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STEP 2005 Solutions Thread

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    (Original post by SimonM)
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    STEP I Q14

    solution
    Note that P(0\leq X \leq x) = k(1-e^{-x}) \implies \displaystyle\int_0^x f(t)dt = k(1-e^{-x}), where f is the PDF for the distribution of X for X>0.

    Differentiating the above:
    \implies \dfrac{d}{dx}\left[\displaystyle\int_0^x f(t)dt\right] = f(x) = ke^{-x}

    Hence, the PDF for X is given by:
    f(x) = \begin{Bmatrix} ke^{-x} & \mathrm{for} & 0 \leq x < \infty \\m & \mathrm{for} & x=-1 \\0 & \mathrm{otherwise}
    (i)
    Note that \displaystyle\int_0^{\infty} ke^{-x} dx = 1-m

    \Leftrightarrow k\left(0-(-1)\right)=1-m

    \Leftrightarrow \boxed{k=1-m}

    (ii)
    Note that E(X) = XP(X=-1) + \displaystyle\int_0^{\infty} (1-m)xe^{-x}dx

    Integrating the second term by parts with u=x, \frac{dv}{dx}=e^{-x}:

    \implies E(X)=-m + (1-m)\left(\left[-xe^{-x}\right]^{\infty}_0 + \displaystyle\int^{\infty}_0 e^{-x}dx \right)

    =-m +(1-m)\left(0 +\left[-e^{-x}\right]^{\infty}_0\right)

    =-m + (1-m)

    =\boxed{1-2m}, as required.

    (iii)
    Variance
    Note that Var(X) = \left(X^2P(X=-1) + \displaystyle\int_0^{\infty} (1-m)x^2e^{-x}dx \right) - \left[E(X)\right]^2

    =m + (1-m)\displaystyle\int_0^{\infty} x^2e^{-x}dx - (1-2m)^2

    Integrating the second term by parts with u=x^2, \frac{dv}{dx}=e^{-x}:

    \implies Var(X) = -(4m^2 - 5m+1) + (1-m) \left( \left[-x^2e^{-x}\right]^{\infty}_0 +2\displaystyle\int ^{\infty}_0 xe^{-x}dx \right)

    In part (ii), we had shown that \displaystyle\int ^{\infty}_0 xe^{-x}dx = 1, hence:

    Var(X) = -(4m^2 -5m+1) + (1-m)\left(0+2)

    =\boxed{1+3m-4m^2}

    Median
    Let M be the median of the distribution of X, it follows that:

    \displaystyle\int_0^{M} (1-m)e^{-x} dx = \dfrac{1}{2} - m

    \Leftrightarrow (1-m)(1-e^{-M}) = \dfrac{1-2m}{2} (By using the CDF given at the start)

    \Leftrightarrow 1-e^{-M} = \dfrac{1-2m}{2(1-m)}

    \Leftrightarrow -M = \ln \left(\dfrac{2(1-m) - (1-2m)}{2(1-m)} \right)

    \Leftrightarrow \boxed{M=\ln \left[2(1-m)\right]}

    (iv)
    E\left(\sqrt{|X|}\right) = \sqrt{|X|}P(X=-1) + \displaystyle\int^{\infty}_0 (1-m)\sqrt{x}e^{-x} dx

    Using a substitution of x=y^2 for the integral:

    E\left(\sqrt{|X|}\right) = m + (1-m) \displaystyle\int ^{\infty}_0 \sqrt{y^2}e^{-y^2} \cdot 2ydy

    =m + 2(1-m)\displaystyle\int ^{\infty}_0 y^2e^{-y^2}dy

    Given that \displaystyle\int ^{\infty}_0 y^2e^{-y^2}dy = \dfrac{1}{4}\sqrt{\pi}:

    E\left(\sqrt{|X|}\right) = m + 2(1-m) \times \dfrac{1}{4}\sqrt{\pi}

    =\dfrac{2m + (1-m)\sqrt{\pi}}{2}

    =\boxed{\dfrac{\left(2-\sqrt{\pi}\right)m + \sqrt{\pi}}{2}}
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    STEP II Q7

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    i) Both P and Q perform circles in a counterclockwise direction. P about the k-axis, radius 1 in the i-j plane, Q about a different axis, radius 3

    ii) \mathbf{P} \cdot \mathbf{Q} = \frac{3}{2} \cos t \cos (t + \frac{\pi}{4} ) + 3 \sin t \sin (t + \frac{\pi}{4})
     = \frac{3}{4} \left ( \cos (\frac{\pi}{4}) + \cos ( 2t + \frac{\pi}{4}) \right ) + \frac{3}{2} \left ( \cos (\frac{\pi}{4}) - \cos ( 2t + \frac{\pi}{4}) \right )
     = \frac{9 \sqrt{2}}{8} - \frac{3}{4} \cos ( 2t + \frac{\pi}{4} )

    |\mathbf{P}| = 1, |\mathbf{Q}| = 3 \Rightarrow \cos \theta = \frac{3 \sqrt{2}}{8} - \frac{1}{4} \cos ( 2t + \frac{\pi}{4} ) as required.

