STEP 2005 Solutions Thread

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  1. Unbounded's Avatar
    61 20-06-2010  01:40
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    Re: STEP 2005 Solutions Thread
    (Original post by The Bigtime)
    I also think that the solution to the final part of I question 7 can be simplified further by expanding the sine function.
    Indeed it can: \dfrac{\sin \left( \frac{(2n+1)\pi}{n} \right)}{\sin \left( \frac{\pi}{n} \right)} = \dfrac{\sin \left( 2\pi + \frac{\pi}{n} \right)}{\sin \left( \frac{\pi}{n} \right)} = \dfrac{ \sin \left( \frac{\pi }{n} \right) }{\sin \left( \frac{\pi}{n} \right) } = 1

    I'll edit it now.
  2. Farhan.Hanif93's Avatar
    62 05-11-2010  09:46
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    Re: STEP 2005 Solutions Thread
    (Original post by Sk1lLz)
    STEP I 2005 Question 10:

    First part
    Spoiler:
    Show

    From A colliding with B:

    Let v_A represent the velocity of A after colliding with A.

    1. NEL: e = \frac{v - v_A}{u}

    2. CLM: au = av_A + bv

    \Rightarrow v_A = \frac{au - bv}{a}

    Substitute into 1.:

    3. v = \frac{au(e + 1)}{a + b}

    From B colliding with C:

    Let \omega_B represent the velocity of B after colliding with C.

    4. NEL: e = \frac{\omega - \omega_B}{v}

    5. CLM: bv = b\omega_B + c\omega

    \Rightarrow \omega_B = \frac{bv - c\omega}{b}

    Substitute into 4.:

    6. \omega = \frac{bv(e + 1)}{(b + c)}

    Now, substitute 3. into 6.: \omega = \frac{abu(e + 1)^2}{(a + b)(b + c)}

    i)
    Spoiler:
    Show

    \frac {a}{b} = \frac {b}{c} \Rightarrow b^2 = ac

    \frac {b}{c} = e \Rightarrow b = ce

    \frac {a}{b} = e \Rightarrow a = be

    ab = c^2e^3

    Use 6.: \omega = \frac{c^2e^3u(e + 1)^2}{c^2e^3 + c^2e^2 + c^2e + c^2e^2}

    \omega = ue^2

    For v, use 3.: v= \frac{beu(e + 1)}{b(e + 1)}

    v = ue

    ii)
    Spoiler:
    Show

    \frac{b}{a} = \frac{c}{b} \Rightarrow b^2 = ac

    \frac{c}{b} = e \Rightarrow c = be

    \frac{b}{a} = e \Rightarrow b = ae

    bc = a^2e^3

    Use 6.: \omega = \frac{a^2ue(e + 1)^2}{a^2e + a^2e^3 + a^2e^2 + a^2e^2}

    \omega = u

    Use 3.: v = \frac{au(e + 1)}{a(e + 1)}

    v = u

    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 v_A and w_B but you didn't solve for them. I'll post the remainder of the solution after college.
  3. Farhan.Hanif93's Avatar
    63 06-11-2010  18:21
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    Re: STEP 2005 Solutions Thread
    (Original post by SimonM)
    ...
    STEP II, Q9

    (i)
    Diagram

    Consider the particles individually when they are at the point of slipping upwards.

    Considering A:
    Resolving perpendicular to the slope: R_m - mg\cos \frac{\pi}{6} = 0 \implies R_m = \frac{\sqrt3}{2}mg

    Resolving parallel to the slope: T-\frac{1}{2}mg - \frac{\sqrt3}{6}R_m =0 \implies T=\frac{3}{4}mg by subbing in R_m from above and rearranging.

    Considering B:
    Resolving perpendicular to the slope: R_{2m}- 2mg\cos \frac{\pi}{6} + P \sin \theta =0 \implies R_{2m} = \sqrt 3mg - P\sin \theta

    Resolving parallel to the slope: P\cos \theta - mg - \frac{\sqrt 3}{3}R_{2m} - T =0 \implies \frac{11}{4}mg = P(\cos \theta + \frac{\sqrt 3}{3}\sin \theta ) by subbing in R_{2m} from above and rearranging.

