mreditor16
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HSC 2014 MX2 Marathon (archive) (1 Viewer)
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Kurosaki
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Sy123
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Re: HSC 2014 4U Marathon
So cos(a-b)=1/3
P.s. Sorry for ugly formatting, don't have the time right now to learn latex, maybe later. D:
Squaring both and adding gives you 2*(expanded form of cos(a-b)) + 2 =8/3 (since sin^2a+cos^2a=1) )
So cos(a-b)=1/3
P.s. Sorry for ugly formatting, don't have the time right now to learn latex, maybe later. D:
Kurosaki
True Fail Kid
Re: HSC 2014 4U Marathon
Add them together and simplify, getting
?
Squaring the two equations:
Add them together and simplify, getting
Re: HSC 2014 4U Marathon
noting a^3 = 30+18a, it follows that a^3-18a-30=0.
therefore, P(x)=x^3-18x-30
A more rigorous proof would prove that its the lowest degree. I would probably do something like square it and show you get something you can't cancel with a degree 1 polynomial?
let that cube root thingo be a.
noting a^3 = 30+18a, it follows that a^3-18a-30=0.
therefore, P(x)=x^3-18x-30
A more rigorous proof would prove that its the lowest degree. I would probably do something like square it and show you get something you can't cancel with a degree 1 polynomial?
10dragonsboi
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Re: HSC 2014 4U Marathon
From Newington Trials 2009
From Newington Trials 2009
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aDimitri
i'm the cook
Re: HSC 2014 4U Marathon
 \quad u=(1-lnx) \quad dv = dx \quad I_{n} = [x(1-lnx)^{n}]_{1}^{e} + n\int_{1}^{e}(1-lnx)^{n-1}dx \\ I_{n} = -1 + nI_{n-1} \\ \\ (ii) \quad I_{3} = 6e - 16 \\ \\ (iii) \quad I_{n} = -1 + nI_{n-1} \\ = -1 -n + n(n-1)I_{n-1} \\ = -1 -n -n(n-1) - n(n-1)(n-1)I_{n-3} \\ ... \\ = n!I_{0} - \frac{n!}{n!} - \frac{n!}{(n-1)!} - ... - \frac{n!}{1!} \\ = n!e - \frac{n!}{n!} - \frac{n!}{(n-1)!} - ... - \frac{n!}{1!} - \frac{n!}{0!} \\ \frac{I_{n}}{n!} = e - \sum_{r=0}^{n} \frac{1}{r!} \\ \\ (iv) \quad \text{Consider the graph of (1-lnx)} \\ \text{In the domain that the integral describes, f(x) is continuous and positive between the points (1,1) and (0,0)} \\ \text{Hence, it is clear that for all values n, the area will be a positive number.} \\ I_{n} \geq0 \\ \text{Since the function lies between 1 & 0, raising it to any positive power will cause a decrease.} \\ \text{Hence, the maximum value occurs at the minimum n, n=0. } \\ I_{n} \leq I_{0} = e - 1 \\ 0 \leq I_{n} \leq e-1 )
hahaha got the equation too long message and had to split up part v
 \text{Once again examine the graph. It is clear minimum value for } I_{n} \text{ occurs as } n \to \infty \\ \lim_{n \to \infty} I_{n} = 0. \\ \text{Hence, the desired statement follows quite simply from (iii)} \\ e - \lim_{n \to \infty}\sum_{r=0}^{n} \frac{1}{r!} = 0 \\ \lim_{n \to \infty}\sum_{r=0}^{n} \frac{1}{r!} = e)
Gonna skip lots of lines, especially in the earlier parts, because this is extremely long.
From Newington Trials 2009
hahaha got the equation too long message and had to split up part v
Last edited:
aDimitri
i'm the cook
Re: HSC 2014 4U Marathon
^2 = (7-5)^2 \\ 3-5 = 7-5 \\ 1+2 = 7 )
gg
Got a difficult one for you guys. I saw that it was going to be in carrot's trial.
gg
madharris
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Re: HSC 2014 4U Marathon
It's using 5u reasoning though.
You will come to learn the reason if you pursue maffs
naa the answer is C.
gg
It's using 5u reasoning though.
You will come to learn the reason if you pursue maffs
aDimitri
i'm the cook
Re: HSC 2014 4U Marathon
let x = 2n-1 where n is an integer.
^{2} - 1 = 8M \text{ for all natural numbers n, where M is an integer} \\ \text{Test n=1, statement is true} \\ \text{Assume true for n=k} \\ \text{Prove for n = k+1} \\ LHS = (2(k+1)-1)^{2} - 1 \\ = (2k - 1 + 2)^{2} - 1 \\ =(2k-1)^{2} + 4(2k-1) + 4 - 1 \\ = 8M + 8k = 8(M+k) \\ \text{Therefore, blah blah blah true by mathematical induction})
same shit to prove it for all negative odd but do n = k-1
but x^2 - 1 is an even function so it's sorta redundant to do it for negative numbers too
equivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.
same shit to prove it for all negative odd but do n = k-1
but x^2 - 1 is an even function so it's sorta redundant to do it for negative numbers too
Stygian
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Re: HSC 2014 4U Marathon
1 mod 8
3 mod 8
5 mod 8
7 mod 8
Squaring it becomes
1 mod 8
9 mod 8 or 1 mod 8
25 mod 8 or 1 mod 8
49 mod 8 or 1 mod 8
Subtracting 1 from each gives 0 mod 8 i.e. all are divisible by 8
alternatively u can take the approach that every odd number is congruent toequivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.
same shit to prove it for all negative odd but do n = k-1
but x^2 - 1 is an even function so it's sorta redundant to do it for negative numbers too
1 mod 8
3 mod 8
5 mod 8
7 mod 8
Squaring it becomes
1 mod 8
9 mod 8 or 1 mod 8
25 mod 8 or 1 mod 8
49 mod 8 or 1 mod 8
Subtracting 1 from each gives 0 mod 8 i.e. all are divisible by 8
aDimitri
i'm the cook
Re: HSC 2014 4U Marathon
but it's still not proven for all odd, just shown for the first 4
Stygian
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Re: HSC 2014 4U Marathon
How so? Doesn't every odd number fit into those congruencies listed?
aDimitri
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Stygian
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Re: HSC 2014 4U Marathon
i do think the mod arithmetic approach is better in terms of efficiency but obviously your solution is more within the scope of the course and so better for this thread
haha it's ok
i do think the mod arithmetic approach is better in terms of efficiency but obviously your solution is more within the scope of the course and so better for this thread
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