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    Specimen Paper 4: Question 6.
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    Use AM-GM.
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    First off, we're gonna assume (although it's not explicitly stated) that the "division" only allows positive reals, else the problem is silly.
    Let A be the number we start with.
    Let S=\{ a_1,a_2,...,a_n \} be a set of order n such that \displaystyle\sum_{i=1}^n a_i = A.

    It follows from the AM-GM inequality: \dfrac{1}{n} \displaystyle\sum_{i=1}^n a_i \leq \left( \displaystyle\prod_{i=1}^n a_i \right)^\frac{1}{n} with equality iff all the a_is are equal. (see here if you're unfamiliar with it - you can probably quote it - if not, it's not that tricky to prove) that the product is maximised when a_1=a_2=...=a_n.

    From here we note that we now wish to find the maximum n such that: \left(\dfrac{A}{n} \right)^n is maximised.
    Note now that the function: f(x)=\left( \dfrac{A}{x} \right)^x has a single stationary point, which we now compute:
    f(x)=\left( \dfrac{A}{x} \right)^x=e^{x\ln \left(\frac{A}{x} \right)}

\Rightarrow f'(x)=f(x)(\ln \left( \dfrac{A}{x} \right)-1)

\Rightarrow f'(x)=0 \Leftrightarrow x=\dfrac{A}{e}.

    Thus we see that either: n=\left \lfloor {\dfrac{A}{e}} \right \rfloor or n=\left\lceil { \dfrac{A}{e} } \right\rceil.
    The maximum product is then given by:
    \max \left \{ \left( \dfrac{A}{\lfloor {\frac{A}{e}} \rfloor} \right) ^{\lfloor {\frac{A}{e}} \rfloor}, \left( \dfrac{A}{\lceil {\frac{A}{e}} \rceil} \right)^{\lceil {\frac{A}{e}} \rceil}\right \}
    In the case A=11 it's easy to see that we have n=4, and the maximum product is 2.75^4 which I guess will be around 50 something.
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    Downing Test Question 6 Solution:
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    The answer is 1/75 seconds
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    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.
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    There are more practice questions in the following:

    https://www1.maths.leeds.ac.uk/~read/TQ2.pdf
    https://www1.maths.leeds.ac.uk/~read/TQ3.pdf

    (Mostly the first couple seem to be different, but haven't done a proper read of them. Have fun! )

    If you're done with all of those, then you might want to try:

    https://www1.maths.leeds.ac.uk/~read/TQ1.pdf

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    Trinity Paper 4 - Question 7

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    How can you find a volume without using geometry?


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    I shall assume that the plane cuts at a perpendicular distance a from the centre, otherwise the question doesn't make much sense.

    The volume of the smaller part can be found using the volume integral:

    

\pi \int_{a}^{r} r^2 - x^2 dx



= \dfrac{2\pi r^3}{3} - \pi ar^2 + \dfrac{\pi a^3}{3}

    and since the volume of the whole sphere is 4/3pir^3, the volume of the larger part is

    

\dfrac{2\pi r^3}{3} + \pi ar^2 - \dfrac{\pi a^3}{3}

    As a quick check, plugging in a=0 gives two equal hemispheres, whereas a=r gives a single sphere, as we would expect.

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    Specimen Test 3, Question 4:

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    Think about which flips must be heads, which must be tails and for which it doesn't matter.
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    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}.
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    Specimen Test 3, Question 9:

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    For part (ii) apply \frac{\mathrm d}{\mathrm d x} as you would a matrix.
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    (i) By the chain rule

    \frac{\mathrm d y}{\mathrm d x}=-xe^{-\frac{x^2}{2}}.

    Using the product rule

    \frac{\mathrm d^2 y}{\mathrm d x^2}=-e^{-\frac{x^2}{2}}+x^2e^{-\frac{x^2}{2}}.

    So we have that

    -\frac{\mathrm d^2 y}{\mathrm d x^2}+x^2y=e^{-\frac{x^2}{2}}-x^2e^{-\frac{x^2}{2}}+x^2e^{-\frac{x^2}{2}}=e^{-\frac{x^2}{2}}=y.