    Expressing \frac{\sqrt 3}{3}\sin \theta +\cos \theta} in the form R\sin (\theta + \alpha) gives it to be equivalent to \frac{2}{\sqrt3}\sin (\theta + \tan^{-1} \frac{3}{\sqrt{3}}).

    Note that this implies that P= \frac{11 mg \sqrt {3}}{8 \sin \left( \theta + \tan ^{-1} \frac{3}{\sqrt 3}\right)}, which is minimised when \sin\left(\theta + \tan^{-1} \frac{3}{\sqrt{3}}\right) is maximised i.e. when \theta + \tan^{-1} \frac{3}{\sqrt{3}} = \frac{\pi}{2}.

    Therefore \boxed{P_{min} = \frac{11\sqrt3}{8}mg} and P must act at an angle of \boxed{\frac{2\pi}{3}-\tan ^{-1} \frac{3}{\sqrt 3}} to the horizontal i.e. parallel to the slope.

    (ii)
    If the particles are not sliding down the slope then they are either on the point of slipping down it or moving up it. We've considered the latter in part (i) so we need to find the smallest value of P for which the system is at the point of slipping downwards. If you were to draw a diagram for this, it would be the same as the one for part (i) except the frictional forces will be acting in the opposite direction.

    Considering A:
    Resolving perpendicular to the slope: R_m - mg\cos \frac{\pi}{3} = 0 \implies R_m = \frac{\sqrt3}{2}mg

    Resolving parallel to the slope: T + \frac{\sqrt3}{6}R_m -\frac{1}{2}mg =0 \implies T=\frac{1}{4}mg

    Considering B:
    Resolving perpendicular to the slope: R_{2m} - \sqrt 3mg + P\sin \theta =0 \implies R_{2m}=\sqrt 3mg-P\sin \theta

    Resolving parallel to the slope: P\cos \theta + \frac{\sqrt3}{3}R_{2m} - mg -T=0 \implies P=\dfrac{mg}{4(\cos \theta - \frac{\sqrt 3}{3}\sin \theta)}

    Note that cos \theta - \frac{\sqrt 3}{3}\sin \theta \equiv \frac{2}{\sqrt 3}\cos \left(\theta + \tan ^{-1}\frac{\sqrt3}{3}\right).

    Therefore \boxed{P_{min} = \frac{\sqrt 3}{8}mg} in this case. This is smaller than the answer to part (i), therefore is the correct answer.
    Last edited by Farhan.Hanif93; 07-11-2010 at 02:05.
  4. Farhan.Hanif93's Avatar
    64 06-11-2010  22:58
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    Re: STEP 2005 Solutions Thread
    (Original post by SimonM)
    ...
    That was a long one!
    STEP II, Q11
    First part
    Diagram

    Where F is the tension in the string, A is the object on the slope and B is the hanging object. (I chose not to use the conventional T because this letter is assigned to something else later on)

    Considering B:
    Resolving downwards: m_2g-F=m_2a \implies F=m_2(g-a) (\alpha)

    Considering A
    Resolving perpendicular to the slope: R-m_1g\cos (\arctan \frac{3}{4}) =0 \implies R=m_1g\cos (\arctan \frac{3}{4})

    Resolving parallel to the slope: F-m_1g\sin (\arctan \frac{3}{4}) - \mu R = m_1a (\beta)

    Note that:
    Useful triangle

    Using SOHCAHTOA on this triangle yields that:
    \sin (\arctan \frac{3}{4})= \frac{3}{5}
    \cos (\arctan \frac{3}{4}) = \frac{4}{5}

    Therefore using these results and subbing (\alpha) into (\beta):
    (\beta) : m_1a + m_2a =m_2g - m_1g \implies \boxed{a=\frac{m_2-m_1}{m_2+m_1}g} as required.

    Second part
    Considering the velocity, v_1, of A at the point where the string breaks:

    u=0, t=T, a=\frac{m_2-m_1}{m_2+m_1}g, v=v_1

    Therefore, using v=u+at:

    v_1=\frac{m_2-m_1}{m_2+m_1}gT.