    iii)  \frac{3 \sqrt{2}}{8} - \frac{1}{4} \cos ( 2t + \frac{\pi}{4} ) = \frac{1}{\sqrt{2}} \Rightarrow \cos (2t + \frac{\pi}{4}) = - \frac{1}{\sqrt{2}} Sketching this, the answer is clear.
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    2005 STEP I question 13

     (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 \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}
     (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)}  }
      \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( \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
     \text{and }Pr(Y < 505)=0.6Pr(X_1< \mu+ \frac{1}{2} \sigma)+0.4Pr(X_2< \mu+ \sigma)=0.6b+0.4a
     \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 \leq 505)=0.7 \implies 0.6b+0.4a=0.7 \implies 4a+6b=7
     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}
      \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}
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    2005 STEP II question 13

    (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) \geq1- \lambda \implies p^n \geq 1- \lambda \implies (1+ \lambda)^n \text{e}^{-n \lambda} \geq 1- \lambda
     \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
     \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
     \text{hence, larger value of }n \text{ is approximately } \dfrac{2}{ \lambda}
     (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}
     = \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
      \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 \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)
     \approx 1-n \left( \dfrac{ \lambda^2}{2} \right)\^{n-1} \text{ taking leading term only }
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    2005 STEP II question 14
     \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
     \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)}

     \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}

     \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)}

     \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}

     (i) \text{f}(x)=k[ \phi(x)+2 \text{g}(x)] \text{ and g}(x)= \dfrac{1}{2} \text{ for }0 \leqx \leq2
     \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
     \text{lifted vertically through a distance of } \dfrac{1}{3}\text { (see diagram below)}

     \text{C.D.F is }\left\{\begin{array}{lc}\dfrac{  1}{3}\Phi(x)& \text{ for }x<0\\

\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< \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 )
     = \dfrac{0.9921}{3}+0.6667- \dfrac{0.5}{3}=0.3307+0.6667-0.1667=0.8307
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    2005 STEP III question 10

     (i) \text{ When discs are a distance }2x \text{ apart, the length of elastic is }4(x+r)+2 \pi r
     \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}
     \text{so the force acting on each disc is } \dfrac{(x+r)mg}{3r}-F \text{ acting towards each other}
     \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
     \text {Elastic energy stored in string initially is } \dfrac{1}{2}  \times \dfrac{\pi mg}{12} \times \dfrac{(12r)^2}{2 \pi r}=3mgr
     \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}
     \text{Discs will collide if this is sufficient to overcome work done against friction for both discs}
     \text{hence, if }3mgr- \dfrac{mgr}{3}>2 \times \mu mg \times 2r \implies \mu< \dfrac{2}{3}
     \text{hence, discs will slide but come to rest before colliding if } \dfrac{2}{3}< \mu <1
     (ii) \text{ If discs collide, kinetic energy just before collision will be }3mgr- \dfrac{mgr}{3}-4 \mu mgr
     = \dfrac{8}{3}mgr-4 \mu mgr= \dfrac{4}{3}mgr(2-3 \mu)
     (iii) \text{ Note first that the discs must have collided. So }\mu< \dfrac{2}{3} \implies \mu^2< \dfrac{4}{9}
     \text{for discs not to move when a distance apart of }2x \text{ we must have}
      \dfrac{mg(x+r)}{3r}< \mu mg \implies x< \dfrac{3 \mu mg-mgr}{mg}=(3 \mu -1)r
     \text{ and for discs to come to rest at a distance apart of }2x
     \text{ K.E. = work done against friction + elastic energy gained by string }
     \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}
      \implies \dfrac{2}{3}mgr(2-3 \mu)-2x \mu mg- \dfrac{mg}{3r}(x+r)^2+ \dfrac{mgr}{3}=0
     \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}
     \text {by using the inequality }x<(3 \mu-1)r
      \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
     \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}
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     \text {2005 STEP III question 11}

     \text {Potential energy of system referred to spindle as origin is }
     P=mga \cos \theta-mgb \cos \left( \dfrac{\pi}{3}- \theta \right)-mgc \cos \left( \dfrac{\pi}{3}+ \theta \right)
    =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) \cos \theta-=(b-c) \sqrt3 \sin \theta ]
     \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 \theta-(b-c) \sqrt3 \cos \theta=0 \implies \tan \theta= -\dfrac{(b-c) \sqrt3}{2a-b-c}
     \text {We also have equilibrium if the total moment about the spindle is zero, i.e. if}
     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 (*)}
     \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}
     \text {i.e. 2 equilibrium positions}
     \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]
     \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}
     \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}
     \theta=\pi- \alpha \implies \sin \theta=-\dfrac{(b-c) \sqrt3}{X} \text { and } \cos \theta=-\dfrac{2a-b-c}{X}
     \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(a^2+b^2+c^2)-4(ab+bc+cxa)=4[(a-b)^2+(b-c)^2+(c-a)^2]
     \texzt {i.e. }X=2R \text { where }R \text { is as defined in question}
     \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]
     \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}
     \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
     \text {For complete revolutions the loss of P.E. From maximum to minimum positions must be less than the}
     \text {K.E. at position of stable equilibrium so if angular velocity at this position is } \omega
     \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}
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     \text {2005 STEP III question 13}

