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STEP maths I, II, III 1991 solutions

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Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
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    To be honest I don't think there'll be any demand for a 1990 thread right now. This thread's gone much slower than the 1992 thread has - I think people are losing interest in favour of *gasps* revision. But you're welcome to start one and revive it after the exams if it dies.
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    For what it's worth I've just done 1990 III/1, very tedious at the start but the last parts are interesting. It's very similar to a more recent one, except this time you have to expand up to 9th powers rather than 7.
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    Yeah I actually did that boring demoivre expansion yesterday before going to sleep, it is quite okay. I'm working through II/1 now, seems to be quite GCSE-like quadratics...

    I'm sure a few of us will post soluitins once a thread is running
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    Done III/2 aswell (surprisingly for me...) assuming that my work with matrices is right.
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    (Original post by Speleo)
    Done III/2 aswell (surprisingly for me...) assuming that my work with matrices is right.
    I didn't know you can do this question using matrices. I think this is a very nice question, particularly because I've an elegant solution. III/7 is an absolute torture! And maybe if someone has done III/5, do tell me whether the last part can be done using the previous results. And also whether there's anything wrong with the last part of I/4 if you have done it already.

    EDIT:We're talking about 1990 papers right?
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    (Original post by khaixiang)
    I didn't know you can do this question using matrices. I think this is a very nice question, particularly because I've an elegant solution. III/7 is an absolute torture! And maybe if someone has done III/5, do tell me whether the last part can be done using the previous results. And also whether there's anything wrong with the last part of I/4 if you have done it already.
    For III/5, I think you still need to work out directly that

    (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0 (hope that's right!).

    You can then use the previous result on both products to get the final result. But it's probably easier just to prove it directly using induction.
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    Just did III/5 and III/9

    III/5: I didn't use the previous results for the last part, since it's beyond easy to just do it by induction, unless I'm reading it wrong.

    III/9: Very short and easy for STEP III, especially the old ones...

    If anyone has done III/8 I would be interested how they did the very last bit. I can get to y' = f(x,y) but the integrating factor is of the form e^[lnp(x) + karctanq(x)] which is pretty messy, and I'm not even going to try to force my way to the end of the question using that method.
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    My III/2 solution for the first part (excuse the lack of boldface vectors):
    OA = m[a^3i + a^2j + ak]
    BC = b^3i + b^2j + bk + n[(b^3-c^3)i + (b^2-c^2)j + (b-c)k]

    So we want solutions to:
    m[a^3i + a^2j + ak] = b^3i + b^2j + bk + n[(b^3-c^3)i + (b^2-c^2)j + (b-c)k]
    Equating co-efficients of i, j and k:
    ma^3 = b^3 + nb^3 - nc^3
    ma^2 = b^2 + nb^2 - nc^2
    ma = b + nb - nc

    So:
    \begin{pmatrix}b^3 & b^3 & c^3 \\ b^2 & b^2 & c^2 \\ b & b & c \end{pmatrix}\begin{pmatrix} \frac{1}{m} \\ \frac{n}{m} \\ -\frac{n}{m} \end{pmatrix} = \begin{pmatrix} a^3 \\ a^2 \\ a \end{pmatrix}

    But the determinant of the matrix is 0 (if you work it out you get two terms that are - each other and one term = 0) so there are no solutions and therefore no intersections.

    However, I don't really know anything about matrices and so I'm probably making up that rule about the determinant...

    (My first latex matrix )


    EDIT: Googling for matrix latex only brings up one bdsm side on the first page, I was very surprised.
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    (Original post by Speleo)
    But the determinant of the matrix is 0 (if you work it out you get two terms that are - each other and one term = 0) so there are no solutions and therefore no intersections.

    However, I don't really know anything about matrices and so I'm probably making up that rule about the determinant...

    (My first latex matrix )
    Det=0 implies either infinitely many solutions or no solutions if I am not being silly
    E.g. 4x+5y=20 [1] and 8x+10y=40[2] has infinitely many solutions. To me your working seems okay (although I've not actually looked much at the question yet...).

    Congrats on the Latex

    edit: And maybe we should start that 1990 thread as we seem to be starting to work on them...
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    (Original post by Speleo)
    Just did III/5 and III/9

    III/5: I didn't use the previous results for the last part, since it's beyond easy to just do it by induction, unless I'm reading it wrong.

    III/9: Very short and easy for STEP III, especially the old ones...

    If anyone has done III/8 I would be interested how they did the very last bit. I can get to y' = f(x,y) but the integrating factor is of the form e^[lnp(x) + karctanq(x)] which is pretty messy, and I'm not even going to try to force my way to the end of the question using that method.
    Hmm, you might want to check your working, the integrating factor is just 1-x^3. I must admit that I wasn't very focus when I did this question, I produced 3 wrong integrating factor before I got the right one, and I am pretty sure the integrating factor is just 1-x^3

    How did you prove the iff part, you work from both ways right?
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    (Original post by Speleo)
    Just did III/5 and III/9

    III/5: I didn't use the previous results for the last part, since it's beyond easy to just do it by induction, unless I'm reading it wrong.
    Yeah, agreed, it has become almost trivial now to prove this same relation for 3 different times in 3 different papers!
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    Thanks nota, I guess then I'd have to add to my proof that infinite solutions are only possible if the two lines are the same which implies the four points are collinear, which implies (a^3i + a^2j + ak) = n(b^3i + b^2j + bk), and a little working out reveals the only way that works is if n = 0 or 1, n=0 implies A is the origin, n=1 implies A is B, but the points O, A and B are distinct so the points aren't collinear so there are no solutions.

