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    so there's 0.1 chance of funny face with 1 chance of acceptance
    0.8 chance of normal face with 0.5 chance of acceptance
    0.1 chance of boring face with 0 chance of acceptance (hence number nine got rejected)

    therefore chance of acceptance total is 0.1 from funny face, 0.4 from normal face and 0 from boring face. there's probably a better way to write this but i suck at maths lol

    therefore big george's chance of having a funny face is 0.1/(0.1+0.4) = 0.2

    ------

    second part

    there's now 0.1 chance of funny face (1 acceptance chance) and 0.9 chance of normal face (0.5 acceptance chance)

    so total is 0.1 from funny face and 0.45 from normal face

    so total funny face chance is 0.1/0.55= 2/11 . so yeah it is different

    done. i wish step was all this kind of stuff maybe i'd get in then loooooooooooooooooool
    question 8 test 4
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    Hey guys it seems the links to paper 4 and the maths+physics paper are dead. Any chance someone could post them again?
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    (Original post by error_404_oops)
    Hey guys it seems the links to paper 4 and the maths+physics paper are dead. Any chance someone could post them again?
    They seem to have been taken down by Trinity themselves and as such, I don't think it'd be okay for me to post them up, sorry! You do have 3 other tests which should prove more than enough.
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    (Original post by Zacken)
    They seem to have been taken down by Trinity themselves and as such, I don't think it'd be okay for me to post them up, sorry! You do have 3 other tests which should prove more than enough.
    Or maybe it was an error that they took it down? If you have to material to help others why keep it to yourself? This is the same thing you did with the textbook. I honestly don't know why you all on here are so difficult.
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    (Original post by Zacken)
    They seem to have been taken down by Trinity themselves and as such, I don't think it'd be okay for me to post them up, sorry! You do have 3 other tests which should prove more than enough.
    Afraid of the big bad Trinity mathmo gestapo are we?
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    (Original post by error_404_oops)
    Afraid of the big bad Trinity mathmo gestapo are we?
    I think with his reasoning for not giving the paper to you, he should also delete it from his computer since it used to be online but they removed it and they want no one to see the paper anymore.
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    Did some digging and found these elsewhere on the internet:

    https://www.yumpu.com/en/document/vi...hematics-with-

    https://www.yumpu.com/en/document/vi...mathematics-2-

    Are these the missing papers?
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    (Original post by ServantOfMorgoth)
    I think with his reasoning for not giving the paper to you, he should also delete it from his computer since it used to be online but they removed it and they want no one to see the paper anymore.
    I know haha. I'm not even 'competition' to any of you lot; I'm just a bored PhysNatSci offer holder on an (unplanned) gap year wanting to do some maths!
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    (Original post by error_404_oops)
    I know haha. I'm not even 'competition' to any of you lot; I'm just a bored PhysNatSci offer holder on an (unplanned) gap year wanting to do some maths!
    I'm already at uni but I'm just giving my input where I see injustice.
    If you're that bored why don't you start Vector/Multivariable Calculus? It'll be the most useful thing to do because the theorems are all similar and learning them all in a short space of time makes you confuse them.
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    (Original post by ServantOfMorgoth)
    I'm already at uni but I'm just giving my input where I see injustice.
    If you're that bored why don't you start Vector/Multivariable Calculus? It'll be the most useful thing to do because the theorems are all similar and learning them all in a short space of time makes you confuse them.
    I have been, Riley Hobson & Bence's Mathematical Methods textbook has been an absolute godsend!
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    Btw I've posted some links to what could potentially be the missing papers because I might have found them elsewhere on the internet (or I've found some extra ones!). The TSR mafia are presumably still checking that I've not broken any of their stupid rules...
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    (Original post by error_404_oops)
    I have been, Riley Hobson & Bence's Mathematical Methods textbook has been an absolute godsend!
    Oh well here are some pratice questions I saw on this thread.

