STEP 2013 Solutions
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Slowbro93
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Unofficial STEP Solutions 2013
Here is a thread of all of the STEP Solutions (I, II and III) that have been created by TSR users. I will update (or the relevant mod) when new solutions are posted
STEP I (Sat 25th June 2013)
Question 1: cpdavis
Question 2: mikelbird
Question 3: FJacob
Question 4: Pyoro, cpdavis
Question 5: mikelbird
Question 6: mikelbird
Question 7: Mastermind2
Question 8: metaltron
Question 9: DJMayes
Question 10: DJMayes
Question 11: DJMayes
Question 12: HCIO
Question 13: DFranklin
STEP II (Sat 19th June 2013)
Question 1: mikelbird
Question 2: Smaug123 Alternative to part ii - Fuzzy12345
Question 3: mikelbird Part ii without considering turning points: Nebula
Question 4: jack.hadamard
Question 5: cpdavis
Question 6: metaltron
Question 7: DJMayes
Question 8:MW24595
Question 9: LShirley95
Question 10: DJMayes
Question 11: DJMayes
Question 12: Smaug123
Question 13: Nebula
STEP III (Sat 26th June 2013)
Question 1: Pyoro
Question 2: borealis72
Question 3: jack.hadamard
Question 4: IrrationalNumber, Blutooth
Question 5: jack.hadamard
Question 6: jack.hadamard
Question 7: DJMayes
Question 8: jack.hadamard
Question 9: Brammer
Question 10: DJMayes
Question 11: bananarama2
Question 12: Blutooth, ukdragon37 - only part iic, MAD Phil
Question 13: DFranklin
STEP I Paper: http://www.thestudentroom.co.uk/show...postcount=1205
STEP II Paper: http://www.thestudentroom.co.uk/show...&postcount=671
STEP III Paper: http://www.thestudentroom.co.uk/show...postcount=1384
Here is a thread of all of the STEP Solutions (I, II and III) that have been created by TSR users. I will update (or the relevant mod) when new solutions are posted

STEP I (Sat 25th June 2013)
Question 1: cpdavis
Question 2: mikelbird
Question 3: FJacob
Question 4: Pyoro, cpdavis
Question 5: mikelbird
Question 6: mikelbird
Question 7: Mastermind2
Question 8: metaltron
Question 9: DJMayes
Question 10: DJMayes
Question 11: DJMayes
Question 12: HCIO
Question 13: DFranklin
STEP II (Sat 19th June 2013)
Question 1: mikelbird
Question 2: Smaug123 Alternative to part ii - Fuzzy12345
Question 3: mikelbird Part ii without considering turning points: Nebula
Question 4: jack.hadamard
Question 5: cpdavis
Question 6: metaltron
Question 7: DJMayes
Question 8:MW24595
Question 9: LShirley95
Question 10: DJMayes
Question 11: DJMayes
Question 12: Smaug123
Question 13: Nebula
STEP III (Sat 26th June 2013)
Question 1: Pyoro
Question 2: borealis72
Question 3: jack.hadamard
Question 4: IrrationalNumber, Blutooth
Question 5: jack.hadamard
Question 6: jack.hadamard
Question 7: DJMayes
Question 8: jack.hadamard
Question 9: Brammer
Question 10: DJMayes
Question 11: bananarama2
Question 12: Blutooth, ukdragon37 - only part iic, MAD Phil
Question 13: DFranklin
STEP I Paper: http://www.thestudentroom.co.uk/show...postcount=1205
STEP II Paper: http://www.thestudentroom.co.uk/show...&postcount=671
STEP III Paper: http://www.thestudentroom.co.uk/show...postcount=1384
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LShirley95
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Slowbro93
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Smaug123
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Question 2:
Spoiler:

i) Substitute
:
, as required.
Now,
by the first part; this is
as required.
We then integrate by parts:
, so
, which can easily be seen to reduce to the correct formula.
Since:

then:

Hence:
and the result follows by rearrangement.
ii) We proceed inductively: it is easy to verify that
. Then
as required.
iii)
. Substitute
, so
; then we use
to obtain
.
Then we use the first part:
, which can easily be seen to be
.
Show

i) Substitute


Now,


We then integrate by parts:

