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    (Original post by physicsmaths)
    Yeh hopefully shud be ok for S,1 maybe.
    I did Maths FM Physics last year got A*A*B so resitting physics(need A*) and doin further additional maths(need A*).
    For olympids just art of problem solving mate!
    And I thought I was the only one who used AoPS... Have you tried the Calculus book? It's great .
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    :ciao:

    STEP III 2014 mock:
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    Question 1. 32 min. Made it a little bit harder for myself by trying to do S_x algebra in terms of a, b, and c, but got back on track when I realised that there had to be an easier way.
    Question 2. 23 min. This would have been much tougher had I not seen essentially exactly the same question in a different STEP paper. I don't remember where, but it has come up before. I got a different final result, I think due to a sign error, so I shouldn't lose too much for that.
    Question 3. 31 min. Whatever. Depends on how the examiner is feeling.
    Question 4. 36 min. Didn't finish the last part, so 4 marks certainly gone. Also I did it the most bizarre way with integration by parts when it was simple recognition. Got a < pi/2 right, I'm so proud lol.
    Question 6. 23 min. This wasn't hard *shrugs*
    ... debating for 2 min. should I do the mechanics vectors one or should I stick with pure. *sees question 8* hmm, that starter looks ugly, if I can see what to do by staring at it a little... *jumps straight in*
    Question 8. 27 min. Nice tight and cozy inequality at the end.

    lol.
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    For question 8(i) 1,2013, why is the domain of da infinity to infinity? Thanks!
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    (Original post by Glavien)
    For question 8(i) 1,2013, why is the domain of da infinity to infinity? Thanks!
    da(x)=\sqrt{x^2}=|x|. This clearly has domain (-\infty,+\infty), since domain of a is this and the range of a is a subset of the domain of d. (In fact the range of a is the domain of d.)
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    (Original post by IrrationalRoot)
    da(x)=\sqrt{x^2}=|x|. This clearly has domain (-\infty,+\infty), since domain of a is this and the range of a is a subset of the domain of d. (In fact the range of a is the domain of d.)
    Thanks!
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    (Original post by Glavien)
    For question 8(i) 1,2013, why is the domain of da infinity to infinity? Thanks!
    You could also just argue that in general fg(x) has the domain of g(x). So da has domain of a.
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    (Original post by Zacken)
    You could also just argue that in general fg(x) has the domain of g(x). So da has domain of a.
    So in general fg(x) has the domain of g(x), but what is the domain of f(x) is restricted? Shouldn't the domain of fg(x) be the domain of f(x) as it is a tighter interval? Sorry that may be a bit confusing.


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    (Original post by Glavien)
    So in general fg(x) has the domain of g(x), but what is the domain of f(x) is restricted? Shouldn't the domain of fg(x) be the domain of f(x) as it is a tighter interval? Sorry that may be a bit confusing.


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    No, in a composite function. You feed things (x's) into g(x), not f(x) so your domain is g(x)'s domain in general.

    However, you need to ensure that g(x)'s image fits into f(x)'s domain because you put g(x) into f(x).

    In your case, can you see how a's range fits into d's domain?
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    (Original post by Zacken)
    No, in a composite function. You feed things (x's) into g(x), not f(x) so your domain is g(x)'s domain in general.

    However, you need to ensure that g(x)'s image fits into f(x)'s domain because you put g(x) into f(x).

    In your case, can you see how a's range fits into d's domain?
    a's range is a(x) =>0 which is the same as d's domian so it fits, right?
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    (Original post by Glavien)
    a's range is a(x) =>0 which is the same as d's domian so it fits, right?
    Yes.
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    (Original post by Zacken)
    You could also just argue that in general fg(x) has the domain of g(x). So da has domain of a.
    No in general the domain of fg(x) is the largest subset S of of the domain of g such that the image of S under g is a subset of the domain of f; that's why my explanation wasn't so simple.
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    (Original post by Zacken)
    No, in a composite function. You feed things (x's) into g(x), not f(x) so your domain is g(x)'s domain in general.

