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    Q4:
    first bit
    Write c = cos t, s = sin t, z = c+is.

    Then

    \frac{2z}{z-1} = 2\frac{c+is}{c-1 + is} =2 \frac{(c+is)(c-1-is)}{(c-1)^2+s^2}

     =2 \frac{c^2+s^2 - (c+is)}{c^2+s^2+1-2c} = 2\frac{1-c-is}{2-2c} = 1  - \frac{is}{1-c}\right]

    and so the locus is the line Re z = 1.

    (a)
    Let z = x+iy, then (x-1)^2+y^2+(x+1)^2+y^2 = 4. So 2x^2 +2 + 2y^2 = 4, so x^2+y^2 = 1.

    (b)
    Again let z = x+iy. Then \sqrt{x^2+(y-1)^2} = 3 - \sqrt{x^2+(y+1)^2}. So x^2+(y-1)^2 = 9 + x^2 + (y+1)^2 - 6 \sqrt{x^2+(y+1)^2}

    So 6\sqrt{x^2+(y+1)^2} = 9 + x^2 + (y+1)^2 - x^2 - (y-1)^2 = 9 + 4y

    So 36 (x^2+(y+1)^2} = 81 + 72y + 16y^2

    36x^2 + 36y^2 + 72 y + 36 = 81 + 72y + 16 y^2

    36x^2 + 20 y^2 = 45 (an ellipse. Probably an arithmetic error somewhere, but life is too short...)

    (c)
    We must have tan(arg(z-1)) = tan(arg(z+1)) and so

    Im(z-1)/Re(z-1) = Im(z+1)/Re(z+1). Writing z = x+iy we get

    y/(x-1) = y/(x+1), and so y = 0. So z lies on the real line. For the arguments to be equal, it must lie on same side of -1, 1. So either z<-1 or z>1.
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    (Original post by DFranklin)
    Q4:
    first bit
    Write c = cos t, s = sin t, z = c+is.

    Then

    \frac{2z}{z-1} = 2\frac{c+is}{c-1 + is} =2 \frac{(c+is)(c-1-is)}{(c-1)^2+s^2}

     =2 \frac{c^2+s^2 - (c+is)}{c^2+s^2+1-2c} = 2\frac{1-c-is}{2-2c} = 1  - \frac{is}{1-c}\right]

    and so the locus is the line Re z = 1.

    (a)
    Let z = x+iy, then (x-1)^2+y^2+(x+1)^2+y^2 = 4. So 2x^2 +2 + 2y^2 = 4, so x^2+y^2 = 1.

    (b)
    Again let z = x+iy. Then \sqrt{x^2+(y-1)^2} = 3 - \sqrt{x^2+(y+1)^2}. So x^2+(y-1)^2 = 9 + x^2 + (y+1)^2 - 6 \sqrt{x^2+(y+1)^2}

    So 6\sqrt{x^2+(y+1)^2} = 9 + x^2 + (y+1)^2 - x^2 - (y-1)^2 = 9 + 4y

    So 36 (x^2+(y+1)^2} = 81 + 72y + 16y^2

    36x^2 + 36y^2 + 72 y + 36 = 81 + 72y + 16 y^2

    36x^2 + 20 y^2 = 45 (an ellipse. Probably an arithmetic error somewhere, but life is too short...)

    (c)
    We must have tan(arg(z-1)) = tan(arg(z+1)) and so

    Im(z-1)/Re(z-1) = Im(z+1)/Re(z+1). Writing z = x+iy we get

    y/(x-1) = y/(x+1), and so y = 0. So z lies on the real line. For the arguments to be equal, it must lie on same side of -1, 1. So either z<-1 or z>1.
    For b, is it not sufficient to note that |z + i| + |z - i| = constant => distance from (0,-1) + distance from (0,1) is constant, which defines an ellipse with foci at those points?
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    Generally looks alright, i.e there is nothing which stands out as completely undoable, having said this i haven't done maths for a year and therefore probably would actually struggle to get a solution to the question, and i am comparing it to the advanced higher course so i don't know what current A level is actually like
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    (Original post by Mr M)
    yes s level was used for cambridge (and oxford) entry
    I was talking to my dad about this. He said that many schools, especially state sector, didn't enter candidates for S level papers - I suppose in the same way as many now don't enter for STEP or AEA papers.

    At my dad's school entering S level papers were optional, there was no additional tuition for them, and they were taken by any students who wanted to stretch themselves. It wasn't exclusively about Oxbridge admission though I suppose many who entered for S levels applied for Oxbridge. S levels were around til 1990 apparently.
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    Q5
    first bit
    \sin^2 (n+1)x - \sin^2 nx = (\cos 2(n+1)x - \cos 2n x) / 2

     = \sin [(n+1) x - nx] \sin [(n+1)x + nx] = \sin x \sin(2n+1) x.

    So assume sin x + sin 3x +... + sin (2n-1) x = \sin^2 nx \csc x (obviously true for n = 1).

    By the calc above, \sin(2n+1) x = (\sin^2 (n+1)x  - \sin^2 nx) \csc x) and the result follows by induction. Don't understand what they mean by "find the sum" - isn't that what we've just done?

    (a)
    \sin^2 6x \csc x = sin x + sin 3x + ... + sin 11 x.

    So the integral I = \sum_1^6 \int_0^{2\pi} \sin (2k-1)x \, dx But for any integer N, \int_0^{2\pi} \sin Nx \,dx = 0, and so I = 0.

