STEP Maths I, II, III 2001 Solutions
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SimonM
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#1
(Updated as far as #40) SimonM - 28.03.2009
STEP I:
1: Solution by Glutamic Acid
2: Solution by Aurel-Aqua
3: Solution by Aurel-Aqua
4: Solution by SimonM
5: Solution by SimonM
6: Solution by Unbounded
7: Solution by brianeverit
8: Solution by SimonM
9: Solution by sonofdot
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by Mark13
14: Solution by SimonM
STEP II:
1: Solution by Glutamic Acid
2: Solution by ForGreatJustice
3: Solution by SimonM
4: Solution by Dadeyemi
5: Solution by Glutamic Acid
6: Solution by sonofdot
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by Glutamic Acid
10: Solution to tommm
11: Solution by Glutamic Acid
12: Solution by brianeverit
13: Solution by Aurel-Aqua
14: Solution by Aurel-Aqua
STEP III:
1: Solution by Aurel-Aqua
2: Solution by SimonM
3: Solution by tommm
4: Solution by tommm
5: Solution by SimonM
6: Solution by Dadeyemi
7: Solution by tommm
8: Solution by SimonM
9: Solution by tommm
10: Solution by brianeverit
11: Solution by tommm
12: Solution by brianeverit
13: Solution by Aurel-Aqua
14: Solution by SimonM
Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
STEP I:
1: Solution by Glutamic Acid
2: Solution by Aurel-Aqua
3: Solution by Aurel-Aqua
4: Solution by SimonM
5: Solution by SimonM
6: Solution by Unbounded
7: Solution by brianeverit
8: Solution by SimonM
9: Solution by sonofdot
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by Mark13
14: Solution by SimonM
STEP II:
1: Solution by Glutamic Acid
2: Solution by ForGreatJustice
3: Solution by SimonM
4: Solution by Dadeyemi
5: Solution by Glutamic Acid
6: Solution by sonofdot
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by Glutamic Acid
10: Solution to tommm
11: Solution by Glutamic Acid
12: Solution by brianeverit
13: Solution by Aurel-Aqua
14: Solution by Aurel-Aqua
STEP III:
1: Solution by Aurel-Aqua
2: Solution by SimonM
3: Solution by tommm
4: Solution by tommm
5: Solution by SimonM
6: Solution by Dadeyemi
7: Solution by tommm
8: Solution by SimonM
9: Solution by tommm
10: Solution by brianeverit
11: Solution by tommm
12: Solution by brianeverit
13: Solution by Aurel-Aqua
14: Solution by SimonM
Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
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SimonM
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#2
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions
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Aurel-Aqua
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#3
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#3
I Question 2
Part (i)
,
.
First it might help to find the points of equality: multiplying by x and rearranging, we get
,
. The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for
(the zero due to the asymptote) and
.
Part (ii)
,
.
Because of the condition, squaring both sides still leaves an iff, giving:
. Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again,
, giving
. From this, we can deduce that
and
, but our new condition,
only gives us
as the final solution.
Part (i)


First it might help to find the points of equality: multiplying by x and rearranging, we get




Part (ii)


Because of the condition, squaring both sides still leaves an iff, giving:







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sonofdot
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#4
ForGreatJustice
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#5
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#5
STEP II Question 2
Sketch not included, but pretty simple (y=0 for 0<x<N, and y=1 for N<x<2N)
i)
for
can be split into 3 regions:
for
for
and
for 
So by splitting the summation into these three ranges we get
![\displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k = \displaystyle\sum_{k=1}^{N-1} k - \displaystyle\sum_{k=N}^{2N-1} k + 2N \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k = \displaystyle\sum_{k=1}^{N-1} k - \displaystyle\sum_{k=N}^{2N-1} k + 2N](https://www.thestudentroom.co.uk/latexrender/pictures/fa/faf1e2a37774dd4aa4420d6890a88401.png)



as required
ii)![S_N= \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} 2^{-k} S_N= \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} 2^{-k}](https://www.thestudentroom.co.uk/latexrender/pictures/83/831e3a0ddf541913cb0b3688b094f8f5.png)
Splitting into 3 as before:

Using the formula for the sum of a geometric progression
we get

Which could probably be made simpler, so now we can see as
then
Sketch not included, but pretty simple (y=0 for 0<x<N, and y=1 for N<x<2N)
i)
![[k/N] [k/N]](https://www.thestudentroom.co.uk/latexrender/pictures/5e/5e44befb593e5c352691ba4f7b5a3ee6.png)

![[k/N]=0 [k/N]=0](https://www.thestudentroom.co.uk/latexrender/pictures/86/86075123db3c03cb3443cff2a5486ff6.png)

