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    (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 half-way rigourous answer was extremely tricky to obtain.

    I'm really not sure what they are after on this one.
    I enjoyed doing the question, quite analysis-y, in a certain sense, but formalising it was a real pain and I just couldn't quite finish it off.
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    Test 2, Question 3

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    Consider the relationship between the coefficients of a polynomial and its roots.


    Solution
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    Suppose four points on the curve are collinear along line y=mx+c. Then their x-coordinates are the roots of the equation

    

2x^4 +7x^3 + 3x -5 -mx - c =0



or



2x^4 +7x^3 + (3-m)x - (5+c) = 0

    The sum of the roots of this polynomial is hence -7/2, giving the roots an average, k, of -7/8.


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    (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.
    To confirm other comments: in this scenario momentum is conserved but energy isn't.
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    Specimen Test 2, Question 1:

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    Consider logarithms... more than once
    Solution:
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    Consider (base ten logarithms): \log \log f(n) = 2^n \log 2 + \log \log 2 and \log \log g(n) = 2n + \log 2

    Then it is quite clear that for large n we have \log \log f(n) > \log \log g(n) since \log \log f(n) has exponential growth whilst \log \log g(n) has polynomial (of degree 1) growth.

    Hence, since the function f \colon (0, \infty) \mapsto \mathbb{R} given by f(x) = \log x is a bijective, strictly increasing continuous function, we can assert that \log \log f(n) > \log \log g(n) \Rightarrow f(n) > g(n) for large n.
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    Another solution to 9:
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    Observe first that e^x-Mx is a continuous function.
    Now, e^1-M<0 for M>e, since M is very large, we may assume this holds.
    Also:
    e^M=1+M+\dfrac{M^2}{2}+\dfrac{M^  3}{3}+...> \dfrac{M^2}{2}+\dfrac{M^3}{3}
    For M>\dfrac{3}{2} ,\ \dfrac{M^3}{3}>\dfrac{M^2}{2} \Rightarrow e^M-M^2>0
    Hence by the Intermediate Value Theorem, \exists w \in (1 , M) such that e^w=Mw
    Suppose that such a solution was not unique, then let two such distinct solutions be w_1 and w_2.
    Then: e^{w_1-w_2}=\dfrac{w_1}{w_2}
    \Leftrightarrow w_1-w_2=\log \left( \dfrac{w_1}{w_2} \right)
    w.l.o.g, let w_1>w_2 \Rightarrow w_1-w_2=k \ , k>0
    \Rightarrow k= \log \left( \dfrac{w_1}{w_2} \right)=log \left(1+ \dfrac{k}{w_2} \right)

\Rightarrow e^k=1+\dfrac{k}{w_2}
    But: e^k>1+k , \ w_2>1 \Rightarrow e^k>1+k>1+\dfrac{k}{w_2}
    A contradiction, meaning that w_1=w_2.

    Now if Mw=e^w \Rightarrow w=\log(M)+\log(w)
    \Rightarrow w-\log(w)=\log(M)
    As \log(w)<<w for reasonable sized w which occur when M is large, we have \log(M) \approx w.

    Letting w=\log(M)+y we see that:
    M(\log(M)+y)=Me^y

\Rightarrow \log(M)+y=e^y

\Rightarrow y=\log \left(\log(M)+y \right)
    For large M we may write y \approx \log \left( \log(M) \right).
    \therefore w \approx \log(M)+\log(\log(M)).


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    Test 2, Question 4:

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    Sketch x-1 and (e-1)\log x - find the roots of each and the value of each at x=e to help you sketch it. That should reveal the solutions immediately. For the second graph, solve f'(x) =0 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 points
    Solution:
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    We sketch x-1 and (e-1)\log x. They both have a root at x=1 so that's one solution and to help us sketch the graphs, we look at x=e only to find that's a solution as well. Since our graph clearly depicts only two solutions, all our solutions are given by x=1 , e.

    Sketching the second graph:


    First thing to note it that the domain of our function is x \geq 0.

    1. An easy root (by inspection) is x=e since we get e^e - e^e = 0. (Note that the graph touches the x-axis here, we justify this later)

    2. Next, we look at the y-intercepts: x = 0 yields f(0) = 1.

    3. We can also see that x \to \infty \Rightarrow f(x) \to \infty 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): \displaystyle f'(x) = e^x - \left(\frac{e}{x}\right)x^e = ex^{e-1} so we know that the stationary points are given by f'(x)  = 0 \iff e^x = ex^{e-1} \iff e^{x-1} = x^{e-1} upon which, taking logarithms yields x-1 = (e-1)\ln x so we have  f'(x) = 0 \iff x = 1, e so our root is also a turning point, meaning that it has to touch and not cross the x-axis.