    http://www.thestudentroom.co.uk/show....php?t=3770439

    Also Zacken may be doubly mad now that were not talking about the thread title topic so if you wanna say anything more you should PM me.
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    (Original post by error_404_oops)
    Did some digging and found these elsewhere on the internet:

    https://www.yumpu.com/en/document/vi...hematics-with-

    https://www.yumpu.com/en/document/vi...mathematics-2-

    Are these the missing papers?
    So are these similar to the missing papers? Or better still could someone just repost specimen test iv and the maths+physics one?
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    (Original post by joostan)
    Paper 3 Question 1:
    Spoiler:
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    Proof by picture seems to be the best way to go, without any fancy expressions for the \zeta function.
    A quick sketch should show that:
    \displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx \leq \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^s}.
    We can compute the integral:
    \displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx = -\dfrac{1}{s-1} \left[ \dfrac{1}{x^{s-1}} \right] _1^{\infty}=\dfrac{1}{s-1}
    Also from the sketch we see that:
    \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n^s} \leq \displaystyle\int_2^{\infty} \dfrac{1}{(x-1)^s} \ dx
    Hence:
    \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^s} \leq 1+\displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx
    Observing that 1+\dfrac{1}{s-1}=\dfrac{s}{s-1} gives the result:
    \dfrac{1}{s-1} \leq 1+2^{-s}+3^{-s}+... \leq\dfrac{s}{s-1}

    Paper 3 Question 2:
    Spoiler:
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    I=\displaystyle\int_0^1 \dfrac{dx}{x+\sqrt{1-x^2}} \ dx
    Let x=\sin(\theta)
    \Rightarrow I = \displaystyle\int_0^{\frac{\pi}{  2}} \dfrac{\sin(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta

\theta \mapsto \dfrac{\pi}{2}-\theta

\RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta 

\Rightarrow 2I=\displaystyle\int_0^{\frac{ \pi}{2}} \ d\theta  = \dfrac{\pi}{2}

\therefore I=\dfrac{\pi}{4}.


    Paper 3 Question 6:
    Spoiler:
    Show

    False.
    In decimal expansion, an n digit number N may be written as:
    N=\displaystyle\sum_{r=0}^n a_r 10^r \ , \ a_r \in \{0,1, . . . ,9 \}
    Let x=10^n-1 for some n \in \mathbb{N}.
    Then:
    x^2=(10^n-1)^2=10^{2n}-2(10^n)+1= \left( \displaystyle\sum_{r=n+1}^{2n} 9(10)^r \right) + 8(10^n)+1.
    Certainly then, we have n digits a_r \not= 0,1.
    Hence if n>1000 we have a square number with more than 1000 digits a_r \not= 0,1.

    For Q6 I didn't follow how you got x^2 to equal the last part of this

    x^2=(10^n-1)^2=10^{2n}-2(10^n)+1= \left( \displaystyle\sum_{r=n+1}^{2n} 9(10)^r \right) + 8(10^n)+1[/latex].

    Can anyone explain the last part? Im not sure my decimal expansion understanding is great XD
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    (Original post by LeopardK)
    For Q6 I didn't follow how you got x^2 to equal the last part of this

    x^2=(10^n-1)^2=10^{2n}-2(10^n)+1= \left( \displaystyle\sum_{r=n+1}^{2n} 9(10)^r \right) + 8(10^n)+1[/latex].

    Can anyone explain the last part? Im not sure my decimal expansion understanding is great XD
    Forget the general case for now, and just imagine n=3, say.
    So x=1000-1=999. But then x^2=(1000-1)^2=10^6-2(10)^3+1=998001=9(10^5)+9(10^4)  +8(10^3)+1.

\Rightarrow x^2=\left( \displaystyle\sum_{r=4}^{6} 9(10^r) \right) +8(10^3)+1.

    We can extend this idea to any n \in \mathbb{N}.
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    For Test 1 Q9, would it be much easier to calculate the maximum point of f(x) = Mx - exp(x), and show that the maximum point is positive? Because there is a zero point between 0 and 1, if the maximum point's x coordinate is larger than 1, and its y coordinate is larger than 0 then there must be only 1 solution larger than 1.
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    (Original post by Renzhi10122)
    Specimen 2, q8

    Hint:
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    Find the area of a fifth of the pentagon, namely a triangle of it

    Solution:
    Spoiler:
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    Let AC'=a, C'D'=b and CC'=c.We have <C'AD'=36, <AC'D'=72, <C'OD'=72, <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}
    Is it c = a/(4cos^2(36))
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    (Original post by Euclidean)
    Specimen Paper III, Question 8:
    Spoiler:
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    In a grid of 8 dots, we have an area of 7 x 7 = 49 unit squares

    It follows that there are 49 unit squares (1x1 squares)

    There are 6 double squares on each row (2x2 squares) and hence 6x6 = 36 double squares in the whole grid

    There are 5 triple squares on each row (3x3 squares) and hence 5x5 = 25 triple squares in the whole grid

    There are 4 quadruple squares on each row (4x4 squares) and hence 4x4=16 quadruple squares in the whole grid

