

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STEP III, Question 5

$yz+zx+xy=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\dfrac{(1)^2-2}{2}=-\dfrac{1}{2}$
as required.

$x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=x^2(y+z)+y^2(z+x)+z^2(x+y)= x^2(1-x)+y^2(1-y)+z^2(1-z)=x^2+y^2+z^2-(x^3+y^3+z^3)=2-3=-1$
as required.

$-\frac{1}{2}=-\frac{1}{2} \times 1=(yz+zx+xy)(x+y+z)=xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz=3xyz+x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=3xyz-1[br][br]\Rightarrow xyz=\dfrac{1-\frac{1}{2}}{3}=\dfrac{1}{6}$
as required.

We require

$S_{n+1}=aS_n+bS_{n-1}+cS_{n-2} [br][br]\Leftrightarrow x^{n+1}+y^{n+1}+z^{n+1}=ax^n+ay^n+az^n+bx^{n-1}+by^{n-1}+bz^{n-1}+cx^{n-2}+cy^{n-2}+cz^{n-2} [br][br]\Leftrightarrow x^{n-2}(x^3-ax^2-bx-c)+y^{n-2}(y^3-ay^2-by-c)+z^{n-2}(z^3-az^2-bz-c)=0$.

It therefore suffices to find $a$, $b$ and $c$ such that the cubic $m^3-am^2-bm-c=0$ for $m=x$,$y$,$z$.

Therefore we can take $a=x+y+z=1$, $b=-(yz+zx+xy)=\dfrac{1}{2}$ and $c=xyz=\dfrac{1}{6}$.
(edited 7 years ago)
Can I reserve STEP III Q4?
Original post by EricPiphany
Can I reserve STEP III Q4?

I was gonna do that one but I'm pretty new to Latex and so I'd be terribly slow with all those integrals and stuff so go ahead haha.

I'd like to reserve II Q2 (i.e. I'm doing it now) .

EDIT: Meant to say Q2 sorry.
(edited 7 years ago)
STEP III Q4

(i).
Laplace transform of $\displaystyle e^{-bt}f(t)$ is
Unparseable latex formula:

\displaystyle \int_0^\infty e^{-st}\left(e^{-bt}f(t)\right) \,\mathrm{d}t = \int_0^\infty e^{-(s+b)t}f(t)\right) \,\mathrm{d}t = F(s+b)

.

(ii).
Laplace transform of $\displaystyle f(at)$ is
$\displaystyle \int_0^\infty e^{-st}f(at) \,\mathrm{d}t$, making the substitution $\displaystyle u = at, \mathrm{d}u = a \mathrm{d}t$,
$\displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}u}f(u) \,\mathrm{d}u = \displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}t}f(t) \,\mathrm{d}t = a^{-1}F\left(\frac{s}{a}\right)$.

(iii).
Laplace transform of $\displaystyle f'(t)$ is
$\displaystyle \int_0^\infty e^{-st}f'(t) \,\mathrm{d}t = \left[e^{-st}f(t)\right]_0^\infty + s\int_0^\infty e^{-st}f(t) \,\mathrm{d}t = -f(0)+sF(s)$, using integration by parts.

(iv).
Laplace transform of $\displaystyle \sin(t)$ is
$\displaystyle \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = \left[-e^{-st}\cos(t)\right]_0^\infty -s \int_0^\infty e^{-st}\cos(t) \,\mathrm{d}t =$
$\displaystyle =1 - s \left[e^{-st}\sin(t)\right]_0^\infty -s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = 1- s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t$.
So, $\displaystyle \mathrm{I} = 1-s^2 \mathrm{I}$, so $\displaystyle \mathrm{I}=\frac{1}{1+s^2}$.

Now, $\displaystyle \cos(t)$ is the derivative of $\displaystyle \sin(t)$, so by part (iii), Laplace transform of $\displaystyle \cos(t)$ is $\dfrac{s}{1+s^2}$.
And by part (ii), Laplace transform of $\displaystyle \cos(qt)$ is $\displaystyle q^{-1} \left(\frac{s/q}{1+(s/q)^2}\right) = \frac{s}{s^2+q^2}$.
Thus, by part (i), Laplace transform of $\displaystyle e^{-pt}\cos(qt)$ is $\displaystyle \frac{s+p}{q^2+(s+p)^2}$.

What a mess, feel free to edit and repost.
(edited 7 years ago)
Original post by StrangeBanana
Doing STEP III, Q5

Done a while ago!!
Original post by IrrationalRoot
Done a while ago!!

hmm

I'm tired

EDIT:
I'll come back tomorrow and see if any are still available
(edited 7 years ago)
STEP II, Question 2

(i)

(ii)

(iii)

(iv)

(edited 7 years ago)
Original post by Zacken
STEP II, Question 4

So it must be the case that$p'(x) = k(x-1)^4(x+1)^4 = k(x^2-1)^4$

Just a small typo
(edited 7 years ago)
Original post by IrrationalRoot
...

Would have been simpler to note that $\sin (\pi e^x)$ is a bounded function, so that $a^{\sin (\pi e^x)}$ is maximum when $\sin (\pi e^x)$ is 1, giving rise to $a^1$ and $\sin (\pi e^x) = -1$ giving rise to $a^{-1}$.
Original post by EricPiphany
Just a small typo

I'm being incredibly thick, but I can't see it?
Original post by Zacken
I'm being incredibly thick, but I can't see it?

coz I've tried to correct it
Original post by EricPiphany
coz I've tried to correct it

Ahhh, yeah lol, dumb mistake that one. Thanks!
STEP II, Question 3:

$\displaystyle \tan \left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} = \frac{1 - 2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} = \frac{1 - \sin x}{\cos x}$.

Which simplifies down to $\sec x - \tan x$.

(i)

(ii)

(iii)

(edited 7 years ago)
STEP II Q8

Solution

(edited 7 years ago)