    In this case, A can be modelled as follows:
    Diagram

    Where a_2 is the acceleration of A after the string breaks.

    Resolving perpendicular to the slope: R-\frac{4}{5}m_1g=0 \implies R=\frac{4}{5}m_1g

    Resolving down the slope: \frac{3}{5}m_1g + \mu R = m_1a_2 \implies a_2 = g

    Considering the motion of A between the time where the string breaks and the particle reaches it's maximum height (let this time be denoted by t_1):

    u=v_1, a=-g, v=0, t=t_1

    Therefore using v=u+at:

    t_1=\frac{m_2-m_1}{m_2+m_1}T

    Note that the time taken (after release) to reach maximum height is given by:

    T+t_1 = \boxed{\dfrac{2m_2T}{m_2+m_1}}

    Third part
    Let the maximum displacement of the particle be denoted by d; let the distance between the initial position and the position at which the string breaks be d_1 and let the distance between the string's breaking position and the maximum height of motion be d_2 such that d=d_1+d_2

    Considering upwards motion for the distance d_1:

    u=0, a=\frac{m_2-m_1}{m_2+m_1}g, t=T, s=d_1

    Therefore, using s=ut +\frac{1}{2}at^2:

    d_1 = \dfrac{(m_2-m_1)gT^2}{2(m_2+m_1)}

    Considering upwards motion from d_1 to d_2:

    u=v_1, a=-g, v=0 s=d_2

    Therefore, using v^2=u^2+2as:

    d_2 = \left(\dfrac{m_2-m_1}{\sqrt2(m_2+m_1)} \right)^2gT^2

    \implies d = d_1+d_2 = \dfrac{(m_2-m_1)m_2}{(m_2+m_1)^2}gT^2

    Note that, by resolving forces of the object when it is at it's maximum height, the acceleration down the slope is \frac{g}{5} ms^{-2}.

    Considering motion down the slope from max height to initial position:

    u=0, a=\frac{g}{5}, t=T+t_1, s=d

    Therefore, using s=ut+\frac{1}{2}at^2:

    \implies \dfrac{(m_2-m_1)m_2}{(m_2+m_1)^2}gT^2 = \dfrac{2m^2_2gT^2}{5(m_2+m_1)^2}

    \implies \boxed{m_1:m_2 = 3:5}
    Last edited by Farhan.Hanif93; 07-11-2010 at 04:56.
  5. Farhan.Hanif93's Avatar
    65 10-04-2011  13:07
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    Re: STEP 2005 Solutions Thread
    (Original post by SimonM)
    ...
    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}}
    Last edited by Farhan.Hanif93; 10-04-2011 at 13:09.
  6. SimonM's Avatar
    66 06-06-2011  16:13
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    Re: STEP 2005 Solutions Thread
    STEP II Q7

    Spoiler:
    Show
    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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  7. brianeverit's Avatar
    67 26-06-2011  16:46
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    Re: STEP 2005 Solutions Thread
    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}
  8. brianeverit's Avatar
    68 27-06-2011  10:21
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    Re: STEP 2005 Solutions Thread
    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 }
  9. brianeverit's Avatar
    69 28-06-2011  10:44
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    Re: STEP 2005 Solutions Thread
    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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  10. brianeverit's Avatar
    70 28-06-2011  14:05
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    Re: STEP 2005 Solutions Thread
    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}
    Last edited by brianeverit; 21-07-2011 at 19:45.
  11. brianeverit's Avatar
    71 22-07-2011  10:23
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP 2005 Solutions Thread
    \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}
  12. brianeverit's Avatar
    72 22-07-2011  16:07
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP 2005 Solutions Thread
    \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}
  13. rath90's Avatar
    73 15-03-2013  18:52
    • Junior Member
    • Posts: 63
    Re: STEP 2005 Solutions Thread
    (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 ?
  14. davros's Avatar
    74 15-03-2013  21:58
    • Overlord in Training
    • Location: Skaro
    Re: STEP 2005 Solutions Thread
    (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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