Specimen Paper 4: Question 6.
Hint:Spoiler:.ShowUse AMGM.
Solution:Spoiler:ShowFirst off, we're gonna assume (although it's not explicitly stated) that the "division" only allows positive reals, else the problem is silly.
Let be the number we start with.
Let be a set of order such that .
It follows from the AMGM inequality: with equality iff all the s 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 .
From here we note that we now wish to find the maximum such that: is maximised.
Note now that the function: has a single stationary point, which we now compute:
.
Thus we see that either: or .
The maximum product is then given by:
In the case it's easy to see that we have , and the maximum product is which I guess will be around something.
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 29112015 13:34
Last edited by joostan; 03122015 at 18:38. Reason: Floors and ceilings. 
EmptyMathsBox
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 29112015 14:33
Downing Test Question 6 Solution:
Answer:
Spoiler:Full Solution:ShowThe answer is 1/75 seconds
Spoiler:ShowIf 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 distance. Then as is small the particle moves approximately in a straight line in this period. Also the particle it is chasing has also moved as well but at an angle of 60 degrees from the line of motion of this particle. As all three particles move 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 . Then the time for the particles to move from their new positions to the centre of the triangle is where 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 + time to get from smaller triangle to centre of the triangle and so .
Now if we can find in terms of the we can write in terms of and take limits to find . We can identify a triangle on the diagram with sides of with the angles opposite them respectively where is the angle opposite the side with length (there are three such triangles on the diagram, we can concentrate on any one).
Now by applying the sine rule once we get that . We rearrange this to get . So if we can work out in terms of we can proceed.
To do this we use the sine rule again to get that . After some rearrangement and trigonometric identities (which I will not detail here but involve expanding and finding from ) we get that . We plug this into our first sine rule equation to get that .
Now as by rearranging we get that so or when we substitute for u .When we take the limit as 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 approaches 0, approaches .
I am sure there is a faster and neater solution but this was the one I found anyway.Last edited by EmptyMathsBox; 29112015 at 14:49. 
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 29112015 14:39
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
Spoiler:Show 
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 29112015 16:29
Trinity Paper 4  Question 7
Hint
Spoiler:Show
How can you find a volume without using geometry?
Solution
Spoiler:Show
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:
and since the volume of the whole sphere is 4/3pir^3, the volume of the larger part is
As a quick check, plugging in a=0 gives two equal hemispheres, whereas a=r gives a single sphere, as we would expect.

MadChickenMan
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 01122015 15:46
Specimen Test 3, Question 4:
Hint:Spoiler:Solution:ShowThink about which flips must be heads, which must be tails and for which it doesn't matter. 
MadChickenMan
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Specimen Test 3, Question 9:
Hint:Solution:Spoiler:Show(i) By the chain rule
.
Using the product rule
.
So we have that
.
(ii) As is an operator, we compute the rightmost terms first terms first
.
(iii) Differentitaing
Substituting this value into the equation we have
Using the original equation and differentiating gives
Substituting the values for and in to the new equation
(LaTeXathon over) 
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 03122015 16:10
Posted from TSR Mobile 
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 03122015 16:26
(Original post by joostan)
Specimen Paper 4: Question 6.
Hint:Spoiler:.ShowUse AMGM.
Solution:Spoiler:ShowFirst off, we're gonna assume (although it's not explicitly stated) that the "division" only allows positive reals, else the problem is silly.
Let be the number we start with.
Let be a set of order such that .
It follows from the AMGM inequality: with equality iff all the s 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 .
From here we note that we now wish to find the maximum such that: is maximised.
Note now that the function: has a single stationary point, which we now compute:
.
Thus we see that either: or .
The maximum product is then given by:
In the case it's easy to see that we have , and the maximum product is which I guess will be around something.
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!Last edited by studentro; 03122015 at 17:25. 
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 03122015 18:32
(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!
You're right about the 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 cases to check, particularly if 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 .
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. 
Number Nine
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 03122015 19:35
Downing paper  Question 4
Hint 1:Spoiler:Hint 2:ShowHow can you define the volume of the box?Spoiler:Solution:ShowWhat do you know about minima/ maxima of functions?Spoiler:ShowThe base will have area as 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. To find maximum and minimum points of we can differentiate, leading to . This factorises to as for maxima/ minima. This gives solutions &. If was cut from both sides, this would make the sides of the base length , making this the minimum volume possible. This leaves as the solution for the maximum volume.Last edited by Number Nine; 03122015 at 20:45. 
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 03122015 21:55
I'm a bit iffy about this, but I think it works...
Paper 4, Question 10
Hint
Spoiler:Show
What is the difference in the energy of an object rotating about an axis, compared to an object moving in a straight line?
Solution
Spoiler:Show
By energy, the man who slides (modelled as a particle) will have speed u at the bottom, given by
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:
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 :))

Johann von Gauss
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 04122015 11:40
(Original post by Krollo)
I'm a bit iffy about this, but I think it works...
Paper 4, Question 10
HintSpoiler:Show
What is the difference in the energy of an object rotating about an axis, compared to an object moving in a straight line?
SolutionSpoiler:Show
By energy, the man who slides (modelled as a particle) will have speed u at the bottom, given by
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:
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 :))
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 same translational acceleration accelerated for lower period of time )
P.S. Why you use Tex rather than LaTex? 
physicsmaths
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 04122015 21:05
(Original post by Krollo)
Test 2, Question 3
Hint
Spoiler:Show
Consider the relationship between the coefficients of a polynomial and its roots.
Solution
Spoiler:Show
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 04122015 21:31
(Original post by Johann von Gauss)
That's the answer I got! Krollo agrees with me!
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 same translational acceleration accelerated for lower period of time )
P.S. Why you use Tex rather than LaTex?
In relation to your other point, bad habit. 
AmarPatel98
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 05122015 09:57
(Original post by Krollo)
Test 2, Question 3
HintSpoiler:Show
Consider the relationship between the coefficients of a polynomial and its roots.
SolutionSpoiler:Posted from TSR MobileShow 
physicsmaths
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 05122015 10:09
(Original post by AmarPatel98)
How do you work out the sum of the roots of the polynomial?
Or (xa)(xb)(xc)(xd) 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 (xa)(xb)(xc)(xd) where abcd are the roots. Expand that and eqaute coefficients.
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 05122015 10:16
(Original post by Jordan\)
How did he come up with the polynomial?
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
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Has he just made them up? 
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 05122015 10:32
(Original post by Johann von Gauss)
P.S. Why you use Tex rather than LaTex?
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