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2006 STEP II question 9 \text {2006 STEP II question 9}

The first diagram shows the forces on the ladder when the man is standing a distance x up from the foot\text {The first diagram shows the forces on the ladder when the man is standing a distance }x \text { up from the foot}
Resolving horizontally and vertically F=X and R=(1+k)W \text {Resolving horizontally and vertically } F=X \text { and } R=(1+k)W
Taking moments about foot of ladder 3aWcosθ+xakWcosθ=6aXsinθ \text {Taking moments about foot of ladder } 3aW \cos \theta +xak W \cos \theta =6aX \sin \theta
 so (3+xk)W=6Xtanθ=12X    X=3+xk12(1+k) \text { so } (3+xk)W=6X \tan \theta=12X \implies X= \dfrac{3+xk}{12(1+k)}
ladder does not slip if F<12R=(1+k)W2 \text {ladder does not slip if }F< \dfrac{1}{2}R = \dfrac{(1+k)W}{2}
maximum value of F is when x=6 and F=3+6k12W \text {maximum value of }F \text { is when }x=6 \text { and }F= \dfrac{3+6k}{12}W
i.e. Max F=1+2k4W which is less than (1+k)W2 \text {i.e. Max }F=\dfrac{1+2k}{4}W \text { which is less than }\dfrac{(1+k)W}{2}
so ladder will not slip \text {so ladder will not slip}
(ii) consider now the forces on the table (see second diagram) when it is free to move \text {(ii) consider now the forces on the table (see second diagram) when it is free to move}
Forces on ladder will be as before with k=9 so R=10W and F=3+9x12W \text {Forces on ladder will be as before with }k=9 \text { so }R=10W \text { and }F= \dfrac{3+9x}{12}W
Resolving forces P=F and N=2W+R=12W \text {Resolving forces }P=F \text { and }N=2W+R=12W
if table tilts it will do so about point A so taking moments about A we have  \text {if table tilts it will do so about point A so taking moments about A we have }
12aW4=aF    x=339=113 \dfrac{12aW}{4}=aF \implies x= \dfrac{33}{9}= \dfrac{11}{3}
table slips before man reaches top of ladder if P>13N i.e. If 3+9x12W>4W \text {table slips before man reaches top of ladder if }P> \dfrac{1}{3}N \text { i.e. If } \dfrac{3+9x}{12}W>4W
hence, if 3+9x>48 or x>5 \text {hence, if }3+9x>48 \text { or } x>5
table tilts if x=113 and slips if z>5 hence, tilting occurs first\text {table tilts if }x=\dfrac{11}{3} \text { and slips if }z>5 \text { hence, tilting occurs first}
2006 STEP II question 12 \text {2006 STEP II question 12}

(i) Number of wickets taken by Arthur will have a B(30,136) distribution \text {(i) Number of wickets taken by Arthur will have a B} \left( 30, \dfrac{1}{36} \right) \text { distribution}
by Betty a B(30,125) and by Cuba a B(30,141) \text {by Betty a B} \left(30, \dfrac{1}{25} \right) \text { and by Cuba a B} \left(30, \dfrac{1}{41} \right)
Unparseable latex formula:

\text {Pr(1 wicket }= \text {P}(A1 \cap B0 \cap C0)+ \terxt{P}( A0 \cap B1 \cap C))+ \text {P}(A0 \cap B0 \cap C1)


