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# Trinity Admissions Test Solutions watch

1. (Original post by physicsmaths)
Yh that is what i meant, when the powers are 1/2 and it is 2 sequences holders becomes cauchy schwarz.
I would cautious of such questions if they ask me some crazy indepth questions about Holders inequality ill be like 'sorry i only know the basic proofs mate'. Atleast I am fairly certain I won't need this in my 30 minute test 😂.

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Yeah, I don't even know the proof lol. I've never used it before. This is definitely a hard question though.
Actually, I think I've done a proof for integer p and q, but not for real ones... I guess that would require limits everywhere.
2. (Original post by Renzhi10122)
Yeah, I don't even know the proof lol. I've never used it before. This is definitely a hard question though.
Actually, I think I've done a proof for integer p and q, but not for real ones... I guess that would require limits everywhere.
Yh I thought, Cauchy is normally in R^3 so this is R^N so maybe holders or an argument similar, so I just used the previous and summed to n and used the p+q=PQ! the second was definitely easier I thought! first part although I had something similar ina step paper! here they had not given anything so would have been sad if I did not know jensens inequality.

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3. (Original post by physicsmaths)
Yh I thought, Cauchy is normally in R^3 so this is R^N so maybe holders or an argument similar, so I just used the previous and summed to n and used the p+q=PQ! the second was definitely easier I thought! first part although I had something similar ina step paper! here they had not given anything so would have been sad if I did not know jensens inequality.

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Cauchy is R^3??? It's in R^n. But yeah, you use the previous result. I spent a while on that question, I didn't really think that using jensen's would have been the best method, but I guess that that is what they wanted.
4. (Original post by Renzhi10122)
Cauchy is R^3??? It's in R^n. But yeah, you use the previous result. I spent a while on that question, I didn't really think that using jensen's would have been the best method, but I guess that that is what they wanted.
Yh it is R^n lol I was thinking of R^3 since I had seen some proof of Cauchy in step II 2006 and extended it to R^n but this had nothing to do with this lol. This is what happens when you have a brain fart.
5. (Original post by joostan)
Paper 3 Question 1:
Spoiler:
Show
Proof by picture seems to be the best way to go, without any fancy expressions for the function.
A quick sketch should show that:
.
We can compute the integral:

Also from the sketch we see that:

Hence:

Observing that gives the result:

Paper 3 Question 2:
Spoiler:
Show

Let

Paper 3 Question 6:
Spoiler:
Show
False.
In decimal expansion, an digit number may be written as:

Let for some .
Then:
.
Certainly then, we have digits .
Hence if we have a square number with more than digits .
Shouldn't the numerator be cos and not sin after the u sub?

I also divided the top and bottom by cos to get

1/1+tan and then used t =tan theta/2

Which gave me I=1 :/

What've I done wrong?

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6. (Original post by I am Ace)
Shouldn't the numerator be cos and not sin after the u sub?

I also divided the top and bottom by cos to get

1/1+tan and then used t =tan theta/2

Which gave me I=1 :/

What've I done wrong?

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Na what he has done is right with the numerator, the result actually holds for 1/(1+tan^n(x))
Could you post your working? It might be since x=pi/2 cosx=0, but I am not too sure as the result holds backwards.

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7. (Original post by physicsmaths)
Na what he has done is right with the numerator, the result actually holds for 1/(1+tan^n(x))
Could you post your working? It might be since x=pi/2 cosx=0, but I am not too sure as the result holds backwards.

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Attached Images

8. (Original post by physicsmaths)
Na what he has done is right with the numerator, the result actually holds for 1/(1+tan^n(x))
Could you post your working? It might be since x=pi/2 cosx=0, but I am not too sure as the result holds backwards.

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The bottom 2 are fails I tried to delete

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9. Lol ur gna be so pissed, double angle tan formula is wrong.

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10. (Original post by physicsmaths)
Lol ur gna be so pissed, double angle tan formula is wrong.

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Brutal. It was so pretty as well

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11. (Original post by I am Ace)
Brutal. It was so pretty as well

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Lol, I have had plenty of amazing proofs like yours disproving well known mathematical theorems

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12. (Original post by physicsmaths)
Lol, I have had plenty of amazing proofs like yours disproving well known mathematical theorems

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Well done on spotting it, I'd never have realised! Ben though I've been meaning to fix that plus to a minus on my formula sheet for ages! Finally bit me in the arse

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What did people get for Q4 a,b?

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14. (Original post by physicsmaths)
What did people get for Q4 a,b?

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(a) (1/A)^(1/5)
(b) 1 + (1/2A), -1 + 1/(2A), (-1/A)^(1/3).

(they may accept +-1 for the first two solutions for (b))
15. (Original post by Zacken)
(a) (1/A)^(1/5)
(b) 1 + (1/2A), -1 + 1/(2A), (-1/A)^(1/3).

(they may accept +-1 for the first two solutions for (b))
How did u get to them? I have made a mistake somewhere.....

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16. (Original post by physicsmaths)
How did u get to them? I have made a mistake somewhere.....

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You get after some simplification .

For small A, you know that x has to be huge (from your graph, so x^3 can be thrown away in comparison to x^5) hence x = (1/A)^(1/5).

For big A, you know x lies somewhere between -1 and 0, so x^5 is smaller than x^3 and you can throw that away, hence A = 1/(-x^3) and solve for x, then looking at the graph, you notice solutions at x = +-1 roughly, so use x = +-1 + delta x and then approximate using binomial shiz, etc...

(I just put +-1)
17. (Original post by Zacken)
You get after some simplification .

For small A, you know that x has to be huge (from your graph, so x^3 can be thrown away in comparison to x^5) hence x = (1/A)^(1/5).

For big A, you know x lies somewhere between -1 and 0, so x^5 is smaller than x^3 and you can throw that away, hence A = 1/(-x^3) and solve for x, then looking at the graph, you notice solutions at x = +-1 roughly, so use x = +-1 + delta x and then approximate using binomial shiz, etc...

(I just put +-1)
Well I don't like that question lol. I just said infinity and 0- lol.

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18. (Original post by physicsmaths)
Well I don't like that question lol. I just said infinity and 0- lol.

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I quite enjoyed the question, I enjoyed the paper in general (the second paper, not so much), I think I got 5 full solutions out - but I freaking hated how they put Jensens inequality in the previous question, that was so unfair!
19. (Original post by Zacken)
I quite enjoyed the question, I enjoyed the paper in general (the second paper, not so much), I think I got 5 full solutions out - but I freaking hated how they put Jensens inequality in the previous question, that was so unfair!
Well you should done more olympiad stuff haha, Yh It was a nice paper! Paper 2 was good aswell I thought, I got Q1,2,3,5 and 7(kind of) for 7 ax+B is the remainder as they didn't say that at the start which made me wonder otherwise it does not make sense.

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20. Anyone have the maths and physics paper 1 Q 4 curve sketching solutions?

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