STEP II/III 2014 solutions
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Farhan.Hanif93
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#1
Papers found here. (Credit to Tarquin Digby for those [and Charles98 for his photos])
STEP II:
1: Solution by mikelbird
2: Solution by LightBlueSoldier
3: Solution by Stray
4: Solution by Khallil
5: Solution by Blazy
6: Solution by DFranklin
7: Solution by cliverlong
8: Solution by LightBlueSoldier, Solution by Stray
9: Solution by Brammer
10: Solution by JosephML
11: Solution by Brammer
12: Solution by shamika
13: Solution by DJMayes
STEP III:
1: Solution by jtSketchy
2: Solution by Brammer
3: Solution by edrraa, Solution by mikelbird
4: Solution by Farhan.Hanif93
5: Solution by metaltron
6: Solution by ctrls
7: Solution by Farhan.Hanif93
8: Solution by DFranklin
9: Solution by JosephML
10: Solution by JosephML
11: Solution by Brammer
12: Solution by DJMayes
13: Solution by DJMayes
STEP II:
1: Solution by mikelbird
2: Solution by LightBlueSoldier
3: Solution by Stray
4: Solution by Khallil
5: Solution by Blazy
6: Solution by DFranklin
7: Solution by cliverlong
8: Solution by LightBlueSoldier, Solution by Stray
9: Solution by Brammer
10: Solution by JosephML
11: Solution by Brammer
12: Solution by shamika
13: Solution by DJMayes
STEP III:
1: Solution by jtSketchy
2: Solution by Brammer
3: Solution by edrraa, Solution by mikelbird
4: Solution by Farhan.Hanif93
5: Solution by metaltron
6: Solution by ctrls
7: Solution by Farhan.Hanif93
8: Solution by DFranklin
9: Solution by JosephML
10: Solution by JosephML
11: Solution by Brammer
12: Solution by DJMayes
13: Solution by DJMayes
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JosephML
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#2
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#2
II/10:
Spoiler:
For the first part, get the two equations for
and
and combine them:
and
, giving:

Now we wish to maximize
for a given value of
so we take what is known as the partial derivative of
with respect to
i.e. ignore
and differentiate wrt to
:
.
To get the max, we solve
which gives
. Putting this back into
will give us
:
.
The sketch of this is fairly straight forward with the set of points which can be reached being those with
and
.
Finally the maximum possible distance reached by the particle is when
which gives us
.
Show
For the first part, get the two equations for





Now we wish to maximize







To get the max, we solve





The sketch of this is fairly straight forward with the set of points which can be reached being those with


Finally the maximum possible distance reached by the particle is when


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shamika
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#3
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#3
II/12:
(i)
(ii)
(iii)
(iv)
(v)
(i)
Spoiler:
Show
(ii)
(iii)
Spoiler:
Show
(iv)
Spoiler:
Show
We prove each implication separately. If
, then
, where x is just a dummy variable. This gives
. Subbing this into the formula of h(t), we get h(t) = k, a constant.
On the other hand, if h(t) = k, then if we solve
, which (after separating the variables and integrating to find F(t)), gives
, which differentiates to the required form of f(t).



On the other hand, if h(t) = k, then if we solve


(v)
Spoiler:
Similar to (iii) and (iv), solve
, to give
. Finally, differentiating, we get 
Note: this is the Weibull distribution, which is an important distribution in actuarial science to model very extreme losses. Of course, there was no reason you needed to know that to complere the question. The concept of the hazard function is also a fundamental theoretical actuarial tool, but I've never used it in practice.
Show
Similar to (iii) and (iv), solve



Note: this is the Weibull distribution, which is an important distribution in actuarial science to model very extreme losses. Of course, there was no reason you needed to know that to complere the question. The concept of the hazard function is also a fundamental theoretical actuarial tool, but I've never used it in practice.
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LightBlueSoldier
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#4
JosephML
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#5
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#5
III/9
Spoiler:
We have
,
so differentiating gives us:

and again:
.
For the initial conditions just put in
.
Now we must show that this fits with
:

And so we indeed have
.
Now letting
and
and subbing into
we get:

For
we must have the
component of
to be zero. Hence giving:
.
Similarly writing
in terms of
and
we have:

which then gives

which simplifies to the required result. Combining the two results gives us:

We are given that
. So from the definition of
we have:

And so we get the required
. Finally since
and
are both acute we must have
Show
We have

so differentiating gives us:

and again:

For the initial conditions just put in

Now we must show that this fits with


And so we indeed have

Now letting




For




Similarly writing




which then gives

which simplifies to the required result. Combining the two results gives us:

We are given that



And so we get the required




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user2020user
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#6
31541
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#7
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#7
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Kasel
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#8
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#8
(Original post by JosephML)
III/9
III/9
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ctrls
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#9
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#9
III/6
Spoiler:
By letting
with the given inequality and noting that the inequality
holds for
where
, you get,

Since
Repeating the process by setting
gives the required inequality.
(i) Choosing
, if we can prove that
for all
, then the result follows. To do this, we differentiate it twice, so


Noting that for
and
so it follows that
for this range of
Since
the conditions for the inequality in the original question is met and we can deduce
for the given range of
and the result follows.
(ii) Here you can prove each bound separately. Since for
you can show that
,
and
, we can multiply over the denominators without worrying about flipping the inequality. This gives,

So we set
and prove it is positive in the given domain using the inequality again. Differentiating twice gives,


Noting again the the two functions are positive in the domain and that
the result follows.
Finally for the upper bound, rearranging the inequality as,

And setting
, again it is sufficient to show that
for the given domain to establish the inequality. Differentiating twice,