    Thanks khaixiang, I'll check my working (probably made an algebra mistake early on).
    Py'' + Qy' + Ry = d/dx(py' + qy) iff it = py'' + (p'+q)y' + q'y, which it does, equating derivatives, iff P=p, Q=p'+q, R=q' which is does iff P''=p'', Q'=p''+q', R=q' which it does iff P'' - Q' + R = 0.
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    (Original post by Speleo)

    Thanks khaixiang, I'll check my working (probably made an algebra mistake early on).
    Py'' + Qy' + Ry = d/dx(py' + qy) iff it = py'' + (p'+q)y' + q'y, which it does, equating derivatives, iff P=p, Q=p'+q, R=q' which is does iff P''=p'', Q'=p''+q', R=q' which it does iff P'' - Q' + R = 0.
    I did like this initially but was unsure of the bold part, i.e this: P=p iff P''=p'', since when you integrate back you get constant + linear terms, P+Ax+B=p+Cx+D when you differentiate twice will also give you P''=p'' as well. So can't enforce iff in this situation?
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    (Original post by Speleo)
    Thanks nota, I guess then I'd have to add to my proof that infinite solutions are only possible if the two lines are the same which implies the four points are collinear, which implies (a^3i + a^2j + ak) = n(b^3i + b^2j + bk), and a little working out reveals the only way that works is if n = 0 or 1, n=0 implies A is the origin, n=1 implies A is B, but the points O, A and B are distinct so the points aren't collinear so there are no solutions.
    Yeah, exactly. Was quite clear that that one wouldn't have infinitely many solutions from looking at that they are not multiples of each other.

    Damn the lack of matrices I think they are about the only thing I tend to understand what to do with on step III...
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    (Original post by khaixiang)
    Hmm, you might want to check your working, the integrating factor is just 1-x^3. I must admit that I wasn't very focus when I did this question, I produced 3 wrong integrating factor before I got the right one, and I am pretty sure the integrating factor is just 1-x^3
    I think so too. Funny: all these STEP questions and this is the first one where I've had to use an integrating factor (i.e. look up "integrating factor" on Wiki, 'cos I certainly didn't remember how to do it!).
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    (Original post by khaixiang)
    I did like this initially but was unsure of the bold part, i.e this: P=p iff P''=p'', since when you integrate back you get constant + linear terms, P+Ax+B=p+Cx+D when you differentiate twice will also give you P''=p'' as well. So can't enforce iff in this situation?
    Yeah I agree, the iff chain breaks here. I find in general it is very hard to get a "iff" chain to work directly if one end of the chain is simply proving something exists.

    I can't see any iff chain that works here. Much easier to do separate proofs for each direction.
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    EDIT: On second thoughts, this post is rubbish, ignore it


    Here's my attempt, but I'd be inclined to ignore it and hope for the best

    OK, we've got the only if then, so we need to prove the if.

    If (P+cx+d)'' - (Q+e)' + R = 0, (functions P and Q have no constant or x terms), R = Q' - P''
    So y''(P+cx+d) + y'(Q+e) + y(R) =
    y''(P+cx+d) + y'(Q+e) + y(Q'-P'')
    equals
    y''(P+cx+d) + y'(P'+c+Q+e-P'-c) + y(Q'-P'')
    equals
    P'y' + Py'' + cxy'' + cy' + dy'' + Q'y + Qy' + ey' - P''y - P'y' - cy'
    equals
    d/dx[(P+cx+d)y' + (Q+e-P'-c)y]
    equals
    d/dx[py' + qy]
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    Shall I make the new thread then?
    Or does someone else really really really want to.
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    (Original post by Speleo)
    My III/2 solution for the first part
    Nice way of using matrix to do this question, the last thing on Earth I would think of is matrices.

    O(0,0,0), A(a^3,a^2,a), B(b^3,b^2,b), C(c^3,c^2,c)
    Alternatively, you can note that if a<b,c that is both B and C are further than A from the origin and are located on one side of A, then OA and BC cannot intersect. And if a>b,c then OA and BC cannot intersect as well, by the same argument.

    So we only need to consider when a<b, a>c and a<c, a>b But we actually need to verify that there's no intersection for either case since we can just swap b and c.

    line OA: \frac{x}{a^3}=\frac{y}{a^2}=\fra  c{z}{a}
    line BC: \frac{x-b^3}{b^3-c^3}=\frac{y-b^2}{b^2-c^2}=\frac{z-b}{b-c}

    From the equation of line OA, we have (the simpler one) y=az.
    From equation of line of BC, we have y=z(b+c)-bc (just take the simpler equation)

    Equating them z=\frac{bc}{b+c-a}
    Now if there's intersection between OA and BC, then this value of z must satisfy 0<z<a since the "depth" of line OA only spans from 0 to a (no equality since all points are distinct)

    So if z=\frac{bc}{b+c-a}>a then there's not intersection
    \\ \frac{bc}{b+c-a}>a\\ a^2-a(b+c)+bc>0\\ (a-b)(a-c)>0\\ a<b, a>c
    This is exactly the case we want to prove earlier, so no intersection between OA and BC. I hope I've not done anything silly :p:

    second part seems unusually short and straightforward (I think I've done this in C4), just dot the vectors and show that the dot product is positive isn't it?
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    (Original post by khaixiang)
    Alternatively, you can note that if that is both B and C are further than A from the origin and are located on one side of A, then OA and BC cannot intersect. And if then OA and BC cannot intersect as well, by the same argument.
    Might want to explain more here.

    wrt the second part: yeah, you get cos(AOB) = sqrt[f(x)] so you need to show that f(x) can be any value between 0 and 1 (which it can, it's a quadratc or a quartic, I forget which). You also need to exclude AOB = 0, pi/2. = 0 is easy because the points would be collinear which can't happen (see above), I'm not so sure about pi/2 though, cross product?

    EDIT: I hate vector/geometry questions

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Updated: June 10, 2013
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