     \text {(i) A player can only win exactly £3 by drawing 3 cards in succession showing a number}
     \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
     \text {inclusive is  } \dfrac{w}{w+1} \text { then Pr(3 such numbers in succesion is } \left( \dfrac{w}{w+1} \right)^3
     \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}
     \text {Similarly Pr(winning exactly £}r \text { is } \dfrac{w^r}{(w+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}
     \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 [/latex]
     \text {(ii) In the second situation we need only consider the }w+1 \text { cards since the rest add nothing to the winnings}
     \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}
     \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}
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    (Original post by sonofdot)
    STEP II 2005 Question 8

    [expand=First Part]\displaystyle \frac{dy}{dx} = \frac{x^3 y^2}{(1+x^2)^{5/2}} for x \geq 0 with y=1 when x=0

    \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=x^2 \Rightarrow \frac{du}{dx} = 2x and \frac{dv}{dx} = \frac{x}{(1+x^2)^{5/2}} \Leftarrow v = -\frac{1}{3(1+x^2)^{3/2}}

    \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



\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



\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



\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 \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+x^2 \approx x^2 so (1+x^2)^{3/2} \approx x^3

    \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+\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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    (Original post by rath90)
    Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?
    Standard binomial expansion
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    (Original post by Glutamic Acid)
    II/4:

    1st part

    \tan^{-1} \left( \dfrac{1}{a + b} \right) + \tan^{-1}\left(\dfrac{1}{a + c} \right) = \tan^{-1} \left( \dfrac{1}{a} \right) (*)

    \Leftrightarrow \tan \left[ \tan^{-1} \left( \dfrac{1}{a + b} \right) + \tan^{-1}\left(\dfrac{1}{a + c} \right) \right] = \tan \tan^{-1} \left( \dfrac{1}{a} \right)

    \Leftrightarrow \dfrac{\frac{1}{a + b} + \frac{1}{a + c}}{1 - \frac{1}{a + b} \frac{1}{a +c}} = \dfrac{1}{a}

    \Leftrightarrow \dfrac{2a + b + c}{a^2 + ba + ca + bc - 1} = \dfrac{1}{a}

    Substituting a^2 = bc - 1, LHS = \dfrac{2a + b + c}{2a^2 + ba + ca} = \dfrac{1}{a}, as required. Note, since a, b and c are positive, 1/(a+b), 1/(a+c) and 1/a will all lie between 0 and +infinity, so "implies and is implied by" implication signs can be used.


    2nd part

    Let p + q = "a", s = "b", t = "c" \Rightarrow \tan^{-1} \left( \dfrac{1}{p + q + s} \right) + \tan^{-1}\left(\dfrac{1}{p + q + t}\right) = \tan^{-1} \left( \dfrac{1}{p + q} \right), from (*)

    Let p + r = "a", u = b, v = c \Rightarrow \Rightarrow \tan^{-1} \left( \dfrac{1}{p + r + u} \right) + \tan^{-1}\left(\dfrac{1}{p + r + v}\right) = \tan^{-1} \left( \dfrac{1}{p + r} \right), from (*)

    Let p = "a", b = "q" and r = "c", so \Rightarrow \tan^{-1} \left( \dfrac{1}{p + q} \right) + \tan^{-1}\left(\dfrac{1}{p + r}\right) = \tan^{-1} \left( \dfrac{1}{p} \right), from (*), and we're done.


    3rd part

    We can use the result in the second part by choosing p = 7, q =1, s = 5, t = 13, r = 50, u = 25 and v = 130, and everything falls out neatly.

    For part III does it matter which values you choose?
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    (Original post by Principia)
    For part III does it matter which values you choose?
    In order to use the previous result you must choose values so that p+q+s=13,p+q+r=21,p+r+u=87 ,p+r+v-187 and p=7
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    (Original post by brianeverit)
    In order to use the previous result you must choose values so that p+q+s=13,p+q+r=21,p+r+u=87 ,p+r+v-187 and p=7
    I did that, but can it be any values that meet those criteria? i.e. there are then an infinite amount of solutions.
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    (Original post by Principia)
    I did that, but can it be any values that meet those criteria? i.e. there are then an infinite amount of solutions.
    Yes, the solution is not unique but you must check that your values satisfy all the criteria.

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