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    has it really been dumbed down? or do more people get higher marks / grades in subjects these days as they have greater access to relevant resources meaning the learning process is made more efficient?
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    (Original post by DFranklin)
    Same method for 2(ii) as well (already posted I think).

    for the 3rd bit:
    Spoiler:
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    9 = (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca), so a^2+b^2+c^2 \leq 9 (since a,b,c positive). Since we obtain equality with a = 3, b=c=0, 9 is the maximum range.

    and

     (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2) - 2(ab+bc+ca)

     = 3(a^2+b^2+c^2) - (a+b+c)^2.

    So  3(a^2+b^2+c^2) \ge (a+b+c)^2 = 9, so (a^2+b^2+c^2) \geq 3. Since we have equality with a = b = c = 1, 3 is the minimum value.
    same as me, but I love how you pulled the lower limit! nice!!!
    it never =9 btw, since 0 isn't positive.
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    This isn't an A-level paper, it's an S-level paper (see where it says "special paper"?), which is equivalent to today's AEA (or, more likely, STEP)... and I'd expect to see questions like that on an AEA/STEP, so the comparison doesn't really fit.
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    (Original post by jermaindefoe)
    has it really been dumbed down? or do more people get higher marks / grades in subjects these days as they have greater access to relevant resources meaning the learning process is made more efficient?
    As the post below(I mean above...) me says, it isnt an A level paper so you cant really compare, but I think generally the style of questions in A level papers has been dumbed down over the years.
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    (Original post by nuodai)
    This isn't an A-level paper, it's an S-level paper (see where it says "special paper"?), which is equivalent to today's AEA (or, more likely, STEP)... and I'd expect to see questions like that on an AEA/STEP, so the comparison doesn't really fit.
    I was waiting for somebody to say this - after reading through all the posts...
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    (Original post by wonderwall 124)
    I was waiting for somebody to say this - after reading through all the posts...
    I said it in post 1 so you didn't need to look far!
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    (Original post by Mr M)
    I said it in post 1 so you didn't need to look far!
    Sorry if i was unclear - i was referring more to people who had not noted the information in the original post... The paper is indeed very interesting - thanks for the upload.
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    Is it just me or are there a lot of the two-part questions completely unrelated?
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    (Original post by jermaindefoe)
    has it really been dumbed down? or do more people get higher marks / grades in subjects these days as they have greater access to relevant resources meaning the learning process is made more efficient?
    No, it's been dumbed down. Sorry if that bruises egos of the current generation, but it's the truth.

    I'd say it's actually much less apparent at "the high end": these S-level questions aren't significantly harder than the AEA is (but cover more material). For that matter I don't think FM A-level is that much easier than when I did it.

    But I'd guess the typical student getting a low 'A' in maths now would be lucky to get a 'C' on the equivalent 1970 paper.
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    Would the s-level in collaboration with a-levels have formed offers to study mathematics at cambridge university in 1970 or were there other examination supplements also required?
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    (Original post by wonderwall 124)
    Would the s-level in collaboration with a-levels have formed the offer to cambridge university to study mathematics in 1970 or were there other examination supplements also required?
    yes s level was used for cambridge (and oxford) entry
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    (Original post by Mr M)
    yes s level was used for cambridge (and oxford) entry
    So in comparison to today - was entry more or less difficult?
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    (Original post by wonderwall 124)
    So in comparison to today - was entry more or less difficult?
    I don't know .. ask DFranklin... but very intelligent students would have done well then and now. Weak/average A Level students now would have been ungraded in the past and they wouldn't have found that out until the end of Year 13 as there were no modules.
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    (Original post by Kolya)
    I haven't done any maths for near-on a month, but a rough go gave me this for 6 (ii):

    Spoiler:
    Show
    Because we are contrary, use alternative substitution of x = \cosh ^2 v - 1 (and denote cosh by c and sinh by s).
    Then, leaving sorting out the limits for later:
    \displaystyle \int ^3 _1 \frac{dx}{(x+1)\sqrt{x^2 + x}} = 2 \int \frac{sc dv}{c^2\sqrt{c^4 - c^2}} = 2 \int sech ^2 v \ dv = \frac{s}{c}

    When x=3, we have c = 2, s = 3. When x=1, we have c = \sqrt{2}, s = 1.

    So applying limits to the integral we computed gives us 3 - \frac{2}{\sqrt{2}} = \boxed{3 - \sqrt{2}}
    .
    I get
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     \sqrt{3} - \sqrt{2}
    and wolfram agrees
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    (Original post by Mr M)
    I don't know .. ask DFranklin... but very intelligent students would have done well then and now.
    The earliest Cambridge College Entrance papers I've seen are from about 1983. (I also have some "Further Mathematics" papers from 1980, but in this context FM means "beyond A-level", and answering 3 questions didn't just get you in, it got you a scholarship!).

