STEP Maths I, II, III 1994 Solutions

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  1. insparato's Avatar
    61 20-05-2007  03:03
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    Re: STEP Maths I, II, III 1994 Solutions
    For the second integral.

    make the substitution

    x = tan(theta)

    then you can make the same substitution from the first one and go from there.
    Last edited by insparato; 20-05-2007 at 03:30.
  2. Speleo's Avatar
    62 20-05-2007  03:05
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    Re: STEP Maths I, II, III 1994 Solutions
    Are you sure? Shouldn't it be tan theta?
  3. nota bene's Avatar
    63 20-05-2007  03:05
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    Re: STEP Maths I, II, III 1994 Solutions
    STEP I Q 13

    i) X=X_1+X_2+...+X_n since X_i=1 (and 0 otherwise) if a red is removed and X denotes the number of reds removed.

    ii) E(X_i)=\frac{m}{M}

    iii) E(X)=\frac{nm}{M} This can be seen by thinking it is a mean when we draw n objects of which have a probability p to be red. This p is then determined by (total number of red)/(total number of objects) i.e. the answer in ii)
    edit: As DFranklin points out the following should probably be said to make it formally valid:
    E(X) = \displaystyle\sum_1^n E(X_i) = \displaystyle\sum_1^n \frac{m}{N} = \frac{mn}{N}
    iv) The probability will be (number of way to get our selection)/(total number of selections). The total number of selections of n objects from M is \begin{pmatrix} M \\ n \end{pmatrix}. Now we are concentrating on how we can choose k red from m red. Let k=number of red drawn. Then we can chose \begin{pmatrix} m \\ k \end{pmatrix} and the ways non-reds can be chosen from our sample is naturally \begin{pmatrix} M-m \\ n-k \end{pmatrix} (as M-m is the number of non-reds and n-k is the number of chosen non-reds).

    Then follows P(x=k)=\frac{\begin{pmatrix} m \\ k \end{pmatrix} \begin{pmatrix} M-m \\ n-k \end{pmatrix}}{\begin{pmatrix} M \\ n \end{pmatrix}}

    v) Deduce that \displaystyle\sum_{k=1}^n k\begin{pmatrix} m \\ k \end{pmatrix}\begin{pmatrix} M-m \\ n-k \end{pmatrix}=m\begin{pmatrix} M-1 \\ n-1 \end{pmatrix}
    Uhm, I'm giving in for sleep I think, can't for my life realise how to deduce part v) right now

    (Maybe iii) needs to be far more rigorously proven, if so I'd be happy if someone did it)
    Last edited by nota bene; 22-05-2007 at 16:29.
  4. insparato's Avatar
    64 20-05-2007  03:29
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by Speleo)
    Are you sure? Shouldn't it be tan theta?
    Yes, damn people for being up at this ungodly hour. I was just laying in bed and thinking that i gave the wrong substitution. I did have the tan theta substitution written down on the paper :p:.

    Ah! Frustrating, i was thinking of the 2I thing but i didnt really know whether i could do that. Its just an angle isnt i guess.
    Last edited by insparato; 20-05-2007 at 03:31.
  5. DFranklin's Avatar
    65 20-05-2007  08:32
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by insparato)
    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

    \displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(\frac{2}{1+tan\phi}\right)
    = \int_0^{\pi/4}\ln 2 \,d\phi - \int_0^{\pi/4} \ln(1+\tan \phi) \,d\phi

    The other well known integral where you can play a similar trick is

    \int_0^{\pi/2} \ln \sin \theta \,d\theta

    Spoiler:
    Show
    Prove first that
    \int_0^{\pi/2} \ln \sin \theta \,d\theta = \int_0^{\pi/2} \ln \cos \theta \,d\theta = \int_0^{\pi/2} \ln \sin 2\theta \,d\theta
  6. DFranklin's Avatar
    66 20-05-2007  08:38
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by nota bene)
    iii) E(X)=\frac{nm}{M} This can be seen by thinking it is a mean when we draw n objects of which have a probability p to be red. This p is then determined by (total number of red)/(total number of objects) i.e. the answer in ii)

    (Maybe iii) needs to be far more rigorously proven, if so I'd be happy if someone did it)
    I think you are supposed to use the fact the expectations can be summed, so E(X) = \sum_1^n E(X_i) = \sum_1^n \frac{m}{N} = \frac{mn}{N}

    v) Deduce that \displaystyle\sum_{k=1}^n k\begin{pmatrix} m \\ k \end{pmatrix}\begin{pmatrix} M-m \\ n-k \end{pmatrix}=m\begin{pmatrix} M-1 \\ n-1 \end{pmatrix}
    Uhm, I'm giving in for sleep I think, can't for my life realise how to deduce part v) right now
    Use your expression for P(X=k) to find a different expression for E(X) and then compare with your answer to (iii).
  7. DFranklin's Avatar
    67 20-05-2007  08:54
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by zrancis)
    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:
    I don't think this is right. I think the scenario you are supposed to be considering is "ball falls towards floor, accelerating wagon. Ball hits floor, wagon carries on under it's own momentum, (but now has gravity providing -ve acceleration). Does wagon have enough momentum to hit top?"

