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STEP 2005 Solutions Thread

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Good solution, but how do we know c^2>0? Surely c^2=0 is a possibility?
They told us c does not equal one, so c could equal zero. I believe I have a good argument why c cannot equal zero, but you might like to look for it yourself. (As your solutions show you're a cut above me at this STEP lark, so I'm sure you'll find it, but i'm fairly sure your solution is incomplete).
It doesn't, you need to exclude the possibility that c = 0 seperately
Reply 43
Avatar for miml
miml
17
It's actually quite a nice contradiction..
No problem, I still have one small qualm though. In the case c=0, if a or b=0, then surely there is only one solution, x=0?

Also out of curiosity, how many marks would you (or someone else) estimate your original solution would have got out of 20? I ask because its the kind of overlook I make all the time.
So we are told that, my turn to overlook what the question tells us :o:
When I did the question I used the argument that if c = 0, then we must have ab =0, which is impossible as a and b are both non-zero. I also think it would've been around 2-3 marks.
I also think that the solution to the final part of I question 7 can be simplified further by expanding the sine function.
Original post by Sk1lLz
STEP I 2005 Question 10:

First part

Spoiler


i)

Spoiler


ii)

Spoiler


I don't agree with this. I think you've misread the question. It asks you to find the final velocities of each of the three particles but haven't you just found the final velocity of C and the initial velocity of B following the collision with A?
From your solution, the final velocities of A and B are vAv_A and wBw_B but you didn't solve for them. I'll post the remainder of the solution after college.
Original post by SimonM
...

STEP II, Q9

(i)


(ii)

(edited 13 years ago)
Original post by SimonM
...

That was a long one!
STEP II, Q11

First part


Second part


Third part

(edited 13 years ago)
Original post by SimonM
...

STEP I Q14

solution

(edited 13 years ago)
Reply 52
Avatar for SimonM
SimonM
18
STEP II Q7

Spoiler

2005 STEP I question 13

(a)Pr(μ−12σ≤X≤μ+σ)=Pr(X≤μ+σ)−Pr(X≤μ−12σ)=a−(1−b)=a+b−1 (a) Pr \left( \mu - \frac{1}{2} \sigma \leq X \leq \mu+ \sigma \right)=Pr(X \leq \mu + \sigma)-Pr(X \leq \mu - \frac{1}{2} \sigma) =a-(1-b)=a+b-1
Pr(X≤μ+12σ∣X≥μ−12σ)=Pr(μ−12σ≤X≤μ+12σ)Pr(X≥μ−12σ)=b−(1−b)b=2b−1b Pr(X \leq \mu+ \frac{1}{2} \sigma |X \geq \mu - \frac{1}{2} \sigma)= \dfrac{Pr( \mu- \frac{1}{2} \sigma \leq X \leq \mu+ \frac{1}{2} \sigma)}{Pr(X \geq \mu- \frac {1}{2} \sigma)}= \dfrac{b-(1-b)}{b}= \dfrac {2b-1}{b}
Unparseable latex formula:

(b) (i) \text{If volume of milk is }Y \text{ then we require }Pr(Y>500 | Y<505)= \dfrac{Pr(500<Y<505)}{Pr(Y<505)}}


