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STEP Maths I, II, III 1994 Solutions

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Got a question about Student Finance? Ask the experts this week on TSR! 14-09-2014
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    That works
    (Though for the first part, you can do it backwards.
    \frac{1-\cos{\alpha}}{\sin \alpha} = \frac{1-(1-2\sin^2 \frac{1}{2}\alpha)}{2 \sin{\frac{1}{2}\alpha}{\cos{\fr  ac{1}{2}\alpha}}} = \frac{\sin \frac{1}{2}\alpha}{\cos{\frac{1}  {2}\alpha}}
    Which is a little easier.)
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    (Original post by Rabite)
    That works
    (Though for the first part, you can do it backwards.
    \frac{1-\cos{\alpha}}{\sin \alpha} = \frac{1-(1-2\sin^2 \frac{1}{2}\alpha)}{2 \sin{\frac{1}{2}\alpha}{\cos{\fr  ac{1}{2}\alpha}}} = \frac{\sin \frac{1}{2}\alpha}{\cos{\frac{1}  {2}\alpha}}
    Which is a little easier.)
    Not only easier, but to be honest, I'd be a little antsy about all the square roots in the other method. You really need to prove you don't need to worry about all the implict \pm signs.
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    (Original post by nota bene)
    STEP III Q5

    So we have(1-0)^2f^{2k+3}(0)-(2k+1)^2f^{2k+1}(0)=0. Meaning: f^{2(k+1)+1}(0)=(2k+1)^2f^{2k+1} Where k= 0, 1, 2...
    So the coefficients of the terms are
    a_1=1 \newline a_3=\frac{1}{3!} \newline a_5=\frac{3^2}{5!}=\frac{3}{40} \newline a_7=\frac{5^2\times3^2}{7!}=\fra  c{5\times 3}{7\times6\times4\times2}
    etc. so the MacLaurin expansion is arcsin(x)=x+\frac{1}{6}x^3+\frac  {3}{40}x^5+\frac{5}{196}x^7+...


    To find the power series expansion forg(x)=Ln(\sqrt{\frac{1+x}{1-x}}) we can see it as \frac{1}{2}[Ln(1+x)-Ln(1+(-x))] for which we can use the McLaurin series for Ln(1+x).
    Ln(1+x)= x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{x^n}{n}
    So: \frac{1}{2}[(x - \frac{x^2}{2}+\frac{x^3}{3}+...+  (-1)^{n+1}\frac{x^n}{n})- (-x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{(-x)^n}{n})]
    Even terms cancel and leaves \frac{1}{2}\times 2 (x+\frac{x^3}{3}+\frac{x^5}{5}+.  ..+\frac{x^{2n+1}}{2n+1})
    This means the the power series for g(x) is \displaystyle\sum_{n=0}^{\infty}  \frac{x^{2n+1}}{2n+1} and thus the coefficient is \frac{1}{2n+1}
    Comparing coefficients of the series for f(x) and g(x) we can see that g(x)>f(x). For example the coefficients of x^7 which are\frac{1}{7}>\frac{5\times 3}{7\times6\times4\times2} for g(x) and f(x) respectively.


    Do I need a better justification for the coefficients of g(x) being larger? I find it a bit hard to express a general term for the power series of f(x). (well I can get it off wikipedia obviously but I have no idea how to get there and especially how to see it during a test).
    As an alternative, earlier you've determined that

    f^{2(k+1)+1}(0)=(2k+1)^2f^{2k+1} Where k= 0, 1, 2...
    or
    f^{2(k+1)+1}(0)=(2k+1)(2k+1)f^{2  k+1} Where k= 0, 1, 2...

    Then note that (by comparing the maclaurin series of g(x) that you've found with the g(x)=g(0)+g'(0)x+\frac{g''(0)x^2  }{2!}+\dots:

    g^{2(k+1)+1}(0)=(2k+1)(2k+2)g^{2  k+1} Where k= 0, 1, 2...


    Since g(0)=f(0), it soon follows that for k>1,

    \frac{g^{2k+1}(0)}{(2k+1)!}>\fra  c{f^{2k+1}(0)}{(2k+1)!}

    i.e the coefficients of maclaurin series of g(x) is greater than that of f(x) except for the first term.

    And nota bene, you've done a great job for this question.
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    STEP III
    (ds/dt)^2 + 2gy must always be constant as this equation forms the energy equation (incl. kinetic and GP) multiplied by 2 and divided by m, which is a constant. this equation must be a constant as no external forces are applied to the particle so energy is conserved.

    (ds/dt)^2 + 2gk^(-1)(s/2)^2=c
    so:
    2(ds/dt)(d^2s/dt^2) + (sg/k)(ds/dt)=0
    => 2(d^2s/dt^2) + (sg/k)=0
    => (d^2/dt^2)= -(g/2k)s
    so angular speed= (g/2k)^0.5
    so period= 2pi(2k/g)^(0.5)
    time taken to reach V is 0.25 of period= pi(k/2g)^(0.5)
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    (Original post by khaixiang)
    g^{2(k+1)+1}(0)=(2k+1)(2k+2)g^{2  k+1} Where k= 0, 1, 2...

    |...|

    And nota bene, you've done a great job for this question.
    I knew it was something that I missed! Thanks, that seems to be the 'simple' way of doing it...
    I quite liked this question, it doesn't really have anything complicated in it...:confused: Although it turned out to be a bit long, but not too bad.