![I_n = [x^n \cdot \frac{(1-x)^{n+1}}{n+1}]_0^1 - \int_0^1 -\frac{(1-x)^{n+1}}{n+1} \cdot n x^{n-1} dx I_n = [x^n \cdot \frac{(1-x)^{n+1}}{n+1}]_0^1 - \int_0^1 -\frac{(1-x)^{n+1}}{n+1} \cdot n x^{n-1} dx](https://www.thestudentroom.co.uk/latexrender/pictures/5c/5c32def1c3e1ad923fafe8cbc164a74d.png)
Since:

then:

Hence:

ii) We proceed inductively: it is easy to verify that


iii)





Then we use the first part:


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Fuzzy12345
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DJMayes
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Q7 – STEP II
Part i):
Part ii):
Part iii):
Part i):
Spoiler:
Show
Part ii):
Part iii):
Spoiler:


Now, b is a prime number meaning that
has four possible factors – 1, b,
and
. Let’s examine each of these in turn:
, 
This solves to get
which is not possible as a and b are both positive integers.
, 
, 
, 
This solves for
which is not possible as a and b are both positive integers, yet b squared is greater than b implying that a = 0.
, 
, ![] a = \frac{1}{2}b(b-1) ] a = \frac{1}{2}b(b-1)](https://www.thestudentroom.co.uk/latexrender/pictures/44/44beeb8bd70a6f72ab54935433203944.png)
Now, our final part – we need to find integer solutions of these equations for a, b and c where b is not prime. The easiest place to start is with b and c. From our final solution, we are after a square number which, when doubled, gives us the product of two consecutive numbers. Trying the obvious ones (1-10) and doubling we find that if c = 6, 2c^2 = 72 = 8x9, giving b=8 as a non prime solution, which is what we want. Finally, plugging in b to the equation with a yields a = 28, so a possible set of integer solutions is a=28, b=8, c=6.
Show


Now, b is a prime number meaning that





This solves to get







This solves for




![] a = \frac{1}{2}b(b-1) ] a = \frac{1}{2}b(b-1)](https://www.thestudentroom.co.uk/latexrender/pictures/44/44beeb8bd70a6f72ab54935433203944.png)
Now, our final part – we need to find integer solutions of these equations for a, b and c where b is not prime. The easiest place to start is with b and c. From our final solution, we are after a square number which, when doubled, gives us the product of two consecutive numbers. Trying the obvious ones (1-10) and doubling we find that if c = 6, 2c^2 = 72 = 8x9, giving b=8 as a non prime solution, which is what we want. Finally, plugging in b to the equation with a yields a = 28, so a possible set of integer solutions is a=28, b=8, c=6.
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DJMayes
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DJMayes
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#8
Q11 – STEP II
Part i):
Part ii):
Part i):
Spoiler:
Let A, B and C be the three particles (with mass m), and let the speed of the particles after respective collisions be denoted by
,
and
, respectively. Now, the first collision:
Conservation of Linear Momentum:


Newton’s Law of Restitution:


Adding together, we get:

And subtracting we get:
(Which is the speed we were after!)
Next, we need to look at the second collision, between B and C. You could plug this all in again, but the second collisions is actually identical to the first but with u replaced by
. Plugging this in then immediately yields the speeds of the other two particles.
Now, for a third collision we require
:

Divide across by u and 1-e (Fine as neither are negative or 0) and multiply by 4:

Which then gives the inequality e<1 which we are given, so another collision must occur.
Show
Let A, B and C be the three particles (with mass m), and let the speed of the particles after respective collisions be denoted by



Conservation of Linear Momentum:


Newton’s Law of Restitution:


Adding together, we get:

And subtracting we get:

Next, we need to look at the second collision, between B and C. You could plug this all in again, but the second collisions is actually identical to the first but with u replaced by

Now, for a third collision we require


Divide across by u and 1-e (Fine as neither are negative or 0) and multiply by 4:

Which then gives the inequality e<1 which we are given, so another collision must occur.
Part ii):
Spoiler:
Firstly, we need to consider the third collision. Let Va and Vb be the velocities after the collision, similarly to before:
Conservation of linear momentum:


Newtons Law of restitution:
![e[\frac{1}{2}u(1-e) - \frac{1}{4}u(1-e^2)] = V_b – V_a e[\frac{1}{2}u(1-e) - \frac{1}{4}u(1-e^2)] = V_b – V_a](https://www.thestudentroom.co.uk/latexrender/pictures/de/de2218f1d0abd44671f8b8b66ca341b2.png)
We are only interested in the velocity of B, so add them together:


