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    (Original post by I am Ace)
    Anyone have the maths and physics paper 1 Q 4 curve sketching solutions?


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    2/3 posts above yours.
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    (Original post by I am Ace)
    The bottom 2 are fails I tried to delete


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    The internet never forgets

    Well TSR is weird, anyway

    (Original post by physicsmaths)
    Lol, I have had plenty of amazing proofs like yours disproving well known mathematical theorems


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    I can prove 0=1 bro
    That disproves all of maths in one stroke

    (Original post by Zacken)
    I quite enjoyed the question, I enjoyed the paper in general (the second paper, not so much), I think I got 5 full solutions out - but I freaking hated how they put Jensens inequality in the previous question, that was so unfair! :eek:
    Easy if you are good at Olympiad inequalities, but I agree its unfair to put in the test

    (Bear in mind that people aren't usually expected to get 10/10 questions, your interviewer would guide you through it after the test) I have heard
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    (Original post by physicsmaths)
    https://www1.maths.leeds.ac.uk/~read/TQC1.pdf
    What did people get for Q4 a,b?


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    How did people do 9 and 10? Not happy with my answers
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    (Original post by Johann von Gauss)
    How did people do 9 and 10? Not happy with my answers
    Lol didnt even bother, they looked like stupid questions, too much reading haha.


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    What's the best way to prepare for interview integration Q's?

    I'm thinking MAT or STEP I - those trinity integrals with summations fry my brain and the Hard Integrals Thread is mostly undergrad I believe
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    (Original post by I am Ace)
    What's the best way to prepare for interview integration Q's?

    I'm thinking MAT or STEP I - those trinity integrals with summations fry my brain and the Hard Integrals Thread is mostly undergrad I believe
    Mat will be useless for integration, step is the best bet


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    (Original post by physicsmaths)
    Mat will be useless for integration, step is the best bet


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    I meant to post this on the interview prep not the Trinity solutions, my bad. I was thinking MAT will be shorter but it'll probably be too easy, STEP then?
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    (Original post by I am Ace)
    I meant to post this on the interview prep not the Trinity solutions, my bad. I was thinking MAT will be shorter but it'll probably be too easy, STEP then?
    Mat only has C1-2 knowledge and hardly any integration at all. Especially if ur lookig for change in variabkes and integrals withh substitutions etc mat has nothing on those topicsz


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    (Original post by physicsmaths)
    Mat only has C1-2 knowledge and hardly any integration at all. Especially if ur lookig for change in variabkes and integrals withh substitutions etc mat has nothing on those topicsz


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    Top of the line keyboard you've got there bro
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    (Original post by I am Ace)
    Top of the line keyboard you've got there bro
    Lol its on my phone so i type very fast without even checking.


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    • Thread Starter
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    (Original post by I am Ace)
    What's the best way to prepare for interview integration Q's?

    I'm thinking MAT or STEP I - those trinity integrals with summations fry my brain and the Hard Integrals Thread is mostly undergrad I believe
    Do STEP I 1994 Q8 !
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    (Original post by Jordan\)
    Do STEP I 1994 Q8 !
    *checks spreadsheet to see if done already*

    On it!

    I feel we should make a list of these easier/ shorter ones


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    (Original post by I am Ace)
    *checks spreadsheet to see if done already*

    On it!

    I feel we should make a list of these easier/ shorter ones


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    Have you done the majority of STEP I questions already?

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    (Original post by JakeThomasLee)
    Have you done the majority of STEP I questions already?

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    hahaha
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    (Original post by EmptyMathsBox)
    Downing Test Question 6 Solution:
    Answer:
    Spoiler:
    Show
    The answer is 1/75 seconds
    Full Solution:

    Spoiler:
    Show
    If we look at the diagram there are small triangles inside the big ones. This is a very big hint.Suppose that in a very small time interval each point travels  \Delta x distance. Then as  \Delta x is small the particle moves approximately in a straight line in this period. Also the particle it is chasing has also moved  \Delta x as well but at an angle of 60 degrees from the line of motion of this particle. As all three particles move  \Delta x at 60 degrees to each other the new triangle which is formed is also equilateral and has the same centre as the original triangle (it is just smaller and rotated). Now suppose this triangle has side length  u . Then the time for the particles to move from their new positions to the centre of the triangle is  ut where  t is the total time for the particles to move from their original positions to the centre of the triangle. Thus we have an equation for t as t = time to move  \Delta x + time to get from smaller triangle to centre of the triangle and so t = \frac{\Delta x}{50}+ut .

