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    (Original post by Jordan\)
    No what I'm asking is where did all the numbers come from? The 2, the 7 the 5 etc.
    Has he just made them up?
    (Original post by Trinity 2, Q3)
    Show that if four distinct points of the curve y = 2x^4 + 7x^3 + 3x - 􀀀5 are co-linear then their average x-coordinate is some constant k. Find k.
    The polynomial is given.
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    (Original post by physicsmaths)
    Vietas formulas, it is a common result.
    Or (x-a)(x-b)(x-c)(x-d) where abcd are the roots. Expand that and eqaute coefficients.


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    (Original post by physicsmaths)
    Vietas formulas, it is a common result.
    Or (x-a)(x-b)(x-c)(x-d) where abcd are the roots. Expand that and eqaute coefficients.


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    Wow that is genius! How did you know about this?


    (Original post by Jordan\)
    How did he come up with the polynomial?
    This video will help understand the whole average bit and Vieta's Formula

    https://www.youtube.com/watch?v=9eWyvWq9CCs
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    (Original post by Zacken)
    The polynomial is given.
    Ah from what I remembered of the question is just said something about collinear points or some **** lol
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    (Original post by Zacken)
    The polynomial is given.
    Jordans losing it.


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    (Original post by physicsmaths)
    Jordans losing it.


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    Interview is creeping closer, we're all losing it.
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    (Original post by AmarPatel98)
    Wow that is genius! How did you know about this?




    This video will help understand the whole average bit and Vieta's Formula

    https://www.youtube.com/watch?v=9eWyvWq9CCs
    Just somwthing i have had come up quite alot. It comes handy when solving some hard equations,STEP I Q2 2015 it was very handy. It comes in ocr i think. Nothing genius just equating coefficients.


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    (Original post by Zacken)
    Interview is creeping closer, we're all losing it.
    Lol my interviews far far away! When is yours? How long you in england before. Come london ill show u a good time


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    (Original post by physicsmaths)
    Lol my interviews far far away! When is yours? How long you in england before. Come london ill show u a good time


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    My interview's on the 11th - I've only just got back from a holiday, so prep starts now. I'll be in England one day before the interview and 4 days after. I'll be in London for one day!
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    Interview on Tuesday at Emmanuel College ... pls pls pls pls ask me a question i've already come across
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    Come on guys, one last push, let's try and finish the four main practice tests before we head down :-)

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    (Original post by Krollo)
    Come on guys, one last push, let's try and finish the four main practice tests before we head down :-)

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    I ahve done Q4 of Spec 3 and Q3 of Spec 4. Although I think alot of the front page is not updated since I have seen solutions to many of them before a few pages back.


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    (Original post by joostan)
    Paper 3 Question 2:
    Spoiler:
    Show
    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}.
    Would you mind explaining what the fourth line means where you write \theta \mapsto \dfrac{\pi}{2}-\theta?
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    (Original post by MadChickenMan)
    Specimen Test 3, Question 4:

    Hint:
    Spoiler:
    Show
    Think about which flips must be heads, which must be tails and for which it doesn't matter.
    Solution:
    Spoiler:
    Show
    If s=t+1, the flip at time t+1 must be heads, so the probability, p=\frac{1}{2}.

    If s=t+9, the flips at times t+1 and t+5 must be tails and the flips at times t+6, t+7, t+8 and t+9 must be heads. The remaining flips can be either heads or tails, so the probability, p=(\frac{1}{2})^6=\frac{1}{64}.
    i don't agree with part 1, surely it is P(heads/ 3 heads) which is an 1/8?
    then again if he is right can someone explain since i am **** at probability.
    got it now my division of numbers was wrong
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    (Original post by JakeThomasLee)
    Would you mind explaining what the fourth line means where you write \theta \mapsto \dfrac{\pi}{2}-\theta?
    He's effectively making the substitution u= pi/2 -theta, I think.

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    (Original post by JakeThomasLee)
    Would you mind explaining what the fourth line means where you write \theta \mapsto \dfrac{\pi}{2}-\theta?
    it is just u=pi/2-theta since u and theta are dummies it is just a transformation of the function. which is how joostan has displayed it.
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    (Original post by Krollo)
    He's effectively making the substitution u= pi/2 -theta, I think.

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    (Original post by physicsmaths)
    it is just u=pi/2-theta since u and theta are dummies it is just a transformation of the function. which is how joostan has displayed it.
    Ah I see, thank you!
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    (Original post by JakeThomasLee)
    Ah I see, thank you!
    There are other ways to do it without that swell of course if you aren't used to that sub.
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    (Original post by Jordan\)
    There are other ways to do it without that swell of course if you aren't used to that sub.
    I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

    \RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta
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    (Original post by JakeThomasLee)
    I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

    \RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta
    Perhaps try and multiple both top and bottom by  \sin(\theta )-\cos(\theta) and see if that works?
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    Specimen Paper III, Question 8:
    Spoiler:
    Show



    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

    Modification as of 7/12/16:

    Every square in the grid must lie on one of the edges of a square with sides parallel to the grid. Considering these squares with sides parallel to the grid:

    Take a square X with side length x in the grid where x > 1, such that X has edges parallel to the grid. Considering all squares which have vertices on the edges of X we see that there are (x-1) such squares excluding the square X itself. So each square of side length x for x > 1 in the grid also comes with x-1 corresponding 'slanted' squares. Hence, for the previous sum we have done, to consider all possible squares, we must multiply the number of squares of side length x parallel to the grid by x.

    For a grid of size n, the number of squares is thus given by: \displaystyle S\left(n\right) = \sum_{i=1}^{n} i\left(n-i\right)^{2}, since \left(n-i\right)^{2} returns the number of parallel squares of side length i.

    For a grid of size 8: S\left(n\right) = 49 + 72 + 75 + 64 + 45 + 24 + 7 = 336.



 
 
 
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