Don’t try expand or simplify this yet. For a fourth collision, we require Vb > Vc so 2Vb > Vc:

Now, divide across by 1+e and u and multiply by 4, which are all fine due to non-zero and non-negativity:



This quadratic has roots of
, and looking at the graph of the quadratic (And knowing e cannot be greater than one) we get the inequality 
Show
Firstly, we need to consider the third collision. Let Va and Vb be the velocities after the collision, similarly to before:
Conservation of linear momentum:


Newtons Law of restitution:
![e[\frac{1}{2}u(1-e) - \frac{1}{4}u(1-e^2)] = V_b – V_a e[\frac{1}{2}u(1-e) - \frac{1}{4}u(1-e^2)] = V_b – V_a](https://www.thestudentroom.co.uk/latexrender/pictures/de/de2218f1d0abd44671f8b8b66ca341b2.png)
We are only interested in the velocity of B, so add them together:


Don’t try expand or simplify this yet. For a fourth collision, we require Vb > Vc so 2Vb > Vc:

Now, divide across by 1+e and u and multiply by 4, which are all fine due to non-zero and non-negativity:



This quadratic has roots of


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LShirley95
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Got Q9 written with diagrams, unfortunately don't have a scanner so this is the best I can do:
![Name: 1371760817848.jpg
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![Name: 1371760837517.jpg
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Size: 29.6 KB]()
![Name: 1371760848354.jpg
Views: 2203
Size: 22.9 KB]()
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mimx
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In case anyone wants a higher res version of the (II) paper to work from:
http://www.thestudentroom.co.uk/show...&postcount=829
http://www.thestudentroom.co.uk/show...&postcount=829
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Nebula
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#11
13 (i)
13 (ii)
13 (iii)
Alternate to 3ii without considering turning points:
13 (ii)
Spoiler:
Show
13 (iii)
Spoiler:
A=2n happens after (n*(HT))T or (n*(TH))H.


S=2n happens after (n*(HH))T or (n*(TT))H.


Then we note pq>0 and
and
are not equal to 0 since and have the same sign since
.




A=2n+1 happens after (n*(HT))HH or (n*(TH))TT.


S=2n+1 happens after (n*(HH))HT or (n*(TT))TH.


Then we note pq>0 and
and
are not equal to 0 since and have the same sign since
. (
when m>0)




----
[note: as you may expect, both the odd and even cases have the same result]
Show
A=2n happens after (n*(HT))T or (n*(TH))H.


S=2n happens after (n*(HH))T or (n*(TT))H.


Then we note pq>0 and







A=2n+1 happens after (n*(HT))HH or (n*(TH))TT.


S=2n+1 happens after (n*(HH))HT or (n*(TT))TH.


Then we note pq>0 and








----
[note: as you may expect, both the odd and even cases have the same result]
Alternate to 3ii without considering turning points:
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Smaug123
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Q12:
i)
ii)
Expression for Var(Y):
Do non-zero
exist?
i)
Spoiler:
Show
ii)
Spoiler:
Show
The variance of X is "the mean of the squares - the square of the mean". The mean is
, which squares to
. Note that
, and hence the second term in the expression for the variance is precisely "the square of the mean".
So we want "the mean of the squares" to be
.
Note that the denominator is
.
Now,
.
We can see that a factor
has appeared, so we just need to show that
.
But this is certainly true:
as required.



So we want "the mean of the squares" to be

Note that the denominator is

Now,

We can see that a factor


But this is certainly true:

Expression for Var(Y):
Do non-zero

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metaltron
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Question 6:
Part i)
ii)
iii)
Part i)
ii)
Spoiler:
Show
iii)
Spoiler:
The two required series converge. This can be seen by considering:

From this since the bottom of the fraction is always positive, both the numerator and the LHS have the same sign. It can be shown therefore by computing a few terms that the series with odd n is increasing and the series with even n is decreasing. Hence, the series of odd n is increasing and is bounded above by 2. Also, the negative of the even n sequence is bounded above by -1 and is strictly increasing. It follows that the two series converge.
As the the two required series converge:

Therefore we wish to solve:

Also:



We also have:

so:



We can discard the negative solution since all terms in the series are positive and between 1 and 2 inclusive. Hence both series converge to the above positive fraction. And since both series converge to this value, u_n converges to:

Changing u_1 to 3 does not change this answer.
Show
The two required series converge. This can be seen by considering:

From this since the bottom of the fraction is always positive, both the numerator and the LHS have the same sign. It can be shown therefore by computing a few terms that the series with odd n is increasing and the series with even n is decreasing. Hence, the series of odd n is increasing and is bounded above by 2. Also, the negative of the even n sequence is bounded above by -1 and is strictly increasing. It follows that the two series converge.
As the the two required series converge:

Therefore we wish to solve:

Also:



We also have:

so:



We can discard the negative solution since all terms in the series are positive and between 1 and 2 inclusive. Hence both series converge to the above positive fraction. And since both series converge to this value, u_n converges to:

Changing u_1 to 3 does not change this answer.
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Zorgz
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(Original post by Smaug123)
Q12:
i)
ii)
Expression for Var(Y):
Do non-zero
exist?
Q12:
i)
Spoiler:
Show
ii)
Spoiler:
Show
The variance of X is "the mean of the squares - the square of the mean". The mean is
, which squares to
. Note that
, and hence the second term in the expression for the variance is precisely "the square of the mean".
So we want "the mean of the squares" to be
.
Note that the denominator is
.
Now,
.
We can see that a factor
has appeared, so we just need to show that
.
But this is certainly true:
as required.



So we want "the mean of the squares" to be

Note that the denominator is

Now,

We can see that a factor


But this is certainly true:

Expression for Var(Y):
Do non-zero

I got very similar but ended up having to show that a^2+b^2/(a+b)^2= 1, which of course isn't possible for non zero values. Not sure but think you might have missed out a lambda squared term when cancelling.
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Smaug123
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(Original post by Zorgz)
Not sure but think you might have missed out a lambda squared term when cancelling.
Not sure but think you might have missed out a lambda squared term when cancelling.

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Slowbro93
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#16
Getting ready to type up a solution to question 5
Also going to update OP

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Slowbro93
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MW24595
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Question 8
i- Let
be the area of any rectangle with sides parallel to the co-ordinate axes, with one side on
and another on the y-axis, and with one corner on the curve f(x). It is clear that, the rectangle with the maximum area will be one of this type.
Looking at a rudimentary diagram, it is evident that:

Now, as we're concerned with maximising area, let
be the value of x for which
is greatest. So,

By definition,
is the maximum value of
so,

ii- So, we have,

Now, notice that on integrating, we have a difference between a function of t and a constant. On differentiating this with respect to t, we just get
on the RHS. To see this, let
be the antiderivative of
.
Then,

Now, consider that:

Further, as
is a decreasing function, we have, for all
,

Further, for
, we have,

Therefore, we clearly have,

Now, observe that,

iii-

Now, if order to find
, we simply plug these into the condition for
, namely:

Now,

Similarly, we have,

Now, from the previous part's inequality, we get,
i- Let


Looking at a rudimentary diagram, it is evident that:

Now, as we're concerned with maximising area, let



By definition,



ii- So, we have,

Now, notice that on integrating, we have a difference between a function of t and a constant. On differentiating this with respect to t, we just get



Then,

Now, consider that:

Further, as



Further, for


Therefore, we clearly have,

Now, observe that,

iii-

Now, if order to find



Now,

Similarly, we have,

Now, from the previous part's inequality, we get,

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jack.hadamard
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I had to practice some drawing on a software, so enjoy.
Question 4
Preliminary stuff
First part
Second part (a)
Second part (b)
Question 4
Preliminary stuff
Spoiler:
Show
First part
Spoiler:
Show
Second part (a)
Spoiler:
Show
Second part (b)
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Brister
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#20
(Original post by LShirley95)
Got Q9 written with diagrams, unfortunately don't have a scanner so this is the best I can do:
![Name: 1371760817848.jpg
Views: 3072
Size: 24.2 KB]()
![Name: 1371760837517.jpg
Views: 2610
Size: 29.6 KB]()
![Name: 1371760848354.jpg
Views: 2203
Size: 22.9 KB]()
Posted from TSR Mobile
Got Q9 written with diagrams, unfortunately don't have a scanner so this is the best I can do:
Posted from TSR Mobile
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