    (ii) As \frac{\mathrm d}{\mathrm d x} is an operator, we compute the rightmost terms first terms first

    (\pm \frac{\mathrm d}{\mathrm d x} +x)(\mp \frac{\mathrm d}{\mathrm d x} +x)z=(\pm \frac{\mathrm d}{\mathrm d x} +x)(\mp \frac{\mathrm d z}{\mathrm d x}+xz)=-\frac{\mathrm d^2 z}{\mathrm d x^2}\pm z \pm x\frac{\mathrm d z}{\mathrm d x} \mp x\frac{\mathrm d z}{\mathrm d x} +x^2z \\ =-\frac{\mathrm d^2 z}{\mathrm d x^2}+x^2z \pm z.

    (iii) Differentitaing w

    \frac{\mathrm d w}{\mathrm d x}=-\frac{\mathrm d^2 y}{\mathrm d x^2}+y+x\frac{\mathrm d y}{\mathrm d x}

    \frac{\mathrm d^2 w}{\mathrm d^2 x}=-\frac{\mathrm d^3 y}{\mathrm d x^3}+2\frac{\mathrm d y}{\mathrm d x}+x\frac{\mathrm d^2 y}{\mathrm d x^2}

    Substituting this value into the equation we have

    -\frac{\mathrm d^2 w}{\mathrm d x^2}+x^2w=\frac{\mathrm d^3 y}{\mathrm d x^3}-2\frac{\mathrm d y}{\mathrm d x}-x\frac{\mathrm d^2 y}{\mathrm d x^2}-x^2\frac{\mathrm d y}{\mathrm d x}+x^3y

    Using the original equation and differentiating gives

    \frac{\mathrm d^3 y}{\mathrm d x^3}=x^2\frac{\mathrm d y}{\mathrm d x}-\frac{\mathrm d y}{\mathrm d x}+2xy

    Substituting the values for \frac{\mathrm d^3 y}{\mathrm d x^3} and \frac{\mathrm d^2 y}{\mathrm d x^2} in to the new equation

    x^2\frac{\mathrm d y}{\mathrm d x}-\frac{\mathrm d y}{\mathrm d x}+2xy-2\frac{\mathrm d y}{\mathrm d x}-x^3y+xy-x^2\frac{\mathrm d y}{\mathrm d x}+x^3y=-3\frac{\mathrm d y}{\mathrm d x}+3y=3w

    (LaTeX-athon over)
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    (Original post by Jordan\)
    SPECIMEN TEST 1, QUESTION 5

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    Using the formula  P(A|B) = \frac{P(B|A)P(A)}{ P(B|A)P(A) + P(B|A' )P(A' )} will help.
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    P ( biased | 8 heads ) =  \displaystyle \frac{\frac{1}{129}}{\frac{1}{12  9} + \frac{128}{129} \left ( \frac{1}{2} \right )^8} = \frac{2}{3}

    P ( 'biased | 8 heads) = \frac{1}{3}

    P (heads) =  \frac{1}{3}* \frac{1}{2} + \frac{2}{3} = \frac{5}{6}
    lol first probability question I have ever got right.


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    (Original post by joostan)
    Specimen Paper 4: Question 6.
    Hint:
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    Use AM-GM.
    .
    Solution:
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    First off, we're gonna assume (although it's not explicitly stated) that the "division" only allows positive reals, else the problem is silly.
    Let A be the number we start with.
    Let S=\{ a_1,a_2,...,a_n \} be a set of order n such that \displaystyle\sum_{i=1}^n a_i = A.

    It follows from the AM-GM inequality: \dfrac{1}{n} \displaystyle\sum_{i=1}^n a_i \leq \left( \displaystyle\prod_{i=1}^n a_i \right)^\frac{1}{n} with equality iff all the a_is are equal. (see here if you're unfamiliar with it - you can probably quote it - if not, it's not that tricky to prove) that the product is maximised when a_1=a_2=...=a_n.

    From here we note that we now wish to find the maximum n such that: \left(\dfrac{A}{n} \right)^n is maximised.
    Note now that the function: f(x)=\left( \dfrac{A}{x} \right)^x has a single stationary point, which we now compute:
    f(x)=\left( \dfrac{A}{x} \right)^x=e^{x\ln \left(\frac{A}{x} \right)}

\Rightarrow f'(x)=f(x)(\ln \left( \dfrac{A}{x} \right)-1)

\Rightarrow f'(x)=0 \Leftrightarrow x=\dfrac{A}{e}.