    (Incidentally, the condition M > m sin \theta is basically the condition that the ball is heavy enough to accelerate the wagon in the first place).
  8. Kolya's Avatar
    68 20-05-2007  12:28
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by Speleo)
    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
    Thanks for that.
    Last edited by Kolya; 20-05-2007 at 12:29.
  9. Speleo's Avatar
    69 20-05-2007  13:16
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    Re: STEP Maths I, II, III 1994 Solutions
    No probs.
  10. ad absurdum's Avatar
    70 20-05-2007  13:30
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by Speleo)
    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).
    I suppose I might aswell type the last parts up fully, but I think I done the very last part differently.

    \int_0^1 \frac{ln(1+x)}{1+x^2} dx
    Let x=\tan\theta, so \frac{d\theta}{dx} = \frac{1}{1+x^2}, and the limits go from 0->0 and 1->pi/4, and we have:
    \int_0^1 \frac{ln(1+tan\theta)}{1+x^2} (1+x^2) d\theta
    \int_0^1 ln(1+tan\theta) d\theta = \frac{\pi}{8} ln2

    \int_0^{\frac{\pi}{2}} ln(\frac{1+\sin x}{1 + \cos x}) dx
    =\int_0^{\frac{\pi}{2}} ln(1+sinx) dx - \int_0^{\frac{\pi}{2}} ln(1+cosx) dx
    I thought that here I could have said this was zero, but just to make absolutely sure, for the first interal, let \sin x=\tan\theta, so dx = \frac{\sec^2\theta d\theta}{\sqrt{1-\tan^2\theta}} and the limits go from 0->0 and pi/2->pi/4. For the second integral, let \cos x=\tan\theta, so dx = - \frac{\sec^2\theta d\theta}{\sqrt{1-\tan^2\theta}}, and the limits go from 0->pi/4 and 1->0, giving:
    \int_0^{\frac{\pi}{4}} \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta + \int_{\frac{\pi}{4}} ^0 \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta
    =\int_0^{\frac{\pi}{4}} \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta - \int_0 ^{\frac{\pi}{4}} \frac{ln(1+\tan\theta) \sec^2\theta}{\sqrt{1-\tan^2\theta}} d\theta = 0
  11. zrancis's Avatar
    71 20-05-2007  14:57
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by DFranklin)
    I don't think this is right. I think the scenario you are supposed to be considering is "ball falls towards floor, accelerating wagon. Ball hits floor, wagon carries on under it's own momentum, (but now has gravity providing -ve acceleration). Does wagon have enough momentum to hit top?"

    (Incidentally, the condition M > m sin \theta is basically the condition that the ball is heavy enough to accelerate the wagon in the first place).
    I did think that about part ii, I sort of stumbled through it! I'll try take another look at it today.
  12. Speleo's Avatar
    72 20-05-2007  15:08
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    Re: STEP Maths I, II, III 1994 Solutions
    Last part - easy way.
    I = INT (0 to pi/2) ln(1+sinx) - ln(1+cosx) dx
    Let y = pi/2 - x. sin(pi/2 - y) = cosy; cos(pi/2 - y) = siny.
    dy/dx = -1 and limits swapped, therefore no change in limits overall.
    I = INT (0 to pi/2) ln(1+cosy) - ln(1+siny) dy
    Add:
    2I = INT (0 to pi/2) 0 dx = 0; I = 0.
  13. ad absurdum's Avatar
    73 20-05-2007  16:21
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by Speleo)
    Last part - easy way.
    I = INT (0 to pi/2) ln(1+sinx) - ln(1+cosx) dx
    Let y = pi/2 - x. sin(pi/2 - y) = cosy; cos(pi/2 - y) = siny.
    dy/dx = -1 and limits swapped, therefore no change in limits overall.
    I = INT (0 to pi/2) ln(1+cosy) - ln(1+siny) dy
    Add:
    2I = INT (0 to pi/2) 0 dx = 0; I = 0.
    Yeah that is better I think I was too keen to use the result shown in the first part.

    STEP I question 10

    Let e be the extension of the string
    Resolve vertically: \frac{\lambda e}{l}\cos\theta = mg
    Resolve horizontally: \frac{\lambda e}{l}\sin\theta = mr\omega^2
    But, \sin\theta = \frac{r}{l+e}
    so, \frac{\lambda e r}{l(l+e)} = mr\omega^2
    Rearranging, e = \frac{ml^2\omega^2}{\lambda - ml\omega^2}
    Subsitute into vertical equation and rearrange to get \cos\theta = \frac{gl(\lambda-ml\omega^2)}{\lambda l \omega^2}

