I enjoyed doing the question, quite analysisy, in a certain sense, but formalising it was a real pain and I just couldn't quite finish it off.(Original post by DFranklin)
My recollection of looking at this was that this question was in a completely different league from the other items on the list and even a halfway rigourous answer was extremely tricky to obtain.
I'm really not sure what they are after on this one.
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 26102015 17:29

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 27102015 08:04
Test 2, Question 3
Hint
Spoiler:Show
Consider the relationship between the coefficients of a polynomial and its roots.
Solution
Spoiler:Show
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DFranklin
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 27102015 08:38
(Original post by Gifted)
Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it. 
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 27102015 09:21
Specimen Test 2, Question 1:
Hints:Spoiler:Solution:ShowConsider logarithms... more than onceLast edited by Zacken; 05062016 at 17:25. 
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 27102015 09:29
Another solution to 9:
Spoiler:ShowObserve first that is a continuous function.
Now, for , since is very large, we may assume this holds.
Also:
For
Hence by the Intermediate Value Theorem, such that
Suppose that such a solution was not unique, then let two such distinct solutions be and .
Then:
w.l.o.g, let
But:
A contradiction, meaning that .
Now if
As for reasonable sized which occur when is large, we have .
Letting we see that:
For large we may write .
.

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 27102015 10:38
Test 2, Question 4:
Hints:Spoiler:Solution:ShowSketch and  find the roots of each and the value of each at to help you sketch it. That should reveal the solutions immediately. For the second graph, solve to help you find the stationary points and aid you in graphing, you'll need to make use of your previous result in finding the stationary pointsSpoiler:ShowWe sketch and . They both have a root at so that's one solution and to help us sketch the graphs, we look at only to find that's a solution as well. Since our graph clearly depicts only two solutions, all our solutions are given by .
Sketching the second graph:
First thing to note it that the domain of our function is .
1. An easy root (by inspection) is since we get . (Note that the graph touches the xaxis here, we justify this later)
2. Next, we look at the yintercepts: yields .
3. We can also see that so we know that the tail of the graph will look exponential (since e^x dominates x^e for large x).
4. Stationary points (this is the fun part): so we know that the stationary points are given by upon which, taking logarithms yields so we have so our root is also a turning point, meaning that it has to touch and not cross the xaxis.
Putting all of this together yields: ( is a maximum and a minimum).Last edited by Zacken; 27102015 at 10:40. 
DFranklin
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 27102015 10:54
(Original post by Zacken)
Specimen Test 2, Question 1:
Hints:Spoiler:Solution:ShowConsider logarithms... more than once
Spoiler:Show100^100^n < 128^128^n = 2^(7*2^7n) < 2^2^(7n+3) < 2^2^2^n if 2^n > 7n+3, which is clearly true for n>10 (can prove by induction if you really want to). 
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 27102015 10:59
(Original post by DFranklin)
Alternative:Spoiler:Show100^100^n < 128^128^n = 2^(7*2^7n) < 2^2^(7n+3) < 2^2^2^n if 2^n > 7n+3, which is clearly true for n>10 (can prove by induction if you really want to).
It's a good technique as well, for inequalities that have one quantity being much bigger/smaller than the other, you should make use of all the extra room to your advantage. Although I always have trouble knowing what to do initially that will get me what I want down the road. 
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 27102015 11:41
Specimen Test 1, Question 10:
Hints:Spoiler:Solution Part 1:ShowSpoiler:Solution Part 2:ShowWe have 20 balls in an urn
5 red, 5 green, 5 yellow, 5 blue.
Please note the question states:
'Write down expressions for the probabilities of the following events. (You need not calculate their numerical values.)'
Though I will be giving numerical values to aid the learning process.
Three balls are drawn from the urn in random, therefore let us note that there are possible outcomes.
A) The question specifically asks for exactly one of the balls drawn to be red. However, we've already established that three balls are drawn at random.
We should split it up into reds and nonreds.
We have 5 reds and 15 nonreds.
We want exactly one red.
There are ways to pick exactly one red. (As there are five reds, and we want one)
There are ways to pick the remaining nonreds. (As there are 15 nonreds, and we want two due to the fact that there can only be one red in the three drawn)
Multiply these together and we have ways of picking one red and 2 nonreds.
