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STEP maths I, II, III 1992 solutions

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    (Original post by *bobo*)
    for some reason my computer won't let me access paper III anymore, says its damaged? :confused:
    Odd (do the other papers work?)

    I can email a copy if you think that would help...
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    all the other papers work. strange.
    please do
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    (Original post by *bobo*)
    for some reason my computer won't let me access paper III anymore, says its damaged? :confused:
    It could be your cache that's got damaged. Try clearing it then opening the paper again.
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    STEP II
    11)
    if the purse scrapes the edge of the cliff, it reaches its max point at 0.5d from edge of cliff.
    vertical movement:
    u=VsinA, v=0, a=-g
    => s= (VsinA)^2/(2g)

    horizontl movement after max point:
    v=VcosA, s=0.5d => t=d/(2VcosA)
    vertical movement after max point:
    s=((VsinA)^2/(2g))-h, u=0, a=g, t= d/(2VcosA)

    ((VsinA)^2/(2g))-h= 0.5g(d/(2VcosA))^2

    => 8V^4sin^4A- 8V^2(2gh+V^2)sin^2A + 2g(gd^2 + 8V^2h)=0
    sin^2A= (8V^2(2gh+V^2) +/- (64V^4(V^4+4gh+4g^2h^2)-64gV^4(gd^2+8V^2h))^0.5)/16V^4

    for the purse to be able to be thrown in the situations given, sinA must exist so:
    (V^4 +4ghV^2 +4g^2h^2)>= g^2d^2 + 8ghV^2

    (V^4- 4ghV^2 +4g^2h^2)>=(gd)^2
    (V^2-2gh)>=gd
    V^2>= g(2h +d)
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    STEP I Question 7(This question was partially completed on post #4)

     g(x) = ax + b

     g(0) = b

     g(1) = a+b

     g(n) = an+bn

     ng(1) = an + bn

     g(n) = ng(1) - bn + b

     g(n) = ng(1) -n(g(0)) + g(0)

    if g(1) and g(0) for any given value of n, g(n) is an integer.

     f(x) = Ax^2 + Bx + C

     f(-1) = A - B + C

     f(0) = C

     f(1) = A + B + C

     f(n) = An^2 + Bn + C

     f(n) = \frac{1}{2}n^2(f(-1) + f(1) - 2f(0)) + \frac{1}{2}n(f(1)-f(-1))  + f(0)

    Therefore because f(-1),f(0) and f(1) are integers

    Now for the bit that harder

     f(\alpha) = A\alpha^2 + B\alpha + C

     f(\alpha - 1) = A\alpha^2 - 2A\alpha + A + B\alpha - B + C

     f(\alpha + 1) = A\alpha^2 + 2A\alpha + A + B\alpha + B + C

     f(\alpha + n) = A\alpha^2 - 2A\alpha n + n^2 + Ba + Bn + C

    Take the first term Aalpha^2

     f(\alpha) - f(\alpha+1) + [f(\alpha) - f(\alpha-1)] = A

     A = \frac{1}{2}[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]

     A\alpha^2 = \frac{1}{2}\alpha^2(2f(\alpha) - f(\alpha+1) - f(\alpha-1))

    next term -2A\alpha n

     -2A\alpha n = -\alpha n[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]

    Next two easier terms Ba + C

     B\alpha + C = f(\alpha) - A\alpha^2 = f(\alpha) - \frac{1}{2}\alpha^2[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]

    Next term is Bn

     Bn = \frac{1}{2}n[ (-f(\alpha+1)  - f(\alpha-1)) ] - 2\alpha ( 2f(\alpha) - f(\alpha +1) - f(\alpha -1) )]

    So!

     f(\alpha + n) = A\alpha^2 - 2A\alpha n + n^2 + B\alpha + Bn + C

     =  \frac{1}{2}\alpha^2(2f(\alpha) - f(\alpha+1) - f(\alpha-1)) -\alpha n[2f(\alpha) - f(\alpha+1) - f(\alpha-1)] + n^2 + f(\alpha) - \frac{1}{2}\alpha^2[2f(\alpha) - f(\alpha+1) - f(\alpha-1)] + \frac{1}{2}n[ (-f(\alpha+1)  - f(\alpha-1)) ] - 2\alpha ( 2f(\alpha) - f(\alpha +1) - f(\alpha -1) )]

    As \alpha and all the functions involving alpha are integers therefore f(\alpha + n) must be an integer.

    Id tidy that up too but i really cant be arsed at the moment.

    What a bitch that was. I bet theres a much more elegant way. Ive most definitely made an error in that but the method is there. You just have to find the function of f(\alpha + n) in terms of functions that are known integers.
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    STEP I Question 8

    Calling F(t) the integrated function of f(t)

     \frac{d}{dx}  \int_a^x f(t) \hspace5 dt  = \frac{d}{dx}[F(t) ]_a^x

     = \frac{d}{dx} [ F(x) -  F(a) ]

     = \frac{d}{dx}(F(x)) = f(x)

    (thanks to Billy (GE) for the cleaning up the tizwozz i got into about the third and last part)
    Consider
     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + \text{const} .
    You can just differentiate both sides from here and get
    f'(x) = f(x) \Rightarrow f(x) = Ae^x
    and plug this into the integrals in the second and third part:
     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + 1
     Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt + 1
     Ae^x = Ae^x - A + 1 \Rightarrow A = 1