    Now, the motion is possible only if 0<\cos\theta<1 (if \cos\theta = 1, then the string is vertical and so no angular motion is actually taking place, if \cos\theta \leq 0, then the string is horizontal or above, which is impossible due to the weight of the particle).
    So, \cos\theta > 0 \Rightarrow \lambda > m l \omega^2, so \omega^2 < \frac{\lambda}{ml}
    And, \cos\theta < 1 \Rightarrow gl(\lambda - m l \omega^2) < \lambda l \omega^2, \omega^2(gml^2 + \lambda l) > gl\lambda
    \omega^2 > \frac{g\lambda}{l(\lambda + mg)}
  14. Speleo's Avatar
    74 22-05-2007  14:43
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    Re: STEP Maths I, II, III 1994 Solutions
    Does anyone mind if I start a 1993 thread?
    This one is pretty much dead, and none of the remaining ~5 questions look all that interesting. We're doing the pre-1996 ones pretty much just for our own benefit, anyway.
  15. Rabite's Avatar
    75 22-05-2007  14:56
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    Re: STEP Maths I, II, III 1994 Solutions
    NO! You most certainly shall not! My gosh, the youth these days, trying to take everything in their own hands!
    :p:
    (I'm kidding, just in case you didn't pick that up)

    I mean uh.
    Sure that'd be great
    Let's try to go all the way to 1990. Although the papers get increasingly weird (education changes, so that's to be expected I suppose) as we go back.

    I mean, it's not like starting the 1993 means we are absolutely forbidden to add stuff to this thread, right?
    Actually, I was going to ask if I could start one for 1990 (since I've been at it recently), but I guess it's more rational to go for 1993, since you've brought that up.

    Team maths! Forward!
  16. DFranklin's Avatar
    76 22-05-2007  14:56
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by Speleo)
    Does anyone mind if I start a 1993 thread?
    This one is pretty much dead, and none of the remaining ~5 questions look all that interesting. We're doing the pre-1996 ones pretty much just for our own benefit, anyway.
    Go for it.

    On a slight aside, do you know how to do the \int_0^{\pi/2} \ln \sin x \,dx integral I mentioned? I dare say it won't come up (I've seen it on a STEP paper before but I don't recall which one), but it's a "classic" question that just might pop up again. (And if it does pop up again, it's very easy marks if you've seen it before).
  17. Speleo's Avatar
    77 22-05-2007  15:00
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    Re: STEP Maths I, II, III 1994 Solutions
    I'll make the thread.

    DFranklin, does it involve swapping the sin for cos and then combining the two forms of the integral somehow?
  18. DFranklin's Avatar
    78 22-05-2007  15:04
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    Re: STEP Maths I, II, III 1994 Solutions
    (Original post by Speleo)
    I'll make the thread.

    DFranklin, does it involve swapping the sin for cos and then combining the two forms of the integral somehow?
    Yes, but there's another "trick" as well. I've seen this question in at least 3 exams (S-level, STEP, some uni past paper), and I don't think I've ever seen it without some level of hint, usually asking you to show that

    \int_0^{\pi/2} \ln \sin x \,dx = \int_0^{\pi/2} \ln \cos x \, dx = \int_0^{\pi/2} \ln \sin 2x \, dx
  19. Speleo's Avatar
    79 22-05-2007  15:18
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    Re: STEP Maths I, II, III 1994 Solutions
    I can do it from there, I just need to prove the lnsin2x bit. I'll look at it in a bit
  20. insparato's Avatar
    80 22-05-2007  15:27
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    Re: STEP Maths I, II, III 1994 Solutions
    I had alittle go at this myself

    \displaystyle \int_0^{\frac{\pi}{2}} ln (sin\theta) \hspace5 d\theta

    \theta = \frac{\pi}{2} - \phi

    d\theta = -d\phi

    \displaystyle -\int_{\frac{\pi}{2}}^0 ln (sin \frac{\pi}{2} - \phi ) \hspace5 d\phi

    \displaystyle = \int_0^{\frac{\pi}{2}} ln (sin \frac{\pi}{2}cos\phi - cos\frac{\pi}{2}cos\phi ) \hspace5 d\phi

    \displaystyle \int_0^{\frac{\pi}{2}} ln (cos \phi) d\phi = \int_0^{\frac{\pi}{2}} ln (cos\theta) \hspace5 d\theta

    Second Integral

    \displaystyle \int_0^{\frac{\pi}{2}} ln (cos\theta) d\theta

    \theta = \frac{\pi}{2} - 2\phi

    d\theta = -2d\phi

    \displaystyle -2\int_{\frac{\pi}{4}}^{0}} ln (cos (\frac{\pi}{2}-2\phi)) \hspace5 d\phi

    \displaystyle 2\int_0^{\frac{\pi}{4}} ln (cos \frac{\pi}{2} cos 2\phi + sin \frac{\pi}{2} sin 2\phi ) \hspace5 d\phi

    \displaystyle 2\int_0^{\frac{\pi}{4}} ln (sin2\phi) \hspace5 d\phi

    then im guessing that

    \displaystyle = \int_0^{\frac{\pi}{2}} ln (sin2\theta) \hspace5 d\theta

    Total guess but .
    Last edited by insparato; 22-05-2007 at 15:52.
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