    Thus we see that either: n=\left \lfloor {\dfrac{A}{e}} \right \rfloor or n=\left\lceil { \dfrac{A}{e} } \right\rceil.
    The maximum product is then given by:
    \max \left \{ \left\lfloor {\dfrac{A}{e} } \right\rfloor ^{\lfloor {\frac{A}{e}} \rfloor}, \left\lceil {\dfrac{A}{e}} \right\rceil  ^{\lceil {\frac{A}{e}} \rceil}\right \}
    In the case A=11 it's easy to see that we have n=5, and the maximum product is 2.2^5 which I guess will be around 50 something.
    That's not quite correct. With A = 11, n = 4 yields a larger product than n = 5.
    2.75^4 is roughly 57 which is larger than 2.2^5 which is roughly 52.

    The maximum product is the maximum of {(A/n)^n : n is A/e rounded up/down} rather than {n^n : n is A/e rounded up/down} - the latter (which you've said) leads you to incorrectly always pick n as the round up of A/e

    On a side note, you've done a good job narrowing the problem down, but you haven't given any method to determine which n of the two choices to pick to maximise (A/n)^n: the round up or the round down of A/e.

    Surely once you've established that each a{i} must be equal, narrowing n down to two choices (round up/round down of A/e) is a bit pointless if there's no good way of determining which to pick.

    Why not just say take n in X = {1,2,...,A} (it's easier to justify that n won't be greater than A than to narrow it down as you've done) to be such that (A/n)^n > (A/m)^m for all m in X\{n}? This solution gets stuck in the same way yours - we can't actually pick between these n without a calculator - but it takes a lot less work to get there!
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    (Original post by studentro)
    That's not quite correct. With A = 11, n = 4 yields a larger product than n = 5.
    2.75^4 is roughly 57 which is larger than 2.2^5 which is roughly 52.

    The maximum product is the maximum of {(A/n)^n : n is A/e rounded up/down} rather than {n^n : n is A/e rounded up/down} - the latter (which you've said) leads you to incorrectly always pick n as the round up of A/e

    On a side note, you've done a good job narrowing the problem down, but you haven't given any method to determine which n of the two choices to pick to maximise (A/n)^n: the round up or the round down of A/e.

    Surely once you've established that each a{i} must be equal, narrowing n down to two choices (round up/round down of A/e) is a bit pointless if there's no good way of determining which to pick.

    Why not just say take n in X = {1,2,...,A} (it's easier to justify that n won't be greater than A than to narrow it down as you've done) to be such that (A/n)^n > (A/m)^m for all m in X\{n}? This solution gets stuck in the same way yours - we can't actually pick between these n without a calculator - but it takes a lot less work to get there!
    This is why I need a calculator o.O - been a bit dim somewhere, though I've thrown my working away so I've no idea why I said that 2.2^5>2.75^4, though I strongly suspect it's cos I'm rubbish at arithmetic.

    You're right about the \left(\dfrac{A}{n} \right)^n that's just a typo from being too lazy with copy and pasting latex will edit the post.
    I'd say that narrowing it down to two cases is better than having A cases to check, particularly if A is large, and the amount of work it takes isn't huge, certainly less than it would be to even check each value for say A=100.
    Finding which of the two maximises for any given number is, I imagine, an unnecessary amount of work, if it's even possible. I couldn't be bothered at the time, and I'm not that bothered now.
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    Downing paper - Question 4
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    How can you define the volume of the box?
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    What do you know about minima/ maxima of functions?
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    The base will have area  (1-2x)^2 as  x is the length removed from both sides. As the height of the sides will be x, the volume is x * the area. This means we can define the volume, V as a function of x.  V(x) = x(1-2x)^2 To find maximum and minimum points of  V(x) we can differentiate, leading to  V'(x) = 12x^2 - 8x + 1 . This factorises to  (6x-1)(2x-1) = 0 as  V'(x) = 0 for maxima/ minima. This gives solutions    \frac{1}{2} &\frac{1}{6} . If \frac{1}{2} was cut from both sides, this would make the sides of the base length  0 , making this the minimum volume possible. This leaves x =\frac{1}{6} as the solution for the maximum volume.
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    I'm a bit iffy about this, but I think it works...

    Paper 4, Question 10

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    What is the difference in the energy of an object rotating about an axis, compared to an object moving in a straight line?


    Solution
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    By energy, the man who slides (modelled as a particle) will have speed u at the bottom, given by

    



0.5mu^{2} = mgh

-> u =  \sqrt{2gh}

    where h is the starting height.