    I don't actually think the difficulty has changed that much. I think the CCE questions were harder, but you weren't expected to get as many questions out. Back then, you'd never get your actual marks (it was all entirely internal to the university), so it's very hard to be sure how it worked.

    The two changes I think have happened over the years:

    'A'-levels have got easier. So it's a bigger jump to the standard Cambridge expects. On the other hand, the syllabus has shrunk, so you don't actually have as much ground to cover as in the past.

    The internet has levelled the playing field a lot. Back then, if you were at a typical comp, literally all you'd have to go on is a set of past papers to work through (if you were lucky). Odds were your teachers couldn't do most of the questions, and you were unlikely to know many people in the same boat. I think that gave an enormous advantage to the private schools that had 10+ maths applicants every year. TSR would have been an absolute revelation back then.
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    (Original post by Scipio90)
    1's ok, as a physicist I must respond to 2.i with "draw the graphs, it's obvious",
    that's all i want to do :P - sadly AS maths has made this virtually impossible for my meagre knowledge to attempt properly (that said I will, I just haven't yet)
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    Looking at some of those questions makes me feel incredibly bad at maths :p: that said, i've not actually got a pen and paper out and tried to do them ... yet. lol

    Anyone got access to any typical 'normal' A-level papers from back then?
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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 x^3 - 3ax + b = 0, and we introduce a transformation py = x to convert the equation into the form 4y^3 - 3y = c, as this will let us use a substitution  y = cos \theta to solve the equation to give 3 roots (if we use a hyperbolic substitution we can only get 1 root):

    (py)^3 - 3a(py) + b = 0

    p^3y^3 - 3apy = -b

    We want a p such that p^3/3pa = 4/3, and we find that if we define p = 2\sqrt{a} then our desired ends are achieved:

    (2\sqrt{a} x)^3 - 3a(2\sqrt{a} x) = -b

    or 4y^3 -3y = \dfrac{-b}{2a\sqrt{a}}

    In order such that we can use a trigonometric substitution, we require \dfrac{-b}{2a\sqrt{a}} \leq 1, or b^2 < 4a^3. QED

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

    4(y-a)^3 + 24(y-a)^2 +45 (y-a) + k = 0

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

    4y^3 -3y + (k-2) = 0

    Comparing with our original equation x^3 - 3ax + b = 0 we find that a= 1/4 and b = k-2:

    (k-2)^2 &lt; 4(\frac{1}{4}^3 and solving this inequality for k gives 1<k<3 (not \leq 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: 12y^2 - 3 = 0 or y= -1/2, +1/2.

    When y= +1/2 is a root, k=3, and (2y-1)^2 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 (2y+1)^2 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 :|
    For the first inequality it is perhaps simpler to note that if p and q denotes a turning point then;

    f(p)f(q) < 0 and it falls out rather simply
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    Simpler? simpler?

    I'll have no truck with simplicity thank you very much.
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    Q3
    (i)


    \frac{r^2+r-1}{r^2+r} = 1 - \frac{1}{r(r+1}} = 1 + \frac{1}{r+1} - \frac{1}{r}

    So S = \sum_1^n 1 + \frac{1}{r+1} - \frac{1}{r}

     = n + \sum_2^{n+1} \frac{1}{r} - \sum_1^n \frac{1}{r}

     = n - 1 + \frac{1}{n+1} and the result follows.

    (ii) (a)
    (n-1)^3 = (n-1)(n^2-2n + 1) = (n-1)[n^2 - 2n] + (n-1) = n(n-1)(n-2) + n - 1.

    So for n > 2, \frac{(n-1)^3}{n!} = \frac{1}{(n-3)!} + \frac{1}{(n-1)!} - \frac{1}{n!}

    So \sum_3^\infty \frac{(n-1)^3}{n!} = 1 + 1/2! + 1/3! + ... + 1/2! + 1/3! + ... - (1/3! + 1/4!+...) = e + (e - 2) - (e-5/2) = e + 1/2

    The first two terms are 0 and 1/2, so the final sum is e+1/2+1/2 = e+1

    (ii)(b)
    \frac{1}{2n^2+n} = \frac{1}{n(2n+1)} = \frac{A}{n}+\frac{B}{2n+1}, where A(2n+1)+Bn = 1, So A=1, B=-2 and so \frac{1}{2n^2+n} = \frac{2}{2n}-\frac{2}{2n+1}.

    So S = \sum_1^\infty \frac{1}{2n^2+n} = 2(1/2 - 1/3 + 1/4 - 1/5 + ...)

    But log 2 = 1 -1/2 + 1/3 - 1/4 + ... and so S = 2(1-log 2).
 
 
 
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