Slowbro93
Badges: 20
Rep:
?
#1
Report Thread starter 8 years ago
#1
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
0
reply
DFranklin
Badges: 18
Rep:
?
#2
Report 8 years ago
#2
Bumped as this has just come out of moderation...
0
reply
Slowbro93
Badges: 20
Rep:
?
#3
Report Thread starter 8 years ago
#3
(Original post by DFranklin)
Bumped as this has just come out of moderation...
Thanks
0
reply
Alexxh
Badges: 0
#4
Report 8 years ago
#4
completed 2 questions, another 2 nearly completed (last part didn't work out) and 3 other fragmentary answers :/
is that a 1?
0
reply
SParm
Badges: 8
Rep:
?
#5
Report 8 years ago
#5
Here's my solution for question 8:

Part 1:

Spoiler:
Show
 \beta , \alpha , q > 0\\



\beta - \alpha  > q\\



(\beta - \alpha )^{2} > q^{2}\\



\beta ^{2} + \alpha ^{2} - 2\beta \alpha  > q^{2}\\



\beta ^{2} + \alpha ^{2} - q^{2} - 2\beta\alpha > 0\\



\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0\\

As required.


Part 2:

Spoiler:
Show
u_{n + 1} = \dfrac{u_{n} ^{2} - q^{2}}{u_{n - 1}}\\



u_{n}(u_{n} + u_{n + 2}) = u_{n} ^{2} + u_{n}u_{n + 2}\: (\tau )

Note from the first equation:

u_{n} ^{2} = u_{n - 1}u_{n + 1} + q^{2}

So (\tau) becomes:

u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2})

As the sequence is valid for all n greater than or equal to one, we can say:

u_{n + 2} = \dfrac{u_{n + 1} ^{2} - q^{2}}{u_{n}}

Re-arrange to give:

u_{n}u_{n + 2} + q^{2} = u_{n + 1} ^{2}

Substitute back into (\tau ):

u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2}) = u_{n - 1}u_{n + 1} + u_{n + 1} ^{2}

Factorise this expression:

u_{n - 1}u_{n + 1} + u_{n + 1} ^{2} = u_{n + 1}(u_{n - 1} + u_{n + 1})

As required


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:

u_{n + 1} - pu_{n} + u_{n - 1} = 0

When n = 1

u_{n - 1} = u_{0} = \alpha\\



u_{n} = u_{1} = \beta\\



u_{n + 1} = u_{2}

From the sequence equation:

u_{2} = \dfrac{\beta ^{2} - q^{2}}{\alpha}

Substitute back into the original equation:

u_{n + 1} - pu_{n} + u_{n - 1} = \dfrac{\beta ^{2} - q^{2}}{\alpha} - p\beta + \alpha = 0

Divide by \beta:

\dfrac{\beta ^{2} - q^{2}}{\alpha\beta} - p + \frac{\alpha}{\beta} = 0

Note: \dfrac{\alpha}{\beta} = \dfrac{\alpha ^{2}}{\alpha\beta}

Re-arrange our original equation to get:

\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

\alpha,\: \beta \neq 0\\



\alpha,\: \beta,\: q 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 \alpha,\: \beta,\: q, 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:
Show
\beta > \alpha + q\\



\beta - \alpha > q

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

\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0

Note that:

\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

Re-arrange previous inequality and we get:

p > 2, so p = 2 + \mu where \mu > 0

So:

u_{n + 1} - pu_{n} + u_{n - 1} = u_{n + 1} - (2 + \mu)u_{n} + u_{n - 1} = 0

Re-arrange this expression:

u_{n + 1} - u_{n} = (1 + \mu)u_{n} - u_{n - 1}

We can see from this that:

u_{n + 1} - u_{n} > u_{n} - u_{n - 1}

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

u_{1} - u_{0} = \beta - \alpha > q > 0

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 \beta = \alpha + q, p = 2, so:

u_{n + 1} - u_{n} = u_{n} - u_{n - 1}, and we can see the sequence increases constantly.
3
reply
Farhan.Hanif93
Badges: 18
Rep:
?
#6
Report 8 years ago
#6
(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.
0
reply
Dirac Spinor
Badges: 10
Rep:
?
#7
Report 8 years ago
#7
Did anyone do 7? Found it kind of strange
0
reply
TheMagicMan
Badges: 15
Rep:
?
#8
Report 8 years ago
#8
(Original post by SParm)
Here's my solution for question 8:

Part 1:

Spoiler:
Show
 \beta , \alpha , q > 0\\



\beta - \alpha  > q\\



(\beta - \alpha )^{2} > q^{2}\\



\beta ^{2} + \alpha ^{2} - 2\beta \alpha  > q^{2}\\



\beta ^{2} + \alpha ^{2} - q^{2} - 2\beta\alpha > 0\\



\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0\\

As required.