Let X1 be the volume in a skimmed milk carton and X2 that in a full fat carton \text{Let }X_1 \text{ be the volume in a skimmed milk carton and }X_2 \text{ that in a full fat carton}
Pr(500<Y<505)=0.6Pr(μ<X1<μ+12σ)+0.4Pr(μ+12σ<X2<μ+σ) Pr(500<Y<505)=0.6Pr( \mu<X_1< \mu+ \frac{1}{2} \sigma)+0.4Pr( \mu+ \frac{1}{2} \sigma<X_2< \mu+ \sigma)
=0.6(b−0.5)+0.4(a−b)=0.4a+0.2b−0.3 =0.6(b-0.5)+0.4(a-b)=0.4a+0.2b-0.3
and Pr(Y<505)=0.6Pr(X1<μ+12σ)+0.4Pr(X2<μ+σ)=0.6b+0.4a \text{and }Pr(Y < 505)=0.6Pr(X_1< \mu+ \frac{1}{2} \sigma)+0.4Pr(X_2< \mu+ \sigma)=0.6b+0.4a
soPr(Y>500∣Y<505)=0.4a+0.2b−0.30.4a+0.6b=4a+2b−34a+6b \text{so}Pr(Y>500 |Y<505)= \dfrac{0.4a+0.2b-0.3}{0.4a+0.6b}= \dfrac{4a+2b-3}{4a+6b}
(ii)Pr(Y≤505)=0.7  ⟹  0.6b+0.4a=0.7  ⟹  4a+6b=7 (ii) Pr(Y \leq 505)=0.7 \implies 0.6b+0.4a=0.7 \implies 4a+6b=7
Pr(full fat milk∣Y>495)=13  ⟹  Pr(X2≥495)Pr(Y≥495)=0.4×0.50.6Pr(X1≥μ−12σ)+0.4Pr(X2≥μ)=0.20.6b+0.2 Pr( \text{full fat milk} |Y>495)= \dfrac{1}{3} \implies \dfrac{Pr(X_2 \geq 495)}{Pr(Y \geq 495)}= \dfrac{0.4 \times 0.5}{0.6Pr(X_1 \geq \mu- \frac{1}{2} \sigma)+0.4Pr(X_2 \geq \mu)}= \dfrac {0.2}{0.6b+0.2}
so 0.20.6b+0.2=13  ⟹  0.6=0.6b+0.2  ⟹  b=0.40.6=23 so a=14(7−6b)=34 \text{so } \dfrac {0.2}{0.6b+0.2}= \dfrac {1}{3} \implies 0.6=0.6b+0.2 \implies b= \dfrac {0.4}{0.6} = \frac {2}{3} \text{ so } a= \dfrac {1}{4}(7-6b)= \dfrac {3}{4}
2005 STEP II question 13

(i)q=1−p=1−(1+λ)e−λ=1−(1+λ)(1−λ+λ22!−O(λ3))=1−1+12λ2+O(λ3)≈12λ2(i) q=1-p=1-(1+ \lambda) \text{e}^{- \lambda}=1-(1+ \lambda) \left(1- \lambda+ \dfrac{ \lambda^2}{2!}-O( \lambda^3) \right) =1-1+ \dfrac{1}{2} \lambda^2+O( \lambda^3) \approx \dfrac{1}{2} \lambda^2
(ii)Pr(Y=n)≥1−λ  ⟹  pn≥1−λ  ⟹  (1+λ)ne−nλ≥1−λ (ii) Pr(Y=n) \geq1- \lambda \implies p^n \geq 1- \lambda \implies (1+ \lambda)^n \text{e}^{-n \lambda} \geq 1- \lambda
i.e. (1+nλ+n(n−1)2λ2+O(λ3))(1−nλ+12n2λ2+O(λ3))≥1−λ \text{i.e. } \left(1+n \lambda+ \dfrac{n(n-1)}{2} \lambda^2+O( \lambda^3) \right) \left(1-n \lambda+ \dfrac{1}{2}n^2 \lambda^2+O( \lambda^3) \right) \geq1- \lambda
  ⟹  1+(−n2+n(n−1)2+n22)λ2+O(λ3)≥1−λ  ⟹  1−nλ22≥1−λ  ⟹  12nλ≤1 \implies1+ \left(-n^2+ \dfrac{n(n-1)}{2}+\dfrac{n^2}{2} \right) \lambda^2+O( \lambda^3) \geq 1- \lambda \implies1- \dfrac{n \lambda^2}{2} \geq 1- \lambda \implies\dfrac{1}{2}n \lambda \leq1
hence, larger value of n is approximately 2λ \text{hence, larger value of }n \text{ is approximately } \dfrac{2}{ \lambda}
(ii)Pr(Y>1∣Y>0)=Pr(Y>1)Pr(Y>0)=1−P(0)+P(1)1−P(0)=1−qn−npqn−11−qn (ii) Pr(Y>1 |Y>0)= \dfrac{Pr(Y>1)}{Pr(Y>0)}= \dfrac{1-P(0)+P(1)}{1-P(0)}= \dfrac{1-q^n-npq^{n-1}}{1-q^n}
=1−(λ22)n−np(λ22)n−11−(λ22)n since q≈12λ2 = \dfrac{1-( \frac{ \lambda^ 2}{2})^n-np( \frac{ \lambda^2}{2})^{n-1}}{1-( \frac{ \lambda^2}{2})}^n \text{ since }q \approx \dfrac{1}{2} \lambda^2
Unparseable latex formula:

