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    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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    Wow, thanks for posting.

    Really does show how the syllabus has changed. Sure, A level Maths can be difficult. But compared to 1970, its been dumbed down greatly.. =|
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    Wow nice nice post dude,
    sure a lot of people will be interested in this!

    Think i might give it a go see what i get, you get mark scheme?
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    This is pretty tough for A level standard. Guess it really was harder back in the day
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    (Original post by gummers)
    Wow nice nice post dude,
    sure a lot of people will be interested in this!

    Think i might give it a go see what i get, you get mark scheme?
    No mark scheme I'm afraid. I'm sure the TSR team can create one. Use spoilers though!
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    (Original post by coldfusion)
    This is pretty tough for A level standard. Guess it really was harder back in the day
    (Original post by Shayke)
    Wow, thanks for posting.

    Really does show how the syllabus has changed. Sure, A level Maths can be difficult. But compared to 1970, its been dumbed down greatly.. =|
    This isn't a-level standard, this is more comparable to STEP.
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    mark schemes?

    edit: oh well. I would post what I think are the correct answers to the ones I can do but I don't want to get them wrong and look stupid lol
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    2. looks easy
    Spoiler:
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    Base case 1=1

    Inductive hypothesis n=k
    e^k \geq k+1

    to deduce n=k+1
    e\times e^k=e^{k+1} \geq ek + e \geq k+e \geq k+2= (k+1) + 1 (multiplying by e, and using obvious facts like e>2)


    Wait a minute I will post the next one. For all values, looks like they don't want induction, gay. Power series I'm thinking. Yeah, that could work.
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    It doesn't say integer values of x simp.
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    (Original post by around)
    This isn't a-level standard, this is more comparable to STEP.
    Yeah looking at it now it does seem like STEP II standard
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    Rep to anyone who can answer all them questions!
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    Yes I did realize that before you said that.:o:
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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
    which is obvious because I said so

    Well, when x=0 then 0=0 as required

    if x>0 then cleary, the sum of positive numbers must be positive

    if x<0 ?
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    Dibs on question 1.
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    Question 1 the first part, looks weird. Maybe, I should read about depressed cubes.

    Although, the next part seems to be able to be done by just using the old sum of roots method.
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    I wouldn't be able to answer it. I wouldn't be able to answer one from nowadays either.
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    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

    for ii
    a = 3-2*(epsilon)
    b = c = epsilon
    so for maxima: (a^2+b^2+c^2) < 9
    and for minima: (1+1+1) = 3
    so 3<or=(a^2+b^2+c^2)<9
    I'm geussing that derivatives must be used to show that these are the maxima/minima but I cba
    something like that methinks
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    (Original post by Simplicity)
    Okay 2. the correct way
    Spoiler:
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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
    which is obvious because I said so

    Well, when x=0 then 0=0 as required

    if x>0 then cleary, the sum of positive numbers must be positive

    if x<0 ?
    taylor/maclauren series aren't on A-Level...
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    (Original post by jj193)
    taylor/maclauren series aren't on A-Level...
    they were back then! but that isn't the approach expected ... i looked at gradients too
 
 
 
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