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    Do you know if it was standard to do all the papers? They seem to equate roughly to "Paper 1/3" = Maths, "Paper 2/4" = Further Maths.

    Certainly brutal compared with todays exams though.

    Edit: Oops - missed that you'd posted FM ones as well...
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    (Original post by DFranklin)
    Do you know if it was standard to do all the papers? They seem to equate roughly to "Paper 1/3" = Maths, "Paper 2/4" = Further Maths.

    Certainly brutal compared with todays exams though.

    Edit: Oops - missed that you'd posted FM ones as well...
    I think these were 3 separate qualifications, Pure Mathematics, Applied Mathematics (Mechanics might have been a better title!) and Further Mathematics.

    Later on, you were able to mix and match (do two papers out of four) and attain qualifications such as A Level Mathematics (Pure & Applied) but I suspect this wasn't the case then.
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    those normal a-level ones don't look too bad, but certainly much harder than now and you need some degree stuff to answer some of them.

    though this may be interesting, it's gcse (o-level) so shouldn't be too taxing, but some of the geometry ones are difficult just because it is taught very little now.

    http://news.bbc.co.uk/1/shared/bsp/h...athematics.pdf
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    (Original post by around)
    Ok, here goes my attempt:

    the slings and arrows of outrageous fortune


    The equation we're given is , and we introduce a transformation to convert the equation into the form , as this will let us use a substitution to solve the equation to give 3 roots (if we use a hyperbolic substitution we can only get 1 root):





    We want a p such that , and we find that if we define then our desired ends are achieved:



    or

    In order such that we can use a trigonometric substitution, we require , or . QED

    For the second part, let us introduce a linear shift of the form to get rid of that pesky quadratic term:



    Expanding everything out, we find that if we take a=2 then our ends are achieved:



    Comparing with our original equation we find that a= 1/4 and b = k-2:

    and solving this inequality for k gives 1<k<3 (not because at the extreme ends of our interval we get a repeated root and the question asks for 3 roots).

    Differentiate the equation, and note that a repeated root is when f'(x) and f(x) both equal zero: or y= -1/2, +1/2.

    When y= +1/2 is a root, k=3, and is a factor. Dividing it out, we see that the other factor is (y+1), and hence y = 1/2 or -1, and since y-2 = x our roots are x=-3/2 or -3.


    When y= -1/2 is a root, k=1, and is a factor. Dividing it out, we see that the other factor is (y-1), and hence y = -1/2 or +1, and since y-2 = x our roots are x=-5/2 or -1.



    my method for the first part didn't help with the second :|
    There is a much simpler way!

    Spoiler:
    Show
    f(x)=x^3-3ax+b\newline f'(x)=3(x^2-a) \Rightarrow \mathrm{turning\ points\ are\ at\ }\pm\sqrt a\newline\mathrm{a\ sketch\ shows\ that\ for\ 3\ real\ roots,\ } \newline f(-\sqrt a)\geq 0 \geq f(\sqrt a)
    Splitting up this inequality leads to the desired result, and the same method can be used for part two!
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    Is it strange that part of me now wishes I grew up in the 1970s? :rolleyes: Those maths papers just look so much more... fun than we get today... :o:

    What's really shocking to me is that there was so much more content in the normal a-level maths back then. How did they fit it all in - were people more intelligent, or was school just loads harder or what?
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    :ditto: I think I love this thread...
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    (Original post by sonofdot)
    Is it strange that part of me now wishes I grew up in the 1970s? :rolleyes: Those maths papers just look so much more... fun than we get today... :o:

    What's really shocking to me is that there was so much more content in the normal a-level maths back then. How did they fit it all in - were people more intelligent, or was school just loads harder or what?
    Much less time wasted due to students fooling around. Hardly any banter or answering back. Just a lot of getting on with it. Remember pupils were coming from a position of having more knowledge already as O Level (taken at age 16) included most of the AS mathematics content.
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    What a 14 year old school leaver could do 100 years ago.