\text{hence, }Pr(Y>1 |Y>0)=1- \dfrac{np( \frac{ \lambda^2}{2})^{n-1}}{1-( \frac{\l\mbda^2}{2})^n}=1-np \left( \dfrac{ \lambda^2}{2} \right)^{n-1} \left(1- \left( \dfrac{ \lambda^2}{2} \right)^n \right)^{-1}


=1−np(λ22)n−1(1+(λ22)n… )≈1−n(1+λ)(1−λ+12λ2+O(λ3))(λ22)n−1(1+(λ22)n… ) =1-np \left( \dfrac{ \lambda^2}{2} \right)^{n-1} \left(1+ \left( \dfrac{ \lambda^2}{2} \right)^n \dots \right) \approx 1-n(1+ \lambda) \left(1- \lambda+ \dfrac{1}{2} \lambda^2+O( \lambda^3) \right) \left( \dfrac{ \lambda^2}{2} \right)^{n-1} \left(1+ \left( \dfrac{ \lambda^2}{2} \right)^n \dots \right)
≈1−n(λ22)n−1ˆ taking leading term only  \approx 1-n \left( \dfrac{ \lambda^2}{2} \right)\^{n-1} \text{ taking leading term only }
2005 STEP II question 14
μ=∫−∞∞xf(x)dx=k∫−∞∞xϕ(x)dx+kλ∫0λxλdx=0+kλ[x22λ]0λ=12kλ2 \mu= \displaystyle \int_{-\infty}^{\infty} x \text{f}(x)dx=k \int_{-\infty}^{\infty} x \phi(x) dx+k \lambda \int_0^{ \lambda} \dfrac{x}{ \lambda}dx=0+k \lambda \left[ \dfrac{x^2}{2 \lambda} \right]_0^{ \lambda}= \dfrac{1}{2}k \lambda^2
but ∫−∞∞f(x)dx=1  ⟹  k∫−∞∞ϕ(x)dx+kλ∫0λ1λdx=1  ⟹  k+kλ=1  ⟹  k=11+λ hence μ=λ22(1+λ) \text{but } \displaystyle \int_{-\infty}^{\infty} \text{f}(x) dx=1 \implies k \int_{-\infty}^{\infty} \phi(x)dx+k \lambda \int_0^{\lambda} \dfrac{1}{ \lambda} dx=1 \implies k+k \lambda=1 \implies k= \dfrac{1}{1+ \lambda} \text { hence } \mu= \dfrac{\lambda^2}{2(1+ \lambda)}

E[X2]=∫−∞∞x2f(x)dx=k∫−∞∞x2ϕ(x)dx+kλ∫0λx2λdx=k+kλλ33λ=k(1+λ23)=k(3+λ2)3 \text{E}[X^2]= \displaystyle \int_{-\infty}^{\infty} x^2 \text{f}(x)dx=k \int_{-\infty}^{\infty}x^2 \phi(x)dx+k \lambda \int_0^{\lambda} \dfrac{x^2}{\lambda}dx=k+k \lambda \dfrac{\lambda^3}{3 \lambda}=k \left(1+ \dfrac{\lambda^2}{3} \right)= \dfrac{k(3+ \lambda^2)}{3}

so Var(X)=3+λ33(1+λ)−(λ22(1+λ))2=4(3+λ3)(1+λ)−3λ412(1+λ)2)=λ4+4λ3+12λ+1212(1+λ)2) \text{so Var}(X)= \dfrac{3+\lambda^3}{3(1+\lambda)}-\left( \dfrac{\lambda^2}{2(1+\lambda) } \right)^2= \dfrac{4(3+\lambda^3)(1+\lambda)-3 \lambda^4}{12(1+\lambda)^2)}= \dfrac{\lambda^4+4 \lambda^3+12 \lambda+12}{12(1+ \lambda)^2)}

