The Student Room Group

Scroll to see replies

Reply 1
STEP II, Question 1

First part



Second part



Third part

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

First part



Second part



Third part

(edited 7 years ago)
Reply 3
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
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
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
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
Original post by EricPiphany
coz I've tried to correct it


Ahhh, yeah lol, dumb mistake that one. Thanks!
Reply 18
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)

Quick Reply

Latest

Trending

Trending