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STEP Maths I, II, III 1994 Solutions

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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.)

    (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.

    (Original post by nota bene)

    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...
    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.

    (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
    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)

    (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

    yes just had to go for my tea thoguh

    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.

    9) centre of mass from B is a distance (2a/3)
    let A be the angle between OA and OB

    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)

    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

    questions 9 and 10
    Attached Files
  1. File Type: doc STEPII q9and10.doc (626.0 KB, 279 views)

    q 11
    Attached Files
  2. File Type: doc Doc4.doc (508.0 KB, 206 views)

    11 cont. couldnt upload full attachment
    Attached Files
  3. File Type: doc Doc5.doc (681.0 KB, 159 views)

    and 3rd part to 11
    Attached Files
  4. File Type: doc Doc6.doc (509.0 KB, 148 views)

    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}

    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.

    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.

    STEP I - Question 11


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


    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:


    and to the ball:


    Adding these gives:




    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:




    As required.


    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:






    Provided sin\theta\ge0


    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.

    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?

    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)

    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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