![[k/N]=1 [k/N]=1](https://www.thestudentroom.co.uk/latexrender/pictures/b5/b5a484c24ac8e53110e5da99a5c5ae2f.png)

![[k/N]=2 [k/N]=2](https://www.thestudentroom.co.uk/latexrender/pictures/ee/eed826c616b841dc0cfc8f7d38b08ab4.png)

So by splitting the summation into these three ranges we get
![\displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k = \displaystyle\sum_{k=1}^{N-1} k - \displaystyle\sum_{k=N}^{2N-1} k + 2N \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k = \displaystyle\sum_{k=1}^{N-1} k - \displaystyle\sum_{k=N}^{2N-1} k + 2N](https://www.thestudentroom.co.uk/latexrender/pictures/fa/faf1e2a37774dd4aa4420d6890a88401.png)



![\displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k =2N-N^2 \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k =2N-N^2](https://www.thestudentroom.co.uk/latexrender/pictures/7f/7f5da0b41e945927bd1ac0bb30195a5b.png)
ii)
![S_N= \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} 2^{-k} S_N= \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} 2^{-k}](https://www.thestudentroom.co.uk/latexrender/pictures/83/831e3a0ddf541913cb0b3688b094f8f5.png)
Splitting into 3 as before:

Using the formula for the sum of a geometric progression



Which could probably be made simpler, so now we can see as


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Aurel-Aqua
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I Question 2
,
. Well, we sketch the graph to show the two stationary points and crossing the x-axis at
,
,
. We can conclude the two stationary points exist, so there must be a real x-value for them to be found.
.
Since the two stationary points are distinct and real, by
, we get
.
Considering
, to have three roots, there must be a real solution to the quadratic, giving
. This implies that the two other roots are
and
, and
. Letting
,
,
in any order (since they can be changed around due to triple-symmetry ). Knowing that
and
. Substituting into our previous result, we obtain:
, as required. We can show that it is not necessary by letting
,
,
.






Since the two stationary points are distinct and real, by


Considering














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Mark13
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#7
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#7
STEP I - Q13
Given that plate is broken, the probability that it is broken by the mathematician is 0.25.
Therefore, the number of plates,
, broken by the mathematician can be represented by the following binomial distribution:
~
We need to find
when n=5.



Given that plate is broken, the probability that it is broken by the mathematician is 0.25.
Therefore, the number of plates,



We need to find





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qgujxj39
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#8
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#8
I shall get going. 
Meikleriggs only has solutions since 2004 now

(Original post by SimonM)
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions

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Glutamic Acid
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#9
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#9
II/11:
Highest point of trajectory when v = 0, so 0 = v - gt and t = v/g. The horizontal distance travelled will be uv/g at this time, and the vertical will be
.
P will hit the ground when its vertical displacement is
, so one must solve the quadratic equation
giving
(using the quadratic formula). The horizontal distance travelled in this time will be
.
Adding the distances:
, as required.
If R > D then

, as required.
Highest point of trajectory when v = 0, so 0 = v - gt and t = v/g. The horizontal distance travelled will be uv/g at this time, and the vertical will be

P will hit the ground when its vertical displacement is




Adding the distances:

If R > D then



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Aurel-Aqua
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#10
III Question 1


Case when
:
.
Differentiating the given relation, we get:

.
Since it is true for n = 0, and it is also true for n + 1, by mathematical induction we conclude that it must also be true for n = 1, n = 2, and so forth for all positive values of n.
Substituting
into the relation, we get
. Thus
, all even powers are zero.
.
Maclaurin series with terms of even powers equal to zero is
, so
.


Case when


Differentiating the given relation, we get:


Since it is true for n = 0, and it is also true for n + 1, by mathematical induction we conclude that it must also be true for n = 1, n = 2, and so forth for all positive values of n.
Substituting




Maclaurin series with terms of even powers equal to zero is


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DFranklin
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#11
(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).
On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).
I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.
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SimonM
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#12
Question 4, STEP I 2001
Spoiler:
Show
Consider
by De Moivre's Theorem.
Therefore
Dividing top and bottom by
yields:

Since
, 
Therefore
Therefore

The range of
is
. Considering that as the domain of
, there are three places where
.
is periodic with period of 
Therefore
n =0,1,2
So
The range of
is
. This gives cos a smaller range between
, so there will be two solutions, on negative and one positive, that is


Therefore

Dividing top and bottom by


Since


Therefore

Therefore


The range of

![[0,3\pi] [0,3\pi]](https://www.thestudentroom.co.uk/latexrender/pictures/d8/d8dc502800a49d63f9d95f3bfa304f9a.png)