=3036×(3536)29×(2425)30×(4041)30+3025×(2425)29×(3536)30×(4041)30 = \dfrac{30}{36} \times \left( \dfrac{35}{36} \right)^{29} \times \left( \dfrac{24}{25} \right)^{30} \times \left( \dfrac{40}{41} \right)^{30}+ \dfrac {30}{25} \times \left( \dfrac{24}{25} \right)^{29} \times \left( \dfrac{35}{36} \right)^{30} \times \left( \dfrac{40}{41} \right)^{30}
+3041×(4041)29×(2425)30×(3536)30+ \dfrac{30}{41} \times \left( \dfrac{40}{41} \right)^{29} \times \left( \dfrac{24}{25} \right)^{30} \times \left( \dfrac{35}{36} \right)^{30}
=(35×24×40)30(36×25×41)30(30×35+25×24+41×40) = \dfrac{(35 \times 24 \times 40)^{30}}{(36 \times 25 \times 41)^{30} (30 \times 35 +25 \times 24+ 41 \times 40) }
Pr(Arthur took the one wicket ) is then \text {Pr(Arthur took the one wicket ) is then }
P(A1B0C0)P(A1B0C0)+P(A0B1C0)+P(A0B0C1)) \dfrac { \text{P}(A1 \cap B0 \cap C0)}{ \text {P}(A1 \cap B0 \cap C0)+ \text {P} (A0 \cap B1 \cap C0)+ \text {P} ( A0 \cap B0 \cap C1))}
=3036×(3536)29×(2425)30×(4041)303036×(3536)29×(2425)30×(4041)30+3025×(2425)29×(3536)30×(4041)30+3041×(4041)29×(2425)30×(3536)30 = \dfrac{\frac {30}{36} \times \left( \frac{35}{36} \right)^{29} \times \left( \frac{24}{25} \right)^{30} \times \left( \frac {40}{41}\right)^{30}}{\frac{30}{36} \times \left( \frac{35}{36} \right)^{29} \times \left( \frac {24}{25} \right)^{30} \times \left( \frac {40}{41} \right)^{30}+\frac{30}{25} \times \left( \frac{24}{25} \right)^{29} \times \left( \frac {35}{36} \right)^{30} \times \left( \frac {40}{41} \right)^{30}+ \frac{30}{41} \times \left( \frac{40}{41} \right)^{29} \times \left( \frac {24}{25} \right)^{30} \times \left( \frac {35}{36} \right)^{30}}
=30353035+3024+3040+135135+124+140=11+3524+78=11+3524+78=124+35+2124=2480=310 = \dfrac{ \frac{30}{35}}{ \frac{30}{35}+ \frac{30}{24}+ \frac {30}{40}} + \dfrac{ \frac {1}{35}}{ \frac {1}{35}+ \frac {1}{24} + \frac {1}{40}}= \dfrac {1}{1+ \frac {35}{24}+ \frac {7}{8}}= \dfrac{1}{ 1+ \frac{35}{24} + \frac {7}{8}}= \dfrac{1}{ \frac {24+35+21}{24}}= \dfrac{24}{80}= \dfrac{3}{10}
(ii) Average number of wickets taken is 30×(136+125+141)=56+65+3041=6130+3041 \text {(ii) Average number of wickets taken is }30 \times \left( \dfrac {1}{36}+ \dfrac{1}{25}+ \dfrac{1}{41} \right)= \dfrac {5}{6}+ \dfrac{6}{5}+ \dfrac{30}{41}= \dfrac{61}{30}+ \dfrac{30}{41}
=61×41+30230×41=34011230=3 to nearet whole number = \dfrac{61 \times 41+30^2}{30 \times 41}= \dfrac{3401}{1230} = 3 \text { to nearet whole number}
(iii) Assuming the distribution B(90,390) for the number of wickets taken which, since  \text {(iii) Assuming the distribution B} \left(90, \dfrac {3}{90} \right) \text { for the number of wickets taken which, since }
n is large and p is small may be approximated by the Poisson distribution with mean 3 n \text { is large and }p \text { is small may be approximated by the Poisson distribution with mean 3}
 P(at least 5 wickets is given by \text { P(at least 5 wickets is given by}
1[p0+p1+p2+p3+p4]=1e3(1+3+92+276+8124) 1-[p_0+p_1+p_2+p_3+p_4]=1- \text{e}^{-3} \left( 1+3+ \dfrac {9}{2} + \dfrac {27}{6} + \dfrac {81}{24} \right)
=1120(1+3+4.5+4.5+3.375)=116.375P20=1131160=2916015=1- \dfrac{1}{20}(1+3+4.5+4.5+3.375)=1-\dfrac{16.375}{P20}=1- \dfrac{131}{160}= \dfrac{29}{160} \approx \dfrac{1}{5}
alkternatively use tables of cumulative probability \text {alkternatively use tables of cumulative probability}
2006 STEP II question 13 \text {2006 STEP II question 13}

There are 24 possible sequences of 4 ice creams as follows:-\text {There are 24 possible sequences of 4 ice creams as follows:-}
Unparseable latex formula:

\begin{array}{c,c,c,c} 1234 & 2134 & 3124 & 4123 \\ 1243 & 2143 & 3142 & 4132 \\ 1324 & 2314 & 3214 & 4213 \\ 1342 & 2314 & 3241 & 4231 \\ 1423 & 2413 & 3412 & 4312 \\ 1432 & 2431 & 3421 & 4321 \end{array}



Of these the 1 will be chosen if the first one is 2,3 or 4 and the 1 is the first number smaller than this \text {Of these the 1 will be chosen if the first one is 2,3 or 4 and the 1 is the first number smaller than this}
from then above list we see that this occurs 11 times so P4(1)=1124 \text {from then above list we see that this occurs 11 times so P}_4(1)= \dfrac{11}{24}
number 2 is chosen if the first number is 3 and the 2 comes before 1 or if the first number is 4 and \text {number 2 is chosen if the first number is 3 and the 2 comes before 1 or if the first number is 4 and}
the 2 comes before either 1 or 3 or if the fiurst number is 1 and the last number is 2 \text {the 2 comes before either 1 or 3 or if the fiurst number is 1 and the last number is 2}
again from the above list this occurs 7 times so P4(2)=724 \text {again from the above list this occurs 7 times so P}_4(2)= \dfrac {7}{24}
3 is chosen if the first number is 4 and the second one is 3 or if the sequence is 1243 \text {3 is chosen if the first number is 4 and the second one is 3 or if the sequence is 1243}
so P4(3)=424=16 \text {so P}_4(3)= \dfrac {4}{24}= \dfrac {1}{6}
whilst 4 is chosen only if the sequence is 1234 or 1324 so P4(4)=224=112 \text {whilst 4 is chosen only if the sequence is 1234 or 1324 so P}_4(4)= \dfrac{2}{24} = \dfrac {1}{12}

With n cones to choose from, if the first is 1 then number 1 cannot be chosen so P1(1)=0 \text {With }n \text { cones to choose from, if the first is 1 then number 1 cannot be chosen so P}_1(1)=0
If I am offered the kth biggest first then P(I choose the biggest) is 1n1 \text {If I am offered the }k^{th} \text { biggest first then P(I choose the biggest) is }\dfrac {1}{n-1}
hence, Pn(1)=1nr=1n11r \text {hence, P}_n(1)= \dfrac{1}{n} \displaystyle \sum _{r=1}^{n-1} \dfrac {1}{r}
2006 STEP III question 9 \text {2006 STEP III question 9}