    There are 3 quintuple squares on each row (5x5 squares) and hence 3x3=9 quintuple squares in the whole grid

    There are 2 sixfold squares on each row (6x6 squares) and hence 2x2=4 sixfold squares in the whole grid

    There is 1 sevenfold square in the grid only

    Hence there are 49+36+25+16+9+4+1 = 140 squares in the grid who have sides parallel to the dots in the grid

    It's clear that this may be applicable to any grid of n dots such that the number of squares with sides parallel to the dots is equal to the sum of all square numbers less than n

    Assuming a square must have dots for all four of it's vertices:

    Considering diagonal squares:

    When I say a 1x1 diagonal square I mean a \sqrt{2}x\sqrt{2} square but have written it this way for simplicity.

    1x1 diagonal squares take up a 2x2 parallel area in the grid. Hence there are the same number of 1x1 diagonal squares as 2x2 parallel squares in the grid = 36

    2x2 diagonal squares take up a 4x4 parallel area in the grid. Hence there are the same number of 2x2 diagonal squares in the grid as 4x4 parallel squares = 16

    3x3 diagonal squares take up a 6x6 parallel area in the grid. Hence there are the same number of 3x3 diagonal squares in the grid as 6x6 parallel squares = 4

    There can be no 4x4 diagonal squares contained in the grid.

    Therefore there are 140 + 36 + 16 + 4 = 196 total squares
    You did not count all the squares. Sorry for responding to an old thread but this error must be fixed.

    It is easy to prove that the vertices of ANY square themselves lie on one of the original 140 squares which are parallel to the axes.

    In fact, taking any one of those 140 squares, The number of squares that can be made such that its vertices lie of the original square is equal to the side length of the original square (Try it, too lazy to provide diagram).

    Since the solution to part (i) for a lattice with x^2 points is the sum of the squares from 1 to (x-1), the solution to the second part is the sum of the CUBES from 1 to (x-1). This comes out to be 784 for x = 8.
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    (Original post by YourBoiTudor)
    You did not count all the squares. Sorry for responding to an old thread but this error must be fixed.

    It is easy to prove that the vertices of ANY square themselves lie on one of the original 140 squares which are parallel to the axes.

    In fact, taking any one of those 140 squares, The number of squares that can be made such that its vertices lie of the original square is equal to the side length of the original square (Try it, too lazy to provide diagram).

    Since the solution to part (i) for a lattice with x^2 points is the sum of the squares from 1 to (x-1), the solution to the second part is the sum of the CUBES from 1 to (x-1). This comes out to be 784 for x = 8.
    Hi

    Thanks for your response, I hadn't realised that my attempt was flagged as the solution but I have amended it now. As far as I am aware, the number of squares in the grid of size 8 is not the sum of the cubes from 1 to (8-1) inclusive, and I will explain why by simplifying the problem to a grid of size 3.

    Considering a grid of size 3, there are 4 + 1 = 5 squares which are parallel to the grid by part (i) but the square of size 2 can be rotated to give another square inside itself with vertices along it's original edges. In the larger grid, it is clear that for each square of size n by n in the grid itself, there are n squares which have vertices along it's edges as you claim. However, this factor of n only applies to the number of squares of size n which isn't n^{2}. The number of squares of size n is given by (x-n)^{2} where x is the grid size.

    So \displaystyle S\left(x\right) = \sum_{i=1}^{x} i\left(x-i\right)^{2} represents the number of squares in the grid of size x.
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    (Original post by Euclidean)
    Hi

    Thanks for your response, I hadn't realised that my attempt was flagged as the solution but I have amended it now. As far as I am aware, the number of squares in the grid of size 8 is not the sum of the cubes from 1 to (8-1) inclusive, and I will explain why by simplifying the problem to a grid of size 3.

    Considering a grid of size 3, there are 4 + 1 = 5 squares which are parallel to the grid by part (i) but the square of size 2 can be rotated to give another square inside itself with vertices along it's original edges. In the larger grid, it is clear that for each square of size n by n in the grid itself, there are n squares which have vertices along it's edges as you claim. However, this factor of n only applies to the number of squares of size n which isn't n^{2}. The number of squares of size n is given by (x-n)^{2} where x is the grid size.

    So \displaystyle S\left(x\right) = \sum_{i=1}^{x} i\left(x-i\right)^{2} represents the number of squares in the grid of size x.
    You are absolutely right!
    Sorry, this is what happens when I try to do everything in my head, it appears I multiplied the factors by the wrong squares.
 
 
 
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