# STEP II 2012 discussion thread

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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

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

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(Original post by

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

**DFranklin**)Bumped as this has just come out of moderation...

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#4

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

is that a 1?

is that a 1?

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#5

Here's my solution for question 8:

Part 1:

Part 2:

Part 3:

Part 4:

Part 1:

Part 2:

Spoiler:

Show

Part 3:

Spoiler:

Show

For this part; we can simply find the number p for which this equation works. If p is a real number; then clearly the equation will work. So:

When n = 1

From the sequence equation:

Substitute back into the original equation:

Divide by :

Note:

Re-arrange our original equation to get:

are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

(I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of , so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)

When n = 1

From the sequence equation:

Substitute back into the original equation:

Divide by :

Note:

Re-arrange our original equation to get:

are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

(I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of , so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)

Part 4:

Spoiler:

So we can immediately see from the first part of the question:

Note that:

Re-arrange previous inequality and we get:

, so where

So:

Re-arrange this expression:

We can see from this that:

And if we consider the difference of the first two terms, u_{0} and u_{1}, we get:

The first difference is greater than 0 and the subsequent differences are greater than the previous differences so the differences between two adjacent terms is always greater than 0, and so the sequence is strictly increasing, as required.

When , so:

, and we can see the sequence increases constantly.

Show

So we can immediately see from the first part of the question:

Note that:

Re-arrange previous inequality and we get:

, so where

So:

Re-arrange this expression:

We can see from this that:

And if we consider the difference of the first two terms, u_{0} and u_{1}, we get:

The first difference is greater than 0 and the subsequent differences are greater than the previous differences so the differences between two adjacent terms is always greater than 0, and so the sequence is strictly increasing, as required.

When , so:

, and we can see the sequence increases constantly.

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#6

(Original post by

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

**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

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#8

(Original post by

Here's my solution for question 8:

Part 1:

Part 2:

Part 3:

Part 4:

**SParm**)Here's my solution for question 8:

Part 1:

Part 2:

Spoiler:

Show

Part 3:

Spoiler:

Show

For this part; we can simply find the number p for which this equation works. If p is a real number; then clearly the equation will work. So:

When n = 1

From the sequence equation:

Substitute back into the original equation:

Divide by :

Note:

Re-arrange our original equation to get:

are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

(I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of , so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)

When n = 1

From the sequence equation:

Substitute back into the original equation:

Divide by :

Note:

Re-arrange our original equation to get:

are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

(I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of , so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)

Part 4:

Spoiler:

So we can immediately see from the first part of the question:

Note that:

Re-arrange previous inequality and we get:

, so where

So:

Re-arrange this expression:

We can see from this that:

And if we consider the difference of the first two terms, u_{0} and u_{1}, we get:

The first difference is greater than 0 and the subsequent differences are greater than the previous differences so the differences between two adjacent terms is always greater than 0, and so the sequence is strictly increasing, as required.

When , so:

, and we can see the sequence increases constantly.

Show

So we can immediately see from the first part of the question:

Note that:

Re-arrange previous inequality and we get:

, so where

So:

Re-arrange this expression:

We can see from this that:

And if we consider the difference of the first two terms, u_{0} and u_{1}, we get:

The first difference is greater than 0 and the subsequent differences are greater than the previous differences so the differences between two adjacent terms is always greater than 0, and so the sequence is strictly increasing, as required.

When , so:

, and we can see the sequence increases constantly.

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#10

(Original post by

Did anyone do 7? Found it kind of strange

**ben-smith**)Did anyone do 7? Found it kind of strange

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#11

(Original post by

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

**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

**way**too simple.

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#13

(Original post by

Did anyone do 7? Found it kind of strange

**ben-smith**)Did anyone do 7? Found it kind of strange

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#15

(Original post by

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

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

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#16

(Original post by

Responding to you here, I believe you were supposed to consider from the start.

**Farhan.Hanif93**)Responding to you here, I believe you were supposed to consider from the start.

I got

so

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)

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#17

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#18

(Original post by

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

**Peter8837**)What was the coeff of x^24 in Q1 part (i)?

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#19

(Original post by

Wasn't that a show that question? 55 wasn't it?

**TheMagicMan**)Wasn't that a show that question? 55 wasn't it?

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