    *bobo* are you going to add the finishing off on that question? (just looked and there is something with "describe the motion" of the m and m\alpha thingies...). Good job either way
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    yes just had to go for my tea thoguh
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    if the particles are perfectly elastic then no energy should be lost in the collision, hence the total energy of the particles must be equal to 2ghk(1 + alpha^2). This was deduced by forming the energy equations of each particle and the constants.
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    STEP II
    9) centre of mass from B is a distance (2a/3)
    let A be the angle between OA and OB
    bsinA/sin(90-0.5A)=2a
    =>bcos0.5Asin0.5A=acos0.5A
    =>sin0.5A=a/b

    O must be directly above centre of mass so:
    bsin(90-0.5A)/costheta + (2a/3)sintheta= bsin(90-0.5A+ theta)
    =>(b^2- a^2)^0.5/costheta +(2a/3)sintheta= b(cos0.5Acostheta + sin0.5Asintheta)
    =(b^2-a^2)^0.5costheta + asintheta
    => (a/3)sintheta= (b^2-a^2)^0.5sin^2theta/costheta
    =>tan theta= a/(3(b^2-a^2)^0.5)
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    moments aboutB:
    mgcos theta= T(b^2 -a^2)^0.5/b

    sin theta= costheta a/ (3(b^2-a^2)^0.5
    1-cos^2theta= (acos^2theta)^2/ (9(b^2-a^2)

    => 1- T^2(b^2-a^2)/(mgb)^2= a^2T^2/9(mgb)^2
    =>T^2= 9(mbg)^2/(9b^2-8a^2)
    T= 3mgb/(9b^2-8a^2)^0.5
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    STEP II
    questions 9 and 10
    Attached Files
  1. File Type: doc STEPII q9and10.doc (626.0 KB, 163 views)
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    q 11
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  2. File Type: doc Doc4.doc (508.0 KB, 124 views)
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    11 cont. couldnt upload full attachment
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  3. File Type: doc Doc5.doc (681.0 KB, 83 views)
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    and 3rd part to 11
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  4. File Type: doc Doc6.doc (509.0 KB, 78 views)
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    STEP I Q 12

    i) \frac{2}{28}=\frac{1}{14}

    ii) \frac{1}{28}\times1\times1+\frac  {27}{28}\times \frac{1}{27}\times1+\frac{27}{28  }\times\frac{26}{27}\times1=\fra  c{3}{28}

    iii) 1-(probability of failure) Probability of failure: \frac{26}{28}\times\frac{25}{27}  \frac{24}{26}= \frac{50}{63} So the answer is \frac{13}{63}

    iv) 1\times \frac{1}{27}\frac{26}{26}+1\time  s \frac{26}{27} \times\frac{1}{26}=\frac{2}{27}

    v) P(Newnham+New Hall|Newnham)=\frac{P(N\cap NH)}{P(N)} Then we know P(N) from ii) and P(N\cap NH)=\frac{26}{28}\frac{2}{27}\fr  ac{1}{26}+\frac{2}{28}\frac{26}{  27}\frac{1}{26}+\frac{2}{28}\fra  c{1}{27}\frac{26}{26}=\frac{6}{2  7\times28}
    From that follows \frac{P(N\cap NH)}{P(N)}=\frac{\frac{6}{27\tim  es28}}{\frac{3}{28}}=\frac{2}{27  }

    vi) Same as above, both probabilities double and hence cancel.

    vii) P(two|one)=\frac{P(one\cap \textbb{the other})}{P(one)} Now we already have the probabilities needed, from the answer in iii) and the \frac{6}{27\times28} from v). So \frac{\frac{6}{27\times28}}{\fra  c{13}{63}}=\frac{1}{26}
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    STEP II question 1

    i)work mod 10. So the possible residues of n mod10 are 1,3,7,9. It suffices to show that there exists m mod10 such that n*m=1 (mod10) for each residue of n, so taking m=1,7,3,9 respectively gives the required result.

    ii)If n ends in 1,3,7 or 9 we have shown we can multiply by some m to give a number ending in 1. So we only have to worry about n ending in 0,2,4,5,6,8,10. Now, clearly our number is divisible by 2 or 5, so suppose it is divisible by 2 and 5 a certain number of times each. All we have do is find an m containing the correct number of 2s or 5s such that the product n*m contains the same number of factors of 2 and 5. This means that n*m is divisble by 10 this number of times, and furthermore it is not divisible by 2 or 5 any further and hence the last digit is either 1,3,7 or 9. Applying the first part again gives the required result.

