Mathematics examination paper from 1970

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.

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

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

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 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!

$\frac{x^2}{2!}+ \frac{x^3}{3!} +.... \geq 0$
which is obvious because I said so

A sketch isn't a proof.

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).

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 bitch.
|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....