STEP II 2012 discussion thread

Post all of your STEP II discussions here.

Question Paper

Provided Solutions

Q1 - DFranklin
Q2 - desijut
Q3 - TLRMaths
Q4 - TheMagicMan
Q5 - mikelbird
Q6 Version 1 - ben-smith/Q6 Version 2 - Blutooth
Q7 - Blutooth
Q8 - SParm
Q9 - Tomcrease
Q10 - Farhan.Hanif93
Q11 - DFranklin
Q12 - DFranklin
Q13 - Farhan.Hanif93

Well done everyone, we have a complete set of solutions :biggrin:

(edited 11 years ago)

Scroll to see replies

Reply 1

DFranklin

Bumped as this has just come out of moderation...

Reply 2

Slowbro93

Original post by DFranklin

Bumped as this has just come out of moderation...

Thanks

Reply 3

Alexxh

completed 2 questions, another 2 nearly completed (last part didn't work out) and 3 other fragmentary answers :/
is that a 1?

Reply 4

SParm

Here's my solution for question 8:

Part 1:

Spoiler

Part 2:

Spoiler

Part 3:

Spoiler

Part 4:

Spoiler

(edited 11 years ago)

Reply 5

Farhan.Hanif93

Original post by 8inchestall

question 4/5, i didnt know what it meant by expand in powers of 1/y, i changed the ln|(2+y)/(2-y)| to ln|(1+(2y/2-y))| but had no idea how to go on

Responding to you here, I believe you were supposed to consider

\ln \left(1 + \dfrac{1}{2y}\right) - \ln \left(1 - \dfrac{1}{2y}\right)

from the start.

Reply 6

Dirac Spinor

Did anyone do 7? Found it kind of strange

Reply 7

TheMagicMan

Original post by SParm

Here's my solution for question 8:

Part 1:

Spoiler

Part 2:

Spoiler

Part 3:

Spoiler

Part 4:

Spoiler

aka it's an arithmetic progression<----probably not necessary

Question 3.

Original post by ben-smith

Did anyone do 7? Found it kind of strange

I spent the last 15 minutes or so on it. Didn't manage to do the last part and can't even remember anything about it other than that it was vectors so I hated it

Reply 10

Dirac Spinor

Original post by TheMagicMan

I spent the last 15 minutes or so on it. Didn't manage to do the last part and can't even remember anything about it other than that it was vectors so I hated it

I have solution to the last part but it seems way too simple.

Reply 11

TRLMaths

2012 step ii.

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Reply 12

Farhan.Hanif93

Original post by ben-smith

Did anyone do 7? Found it kind of strange

I saw vectors, and decided it wasn't worth my time. How did you get on?

Reply 13

Peter8837

What was the coeff of x^24 in Q1 part (i)?

Reply 14

SParm

Original post by TheMagicMan

aka it's an arithmetic progression<----probably not necessary

THAT'S THE ONE :biggrin:

!!!! Forgot the name in the heat of the exam.

Reply 15

TheMagicMan

Original post by Farhan.Hanif93

Responding to you here, I believe you were supposed to consider

\ln \left(1 + \dfrac{1}{2y}\right) - \ln \left(1 - \dfrac{1}{2y}\right)

from the start.

I wasn't really sure what they wanted in the last part to be honest

I got

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

1 < 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)