You can let someone else take control of the OP whenever.
STEP I:
1: Solution by Unbounded
2: Solution by nuodai
3: Solution by nuodai
4: Solution by Unbounded
5: Solution by darkness9999
6: Solution by SimonM
7: Solution by SimonM
8: Solution by Glutamic Acid
9: Solution by Unbounded
10: Solution by SimonM
11: Solution by brianeverit
12: Solution by Farhan.Hanif93
13: Solution by AurelAqua
14: Solution by AurelAqua
STEP II:
1: Solution by DeanK22
2: Solution by DeanK22
3: Solution by SimonM
4: Solution by sonofdot
5: Solution by Daniel Freedman
6: Solution by Elongar
7: Solution by SimonM
8: Solution by Daniel Freedman
9: Solution by brianeverit
10: Solution by Farhan.Hanif93
11: Solution by Farhan.Hanif93
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by Farhan.Hanif93
STEP III:
1: Solution by Daniel Freedman
2: Solution by Daniel Freedman
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Elongar
6: Solution by Anonymous
7: Solution by tommm
8: Solution by DeanK22
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
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STEP 2006 Solutions Thread

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 1
 05052009 21:31
Last edited by SimonM; 04082011 at 21:19.Post rating:1 
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 2
 05052009 22:17
STEP III 2006, Question 2
Last edited by Daniel Freedman; 06052009 at 01:37. 
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 3
 05052009 22:40
STEP III 2006, Question 3
Spoiler:Show
i) tan is an odd function ie tan(x) =  tan(x) and therefore can have no even powers of x in its series expansion
Prove
as required.
ii)
as required.
iii)
which is true by Pythagoras'.
EDIT: There was no point in factorising this way. I did the question whilst typing it up, and it seemed an obvious thing to do.
Equating coefficients gives:
as required.
Last edited by Daniel Freedman; 05052012 at 12:06.Post rating:2 
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 4
 06052009 02:12
STEP III 2006, Question 1
Spoiler:Show
Therefore the curve has vertical asymptotes with equations and .
The curve has roots .
As , so the curve has an oblique asymptote with equation .
As . This is the equation of the tangent to the curve at the origin (can be verified by differentiating)
[SKETCH ATTACHED]
i)
[SKETCH ATTACHED]
The graph shows that the two curves have three points of intersection, which means the above equation has three real roots
ii)
[SKETCH ATTACHED]
The graph shows the two curves only have one point of intersection (using the fact that the line has the same gradient as the tangent to the curve at the origin). This means the above equation has one real root.
iii)
[SKETCH ATTACHED]
The graph shows the curves have six points of intersection, which means the above equation has six real roots (the curves don't meet again because there are only two solutions ofLast edited by Daniel Freedman; 06052009 at 02:21. 
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 5
 10052009 02:50
STEP III 8
We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows;
We will now prove property 1
Spoiler:Show
We now prove property 2 is met by this operation
We prove property three is met by the above operation.
We now prove the final property;
We now need to show that delta (the operation given in the question) returns nx^{n1} after being applied to x^n.
Spoiler:ShowWe proceed by induction.
Assume that for some k in N that delta(x^k) = kx^(k1). Assume that the result holds for k + 1.
P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains N. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.
As requiredLast edited by Oh I Really Don't Care; 10052009 at 03:49.Post rating:3 
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 10052009 09:38

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 7
 10052009 11:01
(Original post by DeanK22)
STEP III 8
We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows;
We will now prove property 1
Spoiler:Show
We now prove property 2 is met by this operation
We prove property three is met by the above operation.
We now prove the final property;
We now need to show that delta (the operation given in the question) returns nx^{n1} after being applied to x^n.
Spoiler:ShowWe proceed by induction.
Assume that for some k in N that delta(x^k) = kx^(k1). Assume that the result holds for k + 1.
P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains N. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.
As required 
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 8
 10052009 13:08
STEP III 2006, Question 4
Last edited by Dadeyemi; 10052009 at 13:13. 
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 10052009 14:21
2006 I Question 13
Part (i)
The number of diamonds in one kilogram has a Poisson distribution of . We find that per 100 grams, . For 100T grams, . is the distribution of scores of the die.
Drawing 1, we have .
Similarly,
(recognising a finite geometric sum).
Part (i)  expectation
Part (ii)
This time, is a geometric distribution with . Thus, .
This time, our probability is:
, as required.
Part (ii)  expectation
.
Consider: .
Thus, . 
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 10
 10052009 14:40
2006 I Question 14
Either I missed something crucial, or this question is incredibly easy. I'll write up the incrediblyeasy interpretation of this question:
Part (i)
This is a geometric progression, where , where , , so .
Differentiating, we get: . Equating to zero, . We know that this is the maximum as r tends to infinity, P tends to zero.
Part (ii)
. Clearly maximum at n = 1.
Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.
Edit: see GHOSH5's post here: http://www.thestudentroom.co.uk/show...3&postcount=41Last edited by AurelAqua; 13062009 at 10:53. 