    Putting all of this together yields: (x=1 is a maximum and x=e a minimum).
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    (Original post by Zacken)
    Specimen Test 2, Question 1:

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    Consider logarithms... more than once
    Solution:
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    Consider (base ten logarithms): \log \log f(n) = 2^n \log 2 + \log \log 2 and \log \log g(n) = 10n + 1

    Then it is quite clear that for large n we have \log \log f(n) > \log \log g(n) since \log \log f(n) has exponential growth whilst \log \log g(n) has polynomial (of degree 1) growth.

    Hence, since the function f \colon [0, \infty) \mapsto \mathbb{R} given by f(x) = \log x is a bijective, strictly increasing continuous function, we can assert that \log \log f(n) > \log \log g(n) \Rightarrow f(n) > g(n) for large n.
    Alternative:

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    100^100^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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    (Original post by DFranklin)
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    100^100^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).
    That's interesting, thanks for that.

    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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    Specimen Test 1, Question 10:

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    Binomial coefficients crop up a lot in this question (^{n}\mathrm{C}_k)
    Don't use specific colours in your question, other than the colours it asks for. May cause confusion.
    Solution Part 1:
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    We 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 ^{20}\mathrm{C}_3 possible outcomes. (1140)

    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 non-reds.
    We have 5 reds and 15 non-reds.
    We want exactly one red.

    \Rightarrow P(exactly\ one\ red)

    There are ^{5}\mathrm{C}_1 ways to pick exactly one red. =5 (As there are five reds, and we want one)

    There are ^{15}\mathrm{C}_2 ways to pick the remaining non-reds. =105 (As there are 15 non-reds, 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 525 ways of picking one red and 2 non-reds.

    We return to the following statement

    P(exactly\ one\ red)\ =\ \dfrac{no.\ of\ ways\ we\ can\ pick\ one\ red\ and\ two\ non-reds}{total\ no.\ of\ ways\ we\ can\ draw\ three\ balls}

    Therefore P(exactly\ one\ red) = \dfrac{525}{1140} = \dfrac{35}{76} = 0.46\ (2\ dp.)
    Solution Part 2:
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    We 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 ^{20}\mathrm{C}_3 possible outcomes. (1140)


    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 ^{4}\mathrm{C}_3 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; ^{5}\mathrm{C}_1
    Colour 2: there are only 5 of these coloured balls; ^{5}\mathrm{C}_1
    Colour 3: there are only 5 of these coloured balls; ^{5}\mathrm{C}_1

    We can 'ignore' the last 'colour' as we eliminated this colour from the process the moment we did ^{4}\mathrm{C}_3

    ^{5}\mathrm{C}_1 = 5

    \therefore 5^3 = 125 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 (500)

    P(three\ different\ colours) = \dfrac{no.\ of\ ways\ we\ could\ draw\ three\ different\ colours}{total\ no.\ of\ ways\ we\ can\ draw\ three\ balls}

    \therefore P(three\ different\ colours) = \dfrac{500}{1140} = \dfrac{25}{57} = 0.\dot{4}4\ (2\ dp.)
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    (Original post by edothero)
    Specimen Test 1, Question 10:
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    C) ... is quite the question :creep:
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    Conditions 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 "non-blue")
    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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    Test 2, Question 2:

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    Split the integral into \displaystyle I_n = \int_0^1 \frac{x^2 \cdot x^{n-2}}{\sqrt{x^3 + 1}} \, \mathrm{d}x then use IBP with u = x^{n-2}.


    Solution:

    Part 1:
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    We split the integral into \displaystyle I_n = \int_0^1 \frac{x^2 \cdot x^{n-2}}{\sqrt{x^3 + 1}} \, \mathrm{d}x then use IBP with u = x^{n-2} and \mathrm{d}v = \frac{x^2}{\sqrt{x^3 + 1}}. This yields \displaystyle v = \frac{2\sqrt{x^3 + 1}}{3} and u = (n-2)x^{n-2}, so we get:

    \displaystyle I_n = \bigg[\frac{2}{3}x^{n-2}\sqrt{x^3 + 1}\bigg]_0^1 - \frac{2(n-2)}{3}\int_0^1 x^{n-3}\sqrt{x^3 + 1} \, \mathrm{d}x. This gives (by multiplying the integrand by \frac{\sqrt{x^3+1}}{\sqrt{x^3 + 1}}:

    \displaystyle 3I_n = 2\sqrt{2} - 2(n-2)\int_0^1 \frac{x^n}{\sqrt{x^3 + 1}} \, \mathrm{d}x - 2(n-2)\int_0^1 \frac{x^{n-3}}{\sqrt{x^3+1}} \, \mathrm{d}x which yields:

    \displaystyle 3I_n = 2\sqrt{2} - 2(n-2)I_n - 2(n-2)I_{n-3} \Rightarrow (2n-1)I_n + 2(n-2)I_{n-3} = 2\sqrt{2}.