We return to the following statement
ThereforeSpoiler:ShowWe have 20 balls in an urn
5 red, 5 green, 5 yellow, 5 blue.
Please note the question states:'Write down expressions for the probabilities of the following events. (You need not calculate their numerical values.)'
Though I will be giving numerical values to aid the learning process
Three balls are drawn from the urn in random, therefore let us note that there are possible outcomes.
B) This one uses the same concept as the previous question, however, instead of having one exact colour, we now have 3 unknown balls. Only thing we know is that they should be different colours from eachother.
We have 4 colours, and we want only 3 different colours.(As three are drawn randomly)
There are ways of picking 3 different colours. (4 ways)
Lets not make the mistake of using specific colours in our question (like red, blue, etc.), it'll only make it more confusing and difficult.
Colour 1: there are only 5 of these coloured balls;
Colour 2: there are only 5 of these coloured balls;
Colour 3: there are only 5 of these coloured balls;
We can 'ignore' the last 'colour' as we eliminated this colour from the process the moment we did
ways
If there are 4 ways of picking 3 different colours, and 125 ways of picking 1 ball from each of those colours, this would mean there are 125 x 4 ways to get three different colours
Last edited by edothero; 27102015 at 16:10. 
DFranklin
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 27102015 12:20
(Original post by edothero)
Specimen Test 1, Question 10:
Spoiler:ShowC) ... is quite the questionSpoiler:ShowConditions are met if we have 2 blue, 3 blue, or 1 blue and 0 yellow.
p(2 blue) = 3 x 5/20 * 4/19 * 15/18 = 3 /19 * 15/18 = 5/38 (multiplying by 3 because subsequent calculation assumes BB are first 2 drawn, so need to allow for xBB, BxB where x is "nonblue")
p(3 blue) = 5/20 * 4/19 * 3/18 = 3/(19*18) = 1 / (6 * 19)
p(1 blue, 0 yellow) = 3 * 5/20 * 10/19 * 9/18 = 3/4 * 10/19 * 1/2 = 15 / (4*19) (again multiply by 3 because of 3 positions in which B can be drawn).
Answer is sum of these. Common denominator is 12 * 19 = 228, so we have (30+2+45)/228 = 77 /228. 
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 27102015 12:25
Test 2, Question 2:
Hints:
Solution:
Part 1:Spoiler:Show
Part 2: 
username1763791
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 27102015 13:46
Specimen Test 2, Question 6
Hint:Solution 
First Part:Second Part:Spoiler:Show
Using DeMoivre
From double angle
OR, you can use the first part and say that
and work from there, which should be easy enough if you have done the first part and you end up with the same cubic either way.
Let
So,
By inspection we have
Then, by either using a calculator , or by going the long way and finding
We get the final answers of
Where, in the question,
However, I know that Zacken and Renzhi10122 have some nicer methods for this last bit.
Last edited by username1763791; 27102015 at 13:51. 
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 27102015 15:11
Specimen 2, Q7
Hint
Spoiler:Show
(x1) is a factor of x^n 1 always.
When is (x+1) a factor of x^n+1?
Solution
Spoiler:Show
Unless otherwise stated, assume all new constants are positive integers.
Suppose p = a^n  1.
Then p = (a1)(a^n1 + ... + 1)
So p is composite unless a1 = 1 (as the second bracket is greater or equal to the first), so a=2.
So p = 2^n  1.
Suppose n is composite, i.e. n=cd, with neither c or d less than 2.
Then p =l 2^(cd) 1 = (2^c)^d 1 = (2^c 1)(2^(cdc) + ... +1)
As c>1, p is thus composite. So n cannot be composite  in other words, n is prime.
  
Let q = a^n + 1. We seek the conditions in which q is prime.
If a is 1, then q is clearly always prime.
If a is an odd integer greater than 1, then a^n is odd, so q is even (and not 2) hence it is never prime in this case.
If a is an even integer, then if n is odd and greater than 1, q=(a+1)(a^n1  ... +1). q is thus never prime as a+1 > 1.
If a is an even integer, then if n is even but has an odd factor r where r>1, and n=rs, then q = a^(rs) +1 = (a^s)^r + 1 = (a^s+1)(a^rss... +1) and so q is never prime.
