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STEP Maths II, III 2009 Solutions

(Updated as far as #57)

STEP II:
1: Solution by Zacken
2: Solution by IrrationalRoot
3: Soluions by Zacken
4: Solution by Zacken
5: Solution by Zacken
6: Solution by Zacken
7: Solution by B_9710
8: Solution by Farhan.Hanif93
9: Solution by A-cute-Angle
10: Solution by Zacken
11: Solution by The-Spartan
12: Solution by Farhan.Hanif3
13: Solution by Farhan.Hanif3

STEP III:
1: Solution by IrrationalRoot
2: Solution by Farhan.Hanif93
3: Solution by Zacken
4: Solution by EricPiphany
5: Solution by IrrationalRoot
6: Solution by Farhan.Hanif93
7: Solution by Farhan.Hanif93
8: Solution by IrrationalRoot
9: Solution by Farhan.Hanif93
10: Solution by Farhan.Hanif93
11: Solution by Farhan.Hanif93
12: Solution by Zacken
13: Solution by Farhan.Hanif3

Solutions written by TSR members:
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(edited 7 years ago)

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Reply 1
Avatar for Zacken
Zacken
OP
22
STEP II, Question 1

First part



Second part



Third part

(edited 7 years ago)
Reply 2
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Zacken
OP
22
STEP II, Question 4

First part



Second part



Third part

(edited 7 years ago)
Reply 3
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Zacken
OP
22
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)
Reply 5
Avatar for Krollo
Krollo
17
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:
1: Solution by IrrationalRoot
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007 - ...


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

Posted from TSR Mobile
STEP III, Question 5

yz+zx+xy=(x+y+z)2−(x2+y2+z2)2=(1)2−22=−12yz+zx+xy=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\dfrac{(1)^2-2}{2}=-\dfrac{1}{2}
as required.

x2y+x2z+y2z+y2x+z2x+z2y=x2(y+z)+y2(z+x)+z2(x+y)=x2(1−x)+y2(1−y)+z2(1−z)=x2+y2+z2−(x3+y3+z3)=2−3=−1x^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.

−12=−12×1=(yz+zx+xy)(x+y+z)=xyz+y2z+yz2+x2z+xyz+xz2+x2y+xy2+xyz=3xyz+x2y+x2z+y2z+y2x+z2x+z2y=3xyz−1[br][br]⇒xyz=1−123=16-\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

Sn+1=aSn+bSn−1+cSn−2[br][br]⇔xn+1+yn+1+zn+1=axn+ayn+azn+bxn−1+byn−1+bzn−1+cxn−2+cyn−2+czn−2[br][br]⇔xn−2(x3−ax2−bx−c)+yn−2(y3−ay2−by−c)+zn−2(z3−az2−bz−c)=0S_{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 aa, bb and cc such that the cubic m3−am2−bm−c=0m^3-am^2-bm-c=0 for m=xm=x,yy,zz.

Therefore we can take a=x+y+z=1a=x+y+z=1, b=−(yz+zx+xy)=12b=-(yz+zx+xy)=\dfrac{1}{2} and c=xyz=16c=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) :smile:.

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

(i).
Laplace transform of e−btf(t)\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 f(at)\displaystyle f(at) is
∫0∞e−stf(at) dt \displaystyle \int_0^\infty e^{-st}f(at) \,\mathrm{d}t , making the substitution u=at,du=adt \displaystyle u = at, \mathrm{d}u = a \mathrm{d}t,
a−1∫0∞e−sauf(u) du=a−1∫0∞e−satf(t) dt=a−1F(sa) \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 f′(t) \displaystyle f'(t) is
∫0∞e−stf′(t) dt=[e−stf(t)]0∞+s∫0∞e−stf(t) dt=−f(0)+sF(s) \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 sin⁡(t) \displaystyle \sin(t) is
∫0∞e−stsin⁡(t) dt=[−e−stcos⁡(t)]0∞−s∫0∞e−stcos⁡(t) dt= \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 =
=1−s[e−stsin⁡(t)]0∞−s2∫0∞e−stsin⁡(t) dt=1−s2∫0∞e−stsin⁡(t) dt\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, I=1−s2I\displaystyle \mathrm{I} = 1-s^2 \mathrm{I}, so I=11+s2\displaystyle \mathrm{I}=\frac{1}{1+s^2}.

Now, cos⁡(t)\displaystyle \cos(t) is the derivative of sin⁡(t)\displaystyle \sin(t), so by part (iii), Laplace transform of cos⁡(t) \displaystyle \cos(t) is s1+s2\dfrac{s}{1+s^2}.
And by part (ii), Laplace transform of cos⁡(qt)\displaystyle \cos(qt) is q−1(s/q1+(s/q)2)=ss2+q2\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 e−ptcos⁡(qt)\displaystyle e^{-pt}\cos(qt) is s+pq2+(s+p)2\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 thatp′(x)=k(x−1)4(x+1)4=k(x2−1)4p'(x) = k(x-1)^4(x+1)^4 = k(x^2-1)^4



Just a small typo
(edited 7 years ago)
Reply 14
Avatar for Zacken
Zacken
OP
22
Original post by IrrationalRoot
...


Would have been simpler to note that sin⁡(πex)\sin (\pi e^x) is a bounded function, so that asin⁡(πex)a^{\sin (\pi e^x)} is maximum when sin⁡(πex)\sin (\pi e^x) is 1, giving rise to a1a^1 and sin⁡(πex)=−1\sin (\pi e^x) = -1 giving rise to a−1a^{-1}.
Reply 15
Avatar for Zacken
Zacken
OP
22
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
Reply 17
Avatar for Zacken
Zacken
OP
22
Original post by EricPiphany
coz I've tried to correct it


Ahhh, yeah lol, dumb mistake that one. Thanks!
Reply 18
Avatar for Zacken
Zacken
OP
22
STEP II, Question 3:

tan⁡(π4−x2)=1−tan⁡x21+tan⁡x2=cos⁡x2−sin⁡x2sin⁡x2+cos⁡x2=1−2sin⁡x2cos⁡x2cos⁡2x2−sin⁡2x2=1−sin⁡xcos⁡x\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\sec x - \tan x.

(i)



(ii)



(iii)

(edited 7 years ago)
STEP II Q8

Solution

(edited 7 years ago)

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