Unparseable latex formula:

\text{Now if } \lambda=2 \text{ then }k=\dfrac{1}{3}, \mu= \dfrac{2}{3} \text{ and } \sigma^2= \drfrac{16+32+24+12}{108}= \dfrac{84}{108}= \dfrac{7}{9}



Unparseable latex formula:

(i) \text{f}(x)=k[ \phi(x)+2 \text{g}(x)] \text{ and g}(x)= \dfrac{1}{2} \text{ for }0 \leqx \leq2


so the graph is the standard normal curve y=13ϕ(x) with the portion from x=0 to x=2 \text{so the graph is the standard normal curve }y= \dfrac{1}{3} \phi(x) \text{ with the portion from }x=0 \text{ to }x=2
lifted vertically through a distance of 13 (see diagram below) \text{lifted vertically through a distance of } \dfrac{1}{3}\text { (see diagram below)}

Unparseable latex formula:

\text{C.D.F is }\left\{\begin{array}{lc}\dfrac{1}{3}\Phi(x)& \text{ for }x<0\\[br]\dfrac{1}{3}\Phi(x)+ \dffrac{x}{3} & \text{ for} 0 \leqx \leq2 \\ \dfrac{1}{3}\Phi(x)+ \dfrac{2}{3} & \text{for }x>2 \end{array}



(iii)P(0<X<μ+2σ)=13Φ(23+237)+23−13Φ(0)( since 23+237>2) (iii) P(0<X< \mu+2 \sigma)= \dfrac{1}{3}\Phi \left(\dfrac{2}{3}+ \dfrac{2}{3} \sqrt7 \right)+ \dfrac{2}{3}- \dfrac{1}{3} \Phi(0) \text{( since } \dfrac{2}{3}+ \dfrac{2}{3} \sqrt7>2 )
=0.99213+0.6667−0.53=0.3307+0.6667−0.1667=0.8307 = \dfrac{0.9921}{3}+0.6667- \dfrac{0.5}{3}=0.3307+0.6667-0.1667=0.8307
2005.II.14.jpg9.3KB
2005 STEP III question 10