    The log that rolls will have both rotational and translational kinetic energy at the bottom; suppose it has translational speed v at the bottom. Its angular speed at the bottom will thus be v/r, where r is the radius of the cylinder. Its energy equation is thus:

    

mgh = 0.5mv^2 + 0.5(0.5mr^2)*v^2/r^2

gh = v^2(0.5+0.25)



v = \sqrt{\dfrac{4gh}{3}}

    where the moment of inertia of the cylinder will be given by 0.5mr^2, as may be obtained by the 'stretch rule' as applied to a disc.

    As 4/3 < 2, we see that the log will have a lower speed than the man at the bottom of the slope. The ratio of the two speeds will be root(3/2), that is the speed of the man will be about 1.2 times the speed of the log.

    (As I said, not too sure about this - I welcome any admonishments :-))
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    (Original post by Krollo)
    I'm a bit iffy about this, but I think it works...

    Paper 4, Question 10

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    What is the difference in the energy of an object rotating about an axis, compared to an object moving in a straight line?

    Solution
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    By energy, the man who slides (modelled as a particle) will have speed u at the bottom, given by

    



0.5mu^{2} = mgh

-&gt; u =  \sqrt{2gh}

    where h is the starting height.

    The log that rolls will have both rotational and translational kinetic energy at the bottom; suppose it has translational speed v at the bottom. Its angular speed at the bottom will thus be v/r, where r is the radius of the cylinder. Its energy equation is thus:

    

mgh = 0.5mv^2 + 0.5(0.5mr^2)*v^2/r^2

gh = v^2(0.5+0.25)



v = \sqrt{\dfrac{4gh}{3}}

    where the moment of inertia of the cylinder will be given by 0.5mr^2, as may be obtained by the 'stretch rule' as applied to a disc.

    As 4/3 < 2, we see that the log will have a lower speed than the man at the bottom of the slope. The ratio of the two speeds will be root(3/2), that is the speed of the man will be about 1.2 times the speed of the log.

    (As I said, not too sure about this - I welcome any admonishments :-))
    That's the answer I got! Krollo agrees with me! :ahee:
    Though you haven't answered this question:

    Assuming both startat the top with zero speed, and that friction plays a negligible role in the second case, which will get to the bottom faster, the man who slides or the log that rolls?

    (Log, obviously, since lower translational speed when reaches bottom \wedge same translational acceleration \implies accelerated for lower period of time )

    P.S. Why you use Tex rather than LaTex?
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    (Original post by Krollo)
    Test 2, Question 3

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


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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.


    Posted from TSR Mobile
    Precisely how I done it. Very nice question.


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    (Original post by Johann von Gauss)
    That's the answer I got! Krollo agrees with me! :ahee:
    Though you haven't answered this question:

    Assuming both startat the top with zero speed, and that friction plays a negligible role in the second case, which will get to the bottom faster, the man who slides or the log that rolls?

    (Log, obviously, since lower translational speed when reaches bottom \wedge same translational acceleration \implies accelerated for lower period of time )

    P.S. Why you use Tex rather than LaTex?
    Sorry, that was just me being dopey and interpreting the question as which one reached the bottom with the greater speed (the question seems a tad ambiguous, though I'm fairly sure you're right on this). I agree with your logic.

    In relation to your other point, bad habit.
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    (Original post by Krollo)
    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.

    Posted from TSR Mobile
    How do you work out the sum of the roots of the polynomial?
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    (Original post by AmarPatel98)
    How do you work out the sum of the roots of the polynomial?
    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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    How did he come up with the polynomial?
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    (Original post by Jordan\)
    How did he come up with the polynomial?
    If points are collinear they lie on a line.
    So that means there are four solutions to that polynomial=mx+c
    Solving for 0 we get a new polynomial but see that sum of the roots is unaffected by the change in in polynomial as coefficient of x^3 is unchanged so -2(sum)=7 sum=-7/2
    We require k=sum/4=-7/8


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    (Original post by physicsmaths)
    If points are collinear they lie on a line.
    So that means there are four solutions to that polynomial=mx+c
    Solving for 0 we get a new polynomial but see that sum of the roots is unaffected by the change in in polynomial as coefficient of x^3 is unchanged so -2(sum)=7 sum=-7/2
    We require k=sum/4=-7/8


    Posted from TSR Mobile
    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?
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    (Original post by Johann von Gauss)
    P.S. Why you use Tex rather than LaTex?
    Easier to type on this site + barely any typesetting difference. (on this site)
 
 
 
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