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    (Original post by souktik)
    You mean Lagrange, right?
    No, this is Leibniz notation. Lagrange was a century later.

    EDIT: got that totally wrong way around
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    Also, what would happen if you put the right answer in the rough working but ticked the incorrect box?
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    (Original post by RichE)
    No, this is Leibniz notation. Lagrange was a century later.
    Could you please solve question H? Thank you~
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    (Original post by yl95)
    Also, what would happen if you put the right answer in the rough working but ticked the incorrect box?
    You'd get 0 marks for that question
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    (Original post by J10)
    You'd get 0 marks for that question
    Well, yes, but would the Oxford people take that into account?
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    (Original post by RichE)
    No, this is Leibniz notation. Lagrange was a century later.
    Oh, sorry, I was mistaken, then. I used to think that d/dx is Leibniz, f'(x) is Lagrange and f(x) with a dot above is Newton. The first two are actually the other way around, I'm guessing from your post?

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    (Original post by yl95)
    Well, yes, but would the Oxford people take that into account?
    They would probably just mark whatever you put in the boxes, and just ignore the questions and rough work.
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    (Original post by souktik)
    Oh, sorry, I was mistaken, then. I used to think that d/dx is Leibniz, f'(x) is Lagrange and f(x) with a dot above is Newton. The first two are actually the other way around, I'm guessing from your post?

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    dy/dx is Leibniz, f'(x) is Lagrange and the dots are Newton, so you are correct.
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    (Original post by Tarquin Digby)
    There are 2 correct answers to A, (a) and (b).
    No. The only correct answer is (a). (b) implies a = -3 does not give distinct real roots which is incorrect.
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    (Original post by DFranklin)
    I'm interested to see what RichE has to say about it.
    As you persist I'll make comment, though I wasn't planning to get involved with the discussion. I obviously don't see this as a subtle matter of notation but rather something more crucial about understanding what functions and variables are, and so something worth setting a question on.

    I expected reasoning something along the following lines. We have

    f'(x) = g(x+1)
    g'(x) = h(x-1).

    If you differentiate the first identity then we get

    f''(x) = g'(x+1)
    g'(x) = h(x-1)

    Hence f''(x) = g'(x+1) = h(x+1-1) = h(x).

    So we've shown f'' = h as a function and so f''(2x) = h(2x).

    That was the expected line of thinking for the question.

    More generally we might have:

    a(x) = b(2x).

    Differentiating with respect to x twice we get

    a''(x) = 4 b''(2x).

    All this is relatively standard so I don't really see that there's much of an excuse for confusing a''(x) and b''(2x) as some people are claiming.
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    (Original post by J10)
    They would probably just mark whatever you put in the boxes, and just ignore the questions and rough work.
    Hm, fair enough. I can't remember what I put for my answers now...
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    (Original post by Tarquin Digby)
    There are 2 correct answers to A, (a) and (b).
    No, you should solve all the values of a
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    (Original post by souktik)
    Oh, sorry, I was mistaken, then. I used to think that d/dx is Leibniz, f'(x) is Lagrange and f(x) with a dot above is Newton. The first two are actually the other way around, I'm guessing from your post?

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    (Original post by ZafarS)
    dy/dx is Leibniz, f'(x) is Lagrange and the dots are Newton, so you are correct.
    Apologies - anyway I meant to say f'(x) is Newton, as I thought it was his notation. You're of course right that dy/dx was Leibniz's

    But as you say it is indeed Lagrange's it seems
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    (Original post by yxcai)
    No, you should solve all the values of a
    What? No. It's a) because there must be DISTINCT roots. b^2-4ac therefore must always be GREATER than 0, which happens for all BUT a=2.
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    (Original post by yl95)
    What? No. It's a) because there must be DISTINCT roots. b^2-4ac therefore must always be GREATER than 0, which happens for all BUT a=2.
    I mean we should solve all the values of a rather than only consider a>2...
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    Anyone solve question H?
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    (Original post by yxcai)
    I mean we should solve all the values of a rather than only consider a>2...
    What do you mean by 'solve all the values of a'?
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    (Original post by RichE)
    As you persist I'll make comment, though I wasn't planning to get involved with the discussion. I obviously don't see this as a subtle matter of notation but rather something more crucial about understanding what functions and variables are, and so something worth setting a question on.

    I expected reasoning something along the following lines. We have

    f'(x) = g(x+1)
    g'(x) = h(x-1).

    If you differentiate the first identity then we get

    f''(x) = g'(x+1)
    g'(x) = h(x-1)

    Hence f''(x) = g'(x+1) = h(x+1-1) = h(x).

    So we've shown f'' = h as a function and so f''(2x) = h(2x).

    That was the expected line of thinking for the question.

    More generally we might have:

    a(x) = b(2x).

    Differentiating with respect to x twice we get

    a''(x) = 4 b''(2x).

    All this is relatively standard so I don't really see that there's much of an excuse for confusing a''(x) and b''(2x) as some people are claiming.
    Yeah, I get that, but in my opinion the notation was a bit confusing for those who tend to use Leibniz when they deal with functions of the form a(c(x)) as opposed to a(x). I interpreted f''(2x) as taking f(2x) and differentiating wrt x twice. I hadn't dealt with anything of the form f''(c(x)) before, so it was as good an interpretation as any to me.

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    (Original post by yl95)
    What do you mean by 'solve all the values of a'?
    for b, it only considers a>2, it doesn't consider a<2
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    (Original post by yxcai)
    for b, it only considers a>2, it doesn't consider a<2
    Ah, yes.
 
 
 
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