    iii)Suppose n is a k-digit number as described. Using the number m=900..008000....00020..0010, where there are at least k+1 zeros between the adjacent 9,8,..,2,1 will work.
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    STEP II question 6

    Proof by induction. For n=1:
    RHS =\frac{1}{2}\cot\frac{\theta}{2} - \cot\theta
    \frac{1}{2} \cot\frac{\theta}{2} - \frac{1-\tan^2\frac{\theta}{2}}{\tan\fra  c{theta}{2}}
     \frac{1}{2} \tan\frac{theta}{2} = LHS, as required

    Assume the given formula holds. So, for n+1,
    \frac{1}{2} \tan\frac{\theta}{2} + \frac{1}{2^2} \tan\frac{\theta}{2^2} + ... + \frac{1}{2^{n+1}} \tan\frac{\theta}{2^{n+1}}}
    = \frac{1}{2^n}}\cot\frac{\theta}{  2^n} - \cot\theta + \frac{1}{2^{n+1}}} \tan\frac{\theta}{2^{n+1}}
    = \frac{1}{2^n}(\frac{1-tan^2\frac{\theta}{2^{n+1}}}{2ta  n\frac{\theta}{2^{n+1}}} + \frac{1}{2}\tan\frac{\theta}{2^{  n+1}}}) - \cot\theta
    =\frac{1}{2^{n+1}}\cot\frac{\the  ta}{2^{n+1}}} - \cot\theta, completing the proof by induction.

    For the last part, note that as n-> infinity, the angle we are taking the cot of gets very small, and as sin(theta)-->theta and cos(theta)-->1, we can write the RHS for large n as:
    \frac{1}{2^n} \frac{2^n}{\theta} - \cot\theta = \frac{1}{\theta} - \cot\theta, as required.
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    STEP I - Question 11

    i)

    To ensure the ball hits the ground, it's mass must be greater than the parallel component of the wagon:

    M>mgsin\theta

    By considering both bodies separately, there will be a tension pulling the wagon up the slope and a tension resisting the ball from falling, by applying N2L, firstly to the wagon:

    T-mgsin\theta=ma_c

    and to the ball:

    Mg-T=Ma_c

    Adding these gives:

    Mg-mgsin\theta=a_c(M+m)

    So:

    a_c=\frac{g(M-msin\theta)}{M+m}

    By taking the length of the slope to be: s=dsin\theta
    And applying the fact the system starts from rest,
    Suvat can be applied:

    v^2=u^2+2a_cs

    v^2=0+2\frac{g(M-msin\theta)}{M+m}dsin\theta

    v^2=\frac{2g(M-msin\theta)dsin\theta}{M+m}

    As required.

    ii)

    If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:

    v^2\le0

    hence:

    \frac{2g(M-msin\theta)dsin\theta}{M+m}\le0

    sin\theta(M-msin\theta)\le0

    So:

    Provided sin\theta\ge0

    M-msin\theta\le0

    QED (I hope)

    I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great.
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    I would be interested if anybody could show how to do STEP I Q8. I get the impression that you must use the rather-too-coincidental limits to evaluate the integrals, but I cannot see how. Can anybody help?
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    Hmm i cant see where to go on this the change in variable doesnt really help at all.

     \displaystyle \int_0^{\frac{\pi}{4}} ln (1+tan\theta) \hspace5 d\theta

     \theta = \frac{\pi}{4} - \phi

     \frac{d\theta}{d\phi} = -1

     d\theta = -d\phi

     \displaystyle -\int_{\frac{\pi}{4}}^0 ln (1 + tan (\frac{\pi}{4} - \phi) ) \hspace5 d\phi

     \displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(1 + \frac{tan \frac{\pi}{4} - tan\phi}{1 + tan\frac{\pi}{4}tan\phi}\right)

     \displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(\frac{1 + tan\phi + 1 - tan\phi}{1 + tan\phi}\right)

     \displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(\frac{2}{1+tan\phi}\right)

     \displaystyle \int_{\frac{\pi}{4}}^0 ln \left(\frac{1+tan\phi}{2}\right)
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    I = INT (0 to pi/4) ln(1+tanx) dx
    let y = pi/4 - x
    dy/dx = -1

    -> - INT (pi/4 to 0) ln(1+tan(pi/4 - y)) dy
    = INT (0 to pi/4) ln(1+ tan(pi/4 - y)) dy
    = INT ln(1 + [1-tany/1+tany]) dy
    = INT ln(2/1+tany) dy
    I = INT ln(2/1+tanx) dx
    Add original integral:
    2I = INT (0 to pi/4) ln(1+tanx) + ln(2/1+tanx) dx
    = INT ln2 dx
    I = (1/2) INT (0 to pi/4) ln2 dx
    = (1/2)(pi/4)ln2

    = pi.ln2/8


    Tell me if you want me to do the other parts too.

    EDIT: glancing at the other parts, the second one looks like x = tanu transforms it to the original integral or very close, the last one is a y = pi/2 - x substitution, noting that sin and cos can be swapped over with the given limits, and applying a similar trick to get a simple integral (adding the original integral).

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