HSC 2014 MX2 Marathon (archive) (2 Viewers)

mreditor16 · Aug 13, 2014

Re: HSC 2014 4U Marathon

A q from this year's Baulko 4u trial

Sy123 · Aug 13, 2014

Re: HSC 2014 4U Marathon

$\\ $Given$ \ \sin \alpha + \sin \beta = \sqrt{\frac{5}{3}} \ , \ \cos \alpha + \cos \beta = 1 \\ \\ $Find$ \ \cos (\alpha - \beta)$

Kurosaki · Aug 13, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Given$ \ \sin \alpha + \sin \beta = \sqrt{\frac{5}{3}} \ , \ \cos \alpha + \cos \beta = 1 \\ \\ $Show that$ \ \cos (\alpha - \beta)$

Show that equals what sorry?

Sy123 · Aug 13, 2014

Re: HSC 2014 4U Marathon

Kurosaki said:
Show that equals what sorry?

Sorry that was a 'Find' question, not 'Show that', edited

Tugga · Aug 13, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Given$ \ \sin \alpha + \sin \beta = \sqrt{\frac{5}{3}} \ , \ \cos \alpha + \cos \beta = 1 \\ \\ $Find$ \ \cos (\alpha - \beta)$

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 · Aug 13, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Given$ \ \sin \alpha + \sin \beta = \sqrt{\frac{5}{3}} \ , \ \cos \alpha + \cos \beta = 1 \\ \\ $Find$ \ \cos (\alpha - \beta)$

Squaring the two equations:
$\sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta=\frac{5}{3} \\ \\ \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta =1$

Add them together and simplify, getting $\frac{1}{3}$ ?

Tugga · Aug 13, 2014

Re: HSC 2014 4U Marathon

mreditor16 said:
A q from this year's Baulko 4u trial

screenshot tool

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 · Aug 13, 2014

Re: HSC 2014 4U Marathon

From Newington Trials 2009

aDimitri · Aug 14, 2014

Re: HSC 2014 4U Marathon

10dragonsboi said:
From Newington Trials 2009

Gonna skip lots of lines, especially in the earlier parts, because this is extremely long.
$\\ (i) \quad u=(1-lnx) \quad dv = dx \quad I_{n} = [x(1-lnx)^{n}]_{1}^{e} + n\int_{1}^{e}(1-lnx)^{n-1}dx \\ I_{n} = -1 + nI_{n-1} \\ \\ (ii) \quad I_{3} = 6e - 16 \\ \\ (iii) \quad I_{n} = -1 + nI_{n-1} \\ = -1 -n + n(n-1)I_{n-1} \\ = -1 -n -n(n-1) - n(n-1)(n-1)I_{n-3} \\ ... \\ = n!I_{0} - \frac{n!}{n!} - \frac{n!}{(n-1)!} - ... - \frac{n!}{1!} \\ = n!e - \frac{n!}{n!} - \frac{n!}{(n-1)!} - ... - \frac{n!}{1!} - \frac{n!}{0!} \\ \frac{I_{n}}{n!} = e - \sum_{r=0}^{n} \frac{1}{r!} \\ \\ (iv) \quad \text{Consider the graph of (1-lnx)} \\ \text{In the domain that the integral describes, f(x) is continuous and positive between the points (1,1) and (0,0)} \\ \text{Hence, it is clear that for all values n, the area will be a positive number.} \\ I_{n} \geq0 \\ \text{Since the function lies between 1 & 0, raising it to any positive power will cause a decrease.} \\ \text{Hence, the maximum value occurs at the minimum n, n=0. } \\ I_{n} \leq I_{0} = e - 1 \\ 0 \leq I_{n} \leq e-1$

hahaha got the equation too long message and had to split up part v

$\\ (v) \text{Once again examine the graph. It is clear minimum value for } I_{n} \text{ occurs as } n \to \infty \\ \lim_{n \to \infty} I_{n} = 0. \\ \text{Hence, the desired statement follows quite simply from (iii)} \\ e - \lim_{n \to \infty}\sum_{r=0}^{n} \frac{1}{r!} = 0 \\ \lim_{n \to \infty}\sum_{r=0}^{n} \frac{1}{r!} = e$

madharris · Aug 14, 2014

Re: HSC 2014 4U Marathon

Got a difficult one for you guys. I saw that it was going to be in carrot's trial.

$1+2$

aDimitri · Aug 14, 2014

Re: HSC 2014 4U Marathon

madharris said:
Got a difficult one for you guys. I saw that it was going to be in carrot's trial.

$1+2$

$4=4 \\ -21 = -21 \\ 9-30 = 49-70 \\ 9-30+25 = 49-70+25 \\ (3-5)^2 = (7-5)^2 \\ 3-5 = 7-5 \\ 1+2 = 7$
gg

madharris · Aug 14, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
$4=4 \\ -21 = -21 \\ 9-30 = 49-70 \\ 9-30+25 = 49-70+25 \\ (3-5)^2 = (7-5)^2 \\ 3-5 = 7-5 \\ 1+2 = 7$
gg

naa the answer is C.
It's using 5u reasoning though.
You will come to learn the reason if you pursue maffs

Sy123 · Aug 14, 2014

Re: HSC 2014 4U Marathon

$\\ $Show that if$ \ x \ $is odd, then$ \ x^2-1 \ $is divisible by$ \ 8$

aDimitri · Aug 14, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Show that if$ \ x \ $is odd, then$ \ x^2-1 \ $is divisible by$ \ 8$

equivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.

$\text{It is required to prove that }(2n-1)^{2} - 1 = 8M \text{ for all natural numbers n, where M is an integer} \\ \text{Test n=1, statement is true} \\ \text{Assume true for n=k} \\ \text{Prove for n = k+1} \\ LHS = (2(k+1)-1)^{2} - 1 \\ = (2k - 1 + 2)^{2} - 1 \\ =(2k-1)^{2} + 4(2k-1) + 4 - 1 \\ = 8M + 8k = 8(M+k) \\ \text{Therefore, blah blah blah true by mathematical induction}$

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

Stygian · Aug 14, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
equivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.

$\text{It is required to prove that }(2n-1)^{2} - 1 = 8M \text{ for all natural numbers n, where M is an integer} \\ \text{Test n=1, statement is true} \\ \text{Assume true for n=k} \\ \text{Prove for n = k+1} \\ LHS = (2(k+1)-1)^{2} - 1 \\ = (2k - 1 + 2)^{2} - 1 \\ =(2k-1)^{2} + 4(2k-1) + 4 - 1 \\ = 8M + 8k = 8(M+k) \\ \text{Therefore, blah blah blah true by mathematical induction}$

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

alternatively u can take the approach that every odd number is congruent to

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 · Aug 14, 2014

Re: HSC 2014 4U Marathon

Stygian said:
alternatively u can take the approach that every odd number is congruent to

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

but it's still not proven for all odd, just shown for the first 4

Stygian · Aug 14, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
but it's still not proven for all odd, just shown for the first 4

How so? Doesn't every odd number fit into those congruencies listed?

aDimitri · Aug 14, 2014

Re: HSC 2014 4U Marathon

Stygian said:
How so? Doesn't every odd number fit into those congruencies listed?

oh true yeah.
derp moment T_T

Stygian · Aug 14, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
oh true yeah.
derp moment T_T

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

Stygian · Aug 14, 2014

Re: HSC 2014 4U Marathon

(also im having a brain fart on that problem you posted in the Advanced thread, i can't seem to think of how to approach it)

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