     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt
     Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt
     Ae^x = Ae^x - A \Rightarrow A=0

    Just attempting the final and 4th part now.
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    (Original post by insparato)
    STEP I Question 8

     \frac{d}{dx} \displaystyle \int_a^x f(t) \hspace5 dt  = \frac{d}{dx}[ \int f(t) ]_a^x

     = \frac{d}{dx} [\int f(x) - \int f(a) ]

     = \frac{d}{dx}(\int f(x)) = f(x)

    if

     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + 1
    Consider
     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + \text{const} .
    You can just differentiate both sides from here and get
    f'(x) = f(x) \Rightarrow f(x) = Ae^x
    and plug this into the integrals in the second and third part:
     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + 1
     Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt + 1
     Ae^x = Ae^x - A + 1 \Rightarrow A = 1

     f(x) = \displaystyle \int_0^x f(t) \hspace5 dt
     Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt
     Ae^x = Ae^x - A \Rightarrow A=0.
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    I was mucking about with this yesterday but i must of got myself in a twist becaues i did get A = 1, or something of this nature but didnt know where it led to.

    Yes that seems reasonable.

    Argh i just realised i had a stupid idea that f(x) was on both sides on third equation which obviously doesnt make sense.
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    (Original post by insparato)
    What a bitch that was. I bet theres a much more elegant way.
    There is. (I posted it earlier, actually).

    Put g(x) = f(x+\alpha). Then g is a quadratic, and g(-1),g(0) and g(1) are all integers. So g(n) is an integer for all n. Therefore so is f(n+\alpha).
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    :rolleyes: Nevermind, i thought nobody had finished that last part off as it was up as incomplete.
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    To my knowledge no one had completed it. Which post did you complete it in, David?
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    (Original post by generalebriety)
    To my knowledge no one had completed it. Which post did you complete it in, David?
    I gave a (spoilered) hint in post #40. Looking back at it, it's not actually right (sign of alpha is wrong), but it should have been enough I think.
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    (Original post by insparato)
     f(x) = Ax^2 + Bx + C

     f(-1) = A - B + C

     f(0) = C

     f(1) = A + B + C

     f(n) = An^2 + Bn + C

     f(n) = n^2(f(-1) + f(1) - 2f(0)) + \frac{1}{2}n(f(1)-f(-1))  + f(0)
    Just a slight pedantic note about A:

    f(-1) + f(1) = A - B + C + A + B + C = 2A + 2C
    f(-1) + f(1) - 2f(0) = 2A + 2C - 2C = 2A

    So what you've actually got there is 2An^2 not An^2. You'll have to divide by 2. I'm just saying this in case in a few years time someone sees your solution and thinks their one is wrong.
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    Ooh, ok, so no one has actually completed it yet? :p:
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    (Original post by generalebriety)
    Ooh, ok, so no one has actually completed it yet? :p:
    What, you mean other than insparato's solution about 10 posts earlier and my shorter solution for the end bit? (Or are we talking about different questions? For clarity, I am talking about STEP I, Q7).
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    (Original post by DFranklin)
    What, you mean other than insparato's solution about 10 posts earlier and my shorter solution for the end bit? (Or are we talking about different questions? For clarity, I am talking about STEP I, Q7).
    As I understood, Aaron's solution had a mistake (pointed out by DeathAwaitsU) and yours was just a hint. I don't know though, you tell me.
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    It's solved. I was just pointing out a minor mistake and David was explaining a more elegant solution. Regardless, Insparato finished off the bit I didn't do so the question is complete (assuming his solution was correct - I don't have the strength right now to read it).
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    Ok, well, I'll leave the link up there and Aaron is free to correct his minor mistake if he wants. I shall add a link to David's post and everyone can then fight it out amongst themselves.
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    (Original post by generalebriety)
    Ok, well, I'll leave the link up there and Aaron is free to correct his minor mistake if he wants. I shall add a link to David's post and everyone can then fight it out amongst themselves.
    Yet another solution:

    g(n) = g(0)+n(g(1)-g(0)), so g(n) is always an integer.

    Now set g(n) = f(n)-f(n-1) = A(2n-1)+B. g(0) and g(1) are both integers and so g(n) is always an integer. Since f(0) is an integer it follows that f(n) is always an integer. (This is obvious, but if you want to prove it, suppose not, and let n be the integer of smallest modulus such that f(n) is not an integer. We know f(0) is an integer, so n =/= 0. If n > 0, we have f(n-1) is an integer, and so f(n) = f(n-1)+g(n) is an integer. If n < 0 we have f(n+1) is an integer and so f(n) = f(n+1)-g(n+1) is an integer).

    Finally, consider h(x) = f(x+\alpha). Since h is a quadratic and h(-1),h(0),h(1) are all integers, h is an integer for all n. Therefore f(x+\alpha) is an integer for all n.
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    (Original post by generalebriety)
    Ok, well, I'll leave the link up there and Aaron is free to correct his minor mistake if he wants. I shall add a link to David's post and everyone can then fight it out amongst themselves.
    Handbags at dawn?

    Yeah ill correct it, i should really start writing these question on paper instead. Although its not elegant.

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