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    (Original post by Mr M)
    I thought some of you might enjoy having a go at the A Level Special Paper from 1970. This was the equivalent of the AEA and was designed for the top 15% of A Level candidates. You had to answer 8 questions out of 10 in 3 hours.

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    thank you Mr M. where did you find these?!?! they're so difficult, makes me realise how lucky i am to be sitting maths nowadays! :p:

    did you sit these ones back in the day?
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    (Original post by jj193)
    2
    a)
    e^x >/= x+1
    sketch y=x+1
    sketch y=e^x
    intercept at (0,1) since e^0 = 0+1
    for x<0 d(e^x)/dx < 1 hence for all x<0 e^x is always grater than y=x+1
    for x>0 d(e^x)/dx >1 hence for all x>0 e^x is always greater than y=x+1

    don't know if it counts as proof...?
    same idea for part B)
    but again, will only suffice if my method of 'proof' counts
    A sketch isn't a proof.

    (Original post by Anon the 7th)
    thank you Mr M. where did you find these?!?! they're so difficult, makes me realise how lucky i am to be sitting maths nowadays
    To be fair this is sort of meant to be like a sort of STEP and it has been noted that the difficulty level is similar to STEP 2.
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    And they wonder why more people are getting A's.
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    (Original post by Anon the 7th)
    thank you Mr M. where did you find these?!?! they're so difficult, makes me realise how lucky i am to be sitting maths nowadays! :p:

    did you sit these ones back in the day?
    I have the other six (count 'em) 1970 A Level papers too. This is the hardest one. Graham at Edexcel has a library of old papers.

    I took my A Levels in 1986 and this is more difficult than my papers. I think the time pressure is the problem. I can answer most of the questions but I couldn't knock out 8 good answers in 3 hours!
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    I have question 1, give me some time to latex it up.
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    1's ok, as a physicist I must respond to 2.i with "draw the graphs, it's obvious", but it's not that bad to do properly, 2ii ok unless I've made a mistake.

    Anyone got a nice method for 2.ii? Think I've got it, but my solution is pretty ugly for what's quite a nice problem.

    Not too evil based on those, but definitely harder than current A-level. Obviously you need to compare it to the AEA though, which I can't really comment on since I've only seen one AEA paper.
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    (Original post by Mr M)
    I have the other six (count 'em) 1970 A Level papers too. This is the hardest one. Graham at Edexcel has a library of old papers.

    I took my A Levels in 1984 and this is more difficult than my papers. I think the time pressure is the problem. I can answer most of the questions but I couldn't knock out 8 good answers in 3 hours!
    jeeze, they must be truly difficult
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    (Original post by Simplicity)
    \frac{x^2}{2!}+ \frac{x^3}{3!} +.... \geq 0
    which is obvious because I said so
    For positive values of x it's obvious. Can you prove it for negative values of x?
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    I might have given this a go 2 months ago, but I think I've forgotten everything! :O xD

    It does look quite hard, I reckon I could give it a good go though!
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    That seems to go up to about the same level as MEI FP2, from my semi-educated comparison...
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    To be honest, this isn't nearly as difficult as I expected. Not sure I'd even say it's harder than the current AEA (although you're expected to know more material).
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    Okay 2. the correct way
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    e^x=1+x+\frac{x^2}{2!}+... \geq x+1
    then
    \frac{x^2}{2!}+ \frac{x^3}{3!} +.... \geq 0(1)
    which is obvious because I said so

    Well, when x=0 then 0=0 as required(using (1))

    if x>0 then cleary, the sum of positive numbers must be positive (using (1) )

    if x<0 use the fact that they intersect at e^x=x+1 (0,0) and gradient of x+1 is 1 and e^x is e^x, which is smaller then 1 for x<0. So as x<0 e^x tends to 0 and x+1 tends to minus infinity.
    So e^x&gt;x+1 when x<0.

    So in each case e^x \geq x+1
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    I haven't done any maths for near-on a month, but a rough go gave me this for 6 (ii):

    Spoiler:
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    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}}
    .
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    Incidentally, 2(i) is pretty easy if you do it the right way:

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    Consider f(x) = e^x - x - 1. Then f'(x) = e^x - 1. We know that f(0) = 0; we also know that e^x > 1 for x > 0 and e^x < 1 for x < 0, so f'(x) > 0 (and so f is increasing) for x > 0 and f'(x) < 0 (and so f is decreasing) for x < 0. So f starts off above the axis, then decreases until it reaches the axis, then increases again, so f(x) >= 0 for all x.
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    I found this fairly easy :/
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    (Original post by Simplicity)
    A sketch isn't a proof.
    yeah but it's backed up by the fact that they intercept at x=0 and that the gradients are such that e^x 'pulls away above' from y=x+1
    (also Mr M did something similar)

    i tried you're method and it's fine for -1< x <0 since |x^n|>|x^(n+1)|
    since 1>|x|
    then for -infinity< x < -1 it looks a complete *****.
    |x^n|<|x^(n+1)| since 1<|x|
    maybe add the second and third (so the first is negative(it's further from the convergence) and hence we'll get an estimate that is weighted closer to the lower end - to be on the safe side), take an average. all 'pairs' are converging about the same average and the average can't be lower. the average is positive.
    trying this:
    (1/3!)*(1/4)*(x(x^3)+4(x^3)) >/= 0 <=> x </= -4
    i can only think that exetending this will eventually get it up to -1. though It's probably assymptotic to -1....
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    Ok, here goes my attempt at question 1:

    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 &lt; 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-26) = 0

    Comparing with our original equation x^3 - 3ax + b = 0 we find that a= 1/4 and b = \frac{k-26}{4}:

    \frac{k-26}{4}^2 &lt; 4(\frac{1}{4})^3 and solving this inequality for k gives 25<k<27 (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=27, 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=25, 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 :|

    also i made a mistake which has now been corrected
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    (Original post by DFranklin)
    To be honest, this isn't nearly as difficult as I expected. Not sure I'd even say it's harder than the current AEA (although you're expected to know more material).
    Interestingly, like the current AEA, this exam was supposed to only test the basic A Level knowledge and did not require students to have taken A Level Further Maths. This illustrates how much the A Level core content has reduced over the years.
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    (Original post by jj193)
    yeah but it's backed up by the fact that they intercept at x=0 and that the gradients are such that e^x 'pulls away above' from y=x+1
    (also Mr M did something similar)

    i tried you're method and it's fine for -1< x <0 since |x^n|>|x^(n+1)|
    since 1>|x|
    then for -infinity< x < -1 it looks a complete *****.
    |x^n|<|x^(n+1)| since 1<|x|
    maybe add the second and third (so the first is negative(it's further from the convergence) and hence we'll get an estimate that is weighted closer to the lower end - to be on the safe side), take an average. all 'pairs' are converging about the same average and the average can't be lower. the average is positive.
    trying this:
    (1/3!)*(1/4)*(x(x^3)+4(x^3)) >/= 0 <=> x </= -4
    i can only think that exetending this will eventually get it up to -1. though It's probably assymptotic to -1....
    See generalebriety post.

    Again, a sketch isn't a proof. Although, looking at gradients is the sane way to go.
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    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.
 
 
 
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