Mathematics Optional mains 2020
Is there any group for Mathematics Optional?
If not, then please use this thread to get together and save each other's time.
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@Howzthejoshokay
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This question came in 2013. Can anyone tell the answer to the last part?
https://t.me/joinchat/NeGZBRrgSg4Qmheo410vYQ
Hooke's law of elasticity in Rectilinear motion chapter... Is it in the syllabus??? I skipped it last year due to paucity of time..
Hooke's law of elasticity in Rectilinear motion chapter... Is it in the syllabus??? I skipped it last year due to paucity of time..
In syllabus
@DramaticAdmiral Solution. If σ ∈ S10, we can express σ as a product of disjoint cycles
σ1 · · · σt of lengths k1, k2, . . . , kt (ki ≥ 2) where k1 + · · · kt ≤ 10. The
order of σ is then lcm(k1, . . . , kt). Thus we have to find the maximum
value of lcm(k1, . . . , kt) over all sets of integers {k1, . . . , kt} satisfying
ki ≥ 2 and k1 + . . . + kt ≤ 10.
We may as well assume the ki
’s are distinct from each other (since
repeating one of them doesn’t alter the lcm). This narrows the search
considerably. We list the possibilities for k1 + · · · kt and the corre-
sponding lcm:
2 + 3 + 4, lcm(2, 3, 4) = 12
2 + 3 + 5, lcm(2, 3, 5) = 30
2 + 4, lcm(2, 4) = 4
2 + 5, lcm(2, 5) = 10
2 + 6, lcm(2, 6) = 12
2 + 7, lcm(2, 7) = 14
2 + 8, lcm(2, 8) = 8
3 + 4, lcm(3, 4) = 12
3 + 5, lcm(3, 5) = 15
3 + 6, lcm(3, 6) = 6
3 + 7, lcm(3, 7) = 21
4 + 5, lcm(4, 5) = 20
4 + 6, lcm(4, 6) = 12
We see that the highest possible order is 30 and this is achieved by the
product of a 2-cycle, a 3-cycle and a 5-cycle; eg. (12)(345)(678910). This is half solution as example you may provide with the help of this.
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I was also able to do till here. Can you try calculating number of such elements of order 30? I am a bit weak with permutations and combinations@DramaticAdmiral Solution. If σ ∈ S10, we can express σ as a product of disjoint cyclesσ1 · · · σt of lengths k1, k2, . . . , kt (ki ≥ 2) where k1 + · · · kt ≤ 10. Theorder of σ is then lcm(k1, . . . , kt). Thus we have to find the maximumvalue of lcm(k1, . . . , kt) over all sets of integers {k1, . . . , kt} satisfyingki ≥ 2 and k1 + . . . + kt ≤ 10.We may as well assume the ki’s are distinct from each other (sincerepeating one of them doesn’t alter the lcm). This narrows the searchconsiderably. We list the possibilities for k1 + · · · kt and the corre-sponding lcm:2 + 3 + 4, lcm(2, 3, 4) = 122 + 3 + 5, lcm(2, 3, 5) = 302 + 4, lcm(2, 4) = 42 + 5, lcm(2, 5) = 102 + 6, lcm(2, 6) = 122 + 7, lcm(2, 7) = 142 + 8, lcm(2, 8) = 83 + 4, lcm(3, 4) = 123 + 5, lcm(3, 5) = 153 + 6, lcm(3, 6) = 63 + 7, lcm(3, 7) = 214 + 5, lcm(4, 5) = 204 + 6, lcm(4, 6) = 12We see that the highest possible order is 30 and this is achieved by theproduct of a 2-cycle, a 3-cycle and a 5-cycle; eg. (12)(345)(678910). This is half solution as example you may provide with the help of this.
@DramaticAdmiral I have no expertise in mathematics. I am trying to develop this. I was fond of mathematics many years ago. But I want to appear in upsc 2021 so I thought to go with mathematics. During this watching some lecture this question appeared perhaps in IGNOU ba or ma math text book. So I put it here. I am studying math from basics at graduation level.
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