Part 2:

Spoiler:
Show
u_{n + 1} = \dfrac{u_{n} ^{2} - q^{2}}{u_{n - 1}}\\



u_{n}(u_{n} + u_{n + 2}) = u_{n} ^{2} + u_{n}u_{n + 2}\: (\tau )

Note from the first equation:

u_{n} ^{2} = u_{n - 1}u_{n + 1} + q^{2}

So (\tau) becomes:

u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2})

As the sequence is valid for all n greater than or equal to one, we can say:

u_{n + 2} = \dfrac{u_{n + 1} ^{2} - q^{2}}{u_{n}}

Re-arrange to give:

u_{n}u_{n + 2} + q^{2} = u_{n + 1} ^{2}

Substitute back into (\tau ):

u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2}) = u_{n - 1}u_{n + 1} + u_{n + 1} ^{2}

Factorise this expression:

u_{n - 1}u_{n + 1} + u_{n + 1} ^{2} = u_{n + 1}(u_{n - 1} + u_{n - 1})

As required


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:

u_{n + 1} - pu_{n} + u_{n - 1} = 0

When n = 1

u_{n - 1} = u_{0} = \alpha\\



u_{n} = u_{1} = \beta\\



u_{n + 1} = u_{2}

From the sequence equation:

u_{2} = \dfrac{\beta ^{2} - q^{2}}{\alpha}

Substitute back into the original equation:

u_{n + 1} - pu_{n} + u_{n - 1} = \dfrac{\beta ^{2} - q^{2}}{\alpha} - p\beta + \alpha = 0

Divide by \beta:

\dfrac{\beta ^{2} - q^{2}}{\alpha\beta} - p + \frac{\alpha}{\beta} = 0

Note: \dfrac{\alpha}{\beta} = \dfrac{\alpha ^{2}}{\alpha\beta}

Re-arrange our original equation to get:

\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

\alpha,\: \beta \neq 0\\



\alpha,\: \beta,\: q 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 \alpha,\: \beta,\: q, 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:
Show
\beta > \alpha + q\\



\beta - \alpha > q

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

\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0

Note that:

\dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

Re-arrange previous inequality and we get:

p > 2, so p = 2 + \mu where \mu > 0

So:

u_{n + 1} - pu_{n} + u_{n - 1} = u_{n + 1} - (2 + \mu)u_{n} + u_{n - 1} = 0

Re-arrange this expression:

u_{n + 1} - u_{n} = (1 + \mu)u_{n} - u_{n - 1}

We can see from this that:

u_{n + 1} - u_{n} > u_{n} - u_{n - 1}

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

u_{1} - u_{0} = \beta - \alpha > q > 0

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 \beta = \alpha + q, p = 2, so:

u_{n + 1} - u_{n} = u_{n} - u_{n - 1}, and we can see the sequence increases constantly.
aka it's an arithmetic progression<----probably not necessary
0
reply
TRLMaths
Badges: 0
Rep:
?
#9
Report 8 years ago
#9
Question 3.
Attached files
3
reply
TheMagicMan
Badges: 15
Rep:
?
#10
Report 8 years ago
#10
(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
0
reply
Dirac Spinor
Badges: 10
Rep:
?
#11
Report 8 years ago
#11
(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.
0
reply
TRLMaths
Badges: 0
Rep:
?
#12
Report 8 years ago
#12
2012 step ii.
3
reply
Farhan.Hanif93
Badges: 18
Rep:
?
#13
Report 8 years ago
#13
(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?
0
reply
Peter8837
Badges: 1
Rep:
?
#14
Report 8 years ago
#14
What was the coeff of x^24 in Q1 part (i)?
0
reply
SParm
Badges: 8
Rep:
?
#15
Report 8 years ago
#15
(Original post by TheMagicMan)
aka it's an arithmetic progression<----probably not necessary
THAT'S THE ONE !!!! Forgot the name in the heat of the exam.
0
reply
TheMagicMan
Badges: 15
Rep:
?
#16
Report 8 years ago
#16
(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 &lt; e&lt; (1+1/n)^{n+1/2}

so 1 &lt; e(1+1/n)^{-n}&lt; (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)
0
reply
8inchestall
Badges: 9
Rep:
?
#17
Report 8 years ago
#17
(Original post by ben-smith)
I have solution to the last part but it seems way too simple.
q7 after the first part showing OY=bla, how do you get the hence? and also x1.x2 meant dot product or nay ?
0
reply
TheMagicMan
Badges: 15
Rep:
?
#18
Report 8 years ago
#18
(Original post by Peter8837)
What was the coeff of x^24 in Q1 part (i)?
Wasn't that a show that question? 55 wasn't it?
0
reply
Peter8837
Badges: 1
Rep:
?
#19
Report 8 years ago
#19
(Original post by TheMagicMan)
Wasn't that a show that question? 55 wasn't it?
The first part of Q1...
0
reply
DarkMatter_22
Badges: 0
Rep:
?
#20
Report 8 years ago
#20
(Original post by Peter8837)
The first part of Q1...
I think it was 15.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should the school day be extended to help students catch up?

Yes (96)
27.51%
No (253)
72.49%

Watched Threads

View All