Note that this gives the same inequality from part (i), so we can quickly deduce that
for the given domain. Since
we can deduce that
and the result follows.
Note: Most of this is just tedious manipulation to show the initial inequalities are met, so I've omitted some of the details of showing certain functions are positive and that f(0) and f'(0) are equal to zero in the various cases. If there's an faster way to do this question that I've missed, I would be interested to hear about it.
Show
By letting





Since


(i) Choosing





Noting that for








(ii) Here you can prove each bound separately. Since for





So we set



Noting again the the two functions are positive in the domain and that

Finally for the upper bound, rearranging the inequality as,

And setting




Note that this gives the same inequality from part (i), so we can quickly deduce that



Note: Most of this is just tedious manipulation to show the initial inequalities are met, so I've omitted some of the details of showing certain functions are positive and that f(0) and f'(0) are equal to zero in the various cases. If there's an faster way to do this question that I've missed, I would be interested to hear about it.
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JosephML
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#10
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#10
(Original post by Kasel)
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
(Original post by CD315)
I only got up as far as the result involving uksin - how many marks do you think I'd have got?
I only got up as far as the result involving uksin - how many marks do you think I'd have got?
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Kasel
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#11
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#11
(Original post by Khallil)
II/4
II/4
There are a few short proofs but I forgot to include any in my solution.
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Farhan.Hanif93
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#12
STEP III, Q4
It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.
It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.
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user2020user
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#13
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#13
(Original post by Kasel)
...
...


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Kasel
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#14
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#14
(Original post by Khallil)
...
...
and
cot(y)=(1/x)=tan(pi/2 - y)
which gives arctan(x)+arctan(1/x)= y+pi/2-y=pi/2
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user2020user
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#15
DJMayes
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#16
STEP III - Q12
Spoiler:
i)
ii)
iii)
iv):
Show
i)
ii)
Spoiler:
Show
iii)
Spoiler:
Show
iv):
Spoiler:
Show
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Stray
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#17
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#17
STEP II - Q8 (LaTeX)
This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
Spoiler:




If
is a maximum, it must be
and 
Taking first,
:

Divide both sides by the common factor

Giving




This is almost the LHS of what we are required to show, just needs a little fiddling


Now repeat the process for
:

Divide both sides by the common factor

Giving




Which combines with (**) to give the required solution

Considering the possible values of


Either
and
are both integers, or
and
are non-integer values either side of an integer, so if
is an integer, it is either
and
(2 values) or it lies between them (1 value).
For the next parts, we are asked to find either the value of
where there is 1 value, or the larger value if there are 2. This means we can consider the function
to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is
.

i)


ii)


iii)
If we fix
and
, the only part of
that can alter is the denominator containing
, so for large
we want small
, so
is when
. (Zero is not a positive integer).
iv)
For fixed
and (unknown) fixed
, the part of the function varying is:

For positive integer values of
So
giving
LightBlueSoldier points out below that my original
is wrong... of course the constraint isn't nearly as harsh as that, regardless of whether infinity is acceptable...



Show




If



Taking first,


Divide both sides by the common factor

Giving




This is almost the LHS of what we are required to show, just needs a little fiddling


Now repeat the process for


Divide both sides by the common factor

Giving




Which combines with (**) to give the required solution

Considering the possible values of



Either







For the next parts, we are asked to find either the value of




i)


ii)


iii)
If we fix








iv)
For fixed



For positive integer values of

So

giving

LightBlueSoldier points out below that my original




This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
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LightBlueSoldier
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#18
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#18
(Original post by Stray)
STEP II - Q8 (LaTeX)




If
is a maximum, it must be
and 
Taking first,
:

Divide both sides by the common factor

Giving




This is almost the LHS of what we are required to show, just needs a little fiddling


Now repeat the process for
:

Divide both sides by the common factor

Giving




Which combines with (**) to give the required solution

Considering the possible values of


Either
and
are both integers, or
and
are non-integer values either side of an integer, so if
is an integer, it is either
and
(2 values) or it lies between them (1 value).
For the next parts, we are asked to find either the value of
where there is 1 value, or the larger value if there are 2. This means we can consider the function
to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is
.

i)


ii)


iii)
If we fix
and
, the only part of
that can alter is the denominator containing
, so for large
we want small
, so
is when
. (Zero is not a positive integer).
iv)
For fixed
and (unknown) fixed
, the part of the function varying is:

This tends to 1 as b tends to infinity, but will never actually reach 1, so
tends towards but never reaches
so for fixed 

---
This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
STEP II - Q8 (LaTeX)




If



Taking first,


Divide both sides by the common factor

Giving




This is almost the LHS of what we are required to show, just needs a little fiddling


Now repeat the process for


Divide both sides by the common factor

Giving




Which combines with (**) to give the required solution

Considering the possible values of



Either







For the next parts, we are asked to find either the value of




i)


ii)


iii)
If we fix








iv)
For fixed



This tends to 1 as b tends to infinity, but will never actually reach 1, so




---
This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
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Blazy
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#19
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#19
STEP II, Q5: (somebody quickly check through this please though I think it should be fine)
The only real difficulty here was spotting how to get rid of the -4 and -3.
The only real difficulty here was spotting how to get rid of the -4 and -3.
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metaltron
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#20
STEP III Q5
Beginning
Part i)
Part ii)
Beginning
Spoiler:
Show
Part i)
Spoiler:
Show
Part ii)
Spoiler:
By symmetry:




So

So
so diagonals of quadrilateral are perpendicular.
So XYZT is a square if and only if
.
That is
.
Looking at the co-efficient of i (which is complex itself)
which is always true.
Equating the other parts
.
So XYZT is a square if and only if
which is the condition for PQRS to be a parallelogram.
Show
By symmetry:




So


So

So XYZT is a square if and only if

That is

Looking at the co-efficient of i (which is complex itself)

Equating the other parts

So XYZT is a square if and only if

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