System will be in stable equilibrium for all positions of the bead if there is no change in P.E. when \text {System will be in stable equilibrium for all positions of the bead if there is no change in P.E. when}
moved from one position to another \text {moved from one position to another}
Taking horizontal plane through O as zero level for P.E. we have  \text {Taking horizontal plane through O as zero level for P.E. we have }
P.E. of system when bead is at H is mghmg(xh) where x is the length of the string \text {P.E. of system when bead is at H is }mgh-mg(x-h) \text { where }x \text { is the length of the string}
P.E. when bead is at B is mgymg(xr) so change in P.E. is  \text {P.E. when bead is at B is }mgy-mg(x-r) \text { so change in P.E. is }
2mghmgxmgy+mg(xr)=mg(2hyr) 2mgh-mgx-mgy+mg(x-r)=mg(2h-y-r)
So there is no change in P.E. if 2hyr=0 i.e. y+r=2h as required \text {So there is no change in P.E. if }2h-y-r=0 \text { i.e. }y+r=2h \text { as required}
If speed at B is v then v=rθ˙ and speed of P will be r˙ \text {If speed at B is }v \text { then }v=r \dot \theta \text { and speed of P will be } \dot r
θ=sin1(vr)=sin1(2hrr)    θ˙rr˙(2hr)r˙r2×11(2hrr)2=2hr˙rr2(2hr)2 \theta= \sin^{-1} \left( \dfrac{v}{r} \right)= \sin^{-1} \left( \dfrac {2h-r}{r}\right) \implies \dot \theta \dfrac {-r \dot r-(2h-r) \dot r}{r^2} \times \dfrac {1}{ \sqrt{1- \left( \frac{2h-r}{r} \right)^2}}= \dfrac {-2h \dot r}{r \sqrt {r^2-(2h-r)^2}}
i.e. θ˙=2hr˙r4h2+4hr=hr˙rhrh2 as required \text {i.e. } \dot \theta= \dfrac {-2h \dot r}{r \sqrt{-4h^2+4hr}}= -\dfrac {h \dot r}{r \sqrt {hr-h^2}} \text { as required}
P.E. constant      K.E. constant, so v2+(rθ˙)2+v2=V2 where v=r˙ is velocity of P \text {P.E. constant } \implies \text { K.E. constant, so } v^2+ (r \dot \theta)^2+v^2=V^2 \text { where }v= \dot r \text { is velocity of P}
Unparseable latex formula:

\text {i.e. } 2v^2+ \dfrac {r^2h^2v^2}{r^2(hr-h^2)}=V^2 \implies 2v^2 \left(1+ \dfrac{h}{2(r-h)} \right)=V^2 \impolies v^2 \left( \dfrac {2r-h}{r-h} \right)=V^2 n\implies v= V \sqrt {\dfrac{r-h}{2r-h}} \text { as required}

2006 STEP III question 10 \text {2006 STEP III question 10}

By conservation of angular momentum mk2Ω+ma2Ω=mk2ω+mr2ω\text {By conservation of angular momentum } mk^2 \Omega +ma^2 \Omega = mk^2 \omega + mr^2 \omega
hence, ω=Ω(k2+a2)k2+r2 as required\text {hence, } \omega= \dfrac { \Omega(k^2+a^2)}{k^2+r^2} \text { as required}
K.E. in initial position is 12mk2Ω2+12ma2Ω2+12mV2 \text {K.E. in initial position is } \dfrac {1}{2}mk^2 \Omega^2 + \dfrac {1}{2} ma^2 \Omega^2 + \dfrac {1}{2}mV^2
and after time t, K.E. is 12mk2ω2+12mr2ω2+12m(drdt)2 \text {and after time }t, \text { K.E. is } \dfrac {1}{2}mk^2 \omega^2+ \dfrac{1}{2}mr^2 \omega^2 + \dfrac {1}{2} m \left( \dfrac { \text{d}r}{ \text{d}t} \right)^2
so, by conservation of energy, 12mk2ω2+12mr2ω2+12m(drdt)2=12mk2Ω2+12ma2Ω2+12mV2 \text {so, by conservation of energy, } \dfrac {1}{2}mk^2 \omega^2+ \dfrac{1}{2}mr^2 \omega^2 + \dfrac {1}{2} m \left( \dfrac { \text{d}r}{ \text{d}t} \right)^2= \dfrac {1}{2}mk^2 \Omega^2 + \dfrac {1}{2} ma^2 \Omega^2 + \dfrac {1}{2}mV^2
[latex]\implies \left( \dfrac { \text{d}r}{ \text{d}t} \right)^2= \Omega^2(k^2+a^2)+ \dfrac { \Omega^2 a^2(k^2+a^2)}{k^2}- \omega^2(k^2+r^2)
Unparseable latex formula:

= \dfra { \Omega^2k^2(k^2+a^2)(k^2+r^2)+ \Omega^2a^2(k^2+a^2)(k^2+r^2)- \Omega^2k^2(k^2+a^2)^2}{k^2(k^2+r^2)}


=Ω2(k2+a2)(k4+k2r2+a2k2+a2r2k4a2k2)k2(k2+r2)=Ω2k2r2(k2+a2)2k2(k2+r2) as required = \dfrac {\Omega^2(k^2+a^2)(k^4+k^2r^2+a^2k^2+a^2r^2-k^4-a^2k^2)}{k^2(k^2+r^2)}= \dfrac { \Omega^2k^2r^2(k^2+a^2)^2}{k^2(k^2+r^2)} \text { as required}
drdt<0 since particle is moving towards the axis and drdθ=drdt×dtdθ=1Ωdrdt \dfrac {\text{d}r}{ \text{d}t}<0 \text { since particle is moving towards the axis and } \dfrac {\text{d}r}{ \text{d} \theta}= \dfrac {\text{d}r}{\text{d}t} \times \dfrac {\text{d}t}{\text{d} \theta}= \dfrac {1}{\Omega} \dfrac {\text{d}r}{\text{d}t}
so drdθ=Ωr(k2+a2)kk2+ar2×k2+r2Ω(k2+a2)=rk2+r2k    kdrdθ=rk2+r2 \text {so } \dfrac {\text{d}r}{\text{d} \theta}= \dfrac {\Omega r(k^2+a^2)}{k \sqrt {k^2+ar^2}} \times \dfrac {k^2+r^2}{ \Omega(k^2+a^2)}= -\dfrac {r \sqrt {k^2+r^2}}{k} \implies k \dfrac {\text{d}r}{\text{d} \theta}=-r \sqrt{k^2+r^2}
Unparseable latex formula:

\text {putting }u= \dfrac{k}{r} \text { we have } r= \dfrac {k}{u} \implies \dfrac {\text{d}r}{\text{d} \theta}=- \dfrac{k}{u^2} \dfrac {\text{d}u}{\text{d} \theta} \text { so } \dfrac {k^2}{u^2} \dfrac {\text{d}u}{ \text{d} \theta}= \dfrac {k}{u} \sqrtr k^2+ \frac{k^2}{u^2}}= \dfrac {k^2}{u^2} \sqrt{1+u^2}


i.e. dudθ=1+u2    du1+u2=dθ    sinh1u=θ+α or u=sinh(θ+α) \text {i.e. } \dfrac {\text{d}u}{\text{d} \theta}= \sqrt {1+u^2} \implies \int \dfrac {\text{d}u}{ \sqrt{1+u^2}}= \int \text{d} \theta \implies \sinh^{-1} u= \theta+\alpha \text { or } u= \sinh( \theta+\alpha)
hence, r=ksinh(θ+α) or rsinh(θ+α)=k and r=a when θ=0    sinhα=ka \text {hence, }r= \dfrac {k}{\sinh (\theta+\alpha)} \text { or } r \sinh(\theta+\alpha)=k \text { and }r=a \text { when } \theta=0 \implies \sinh \alpha=\dfrac{k}{a}
since we are given that k>0 it follows that r can never be zero, i.e. Particle does not reach the axis\text {since we are given that }k>0 \text { it follows that }r \text { can never be zero, i.e. Particle does not reach the axis}
(edited 12 years ago)
2006 STEP III question 11 \text {2006 STEP III question 11}

After tile falls, by Newton’s second law, if tension in cable is T then for counterweight \text {After tile falls, by Newton's second law, if tension in cable is }T \text { then for counterweight}
MgT=Ma and for lift minus tile T(Mm)g=(Mm)a where a is the acceleration Mg-T=Ma \text { and for lift minus tile }T-(M-m)g=(M-m)a \text { where } a \text { is the acceleration}
so, acceleration of lift is given by mg=(2Mm)a    a=m2Mmg and acceleration of tile is g \text {so, acceleration of lift is given by }mg=(2M-m)a \implies a= \dfrac{m}{2M-m}g \text { and acceleration of tile is }g
if lift rises a distance d while tile falls then tile falls a distance hd \text {if lift rises a distance }d \text { while tile falls then tile falls a distance }h-d
so if velocities of tile and lift just before impact are u(downwards) and v(upwards) respectively then \text {so if velocities of tile and lift just before impact are }u \text {(downwards) and }v \text {(upwards) respectively then}
u2=2g(hd) and v2=2m2Mmgd    u2=2gh2Mmmv2 or u2+2Mmmv2=2gh u^2=2g(h-d) \text { and }v^2= \dfrac{2m}{2M-m}gd \implies u^2=2gh- \dfrac{2M-m}{m} v^2 \text { or } u^2+ \dfrac {2M-m}{m}v^2=2gh
now, momentum is constant =0 throughout so (2Mm)v=mu    m2Mm=θu say \text {now, momentum is constant {=0} throughout so }(2M-m)v=mu \implies \dfrac {m}{2M-m}= \theta u \text { say}
now let velocities after impact be w and x(both upwards) respectively then\text {now let velocities after impact be } w \text { and } x \text {(both upwards) respectively then}
Unparseable latex formula:

(2M-m)x+mw=0 \implies x=- \dfrac {m}{2M-m}w \text { or } x=- \tgheta w


by law of restitution at impact wx=e(u+v)    w(1+θ)=eu(1+θ) so w=eu\text {by law of restitution at impact }w-x=e(u+v) \implies w(1+ \theta)=eu(1+\theta) \text { so } w=eu
change in K.E.=12(2Mm)(v2x2)+12m(u2w2)\text {change in K.E.}= \dfrac {1}{2}(2M-m)(v^2-x^2)+ \dfrac {1}{2}m(u^2-w^2)
=12(2Mm)θ2(u2w2)+12m(u2w2)=12(u2w2)[(2Mm)θ2+m] = \dfrac {1}{2}(2M-m) \theta^2(u^2-w^2)+ \dfrac {1}{2}m(u^2-w^2)= \dfrac {1}{2}(u^2-w^2)[(2M-m) \theta^2+m]
Unparseable latex formula:

= \dfrac {1}{2} (u^2-w^2) \lefty[\dfrac{m^2}{2M-m}+m \right] \text { since } \theta^2= \left( \dfrac {m}{2M-m} \right)^2


=12u2(1e2)2Mm2Mm since w=eu = \dfrac{1}{2}u^2(1-e^2) \dfrac {2Mm}{2M-m} \text { since } w=eu
=12(1e2)2Mm2Mm×2MmMgh since u=2MmMgh from first part of question = \dfrac {1}{2}(1-e^2) \dfrac {2Mm}{2M-m} \times \dfrac {2M-m}{M} gh \text { since }u= \sqrt {\dfrac{2M-m}{M}gh} \text { from first part of question}
=(1e2)mgh as required = (1-e^2)mgh \text { as required}
2006 STEP III question 12 \text {2006 STEP III question 12}