    They would also have had a geometry paper but I don't have a copy of that but it would have required sound understanding of Euclid's Elements. Remember no calculator!

    Arithmetic and Algebra 1909.pdf

    A large proportion of the candidates for this exam would have gone on to die in the First World War.
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    Q6 (i)
    Let I(t) = \int_0^t \sin x e^{-2x} \,dx = [ -\cos x e^{-2x}]_0^t - 2 \int_0^t \cos x e^{-2x}\,dx

     = (1 - \cos t e^{-2t}) - [2 \sin x e^{-2x}]_0^t - 4 \int_0^t \sin x e^{-2x}\,dx

     = 1 - (\cos t + 2\sin t)e^{-2t} - 4 I(t)

    Deduce I(t) = \frac{1}{5} - \frac{1}{5} (\cos t + 2 \sin t}e^{-2t})

    Then I(2n\pi) = \frac{1}{5} (1 - e^{-2t}) &lt; \frac{1}{5} and I((2n+1)\pi) = \frac{1}{5} (1+e^{-2t}) &gt; \frac{1}{5} as required.


    (ii)
    x=\frac{1}{y-1}.  \frac{dx}{dy} = -\frac{1}{(y-1)^2}.  x=1\implies y=2, \, x=3 \implies y = 4/3

    x+1 = \frac{y}{y-1}, x^2+x = (x+1)x = \frac{y}{(y-1)^2}

    So the integral becomes \displaystyle \int_{4/3}^2 \frac{1/(y-1)^2}{(y/(y-1)) \sqrt{y/(y-1)^2}}\,dy = \int_{4/3}^2 y^{-3/2}\,dy

    So we get \displaystyle [-2 y^{-1/2}]_{4/3}^2 = -2/\sqrt{2} +2 \sqrt{3/4} = \sqrt{3} - \sqrt{2}
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    its not actually THAT much more difficult, questions are just worded a lot differently, style of questions is different
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    Note: can't face looking at the coord geom questions so if anyone else is feeling brave, go for it...
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    (Original post by 3006)
    its not actually THAT much more difficult, questions are just worded a lot differently, style of questions is different
    Not so. These questions provide a lot less guidance, they require a lot more manipulation of (fairly) complex equations, you're expected to do quite complex trig manipulation without guidance, and you're expected to know a lot more material (complex numbers, conic sections, vectors).

    Without wanting to cause blushes, it's instructive how many of the posted solutions have actually been incorrect.

    [Feel free to prove me wrong by posting solutions to questions 7-10].
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    I did London A level Pure Maths in 1971. I'd love to see the paper again. Have you got a copy of it you could post here Mr M?
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    (Original post by DFranklin)
    Not so. These questions provide a lot less guidance, they require a lot more manipulation of (fairly) complex equations, you're expected to do quite complex trig manipulation without guidance, and you're expected to know a lot more material (complex numbers, conic sections, vectors).

    Without wanting to cause blushes, it's instructive how many of the posted solutions have actually been incorrect.

    [Feel free to prove me wrong by posting solutions to questions 7-10].
    I apologise to my generation in advance.

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    I did my maths A level a year early..and only did further maths at AS level..but some of these questions look famililar..but its been to long since i did any maths...so i'm not even goin to attempt them!
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    (Original post by Elemc)
    I did London A level Pure Maths in 1971. I'd love to see the paper again. Have you got a copy of it you could post here Mr M?
    1971 is missing. My next is 1973.
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    who has done question 1?
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    (Original post by Totally Tom)
    who has done question 1?
    around: http://www.thestudentroom.co.uk/show...8&postcount=37
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    (Original post by DFranklin)
    around: http://www.thestudentroom.co.uk/show...8&postcount=37
    cheers, i don't want to click on an unnumbered spoiler, just incase it spoils it for me
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    Well, if you want something not done yet, there's a distinct lack of sols for Q7-Q10. Q9, Q10 don't look too bad, I confess I'm not really seeing where to start on Q8, and Q7(b) looks horrible. (In my defense, I find coord geometry is one of those things you need to be in practice for to do well).
 
 
 
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