    Now if we can find  u in terms of  \Delta x the we can write  t in terms of  \Delta x and take limits to find  t . We can identify a triangle on the diagram with sides of  u,\Delta x,1- \Delta x with the angles opposite them  60,\Delta \theta,120-\Delta \theta respectively where  \Delta \theta is the angle opposite the side with length  \Delta x (there are three such triangles on the diagram, we can concentrate on any one).

    Now by applying the sine rule once we get that  \frac{u}{sin(60)} = \frac{\Delta x}{sin(\Delta \theta)} . We rearrange this to get  u = \frac{\sqrt{3}\Delta x}{2sin(\Delta \theta)} . So if we can work out  sin(\Delta \theta) in terms of  \Delta x we can proceed.

    To do this we use the sine rule again to get that \frac{\Delta x}{sin(\Delta \theta)} = \frac{1- \Delta x}{sin(120-\Delta \theta)}  . After some rearrangement and trigonometric identities (which I will not detail here but involve expanding  sin(120-\Delta \theta) and finding sin(\Delta \theta) from  cot^2(\Delta \theta)) we get that  sin(\Delta \theta) = \frac{\sqrt{3}\Delta x}{2\sqrt{1-3\Delta x+3(\Delta x)^2}}. We plug this into our first sine rule equation to get that  u = \sqrt{1-3\Delta x+3(\Delta x)^2}.

    Now as t = \frac{\Delta x}{50}+ut by rearranging we get that (1-u)t = \frac{\Delta x}{50} so t = \frac{\Delta x}{50*(1-u)} or when we substitute for u t = \frac{\Delta x}{50(1-\sqrt{1-3\Delta x+3(\Delta x)^2})} .When we take the limit as  \Delta x approaches 0 we find that we have to use L'Hopital's rule (fairly standard and easy to use in this case) to find that as  \Delta x approaches 0,  t approaches  \frac{1}{75}.

    I am sure there is a faster and neater solution but this was the one I found anyway.
    Here is that faster and neater solution.
    Spoiler:
    Show
    For an arbitrary vertex of the triangle, consider the radial frame of reference, from which the vertex is moving at an angle \frac{\pi}{6}

    Thus the radial speed is 50\cos\frac{\pi}{6}=25\sqrt{3}

    The initial radial length was \frac{1}{\sqrt{3}} using elementary facts about equilateral triangles. The time in question then T=\frac{ \frac{1}{ \sqrt{3}}}{25\sqrt{3}}=\frac{1}{  75} Just as before
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    ^^ try and keep this thread on topic, discussing STEP isn't really conducive. :-)
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    (Original post by Zacken)
    ^^ try and keep this thread on topic, discussing STEP isn't really conducive. :-)
    You are not the boss of me.
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    (Original post by Number Nine)
    You are not the boss of me.
    Rep me, betch
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    (Original post by Zacken)
    Rep me, betch
    get off this thread you didnt even apply to trinity admissions test
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    Test 4, Question 3:

    Let r = a + b\sqrt{2} and s = c + d\sqrt{2} so rs = (ac + 2bd)+\sqrt{2}(ad+bc)

    Hence: N(rs) = (ac+2bd)^2 - 2(ad+bc)^2 = (a^2 - 2b^2)(c^2 -2d^2) = N(r)N(s).

    We want to show that there are infinitely many solutions to N(r) = a^2 - 2b^2 = \pm 1 (side note: this an easy result from ring theory, you may want to look it up.)

    Look for the first few solutions: a=b=1, a=3, b=2. a=7, b=5 and from this we can intuit that if (a,b) is a solution then so is (a+2b, a+b) (where b < a):

    \displaystyle

\begin{equation*}(a+2b)^2 - 2(a+b)^2 = a^2 + 4ab + 4b^2 - 2a^2 - 4ab - 2b^2 = -(a^2 - 2b^2)\end{equation*}

    So since each solution is greater in value than the previous there are infinitely many of them.
 
 
 
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