    Part 2:
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    n=8 \Rightarrow 15I_8 + 12I_5 = 2\sqrt{2}
    n=5 \Rightarrow 9I_5 + 6I_2 = 2\sqrt{2}, where I_2 is easy to integrate. This gives us:

    \displaystyle I_5 = \frac{4 - 2 \sqrt{2}}{9} and hence \displaystyle I_8 = \frac{14\sqrt{2} - 16}{45}
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    Specimen Test 2, Question 6

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     \cos(n\theta + \theta) = 2\cos{\theta}cos(n\theta) - \cos(n\theta - \theta)
    Solution -

    First Part:
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     \cos{n\theta} = f_{n}(\cos\theta) 



\therefore \cos(n\theta + \theta) = 2\cos{\theta}\cos(n\theta) - \cos(n\theta - \theta)



\Rightarrow \cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta = 2\cos(n\theta)\cos\theta -\cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta



\Rightarrow \cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta = \cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta 



\therefore LHS = RHS

    as required.
    Second Part:
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     \cos{3\theta} + \cos{2\theta} =0

    Using DeMoivre
     \cos{3\theta} = 4\cos^{3}\theta - 3\cos\theta

    From double angle
     \cos{2\theta} = 2\cos^{2}\theta -1

    \therefore 4\cos^{3}\theta - 3\cos\theta + 2\cos^{2}\theta -1 = 0

    OR, you can use the first part and say that

    f_{3} = 2xf_{2}(x)- f(x) 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  \cos\theta = x,

    So,  4x^3 + 2x^2 - 3x -1 = 0

    By inspection we have

     (x + 1)(4x^{2} - 2x -1) = 0

     x = -1

     x = \frac{1\pm\sqrt{5}}{4}

    Then, by either using a calculator , or by going the long way and finding

     \theta = \arccos(x)

    We get the final answers of \theta = \pi, \frac{3\pi}{5}, \frac{\pi}{5}

    Where, in the question,  \theta = \phi

    However, I know that Zacken and Renzhi10122 have some nicer methods for this last bit.



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    Specimen 2, Q7

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    (x-1) is a factor of x^n -1 always.
    When is (x+1) a factor of x^n+1?


    Solution
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    Unless otherwise stated, assume all new constants are positive integers.


    Suppose p = a^n - 1.
    Then p = (a-1)(a^n-1 + ... + 1)

    So p is composite unless a-1 = 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^(cd-c) + ... +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^n-1 - ... +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^rs-s... +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 non-negative 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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    Specimen Test 1, Question 1:

    Solution 1:
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    In a tennis tournament there are 2n participants.
    In the first round, each player plays exactly one, so there are n games.

    There are 2n! ways to order everyone.You would divide this by n! ways to get the total number of combinations that 2n players could be chosen to play n games.

    There are 2^{n} ways to order each of these pairs.

    Therefore you have
    \dfrac{2n!}{n!\times2^{n}}

    You could further split up these factorials like so,
    \dfrac{2n(2n-1)!}{n(n-1)!\times2(2^{n-1})}


    When further simplified, gives you:
    \dfrac{(2n-1)!}{2^{n-1}(n-1)!}
    As required.
    Solution 2:
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    Denote with P(n) the number of pairings you are looking for. Suppose that two players are already coupled. Then the number of couplings is P(n-1). There are ^{2n}\mathrn{C}_{2} possibility to extract 2 players from 2n. Then:

    P(n) = \dfrac{1}{n}\dbinom{2n}{2}P(n-1)

    If we reiterate the previous formula, we have:

     P(n) = \dfrac{1}{n}\dbinom{2n}{2}P(n-1) = \dfrac{1}{n(n-1)}\dbinom{2n}{2}\dbinom{2(n-1)}{2}P(n-2) = \ldots =

    \displaystyle = \dfrac{1}{n!}\prod_{h=0}^{n-2}\dbinom{2(n-h)}{2}P(1) = \dfrac{1}{n!}\prod_{h=0}^{n-2}\dbinom{2(n-h)}{2}.


    where P(1)=1, because when n=1, then there are only 2 players.Then:
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    Specimen 2, q8