If X is the number of potential passengers then XB(1024,0.5)N(512,162) \text {If }X \text { is the number of potential passengers then }X \sim \text {B}(1024,0.5) \approx \text {N}(512,16^2)
Expected lost profit (X>480) without another bus is k=132kP(X=480+k)+32P(X>512) \text {Expected lost profit }(X>480) \text { without another bus is } \displaystyle \sum_{k=1}^{32} k\text{P}(X=480+k)+32 \text{P}(X>512)
Unparseable latex formula:

\text { i.e. } \displaystyle \sum_{k=1}^{32}k \dfrac {1}{16} \phi \left(-2+\dfrac{k}{16} \right)+16 \approx \displaystyle \int_1^{32} \displaystyle \sum_{k=1}^{32} k \dfrac {1}{16} \phi \left(-2+ \dfrtac {k}{16} \tight) \text {d}x+16


Unparseable latex formula:

= \displaystyle \int _0^{32}\dfrac {x}{16} \dfrac {1}{ \sqrt{2 \pi}} \phi \left(-2+ \dfrtac {k}{16} \tight) \text {d}x+16= \displaystyle \int _0^{32}\dfrac {1}{16} \dfrac {x}{ \sqrt{2 \pi}}\text {exp} \left(- \dfrac{1}{2} t^2 \right) \text{d}t+16 \text { where }z=-2+ \dfrac {x}{16}


=032116x2πexp((x32)2512)dx+16 =\displaystyle \int _0^{32}\dfrac {1}{16} \dfrac {x}{ \sqrt{2 \pi}}\text {exp} \left(- \dfrac{(x-32)^2}{512} \right) \text{d}x+16
Unparseable latex formula:

\text {now let }t=\dfrac {x-32}{16} \implies \text {d}t= \dfarc {1}{16} \text {d}x, x=0 \implies t=-2, x=32 \implies t=0 \text { so integral becomes}


2016t+32162πexp(12t2)dt+16 \displaystyle \int_{-2}^0 \dfrac {16t+32}{16 \sqrt{2 \pi}} \text{exp} \left( -\dfrac{1}{2} t^2 \right) \text {d}t+16
=1620t2πexp(12t2)dt+322012πexp(12t2)dt+16=16 \displaystyle \int_{-2}^0 \dfrac {t}{ \sqrt{2 \pi}} \text{exp} \left( -\dfrac{1}{2} t^2 \right) \text {d}t+32\displaystyle \int_{-2}^0 \dfrac {1}{\sqrt{2 \pi}} \text{exp} \left( -\dfrac{1}{2} t^2 \right) \text {d}t +16
=16[12πexp(12t2)]20+32[12Φ(2)]+16=162π(e21)+32Φ(2)16+16 =16 \left[- \dfrac {1}{ \sqrt{2 \pi}} \text{exp} \left(- \dfrac{1}{2} t^2 \right) \right]_{-2}^0 +32 \left[ \dfrac {1}{2} - \Phi(-2) \right]+16= \dfrac{16}{ \sqrt {2 \pi}} ( \text{e}^{-2}-1)+32 \Phi(2)-16+16
In the course of a year the expected loss is 50 times this i.e. 1600Φ(2)8002π(1e2) \text {In the course of a year the expected loss is 50 times this i.e. }1600 \Phi(2)- \dfrac {800}{\sqrt{2\pi}} (1- \text{e}^{-2})
which is thus the maximum tolerable licence fee foe an extra bus (as required) \text {which is thus the maximum tolerable licence fee foe an extra bus (as required)} r
2006 STEP III question 13 \text {2006 STEP III question 13}

There are two possibilities (i) both points lie on the perimeter of the semicircle (first diagram) \text {There are two possibilities (i) both points lie on the perimeter of the semicircle (first diagram)}
(ii) one point is on perimeter and one on diameter (second diagram) \text {(ii) one point is on perimeter and one on diameter (second diagram)}
(i) Let P1 lie within the arc (α,α+δα) and P2 in the arc (θ,θ+δθ) \text {(i) Let P}_1 \text { lie within the arc }(\alpha,\alpha+\delta \alpha) \text { and P}_2 \text { in the arc }(\theta, \theta + \delta \theta)
 Probability that P2 is in this arc is δθπ+2 and area of triangle OP1P2 where O is centre of semicircle \text { Probability that P}_2 \text { is in this arc is } \dfrac{ \delta \theta}{\pi+2} \text { and area of triangle OP}_1 \text {P}_2 \text { where O is centre of semicircle}
Unparseable latex formula:

\text {is } \dfrac{1}{2} \sin ( \alpha- \theta) \ytext { if } \theta< \alpha \text { or } \dfrac {1}{2} \sin ( \theta- \alpha) \text { if } \theta> \alpha


Unparseable latex formula:

\text {so E[Area] when P}_1 \text { is in }( \alpha, \alphga + \delta \alpha) \text { is given by }