(i) When discs are a distance 2x apart, the length of elastic is 4(x+r)+2πr (i) \text{ When discs are a distance }2x \text{ apart, the length of elastic is }4(x+r)+2 \pi r
so tension is πmg×4(x+r)12×2πr i.e. tension in elastic band is (x+r)mg6r \text{so tension is } \dfrac{\pi mg \times 4(x+r)}{12 \times 2 \pi r} \text{ i.e. tension in elastic band is }\dfrac{(x+r)mg}{6r}
so the force acting on each disc is (x+r)mg3r−F acting towards each other \text{so the force acting on each disc is } \dfrac{(x+r)mg}{3r}-F \text{ acting towards each other}
Maximum value of F is μmg, so discs will slide providing (x+r)mg3r>μmg  ⟹  μ=1 when x=2r \text{Maximum value of }F \text{ is } \mu mg \text{, so discs will slide providing } \dfrac{(x+r)mg}{3r}> \mu mg \implies \mu=1 \text{ when }x=2r
Elastic energy stored in string initially is 12×πmg12×(12r)22πr=3mgr \text {Elastic energy stored in string initially is } \dfrac{1}{2} \times \dfrac{\pi mg}{12} \times \dfrac{(12r)^2}{2 \pi r}=3mgr
Elastic energy stored in spring when discs collide is 12×πmg12×(4r)22πr=mgr3 \text{Elastic energy stored in spring when discs collide is } \dfrac{1}{2} \times \dfrac{\pi mg}{12} \times \dfrac{(4r)^2}{2 \pi r}= \dfrac{mgr}{3}
Discs will collide if this is sufficient to overcome work done against friction for both discs \text{Discs will collide if this is sufficient to overcome work done against friction for both discs}
hence, if 3mgr−mgr3>2×μmg×2r  ⟹  μ<23 \text{hence, if }3mgr- \dfrac{mgr}{3}>2 \times \mu mg \times 2r \implies \mu< \dfrac{2}{3}
hence, discs will slide but come to rest before colliding if 23<μ<1 \text{hence, discs will slide but come to rest before colliding if } \dfrac{2}{3}< \mu <1
(ii) If discs collide, kinetic energy just before collision will be 3mgr−mgr3−4μmgr (ii) \text{ If discs collide, kinetic energy just before collision will be }3mgr- \dfrac{mgr}{3}-4 \mu mgr
=83mgr−4μmgr=43mgr(2−3μ) = \dfrac{8}{3}mgr-4 \mu mgr= \dfrac{4}{3}mgr(2-3 \mu)
(iii) Note first that the discs must have collided. So μ<23  ⟹  μ2<49 (iii) \text{ Note first that the discs must have collided. So }\mu< \dfrac{2}{3} \implies \mu^2< \dfrac{4}{9}
for discs not to move when a distance apart of 2x we must have \text{for discs not to move when a distance apart of }2x \text{ we must have}
mg(x+r)3r<μmg  ⟹  x<3μmg−mgrmg=(3μ−1)r \dfrac{mg(x+r)}{3r}< \mu mg \implies x< \dfrac{3 \mu mg-mgr}{mg}=(3 \mu -1)r
 and for discs to come to rest at a distance apart of 2x \text{ and for discs to come to rest at a distance apart of }2x
 K.E. = work done against friction + elastic energy gained by string  \text{ K.E. = work done against friction + elastic energy gained by string }
i.e. 23mgr(2−3μ)=2xμmg+12×πmg12×16(r+x)22πr−mgr3=2xμmg+mg(x+r)23r−mgr3 \text{i.e. } \dfrac{2}{3}mgr(2-3 \mu)=2x \mu mg+ \dfrac{1}{2} \times \dfrac{\pi mg}{12} \times \dfrac{16(r+x)^2}{2 \pi r}- \dfrac{mgr}{3}=2x \mu mg+ \dfrac{mg(x+r)^2}{3r}- \dfrac{mgr}{3}
  ⟹  23mgr(2−3μ)−2xμmg−mg3r(x+r)2+mgr3=0 \implies \dfrac{2}{3}mgr(2-3 \mu)-2x \mu mg- \dfrac{mg}{3r}(x+r)^2+ \dfrac{mgr}{3}=0
  ⟹  23mgr(2−3μ)−2xμmg−mg3r(x+r)2+mgr3>23mgr(2−3μ)−2mgμ(3μ−1)r−mg3(3μr)2r+mgr3 \implies \dfrac{2}{3}mgr(2-3 \mu)-2x \mu mg- \dfrac{mg}{3r}(x+r)^2+ \dfrac{mgr}{3}> \dfrac{2}{3}mgr(2-3 \mu)-2mg \mu(3 \mu-1)r- \dfrac{mg}{3} \dfrac{(3 \mu r)^2}{r}+ \dfrac{mgr}{3}
by using the inequality x<(3μ−1)r \text {by using the inequality }x<(3 \mu-1)r
hence, 23mgr(2−3μ)−2mgμ(3μ−1)r−mg3r(3μ)2+mgfr3<0  ⟹  4r−6μr−18μ2r+6μr−9μ2r+r<0 \text{hence, } \dfrac{2}{3}mgr(2-3 \mu)-2mg \mu(3 \mu-1)r- \dfrac{mg}{3r} (3 \mu)^2+ \dfrac{mgfr}{3}<0 \implies 4r-6 \mu r-18 \mu^2r+6 \mu r-9 \mu^2 r+r<0
  ⟹  5r−27μ2r<0  ⟹  μ2>527 so discs come to rest exactly once if 49>μ2>527 \implies 5r-27 \mu^2r<0 \implies \mu^2> \dfrac{5}{27} \text{ so discs come to rest exactly once if } \dfrac{4}{9}> \mu^2> \dfrac{5}{27}
(edited 12 years ago)
2005 STEP III question 11 \text {2005 STEP III question 11}

