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!(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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did you sit these ones back in the day?

Anon the 7th
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 04082009 20:59

Revision help in partnership with Birmingham City University

Simplicity
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 04082009 21:00
(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
(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 nowadaysLast edited by Simplicity; 04082009 at 21:05. 
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 04082009 21:03
And they wonder why more people are getting A's.

Mr M
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 04082009 21:03
(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!
did you sit these ones back in the day?
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!Last edited by Mr M; 04082009 at 21:07. 
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 04082009 21:04
I have question 1, give me some time to latex it up.

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 04082009 21:06
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 Alevel. 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. 
Anon the 7th
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 04082009 21:07
(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! 
generalebriety
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 04082009 21:10
For positive values of x it's obvious. Can you prove it for negative values of x?

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 04082009 21:10
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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 04082009 21:11
That seems to go up to about the same level as MEI FP2, from my semieducated comparison...

DFranklin
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 04082009 21:14
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).

Simplicity
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 04082009 21:14
Okay 2. the correct way
Spoiler:Show
then
(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 (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 when x<0.
So in each case

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 04082009 21:15
I haven't done any maths for nearon a month, but a rough go gave me this for 6 (ii):
.Last edited by Kolya; 04082009 at 21:20. 
generalebriety
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 04082009 21:17
Incidentally, 2(i) is pretty easy if you do it the right way:
Spoiler:ShowConsider 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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 04082009 21:19
I found this fairly easy :/

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 04082009 21:20
(Original post by Simplicity)
A sketch isn't a proof.
(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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 04082009 21:22
Ok, here goes my attempt at question 1:
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 = \frac{k26}{4}:
and solving this inequality for k gives 25<k<27 (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=27, 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 y2 = x our roots are x=3/2 or 3.
When y= 1/2 is a root, k=25, and is a factor. Dividing it out, we see that the other factor is (y1), and hence y = 1/2 or +1, and since y2 = 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 correctedLast edited by around; 05082009 at 13:30. 
Mr M
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 04082009 21:22
(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). 
Simplicity
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 04082009 21:26
(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....
Again, a sketch isn't a proof. Although, looking at gradients is the sane way to go. 
DFranklin
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 04082009 21:29
Same method for 2(ii) as well (already posted I think).
for the 3rd bit:
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