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# 2 quick questions from Trinity pre-admissions test watch

1. Hi,

these are the questions:

1.) Investigate the integral from 0 to 3 of 1/((1-x)^2) dx
2.) In a tennis tournament there are 2n participants. In the first round of the tournament, each player plays exactly once, so there are n games. Show that th pairings for the first round can be arranged in exactly (2n-1)!/((2^(n-1))*(n-1)!) ways.

So here are my my questions concerning them:

1.) as indefinite integral, i get 1/(1-x). So if i now add the definite integrals with boundaries (0 to a value very shortly below one) and (a value very shortly above one to 3), i get (-0.5 - "infinity" ) + ( "- infinity" -1) which is -infinity. But by looking at the original formula (and at the graphical calculator btw) it is obvious that the result is positive infinity. What is wrong?
2.) by doing the induction "backwards", i see that for possibilities(n)=possibilities(n-1)*(2n+1). However i do not know how to explain this let alone do i have a clue on how to derive the formula had it not been given. Any advice?

Thanks!
2. For 2), have a look at my posts here: http://www.thestudentroom.co.uk/show...6#post15307946. The principle is the same.
3. 1) What's wrong is that the integral definition you're used to isn't well defined over discontinuities, which is why it's spitting out bad answers. I think a sketch is the best way to answer the question.

*Edit*

It's just a play on the classic interview question

.
4. Well i don't have a graphical calculator but the graph of 1/(1-x) looks like 1/x reflected in the y-axis with asymptotes of x=0, x=1, y=1, y=0 and y=-1 right?

And an integral works out the area between to points right, so if you're calculating the area between 0 and 3 then the area is gonna be positive infinity. However, because the graph has a void in it, its essentially positive or negative infinity i would assume. Perhaps this is what the question is wanting u to investigate so to speak IDK, i'm not applying for maths. General engineering ftw , maths degree on a simpler level lol (i am hoping anyways).
5. (Original post by Swayum)
1) What's wrong is that the integral definition you're used to isn't well defined over discontinuities, which is why it's spitting out bad answers. I think a sketch is the best way to answer the question.

*Edit*

It's just a play on the classic interview question

.
Well, this is why I excluded x=1 from my calculation by addint up the integrals as written in my post ((integral from 0 to 0.999999) + (integral from 1.000000...001 to 3)) so why is that still giving the wrong answer?

6. 0.999... = 1
1.000...001 = nonsense

I get that you're trying to argue it like

And finding the limit as epsilon tends to 0 but my point above stands.
7. well, for the tennis question, i started off by saying.

how many pairs can the first pair be made, then times that by how many ways the second pair can be made, etc..

that'll give you a whole bunch of stuff times together.

then just play around with that in terms of factorials and its pretty easy to derive. i doubt you need to rigorously prove it. i mean it said "show that". i pressume "prove that" would imply a more rigorous approach.
8. (Original post by Swayum)
I get that you're trying to argue it like

And finding the limit as epsilon tends to 0 but my point above stands.
Yes, that was exactly my argument. But I do not see why your point stands when one is excluded. The result should be + infinity then, which would be correct (but, sadly, i recieve - inf).

(@ SouthernFreerider: Just playing around with it and using the intuitive argument works indeed pretty well so I suppose that's what they wanted)
9. 2) let P(n) = pairings
well if there are 2 players there is obviously 1 pairing so
isn't it? P(n)= (2n-1)P(n-1) or else P(2)=(2*2+1)P(1)= 5*1 = 5
but P(2) = 3

anyway that means:

P(n) = 1 *3*5*...*(2n-1)
Spoiler:
Show

=(2n-1)!/(2*4*6*...*(2n-2))
=(2n-1)!/(2^(n-1)*1*2*3*...*(n-1))
=(2n-1)!/(2^(n-1).(n-1)!) as required
10. The wordy proof for 2)

There are 2n! ways to order everyone. Pair everyone of in the obvious way.
There are n! ways to order these pairs. There are 2^n ways to order each pair

Therefore there are
11. 2) another wordy one: if you split the players in two groups, there are (2n)Cn ways of doing this then there are n! ways of making pairings between each group but if you were to swap 2 people from each group and then swap their pairing you get the same pairing this mean for each pair their is an arrangement that was double so Pairing (n) = (2n)!/(n!)^2*n!/2^n = (2n)!/(n!*2^n) = (2n-1)!/(n!*2^n) * 2n (2n-1)!/((n-1)!*2^n) * 2n / n = (2n-1)!/(n!*2^(n-1)) * 2/2 = (2n-1)!/(n!*2^(n-1)) as required.

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