Potential energy of system referred to spindle as origin is  \text {Potential energy of system referred to spindle as origin is }
P=mgacos⁡θ−mgbcos⁡(π3−θ)−mgccos⁡(π3+θ) P=mga \cos \theta-mgb \cos \left( \dfrac{\pi}{3}- \theta \right)-mgc \cos \left( \dfrac{\pi}{3}+ \theta \right)
=mg(acos⁡θ−b2cos⁡θ−b32sin⁡θ−c2cos⁡θ+c32sin⁡θ)=12mg[(2a−b−c)]=mg \left(a \cos \theta -\dfrac{b}{2} \cos \theta- \dfrac{b \sqrt3}{2} \sin \theta- \dfrac{c}{2} \cos \theta+ \dfrac{c \sqrt3}{2} \sin \theta \right)= \dfrac{1}{2}mg[(2a-b-c)]
differentiating w.r.t. θ we have Dpdθ=12mg[−(2a−b−c)sin⁡θ−(b−c)3cos⁡θ]=0 (*) \text {differentiating w.r.t. } \theta \text { we have } \dfrac {\text{D}p}{ \text{d} \theta}= \dfrac{1}{2}mg[-(2a-b-c) \sin \theta-(b-c) \sqrt3 \cos \theta]=0 \text { (*)}
−(2a−b−c)sin⁡θ−(b−c)3cos⁡θ=0  ⟹  tan⁡θ=−(b−c)32a−b−c -(2a-b-c) \sin \theta-(b-c) \sqrt3 \cos \theta=0 \implies \tan \theta= -\dfrac{(b-c) \sqrt3}{2a-b-c}
We also have equilibrium if the total moment about the spindle is zero, i.e. if \text {We also have equilibrium if the total moment about the spindle is zero, i.e. if}
mg(asin⁡θ+bsin⁡(π3−θ)−csin⁡(π3+θ))=0 again giving equation (*) mg \left(a \sin \theta+b\sin \left( \dfrac{\pi}{3}- \theta \right)-c \sin \left( \dfrac{\pi}{3}+ \theta \right) \right)=0 \text { again giving equation (*)}
Unparseable latex formula:

\tan \theta= -\dfrac {(b-c) \sqrt3}{2a-b-c} \implies \rthgeta= \pi-\alpha \text { or } 2 \pi- \alpha \text { where } \alpha = \tan^{-1} \left( \dfrac{(b-c) \sqrt3}{2a-b-c}\right) \text { for } 0< \alpha< \dfrac{\pi}{2}


i.e. 2 equilibrium positions \text {i.e. 2 equilibrium positions}
d2Pdθ2=12mg[−(2a−b−c)sin⁡θ−(b−c)3cos⁡θ]=12mg[−(2a−b−c)cos⁡θ+(b−c)3sin⁡θ] \dfrac{\text{d}^2P}{\text{d} \theta^2}= \dfrac {1}{2}mg[-(2a-b-c) \sin \theta-(b-c) \sqrt3 \cos \theta]= \dfrac{1}{2}mg[-(2a-b-c) \cos \theta+(b-c) \sqrt3 \sin \theta]
If θ=π−α then sin⁡θ>0 and cos⁡θ<0  ⟹  d2Pdθ2>0  ⟹   equilibrium stable \text {If } \theta= \pi-\alpha \text { then } \sin \theta >0 \text { and } \cos \theta <0 \implies \dfrac{\text {d}^2 P}{\text{d} \theta^2}>0 \implies \text { equilibrium stable}
θ=2π−α then sin⁡θ<0 and cos⁡θ>0  ⟹  d2Pdθ2<0  ⟹   equilibrium unstable \theta= 2 \pi-\alpha \text { then } \sin \theta <0 \text { and } \cos \theta >0 \implies \dfrac{\text {d}^2 P}{\text{d} \theta^2}<0 \implies \text { equilibrium unstable}
θ=π−α  ⟹  sin⁡θ=−(b−c)3X and cos⁡θ=−2a−b−cX \theta=\pi- \alpha \implies \sin \theta=-\dfrac{(b-c) \sqrt3}{X} \text { and } \cos \theta=-\dfrac{2a-b-c}{X}
similarly θ=2π−α  ⟹  sin⁡θ=−(b−c)3X and cos⁡θ=2a−b−cX where X is the positive root of  \text {similarly } \theta=2 \pi- \alpha \implies \sin \theta=- \dfrac{(b-c) \sqrt3}{X} \text { and } \cos \theta =\dfrac{2a-b-c}{X} \text { where }X \text { is the positive root of }
(2a−b−c)2+3(b−c)2=4(a2+b2+c2)−4(ab+bc+cxa)=4[(a−b)2+(b−c)2+(c−a)2] (2a-b-c)^2+3(b-c)^2=4(a^2+b^2+c^2)-4(ab+bc+cxa)=4[(a-b)^2+(b-c)^2+(c-a)^2]
Unparseable latex formula:

\texzt {i.e. }X=2R \text { where }R \text { is as defined in question}


so minimum p.e. Is 12mg[−(2a−b−c)(2a−b−c2R)−(b−c)3((b−c)32R)] \text {so minimum p.e. Is } \dfrac{1}{2}mg \left[-(2a-b-c) \left( \dfrac{2a-b-c}{2R} \right)-(b-c) \sqrt3 \left( \dfrac{(b-c) \sqrt3}{2R} \right) \right]
i.e. mg4R[−22a−b−c)2−3(b−c)2]=−mg4R×2R2=−mgR and similarly, maximum np.e. is \text {i.e. } \dfrac{mg}{4R} \left[-22a-b-c)^2-3(b-c)^2 \right]=- \dfrac{mg}{4R} \times 2R^2=-mgR \text { and similarly, maximum np.e. is}
12mg[(2a−b−c)(2ab−c2R)+(b−c)3((b−c)32R)]=mgR \dfrac{1}{2}mg \left[(2a-b-c) \left( \dfrac{2a_b-c}{2R} \right)+(b-c) \sqrt3 \left( \dfrac{(b-c) \sqrt3}{2R} \right) \right]=mgR
For complete revolutions the loss of P.E. From maximum to minimum positions must be less than the \text {For complete revolutions the loss of P.E. From maximum to minimum positions must be less than the}
K.E. at position of stable equilibrium so if angular velocity at this position is ω \text {K.E. at position of stable equilibrium so if angular velocity at this position is } \omega
we must have 12m(a2+b2+c2)ω2>2mgR or ω>4gRa2+b2+c2 as required \text {we must have }\dfrac{1}{2}m(a^2+b^2+c^2) \omega^2>2mgR \text { or } \omega>\sqrt{\dfrac{4gR}{a^2+b^2+c^2}} \text { as required}
2005 STEP III question 13 \text {2005 STEP III question 13}