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 11
 10052009 14:42
(Original post by Daniel Freedman)
I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that , using the four rules. Then let , and apply the operation to this function, making use of rules ii) and iii) to show that .
They've said "given that the operator has these properties; show that...".
Dean has said "A particular operator has those properties, and for , we have ...."
But he hasn't shown that is the only possible operator with the properties.
This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised). 
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 12
 10052009 14:52
(Original post by DFranklin)
Actually, it's worse than that.
They've said "given that the operator has these properties; show that...".
Dean has said "A particular operator has those properties, and for , we have ...."
But he hasn't shown that is the only possible operator with the properties.
This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised). 
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 13
 10052009 15:35
Well, that's effectively what the question is asking you to show. I would:
(1) Prove by induction that (using (i) and (iv)).
(2) Then use (iii) to prove
(3) Then use (ii) to prove .
[Which effectively proves uniqueness of for polynomials].
Then simply observe this is the same as the derivative in the case of polynomials. 
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 14
 12052009 13:29
I have different answers to STEP I Q 14 Im not to sure who is wrong...
Part (i)
Spoiler:Show
Part (ii)
Last edited by darkness9999; 12052009 at 20:46. 
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 15
 12052009 16:34
STEP I: Q5
Part (i)
Substitution:
Therefore the integral becomes
Simplifying this further using partial fractions gives us
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\Rightarrow \ln { \frac{u3}{u+3} + C
If we now sub in the substitution we used we get
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.{\ln {\frac{\sqrt {2x+1}3}{\sqrt {2x+1}+3}+C
Part (ii)
I used this substitution:
So the integral becomes:
Simplifying this further using partial fractions gives us
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\Rightarrow \left[\frac{1}{u+1}  \frac{1}{u1} + \ln {\frac{u+1}{u1} \right]^{3}_{2}
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\Rightarrow \left( (\frac{1}{4} \frac{1}{2} + \ln {\frac{4}{2}) (\frac{1}{3} 1 + \ln 3) \right)
Last edited by darkness9999; 12052009 at 16:48. 
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 16
 20052009 00:05
STEP III 2006, Q5
We wish to show that form an equilateral iff
Multiplying by two and factorising, we wish to show that,
holds for all equilateral triangles.
Note that are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is . Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.
Using this, we reconstruct the equation:
or, more simply,
The bracket evaluates to 0, so we are done.
For the next part, we take the roots of the cubic to be , and expand, yielding Vieta's formula:
Algebra shows that:
and the result follows immediately.
Now we write,
The transformation has the effect of rotating our equilateral triangle by the angle , magnifying it by a factor of and translating it through the vector represented by (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle. 
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 17
 29052009 16:11
STEP I 2006 Question 2
Solution
The barn is square, so it doesn't matter which side the goat is on. Because the lengths are of side 2a and the rope is of length 4a, it makes sense that we should consider the case where the rope is between the middle of a side and the edge of the side
In this, let P be the position of the goat, O be the position where the rope is tethered, and ABCD be the barn, with O being along AB and A, B, C, D marking the corners of the barn, going clockwise. Bear in mind that this is much more easily explained with diagrams, so try drawing it if you don't follow.
Let's say that O is located along AB at distance x from A, where 0 < x < a (so that it is between the midpoint of AB and the point A)
The distance OA is , so the distance OA is and so on. This means that if the goat walks as far as it can clockwise, P will be located at a distance from C along the line CD. So, the goat can make:
 A quartercircle with radius CP = x
 A quartercircle with radius BP = 2a + x
 A halfcircle with radius OP = 4a
 A quartercircle with radius AP = 4a  x
 A quartercircle with radius DP = 2a  x
So the area would be:
This therefore gives us the maximum and minimum values. The maximum value must occur when since , and the maximum value must occur when , therefore
Last edited by nuodai; 12062009 at 14:10. 
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 18
 29052009 16:13
RE; Noudai
opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker. 
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 29052009 16:21
(Original post by DeanK22)
RE; Noudai
opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.Post rating:1 
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 20
 29052009 16:34
STEP III 2006 Q7
Spoiler:ShowPart i):
Using the quadratic formula we obtain
If we let , we obtain the differential equation they're asking about, so:
Using the condition that at , we see that the must be a +, hence
Integrating we obtain
and using y = 0 when x = 0 we find that c = 1.
Part ii):
Using the quadratic formula again to find dy/dx, we obtain (apologies for not writing it out in full):
We can integrate the left hand side using recognition: the top is the derivative of the bottom, so:
Now, if the sign were a minus, then the condition x = 0 when y = 0 would be impossible, as this would mean that the arbitrary constant were infinite. We can therefore deduce that the must be a +. Therefore:
Using x = 0 when y = 0, we find that c = ln2. Therefore
, which is the solution of the differential equation.
Writing the cosh in full exponentials and multiplying by 2 gives:
To find the asymptotes, we consider two cases:
1) when , we can disregard and because they are comparatively small, so . Rearranging gives the equation of the asymptote, .
1) when , we can disregard and because they are comparatively small, so . Rearranging gives the equation of the other asymptote, .
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Updated: May 30, 2015
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