This leaves the only possibility as when a is even, and n is even and has no odd factors > 1, i.e. n is a nonnegative integer power of two. Examples of this would be
2^4 +1 =17, 6^1+1 = 7, 6^2 +1 =37.
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 27102015 15:48
Specimen Test 1, Question 1:
Solution 1:Spoiler:Solution 2:ShowIn a tennis tournament there are 2n participants.
In the first round, each player plays exactly one, so there are n games.
There are ways to order everyone.You would divide this by ways to get the total number of combinations that players could be chosen to play games.
There are ways to order each of these pairs.
Therefore you have You could further split up these factorials like so,When further simplified, gives you:As required.
Spoiler:ShowLast edited by edothero; 01112015 at 02:48. 
Renzhi10122
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 27102015 15:53
Specimen 2, q8
Hint:Spoiler:Solution:ShowFind the area of a fifth of the pentagon, namely a triangle of itSpoiler:ShowLet AC'=a, C'D'=b and CC'=c.We have where O is the center of the pentagon. Then giving by double angle formulae. Also, if then . Noting that , we have .
Now, so .
We have so
We (I) know that so so
EDIT: Just realised that I haven't actually finished the question...
Hence, the area of the pentagon isLast edited by Renzhi10122; 09122015 at 08:00. 
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 27102015 16:20
Test 2, Q10
Hints
Spoiler:Show
Might be worth considering energy.
Solution
Spoiler:Show
For the particle to rest, friction must exceed the restoring force. So F>kx, i.e. x< F/k.
For the second part, consider energy. Call the excursion at the start of the halfcycle x0, and the excursion at the end x1. We want to find x0  x1.
Initially, the energy in the spring is 0.5k(x0)^2. It then loses F*(x0+x1) of energy to friction. It finally has 0.5k(x1)^2 of energy.
, as required.
To discuss the motion, you could say that the particle oscillates about the point at which the spring is anchored, with decreasing amplitude as time goes on until it comes to a stop (as soon as the amplitude falls below F/k).
Last edited by Krollo; 27102015 at 16:22. 
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 27102015 16:45
Edit: Stop doing maths and enjoy your vacation!Last edited by Zacken; 27102015 at 16:46. 
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 27102015 17:26
(Original post by Zacken)
My mechanics is a little rusty, but would the 'discuss' basically just be damped SHM?
Edit: Stop doing maths and enjoy your vacation!
And I am enjoying my holiday  I'm doing maths, of course. :) 
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 27102015 17:27
Specimen Paper 2, Question 9:
Caveat  my probability is a bit weak and I'm not convinced about the last part.
Also succinct notation was a pain.Spoiler:ShowLast edited by joostan; 27102015 at 17:29. 
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 27102015 19:18
(Original post by Krollo)
Test 2, Q10
Hints
Spoiler:Show
Might be worth considering energy.
Solution
Spoiler:Show
For the particle to rest, friction must exceed the restoring force. So F>kx, i.e. x< F/k.
For the second part, consider energy. Call the excursion at the start of the halfcycle x0, and the excursion at the end x1. We want to find x0  x1.
Initially, the energy in the spring is 0.5k(x0)^2. It then loses F*(x0+x1) of energy to friction. It finally has 0.5k(x1)^2 of energy.
, as required.
To discuss the motion, you could say that the particle oscillates about the point at which the spring is anchored, with decreasing amplitude as time goes on until it comes to a stop (as soon as the amplitude falls below F/k).
Spoiler:Show
1. I think that the nomotion condition should be
2. If we write down the equations of motion we get:
with corresponding solutions:
with
Now this looks a little odd since there is no indication directly from the solutions that A and B should be decreasing with time, as we would expect for a system losing energy; compare this with the solutions to SHM with damping that is proportional to velocity, where we get a leading exponential term, which allows the amplitude to die away.
Also these solutions suggest that the velocity exactly matches that of SHM without damping, and again we'd expect the velocity to decrease relative to SHM over corresponding points of the cycle, since we're losing energy constantly over the cycle.
So I don't know what to make of this at all. The solutions to the equations of motion don't seem to match what we "know" must be happening; the only explanation that I can see at the moment is that intuitively I'm wrong, and that over every half cycle we get SHM, but somehow we can make an argument the the "constants" A,B must change at the end of every halfcycle, or something like that.
In short, I'm confused.
Last edited by atsruser; 28102015 at 09:54.
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