Therefore

So

The range of




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Glutamic Acid
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#13
II/5:
dy/dx = 0 when x = 0, so there's a stationary point.
, which equals 1 at x = 0, so it's a minimum point. Consider
, letting u = x^2 gives du = 2x dx so it becomes
. From inspection,
. No constant as the curve passes through the origin. At x = 1, y =
and
; negative, so a maximum.
The other stationary point is
as it's an even function (which is to be expected as the derivative is odd).
For C_2:
. At x = 0, this equals 1; so minimum. At x = 1, this equals
, negative, so maximum.
The derivatives of C_1 and C_2 are equal at x = 0 and x = 1. If the derivative of C_2 is greater than C_1 for 0 < x < 1 then it follows that
. This relies on showing that
for 0 < x < 1, which can be shown with a very simple sketch. As e^(x^2) and e^(x^3) < 1 for 0 < x < 1, when taking logarithms the inequality is reversed.
dy/dx = 0 when x = 0, so there's a stationary point.






The other stationary point is

For C_2:


The derivatives of C_1 and C_2 are equal at x = 0 and x = 1. If the derivative of C_2 is greater than C_1 for 0 < x < 1 then it follows that


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qgujxj39
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#14
STEP III 2001, QUESTION 3
Consider the equation

where b and c are real numbers.
(i) Show that the roots of the equation are real and positive if and only if
and
, and sketch the region of the b-c plane in which these conditions hold.
(ii) Sketch the region of the b-c plane in which the roots are real and less than 1 in magnitude.
(Apologies for any dodgy copypasting, notepad is a ***** to work with sometimes.)
Consider the equation

where b and c are real numbers.
(i) Show that the roots of the equation are real and positive if and only if


Spoiler:
let the roots be
then by the quadratic formula:


Clearly
, so both roots are positive iff 
therefore
(1)
If
, then
, which cannot be true if the roots are real. Therefore
(first condition), and thus we can square both sides of equation (1):

or
(second condition)
Finally,
(third condition), otherwise the roots would clearly be complex.
The three conditions obtained are equivalent to those stipulated in the question.
Pathetic sketch, where the curve drawn is the positive half of
and the shaded area underneath (inclusive of the curve itself) is where the conditions hold:
![Image]()
Show
let the roots be

then by the quadratic formula:


Clearly


therefore

If




or

Finally,

The three conditions obtained are equivalent to those stipulated in the question.
Pathetic sketch, where the curve drawn is the positive half of


(ii) Sketch the region of the b-c plane in which the roots are real and less than 1 in magnitude.
Spoiler:


By assumption, we can take
, therefore these conditions become
, as these both imply the other inequalities involved. These lead to:


The right hand sides of these must clearly be positive, therefore
. On this assumption we can square both inequalities, which (after some manipulation) become:


Plotting these on a graph, together with
, we get something looking vaguely like:
![Image]()
Show


By assumption, we can take




The right hand sides of these must clearly be positive, therefore



Plotting these on a graph, together with


(Apologies for any dodgy copypasting, notepad is a ***** to work with sometimes.)
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Glutamic Acid
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#15
(Original post by DFranklin)
It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.
(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).
On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).
I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.
It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.
(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).
On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).
I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.
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SimonM
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#16
STEP I 2001, Question 5
Which happens to be true. (For more info, look into differentiation under the integral sign)
Which happens to be true. (For more info, look into differentiation under the integral sign)
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vahik92
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#17
(Original post by Aurel-Aqua)
I Question 2
Part (i)
,
.
First it might help to find the points of equality: multiplying by x and rearranging, we get
,
. The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for
(the zero due to the asymptote) and
.
Part (ii)
,
.
Because of the condition, squaring both sides still leaves an iff, giving:
. Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again,
, giving
. From this, we can deduce that
and
, but our new condition,
only gives us
as the final solution.
I Question 2
Part (i)


First it might help to find the points of equality: multiplying by x and rearranging, we get




Part (ii)


Because of the condition, squaring both sides still leaves an iff, giving:








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DFranklin
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#18
(Original post by Glutamic Acid)
Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.
Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.
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Aurel-Aqua
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#19
SimonM
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#20
STEP I, Question 14
Spoiler:
Show
Spoiler:
Show
The probability the best amongst the first n candidates is best or second best is the probability it is the best is there + probability that the second best is there - probability that both are there. This is

Let the candidates be
where ability at maths is measured by one's initial.
Possibilities for interviews are

So the probability that the best guy is there is
The probability that one of the best two are there is

Let the candidates be

Possibilities for interviews are

So the probability that the best guy is there is

The probability that one of the best two are there is

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