0αsin(αθ)2(2+π)dθ+απsin(θα)2(2+π)dθ \displaystyle \int_0^\alpha \dfrac{ \sin (\alpha-\theta)}{2(2+\pi)} \text {d} \theta+\displaystyle \int_ \alpha^\pi \dfrac{\sin(\theta-\alpha)}{2(2+\pi)} \text {d} \theta
=12(2+π)([cos(αθ)]0α+[cos(θα)]απ) =\dfrac{1}{2(2+\pi)}\left([\cos(\alpha-\theta)]_0^\alpha+[-\cos(\theta-\alpha)]_\alpha^\pi \right)
=12(2+π)(1cosα+cosα+1)=12+π =\dfrac{1}{2(2+\pi)}(1-\cos \alpha+\cos \alpha+1)= \dfrac {1}{2+\pi}
(ii) Let P1 lie within the range (r,r+δr) measured from centre of semicircle and\text {(ii) Let P}_1 \text { lie within the range }(r,r+ \delta r) \text { measured from centre of semicircle and}
P2 in arc (θ,θ+δθ) as before then P(P2 lies in this arc)=δθ2+π \text{P}_2 \text { in arc }( \theta, \theta+ \delta \theta) \text { as before then P(P}_2 \text { lies in this arc)}= \dfrac{\delta \theta}{2+\pi}
and area of triangle OP1P2 is 12rsinθ \text {and area of triangle OP}_1 \text {P}_2 \text { is } \dfrac {1}{2} |r| \sin \theta
so expected value of area is 0π12(2+π)rsinθdθ=[12(2+π)rcosθ]0π=r2+π \text {so expected value of area is } \displaystyle \int_0^\pi \dfrac{1}{2(2+\pi)}|r| \sin \theta \text {d} \theta= \left[ \dfrac{-1}{2(2+\pi)} |r| \cos \theta \right]_0^\pi= \dfrac{|r|}{2+\pi}
If P1 is on the arc and P2 on the diameter then obviously the result is the same\text {If P}_1 \text { is on the arc and P}_2 \text { on the diameter then obviously the result is the same}
Unparseable latex formula:

\text {hence, E[Area} is } \displaystyle \int \dfrac {1}{(2+\pi)^2} \text{d} \alpha +2 \displaystyle \int_{-1}^1 \dfrac {|r|}{(2+\pi)^2} \text {d}r= \left[ \dfrac{ alpha}{(2+\pi)^2} \right]_0^\pi+4 \left[ \dfrac {r^2}{2(2+\pi)^2} \right]_0^1= \dfrac {\pi+2}{(2+\pi)^2}= \dfrac {1}{2+\pi}

2006 STEP III question 14  \text {2006 STEP III question 14 }

E[X1+bX2]=aE[X1]+bE[X2] and E[X1X2]=E[X1]E[X2] for independent events\text{E}[X_1+bX_2]= a\text{E}[X_1]+b \text{E}[X_2] \text { and E}[X_1X_2]= \text{E}[X_1]\text{E}[X_2] \text { for independent events}
P=2(X1+X2) so E[P]=2E[X1]+2E[X2]=2μ1+2μ2 P=2(X_1+X_2) \text { so E}[P]=2 \text{E}[X_1]+2 \text{E}[X_2]=2 \mu_1+2 \mu_2
E[P2]=E[2(X1+X2))2=E[4X12]+E[8X1X2]+E[4X22]=4E[X12]+8E[X1]E[X2]+4E[X22] \text{E}[P^2]=\text{E}[2(X_1+X_2))^2=\text{E}[4X_1^2]+\text{E}[8X_1X_2]+\text{E}[4X_2^2]=4\text{E}[X_1^2]+8\text{E}[X_1] \text{E}[X_2] +4\text{E}[X_2^2]
E[X12]=μ12+σ12,E[X22]=μ22+σ22 so E[P2]=4(μ12+σ12)+8μ1μ2+4(μ22+σ22) \text{E}[X_1^2]= \mu_1^2+\sigma_1^2, \text{E}[X_2^2]= \mu_2^2+\sigma_2^2\text { so E}[P^2]=4(\mu_1^2+\sigma_1^2)+8\mu_1\mu_2+4(\mu_2^2+\sigma_2^2)
Var[P]=E[P2](E[P])2=4(μ12+σ12)+8μ1μ2+4(μ22+σ22)4(μ1+μ2)2 \text{Var}[P]=\text{E}[P^2]-(\text{E}[P])^2=4(\mu_1^2+\sigma_1^2)+8\mu_1\mu_2+4(\mu_2^2+\sigma_2^2)-4(\mu_1+\mu_2)^2
     standard deviation =2σ12+σ22 \implies \text{ standard deviation }=2 \sqrt{\sigma_1^2+\sigma_2^2}
    E[A2]=(μ12+σ12)(μ22+σ22)\implies \text{E}[A^2]=(\mu_1^2+\sigma_1^2)(\mu_2^2+ \sigma_2^2)
Var[A2]=E[A2](E[A])2=(μ12+σ12)(μ22+σ22)μ12μ22=σ12μ22+μ12σ22+σ12σ22 \text {Var}[A^2]=\text{E}[A^2]-(\text{E}[A])^2=(\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2)-\mu_1^2\mu_2^2=\sigma_1^2 \mu_2^2+\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2
E[PA]=E[(2X1+2X2)X1X2]=2E[X12]E[X2]+2E[X22]E[X1] \text{E}[PA]= \text{E}[(2X_1+2X_2)X_1X_2]=2\text{E}[X_1^2]\text{E}[X_2]+2\text{E}[X_2^2] \text{E}[X_1]
=2(μ12+σ12)μ2+2(μ22+σ22)μ1 =2(\mu_1^2+\sigma_1^2)\mu_2+2(\mu_2^2+\sigma_2^2)\mu_1
and E[P]E[A]=2(μ1+μ2)μ1μ2=2μ12μ2+2μ22μ1 \text {and E}[P] \text{E}[A]=2(\mu_1+\mu_2)\mu_1\mu_2=2\mu_1^2 \mu_2+2 \mu_2^2 \mu_1
Unparseable latex formula:

\text {hence E}[PA]- \text{E}[P]\text{E}[A]=2\sigmna_1^2\mu_2+2\sigma_2^2 \mu_1 \not=0 \text { since } \mu_1,\mu_2>0 \text { and } \sigma_1, \sigma_2 \not=0


 which implies that P and A are not independent \text { which implies that }P \text { and }A \text { are not independent}
if Z=PαA then E[ZA]=E[PAαA2]=E[PA]αE[A2] \text {if }Z=P- \alpha A \text { then E}[ZA]=\text{E}[PA- \alpha A^2]=\text{E}[PA]- \alpha \text{E}[A^2]
=2(μ12+σ12)μ2+2(μ22+σ22)μ1α(σ12μ22+μ12σ22+σ12σ22)=0 only if  =2(\mu_1^2+ \sigma_1^2) \mu_2+2(\mu_2^2+ \sigma_2^2) \mu_1- \alpha (\sigma_1^2 \mu_2^2+\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2)=0 \text { only if }
Unparseable latex formula:

\alpha= \dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2}


Unparseable latex formula:

\text { so if }\alpha \not=\dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2} \text { then }Z \text { and }A \text { are not independent}


Unparseable latex formula:

\text {For the final part we need only consider the exceptional case when } \alpha= \dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2}


Unparseable latex formula:

X1,X2=1 \text { or }3 \implies A=1,3 \text { or }9 \text { so P}(A=1)=\dfrac {1}{4}, \texct{P}(A=3)=\dfrac {1}{2} \text { and P}(A=9)= \dfrac {1}{4}


E[X1]=E[X2]=1×12+3×12=2,Var[X1]=Var[X2]=1×12+9×1222=1 \text {E}[X_1]=\text{E}[X_2]=1 \times \dfrac {1}{2}+3 \times \dfrac {1}{2}=2, \text{Var}[X_1]=\text{Var}[X_2]=1 \times \dfrac {1}{2}+9 \times \dfrac {1}{2}-2^2=1
so μ1=μ2=2 and σ1=σ2=1    α=4+434+4+14=89 \text {so }\mu_1=\mu_2=2 \text { and } \sigma_1= \sigma_2=1 \implies \alpha= \dfrac{4+4}{34+4+14}= \dfrac{8}{9}
A=1    Z=489=289,A=3    Z=883=163 and A=9    Z=128=4 A=1 \implies Z=4- \dfrac {8}{9}= \dfrac {28}{9}, A=3 \implies Z=8- \dfrac {8}{3}= \dfrac {16}{3} \text { and }A=9 \implies Z=12-8=4
Now, for example P(Z=289)=14 and P(Z=289A=3)=0 \text {Now, for example P} \left(Z= \dfrac {28}{9} \right)= \dfrac {1}{4} \text { and P} \left(Z=\dfrac {28}{9}|A=3 \right)=0
so Z and A are not independent for any value of α \text {so }Z \text { and }A \text { are not independent for any value of }\alpha

!Phew Thank goodness that's done. I wouldn't be at all surprised if there are some errors in it. Please let me know if you spot any.
(edited 12 years ago)
Original post by SimonM
...

STEP II Q11

Solution

Reply 70
Original post by SimonM
STEP II, Question 3

Spoiler



To show (N+root(N2-1)k differs from the nearest integer by less than (2N-0.5)-k you show that:

0<(N+root(N2))-1<(N+root(N2-1))-1<(N+root(N2-N+0.25))<1 as -1>-N+0.25 because N>1.25 as N is an integer greater than 1. So

0<(2N)-k<(n+root(N2-1))-k<(2N-0.5)-k<1, then you use the result similar to that in the first part:

(N+root(N2-1))k = M - (N+root(N2-1))-k>M - (2N-0.5)-k where M is a positive integer.
Original post by Elongar
STEP II 2006, Question 6

Spoiler



I have two questions on this solution ---

I. Can someone please explain why it's true that y+z=56 y + z = 56 ?

II. Also, why are we solving for x x here? The question only asks us to show the inequality? It doesn't say anything about solving?

Thank you ---
Original post by adrienne_om
I have two questions on this solution ---

I. Can someone please explain why it's true that y+z=56 y + z = 56 ?

II. Also, why are we solving for x x here? The question only asks us to show the inequality? It doesn't say anything about solving?

Thank you ---


It does! It says "use this result to solve the equation...."
Reply 73
Original post by Elongar
STEP III 2006, Q5

We wish to show that α,β,γ\alpha, \beta, \gamma form an equilateral iff

α2+β2+γ2αββγγα=0\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0

Multiplying by two and factorising, we wish to show that,

(αβ)2+(βγ)2+(γα)2=0\displaystyle (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 0

holds for all equilateral triangles.

Note that αβ,βγ,γα\alpha - \beta, \beta - \gamma, \gamma - \alpha are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is π3\frac{\pi}{3}. Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

Using this, we reconstruct the equation:

(reiθ)2+(rei(θ+π3))2+(rei(θ+2π3))2=0\displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = 0

or, more simply,

r2(e2iθ+e2i(θ+π3)+e2i(θ+2π3))=0\displaystyle r^2 \left( e^{2 i \theta} + e^{2 i (\theta +\frac{\pi}{3})} + e^{2 i (\theta +\frac{2 \pi}{3})} \right) = 0

r2e2iθ(1+eiπ3+ei2π3)=0\displaystyle r^2 e^{2 i \theta} \left( 1 + e^{i \frac{\pi}{3}} + e^{i \frac{2 \pi}{3}} \right) = 0

The bracket evaluates to 0, so we are done.

For the next part, we take the roots of the cubic to be α,β,γ\alpha, \beta, \gamma, and expand, yielding Vieta's formula:

z3(α+β+γ)z2+(αβ+βγ+γα)zαβγ=0\displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0

Algebra shows that:

α2+β2+γ2αββγγα=(α+β+γ)23(αβ+βγ+γα)=a23b=0\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = (\alpha + \beta + \gamma)^2 - 3 (\alpha \beta + \beta \gamma + \gamma \alpha) = a^2 - 3b = 0

and the result follows immediately.

