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Original post by SParm
Here's my solution for question 8:

Part 3:

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Reply 61
Original post by jack.hadamard

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Oh yeah that's true :/...but is there anything inherently wrong with my proof?
Reply 62
Can't wait to bash out some solutions :biggrin:
Original post by TheMagicMan


so 1<e(1+1/n)n<(1+1/n)1/21 < e(1+1/n)^{-n}< (1+1/n)^{1/2}

The RHS has a limit of one so want we want follows by the squeeze theorem...the real question is is there a simpler way (tbh I'm never quite sure how STEp wants you to approach limits)


You could rewrite it as (1+1n)n1+1n\left(1 + \frac{1}{n}\right)^n\sqrt{1 + \frac{1}{n}}, consider it as a product and then mention that it is decreasing.
I don't see anything wrong with mentioning the Squeeze theorem, and it even sounds better.
Original post by SParm
Oh yeah that's true :/...but is there anything inherently wrong with my proof?


I just skimmed it by far, but I would generally avoid commenting on other people's work. :tongue:
Original post by Peter8837
Hmmm.. fair enough. Also, I did the very first part of Q8 just to grab a mark or two (I showed the first inequality) - do you reckon I'd get any marks for that?

There was a lot going on in that question. Probably 1 mark for that inequality at the start (and even then, to get that mark I think they would expect the proof of it to give a short explanation of why squaring doesn't change anything since βα\beta - \alpha must be positive etc).
Reply 66
Original post by jack.hadamard
I just skimmed it by far, but I would generally avoid commenting on other people's work. :tongue:


Fair. I assume you took the exam yesterday then?
Reply 67
Original post by TheJ0ker
Not great but I don't think I have failed, 2 full solutions and 2 half solutions. I didn't have a Cambridge offer anyway so I think my grade on STEP 1 will be easily enough to get into Warwick.


lucky you! some of us with Cam offers only got 3 almost out and a couple of partials :/ heres hoping for low boundaries anyway!
Reply 68
For question 4, you need to consider the expansion:

e^x = 1 + x + (x^2)/2! + (x^3)/3! + ...

Let x = 1/y

e^(1/y) = 1 + 1/y + 1/(2*y^2) + 1/(6*y^3) + 1/(24*y^3) + ...
< 1+ 1/y + 1/(2*y^2) + 1/(4*y^3) + 1/(8*y^) + ...
Note that we have a geometric series with common ratio of 1/2y and first term 1/y. Since y > 1/2, 1/2y < 1. This allows us to use the formula for the sum to infinity
Hence:
e^(1/y) < 1+ 2/(2y-1)
=> e^(1/y) < (2y+1)/(2y-1)
=> 1/y < ln[(2y+1)/(2y-1)], as required :wink:

Also, for the last part, just consider the ratio between the 2 boundaries for e. As n goes to infinity, the boundaries are closer together. e lies in between. We can deduce immediately that e will get closer to (1 + 1/n)^n
(edited 11 years ago)
Reply 69
Just out of interest, I completed 3 questions successfully and about 2/3 of another question. Would you reckon that's enough for a grade 1. Worry about that since I am a Cambridge offer holder too :confused:
Reply 70
Original post by Farhan.Hanif93
There was a lot going on in that question. Probably 1 mark for that inequality at the start (and even then, to get that mark I think they would expect the proof of it to give a short explanation of why squaring doesn't change anything since βα\beta - \alpha must be positive etc).


Thanks. Also, for Q5 part (ii), I drew the graphs right in both cases (I checked with wolfram) but I didn't explicitly solve the cubic I got when trying to find stationary points... I just sort of wrote the stationary points down after drawing the graphs. How many marks would I lose do you think? This question is open to anyone by the way. Thanks.
(edited 11 years ago)
Original post by Peter8837
Thanks. Also, for Q5 part (ii), I drew the graphs right in both cases (I checked with wolfram) but I didn't explicitly solve the cubic I got when trying to find stationary points... I just sort of wrote the stationary points down after drawing the graphs. How many marks would I lose do you think?

Would probably lose nothing more than 4 or 5; possibly less if you've got the correct S.P's too.
Reply 72
DONE 6 on paper, will post the solution soon, but will try some other qs first :biggrin:.
Reply 73
Can someone do a solution to Q5? I'd be interested to see how you would solve that cubic to find turning points!
Reply 74
Original post by Zuzuzu
How'd your teacher find it?


tough
Reply 75
Original post by Peter8837
Can someone do a solution to Q5? I'd be interested to see how you would solve that cubic to find turning points!


I think you can factorise out (x-a)(x-b) and then youre left with something that again factorises. I'll try and do it again after my tea.
Reply 76
Original post by Peter8837
Can someone do a solution to Q5? I'd be interested to see how you would solve that cubic to find turning points!


i guessed it was symmetrical about the midpoint of a,b meaning (a+b)/2 is either a turning point (1st case) or a discontinuity (2nd case)
Reply 77
Original post by Peter8837
Thanks. Also, for Q5 part (ii), I drew the graphs right in both cases (I checked with wolfram) but I didn't explicitly solve the cubic I got when trying to find stationary points... I just sort of wrote the stationary points down after drawing the graphs. How many marks would I lose do you think? This question is open to anyone by the way. Thanks.


I did exactly the same thing as I was hurrying to get out another question in the last half an hour, think one of my stationary points was right but the others weren't. Hoping the majority of the marks were for the graphs as mine were both right. Did you just guess the stationary points? What did you get?
Original post by Nam Nguyen
Just out of interest, I completed 3 questions successfully and about 2/3 of another question. Would you reckon that's enough for a grade 1. Worry about that since I am a Cambridge offer holder too :confused:


Perhaps; perhaps not. Now you have to perform at the top of your game for STEP III without knowing your result in II.

Suppose that at interview, you had been asked 'can you perform under pressure?'. I think you would have found the answer 'yes'. They've thought of that. So they've designed a process that invites you to demonstate it.
Reply 79
Original post by MrDD
i guessed it was symmetrical about the midpoint of a,b meaning (a+b)/2 is either a turning point (1st case) or a discontinuity (2nd case)


Yes, this exactly. I checked my graphs with wolframalpha and they were correct... I just don't know how many marks I'd lose for not 'explicitly' finding the stationary points; but FarhanHanif93 who is quite experienced said 4-5 marks so that seems a good estimate.

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