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    Find the area of a fifth of the pentagon, namely a triangle of it
    Solution:
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    Let AC'=a, C'D'=b and CC'=c.We have &lt;C'AD'=36, &lt;AC'D'=72, &lt;C'OD'=72, &lt;OC'D'=54 where O is the center of the pentagon. Then  1= \frac{1}{2}a^2 \sin36= \frac{1}{2}ab \sin72 giving  a=2b \cos36 by double angle formulae. Also, if  R=[C'OD'] then  R= \frac{1}{2}bc \sin54=\frac{1}{2}c^2 \sin72 . Noting that  \sin54= \cos36 , we have  b=2c \sin36 .
    Now,  c=\frac{b}{2 \cos36}=\frac{a}{4 \sin36 \cos36} = \frac{a}{2 \sin72} so  R= \frac{1}{2}c^2 \sin72=\frac{a^2}{8 \sin72} .
    We have  a^2=\frac{2}{\sin36} so  R= \frac{1}{4 \sin72 \sin36}= \frac{1}{4 \sin^236 \cos36}

    We (I) know that  \cos36=\frac{1+\sqrt5}{4} so  \sin^236=1-\cos^236=1-\frac{6+2\sqrt5}{16}=\frac{5-\sqrt5}{8} so  R=\dfrac{1}{8(\frac{5-\sqrt5}{8})(\frac{1+\sqrt5}{4})}  =\frac{1}{\sqrt5}

    EDIT: Just realised that I haven't actually finished the question...
    Hence, the area of the pentagon is  5R=\sqrt{5}
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    Test 2, Q10

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    Might be worth considering energy.


    Solution
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    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.
    

So k(x0)^2 = 2F(x0+x1) + k(x1)^2

k(x0-x1)(x0+x1) = 2F(x0+x1)

k(x0-x1)=2F

x0-x1 = 2F/k

, 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).
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    (Original post by Krollo)
    Test 2, Q10
    x
    My mechanics is a little rusty, but would the 'discuss' basically just be damped SHM?

    Edit: Stop doing maths and enjoy your vacation!
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    (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!
    Yes, I think, but I can't imagine it would be expected.

    And I am enjoying my holiday - I'm doing maths, of course. :-)
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    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.
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    Notational stuff:
    Let N_x(n,r) be the number of cards of type x i.e x=Q , K.
    Then, assuming n \leq 52:
    P(N_K(n,r)=i)=\dfrac{{4 \choose i}{n -4 \choose r-i}}{{n\choose r}}

    a) P(N_K(52,13)=1)=\dfrac{{4 \choose 1}{48 \choose 12}}{{52 \choose 13}}

    b) P(N_Q(52,13) \geq 2)=1-P(N_Q(52,13) &lt; 2)=1-\dfrac{{4 \choose 1}{48 \choose 12}+{4 \choose 0}{48 \choose 13}}{{52 \choose 13}}

    c) I claim (perhaps erroneously) that:
    P\left( N_K(52,13)=N_Q(52,13) \right) = \displaystyle\sum_{i=0}^4 P(N_Q(52,13)=i)P(N_K(48,13-i)=i)
    i.e I'm saying that we need i Queens from 52 cards and i Kings from 48 cards - the full pack without the Queens, and that these are independent.
    \Rightarrow P\left( N_K(52,13)=N_Q(52,13) \right) =\displaystyle\sum_{i=0}^4 \left( \dfrac{{4 \choose i}{48 \choose 13-i}}{{52 \choose 13}} \cdot \dfrac{{4 \choose i}{44 \choose 13-2i}}{{48 \choose 13-i}} \right)
    Simplifying a little we obtain:
    P\left( N_K(52,13)=N_Q(52,13) \right)=\displaystyle\sum_{i=0}^  4 \left(\dfrac{{4 \choose i}^2{44 \choose 13-2i}}{{52 \choose 13}} \right).
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    (Original post by Krollo)
    Test 2, Q10

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    Might be worth considering energy.


    Solution
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    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.
    

So k(x0)^2 = 2F(x0+x1) + k(x1)^2

k(x0-x1)(x0+x1) = 2F(x0+x1)

k(x0-x1)=2F

x0-x1 = 2F/k

, 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).
    Some further thoughts (I don't have a complete solution):

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    1. I think that the no-motion condition should be |x| &lt; \frac{F}{k}

    2. If we write down the equations of motion we get:

    \dot{x} &gt; 0: m\ddot{x} = -kx -F \Rightarrow \ddot{x}+\frac{k}{m}x = -\frac{F}{m}

    \dot{x} &lt; 0: m\ddot{x} = -kx +F \Rightarrow \ddot{x}+\frac{k}{m}x = \frac{F}{m}

    with corresponding solutions:

    \dot{x} &gt; 0: x =A\cos(\omega t-\phi)-\frac{F}{k}

    \dot{x} &lt; 0: x =B\cos(\omega t-\phi)+\frac{F}{k}

    with \omega^2=\frac{k}{m}

    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 half-cycle, or something like that.

    In short, I'm confused.
 
 
 
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