    However, you need to ensure that g(x)'s image fits into f(x)'s domain because you put g(x) into f(x).

    In your case, can you see how a's range fits into d's domain?
    (Original post by IrrationalRoot)
    No in general the domain fg(x) is the largest subset S of of the domain of g such that the image of S under g is a subset of the domain of f; that's why my explanation wasn't so simple.
    ..
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    (Original post by Zacken)
    ..
    "You could also just argue that in general fg(x) has the domain of g(x). So da has domain of a."
    "so your domain is g(x)'s domain in general."

    were not correct so had to correct them. Anyway, what I said is (somewhat subtly) different to the statement
    'it's the domain of g(x) as long as you make sure the image is part of the domain of f',
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    (Original post by IrrationalRoot)
    "You could also just argue that in general fg(x) has the domain of g(x). So da has domain of a."
    "so your domain is g(x)'s domain in general."

    were not correct so had to correct them. Anyway, what I said is (somewhat subtly) different to the statement
    'it's the domain of g(x) as long as you make sure the image is part of the domain of f',
    Yeah, I agree with all that. I was just trying to go for brevity and informalness. :lol:
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    (Original post by Zacken)
    Yeah, I agree with all that. I was just trying to go for brevity and informalness. :lol:
    No worries, you can probs tell by now that I'm super pedantic with technical stuff .
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    (Original post by IrrationalRoot)
    No worries, you can probs tell by now that I'm super pedantic with technical stuff .
    It's good for keeping me on my toes!
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    STEP II 2014 Q 13:
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    One thing I didn't quite understand about this question. You get the first identity by summing P(X=r) from 0 to infinity (where the terms greater than n+1 are all zero). But the expression for P(X=r) is only necessarily valid for r <= n+1 (and as far as I can see although it is valid for r = n+2 it isn't zero (so it's wrong) for r > n+2), so can it be used to prove the identity?

    While I'm at it, STEP II 2004 Q 13: (back in the good old days when there were three stats questions):

    For the last part -- "Given that the cook stirs 6 sixpences..." -- a simple counting argument (all ways of distributing coins are equally likely, 9 ways in total, one is a success) suggests the answer is 1/9, but that isn't correct. What am I doing wrong?
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    (Original post by sweeneyrod)
    STEP II 2014 Q 13:
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    One thing I didn't quite understand about this question. You get the first identity by summing P(X=r) from 0 to infinity (where the terms greater than n+1 are all zero). But the expression for P(X=r) is only necessarily valid for r <= n+1 (and as far as I can see although it is valid for r = n+2 it isn't zero (so it's wrong) for r > n+2), so can it be used to prove the identity?

    While I'm at it, STEP II 2004 Q 13: (back in the good old days when there were three stats questions):

    For the last part -- "Given that the cook stirs 6 sixpences..." -- a simple counting argument (all ways of distributing coins are equally likely, 9 ways in total, one is a success) suggests the answer is 1/9, but that isn't correct. What am I doing wrong?
    For the 2014 question,

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    The question's ambiguous, but I'm fairly sure they're just summing from 1 to n.


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    physicsmaths,

    How much easier are older papers compared to newer ones? I did 1998 III, and I got 96/120 but not sure if this is S or Grade 1. What do you think might be the grade boundaries for it? You should consider that Q1, 2 and 9 are literally a 10 minute job for current STEP III candidates (not sure about then though, since I don't know if standard M.I techniques were in M5 back in the old days).

    Also, has anyone tried 1998 III (IrrationalRoot since I know you do 90s papers)? We can discuss if you want, I found 3 very difficult and Question 7 I couldn't spot the binomial expansion and how to pretty much integrate b(x) at the end. I did 1, 2, 3, 4, 7, 9, 10.
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    (Original post by Krollo)
    For the 2014 question,
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    The question's ambiguous, but I'm fairly sure they're just summing from 1 to n.

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    In that case, they should've put in the last term! I think you're right, otherwise the sum wouldn't actually be correct, if it was a sum to infinity I don't think it would have the value given.
 
 
 
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