Now we write,

p=Aeiϕp = Ae^{i \phi}

q=B+Ciq = B + Ci

The transformation z=pw+qz = p w + q has the effect of rotating our equilateral triangle by the angle ϕ\phi, magnifying it by a factor of AA and translating it through the vector represented by qq (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.


For the first part it is clear that if the triangle is equilateral, then α2+β2+γ2αββγγα=0\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0. However, I am not convinced that your argument works the other way around. You have shown that it is necessary but not that it is sufficient.

Also,
(reiθ)2+(rei(θ+π3))2+(rei(θ+2π3))2=r2e2iθ(1+ei2π3+ei4π3)=0\displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = r^2 e^{2 i \theta} \left( 1 + e^{i \frac{2 \pi}{3}} + e^{i \frac{4 \pi}{3}} \right) = 0
Here is an alternative solution for Paper III number 5

α,β,γ\alpha,\beta,\gamma represent the vertices of an equilateral triangle if and only if
(βα)=ω(γβ)=ω2(γβ)(\beta-\alpha)=\omega(\gamma-\beta)=\omega^2(\gamma-\beta) where ω=eiπ3\omega=e^{i\frac{\pi}{3}}
i.e[(βα)ω(γβ)][(βα)ω2(γβ)]=0 \mathrm{i.e}[(\beta-\alpha)-\omega(\gamma-\beta)][(\beta-\alpha)-\omega^2(\gamma-\beta)]=0
(βα)2(βα)(γβ)(ω+ω2)+ω3(γβ)2=0\Leftrightarrow(\beta-\alpha)^2-(\beta-\alpha)(\gamma-\beta)(\omega+\omega^2)+\omega^3(\gamma-\beta)^2=0
α2+β22αβ+γβαγβ2+αβ+γ22βγ+β2=0\Leftrightarrow\alpha^2+\beta^2-2\alpha\beta+\gamma\beta-\alpha\gamma-\beta^2+\alpha\beta+\gamma^2-2\beta\gamma+\beta^2=0
since ω+ω2=1\omega+\omega^2=-1 and ω3=1 \omega^3=1
hence, α2+β2+γ2αββγγα=0\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha=0 as required
If α,β,γ\alpha,\beta,\gamma are the roots of z3+az2+bz+c=0z^3+az^2+bz+c=0 then α+β+γ=a\alpha+\beta+\gamma=-a and αβ+βγ+γα=b\alpha\beta+\beta\gamma+\gamma \alpha=b
α2+β2+γ2αββγγα=a22b\Rightarrow\alpha^2+\beta^2+ \gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha=a^2-2b and so
α2+β2+γ2αββγγα=a23b\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha=a^2-3b hence the roots are the vertices of an equilateral triangle if and only if a2=3ba^2=3b
Under the transformation z=pw+qz=pw+q the equation becomes
(pw+q)3+a(pw+q)2+b(pw+q)+c=0(pw+q)^3+a(pw+q)^2+b(pw+q)+c=0
i.e. p3w3+(3p2q+ap2)w2+(3pq2+2apq+bp)w+(q3+aq2+bq+c)=0p^3w^3+(3p^2q+ap^2)w^2+(3pq^2+2apq+bp)w+(q^3+aq^2+bq+c)=0
comparing with w3+Aw2+Bw+C=0w^3+Aw^2+Bw+C=0
A=3q+ap,B=3q2+2aq+bp2 A= \frac {3q+a}{p} ,B=\frac{3q^2+2aq+b}{p^2} and A23B=(3q+a)23(3q2+2aq+b)p2A^2-3B=\frac{(3q+a)^2-3(3q^2+2aq+b)}{p^2}
=9q2+6aq+a29q26aq3bp2=a23bp2=0=\frac{9q^2+6aq+a^2-9q^2-6aq-3b}{p^2}=\frac{a^2-3b}{p^2}=0 since a2=3ba^2=3b and p0p\not=0
hence, roots also represent an equilateral triangle.
Original post by darkness9999
STEP I: Q5

Part (i)



Part (ii)



Hi could someone explain the partial fraction part in part (ii) to me please? I don't understand, shouldn't it be split up to just A/(u+1)^2 and B/(u-1)^2?Where are the C and the D terms coming from, is this a special case or something?
Reply 76
Original post by Generic Name
Hi could someone explain the partial fraction part in part (ii) to me please? I don't understand, shouldn't it be split up to just A/(u+1)^2 and B/(u-1)^2?Where are the C and the D terms coming from, is this a special case or something?


That is how you deal with repeated factors.
Original post by Brister
That is how you deal with repeated factors.


Yeah I just read up on it :sigh: Need to stop forgetting basic stuff -.-
Reply 78
Original post by Generic Name
Yeah I just read up on it :sigh: Need to stop forgetting basic stuff -.-


I have had a little think, and I believe I can justify it.

Suppose you want to decompose
f(x)(ax+b)2 \frac{f(x)}{(ax + b)^2}
where f(x) is a polynomial.

We can write
f(x)=C(ax+b)+Df(x) = C(ax + b) + D
where C is a function of x and D is a constant.

This gives the composition as
Cax+b+D(ax+b)2\frac{C}{ax + b} + \frac{D}{(ax + b)^2}
Original post by Brister
I have had a little think, and I believe I can justify it.

Suppose you want to decompose
f(x)(ax+b)2 \frac{f(x)}{(ax + b)^2}
where f(x) is a polynomial.

We can write
f(x)=C(ax+b)+Df(x) = C(ax + b) + D
where C is a function of x and D is a constant.

This gives the composition as
Cax+b+D(ax+b)2\frac{C}{ax + b} + \frac{D}{(ax + b)^2}


Makes sense :yep:

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