HSC 2017 MX2 Marathon (archive) (2 Viewers)

leehuan · Nov 16, 2016

Re: HSC 2017 4U Marathon

calamebe said:
Yeah you left off the squares. Put the squares on, then use the result from a) and you should be fine.

EDIT: Though I don't see a need to break the terms up as you did in the second line. You can do it easily without breaking it up.

Was wondering where the squares vanished to.

There wasn't a need for it; I did it purely for my visual purposes.

Paradoxica · Nov 16, 2016

Re: HSC 2017 4U Marathon

You really don't need to split based upon parity. It only makes things worse.

You haven't considered

$\binom{n}{k} = \binom{n}{n-k}$

enough.

Side note your conclusion is wrong.

Paradoxica · Nov 16, 2016

Re: HSC 2017 4U Marathon

Here's a nice anecdote.

"One day Gauss' teacher asked his class to add together all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050"

TaxiCab1729 · Nov 17, 2016

Re: HSC 2017 4U Marathon

Damn it...I was doing that

TaxiCab1729 · Nov 17, 2016

Re: HSC 2017 4U Marathon

View attachment Binomial Question.pdf

I think I got it, but it looks dodgy...

seanieg89 · Nov 17, 2016

Re: HSC 2017 4U Marathon

TaxiCab1729 said:
View attachment 33599

I think I got it, but it looks dodgy...

For odd n, I don't think your S_2 is an integer. I think you don't need that subtraction by 1.

Also note that you can evaluate the sums C(n,k)^2 and kC(n,k)^2 combinatorially. (This takes less effort than other methods of computation imo.)

These sums amount to:

A: The number of ways of choosing a team of n people out of 2n people, n of whom are in year 12 and n of whom are in year 11.

This can be done in C(2n,n) ways.

B: Same as above, but also selecting one of the year 12 students to be team captain.

This can be done in nC(2n,n)/2 ways. (The number of ways of choosing a team and specifying a captain is clearly nC(2n,n). By symmetry, this captain will be in year 12 exactly half of the time.)

seanieg89 · Nov 17, 2016

Re: HSC 2017 4U Marathon

Note that the above yields an end answer of something like

$\left(a+\frac{nd}{2}\right)\binom{2n}{n}.$

Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.

calamebe · Nov 17, 2016

Re: HSC 2017 4U Marathon

seanieg89 said:
Note that the above yields an end answer of something like

$\left(a+\frac{nd}{2}\right)\binom{2n}{n}.$

Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.

Yeah that's what I got too.

TaxiCab1729 · Nov 17, 2016

Re: HSC 2017 4U Marathon

Yeah...I realized my mistake. My bad...

mini8658 · Nov 22, 2016

Re: HSC 2017 4U Marathon

For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?

frog1944 · Nov 23, 2016

Re: HSC 2017 4U Marathon

Paradoxica said:
$$\noindent Prove: $ \\\\ \left( x^{n-1} + \frac{1}{x^{n-1}} \right) + 2 \left( x^{n-2} + \frac{1}{x^{n-2}} \right) + \cdots + (n-1) \left( x + \frac{1}{x} \right) + n \equiv \frac{1}{x^{n-1}} \left(\frac{x^n -1}{x-1}\right)^2 \\\\ $Hence, or otherwise, prove: \\\\ $ \cos{(n-1)\theta} + 2\cos{(n-2)\theta} + \cdots + (n-1)\cos{\theta} + \frac{n}{2} \equiv \frac{1}{2} \times \frac{1+\cos{n \theta}}{1+\cos{\theta}}$

Is x a real or complex number?

calamebe · Nov 24, 2016

Re: HSC 2017 4U Marathon

frog1944 said:
Is x a real or complex number?

It doesn't matter for the first proof.

Paradoxica · Nov 24, 2016

Re: HSC 2017 4U Marathon

mini8658 said:
For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?

By inspection.

Drsoccerball · Nov 24, 2016

Re: HSC 2017 4U Marathon

mini8658 said:
For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?

This is true by definition.
$p(x) = (x-x_0)(x-x_1)...(x-x_n) $ Where $ x_i , i = 0,1,2,3...n $ are roots of $ p(x)$

pikachu975 · Nov 24, 2016

Re: HSC 2017 4U Marathon

http://prntscr.com/db7gfc

How would you sketch this? It's a HSC question.

si2136 · Nov 25, 2016

Re: HSC 2017 4U Marathon

pikachu975 said:
http://prntscr.com/db7gfc

How would you sketch this? It's a HSC question.

The first locus is a Half- plane locus. You sketch the original locus and you add in the inequality.

The 2nd locus is an Argand locus. It has the ranges of pi/4 and -pi/4.

Green Yoda · Nov 25, 2016

Re: HSC 2017 4U Marathon

pikachu975 said:
http://prntscr.com/db7gfc

How would you sketch this? It's a HSC question.

The first part is a linear line that is a perpendicular bisector from points that subtend from 0 and 2. (make z=x+iy and you will get a linear eqn)
Second part is just the argument from the origin from -pi/4 to pi on 4

equate the two inequalities

pikachu975 · Nov 25, 2016

Re: HSC 2017 4U Marathon

si2136 said:
The first locus is a Half- plane locus. You sketch the original locus and you add in the inequality.

The 2nd locus is an Argand locus. It has the ranges of pi/4 and -pi/4.

Rathin said:
The first part is a linear line that is a perpendicular bisector from points that subtend from 0 and 2. (make z=x+iy and you will get a linear eqn)
Second part is just the argument from the origin from -pi/4 to pi on 4

equate the two inequalities

Oh yeah i get it now thanks!

seanieg89 · Nov 25, 2016

Re: HSC 2017 4U Marathon

Rathin said:
a linear line

Green Yoda · Nov 25, 2016

Re: HSC 2017 4U Marathon

seanieg89 said:

Had to make sure lol
In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit

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