    (b)
    \sin^4 6x \csc^2 x = (\sin x + \sin 3x + ... + \sin 11 x)^2

    =\sum_{n=1}^6\sum_{m=1}^6 \sin (2n - 1)x \sin (2m -1)x.

    So we want to find \sum_{n=1}^6\sum_{m=1}^6 \int_0^{2\pi}\sin (2n - 1)x \sin (2m -1)x \, dx (*)

    Now consider \int_0^{2\pi} \sin Ax \sin Bx \, dx where A, B are +ve integers.

    Suppose first A=B. Then we have \int_0^{2\pi} \sin^2 Ax \, dx = \int_0^{2\pi} \frac{1}{2} - \frac{1}{2} \cos 2Ax\, dx = \pi

    Suppose instead A \neq B. Then 2 sin Ax sin Bx = cos(A-B)x - cos(A+B) x. Since for any non-zero integer N we have \int_0^{2\pi} \cos Nx \,dx = 0 we must have \int_0^{2\pi} \sin Ax \sin Bx \,dx = 0.

    It follows that in (*) the terms where m = n each add \pi to the answer, and the other terms are all zero. So the answer required is 6\pi.
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    (Original post by Scipio90)
    For b, is it not sufficient to note that |z + i| + |z - i| = constant => distance from (0,-1) + distance from (0,1) is constant, which defines an ellipse with foci at those points?
    Yeah, except I think you'd be still be expected to get to an actual equation, or at least values for major/minor axes (since you're expected to sketch each locus). And I couldn't be bothered to look up the ellipse equations to do that from the "sum of distance from focii" starting point.
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    (Original post by jj193)
    taylor/maclauren series aren't on A-Level...
    Pretty sure I did them for A-Level. Not sure what module, but was not that long ago, 2005/6.

    The paper does not seem that bad, I am just very much out of practice with maths.

    I am a physicist, don't need to do that complicated stuff too often.
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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.
    Would be interesting to see the other ones - I think the S-level/AEA/etc. has actually seen a smaller drop off in difficulty than the bog standard A-level.

    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!
    Yeah, but don't forget you probably only needed 6 questions for a distinction. 8 in 3 hours looks tough, but not impossible; main problem for current generation is relative lack of coord geometry, since the last 4 questions all lean that way.
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    (Original post by JoshJGordon)
    Pretty sure I did them for A-Level. Not sure what module, but was not that long ago, 2005/6.

    The paper does not seem that bad, I am just very much out of practice with maths.

    I am a physicist, don't need to do that complicated stuff too often.
    Did Taylor/Maclaurin expansions in FP3 last year ,not sure if it's been taken out of the spec since then.

    Pretty enlightening post seeing these questions , I could probably manage most of them but only by the fact I've finished my first year reading Mathematical Physics (we do a lot of the same maths modules as the maths students).I'd have definitely needed hints to finish most of those questions if I'd done them a year ago.
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    It says "special" A-level, which equivalent to today's AEA. My dad did one in Chemistry.

    EDIT: looks like someone else beat me to it :p:
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    (Original post by red_roses)
    It says "special" A-level, which equivalent to today's AEA. My dad did one in Chemistry.

    EDIT: looks like someone else beat me to it :p:
    In fairness, the OP said it in the very first post but most people didn't bother reading it
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    (Original post by OrmondDrone)
    In fairness, the OP said it in the very first post but most people didn't bother reading it
    Ahh I read the third post thinking it was still the OP's about it saying how the A-level syllabus had obviously been dumbed down. I fail at speed reading :p:
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    I have quite a few questions from Oxford + Cambridge scholarship papers dated at the very latest 1963. Only problem is that I have no idea how many of them are per paper or what the time limit is. Here's one for now that I have wrote down:

    1. (Oxford.) By means of the substitutions x \pm x^{-1} = t, or otherwise, find:

    \displaystyle \int \frac{x^2 + 1}{(x^4+x^2+1)^2} \, \mathrm{d}x \qquad \int \frac{x^2 - 1}{(x^4+x^2+1)^2} \, \mathrm{d}x  \qquad \int \frac{1}{x^4+x^2+1} \, \mathrm{d}x.

    I'll try and bring some more later today. :p:
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    (Original post by DFranklin)
    Would be interesting to see the other ones - I think the S-level/AEA/etc. has actually seen a smaller drop off in difficulty than the bog standard A-level.
    I will put some more up later.
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    (Original post by -G-a-v-)
    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?
    i have a set of 1970s physics papers that are also a massive shock compared to current A-level papers.... I mean, they are a lot, lot harder
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    (Original post by Mr Nonsense)
    i have a set of 1970s physics papers that are also a massive shock compared to current A-level papers.... I mean, they are a lot, lot harder
    A Level physics is simply embarrassing these days. The problem comes from not requiring A Level physicists to study any mathematics beyond GCSE. I understand and accept the reason (the low number of candidates is a concern) but it means no topic can be examined in any meaningful depth.
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    Here are pdfs of the 1970 Mathematics A Level Pure Maths papers:

    June 1970 Paper 1.pdf

    June 1970 Paper 2.pdf
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    And pdfs of the 1970 Applied Mathematics A Level papers:

    June 1970 Paper 3.pdf

    June 1970 Paper 4.pdf
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    And finally pdfs of the Further Mathematics A Level papers:

    June 1970 Paper 5.pdf

    June 1970 Paper 6.pdf
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    (Original post by Mr Nonsense)
    i have a set of 1970s physics papers that are also a massive shock compared to current A-level papers.... I mean, they are a lot, lot harder
    Is there any chance you could upload them (physics section) ? I'd love to give them a try.
 
 
 
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