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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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Updated: March 15, 2013
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