(i) A player can only win exactly £3 by drawing 3 cards in succession showing a number \text {(i) A player can only win exactly £3 by drawing 3 cards in succession showing a number}
between 1 and w and then drawing a zero with his 4th card, so, since Pr(drawing a number between 1 and w \text {between 1 and }w \text { and then drawing a zero with his 4th card, so, since Pr(drawing a number between 1 and }w
inclusive is ww+1 then Pr(3 such numbers in succesion is (ww+1)3 \text {inclusive is } \dfrac{w}{w+1} \text { then Pr(3 such numbers in succesion is } \left( \dfrac{w}{w+1} \right)^3
Pr)drawing a zero =1w+1 hence, Pr(winningt exactly £3 =(ww+1)3×1w+1=w3(w+1)4 as required \text {Pr)drawing a zero }=\dfrac{1}{w+1} \text { hence, Pr(winningt exactly £3 }= \left( \dfrac{w}{w+1} \right)^3 \times \dfrac{1}{w+1}= \dfrac{w^3}{(w+1)^4} \text { as required}
Similarly Pr(winning exactly £r is wr(w+1)r+1 \text {Similarly Pr(winning exactly £}r \text { is } \dfrac{w^r}{(w+1)^{r+1}}
so expected winnings are ∑r=0∞rwr(w+1)r+1=w(w+1)2×∑r=0∞(ww+1)r−1 \text {so expected winnings are }\displaystyle \sum_{r=0}^ \infty \dfrac{rw^r}{(w+1)^{r+1}}= \dfrac{w}{(w+1)^2} \times \displaystyle \sum_{r=0}^\infty \left(\dfrac{w}{w+1} \right)^{r-1}
and ∑r=0∞r(x)r−1=(1−x)−2 so expected winnings are w(w+1)2 \text {and } \displaystyle \sum_{r=0}^\infty r(x)^{r-1}= (1-x)^{-2} \text { so expected winnings are } \dfrac{w}{(w+1)^2} \times \dfrac{1}{1-\frac{w}{w+1})^2=w
(ii) In the second situation we need only consider the w+1 cards since the rest add nothing to the winnings \text {(ii) In the second situation we need only consider the }w+1 \text { cards since the rest add nothing to the winnings}
so Pr(winning £r is ww+1×w−1w×w−1w−2×⋯×w−r−1w−r−2=1w+1 \text {so Pr(winning £}r \text { is } \dfrac{w}{w+1} \times \dfrac{w-1}{w} \times \dfrac{w-1}{w-2} \times \dots \times \dfrac{w-r-1}{w-r-2}=\dfrac {1}{w+1}
hence, expected winnings are now ∑r=0wrw+1=1w+1×w2(w+1)=w2 \text {hence, expected winnings are now } \displaystyle \sum_{r=0}^w \dfrac{r}{w+1}= \dfrac{1}{w+1} \times \dfrac{w}{2}(w+1)= \dfrac{w}{2}
Reply 59
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rath90
Original post by sonofdot
STEP II 2005 Question 8

dydx=x3y2(1+x2)5/2\displaystyle \frac{dy}{dx} = \frac{x^3 y^2}{(1+x^2)^{5/2}} for x≥0x \geq 0 with y=1 when x=0

⇒∫1y2 dy=∫x3(1+x2)5/2 dx\displaystyle\Rightarrow \int \frac{1}{y^2} \, dy = \int \frac{x^3}{(1+x^2)^{5/2}} \, dx

The RHS can be integrated by parts, with u=x2⇒dudx=2xu=x^2 \Rightarrow \frac{du}{dx} = 2x and dvdx=x(1+x2)5/2⇐v=−13(1+x2)3/2\frac{dv}{dx} = \frac{x}{(1+x^2)^{5/2}} \Leftarrow v = -\frac{1}{3(1+x^2)^{3/2}}

Unparseable latex formula:

\displaystyle\Rightarrow -\frac{1}{y} = - \frac{x^2}{3(1+x^2)^{3/2}} + \int \frac{2x}{3(1+x^2)^{3/2}} \, dx \br [br][br]\br[br][br]\displaystyle\Rightarrow -\frac{1}{y} = - \frac{x^2}{3(1+x^2)^{3/2}} - \frac{2}{3(1+x^2)^{1/2}} + C \br [br][br]\br[br][br]\displaystyle\Rightarrow \frac{1}{y} = \frac{x^2}{3(1+x^2)^{3/2}} + \frac{2(1+x^2)}{3(1+x^2)^{3/2}} + C \br [br][br]\br[br][br]\displaystyle\Rightarrow \frac{1}{y} = \frac{2+3x^2}{3(1+x^2)^{3/2}} + C



Substituting in x=0 and y=1 gives C=1/3, so we get 1y=2+3x23(1+x2)3/2+13\displaystyle\boxed{\frac{1}{y} = \frac{2+3x^2}{3(1+x^2)^{3/2}} + \frac13} as required.

For large x, we can say 1+x2≈x21+x^2 \approx x^2 so (1+x2)3/2≈x3(1+x^2)^{3/2} \approx x^3

∴1y≈x2x3+13≈13(1+3x)\displaystyle\therefore \frac{1}{y} \approx \frac{x^2}{x^3} + \frac13 \approx \frac13 \left(1+\frac{3}{x} \right)

Looking at the first two terms of the expansion of (1+3x)−1(1+\frac{3}{x})^{-1} gives, for large x:


Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?

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