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March 29, 2024, 07:00:43 am

Author Topic: Mathematics Extension 2 Challenge Marathon  (Read 31643 times)  Share 

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RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #15 on: October 05, 2016, 07:27:38 pm »
+3
Jesus Rui that was quick, save some for the students ;)
So not touching perms and combs lel

Eh, I've seen integrals way worse than that, and I couldn't get them out. This one was only barely in my doable margin because of the hint. So I was just like 'why not'.

Paradoxica

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #16 on: October 05, 2016, 07:28:48 pm »
0


Hint: Divide.

RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #17 on: October 06, 2016, 11:34:54 pm »
+1
Here is one extremely brutal way of doing the above question. It overuses partial fractions and throws complex numbers into an integral - something not needed in the HSC, just like my very first question


I considered the hint, but it took me way too long to figure out what happens after the hint is applied, hence all of this.

I won't put up the solution to the 3x quicker method just yet.
Unnecessarily over complicated non-HSC solution

Paradoxica

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #18 on: October 06, 2016, 11:43:30 pm »
0
By dividing throughout by , or otherwise, evaluate:


Mahan

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #19 on: November 08, 2016, 10:49:28 pm »
0
By dividing throughout by , or otherwise, evaluate:



Since, this is a relatively old post, I thought it would be useful to give a solution for it.
This method doesn't use the dividing trick:
before I start the proof it is useful to prove:
by integration by part we get :

let that yields


(by tan^{2}x+1=sec^{2}x)

by back substitution we can write it in terms of x.
the answer is
« Last Edit: November 08, 2016, 10:54:29 pm by Mahan »
Mahan Ghobadi

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RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #20 on: November 08, 2016, 10:56:34 pm »
+3

Mahan

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #21 on: November 08, 2016, 11:03:46 pm »
0


Yes, using the hint makes the question pretty simple.Just for the fun of it, I just presented a different proof. :)
Mahan Ghobadi

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RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #22 on: November 08, 2016, 11:16:36 pm »
+3
« Last Edit: November 08, 2016, 11:24:20 pm by RuiAce »

RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #23 on: November 21, 2016, 11:24:12 am »
+3
Consider an alphabet with n different available letters.

Let P(k) be the number of ways you can use exactly k different letters in an n letter word.

i) Explain why




ii) Show that



Let the Score of a word, X, be defined as 1/(1+ρ(X)), where ρ(X) is the number of letters that were not used by the word X.

iii) Show that the sum of all the Scores, S, over all possible n letter words, is given by:




iv) Hence, evaluate S in closed form.
To help explain part (i) (because I took a while understanding what was going on)
Explanation
Consider a three letter alphabet: {a, b, c}
Then, P(1) is the number of ways we can make three letter words, out of just one letter of the alphabet. If that letter was a, then aaa is the only word.
P(2) is the number of ways we can make three letter words, out of any two letters of the alphabet. If the letters are a and b, then the words are:
aab, aba, baa, bba, bab, abb
P(3) is the number of ways we can make three letter words using all the letters. So that's the easy 3!

The point of introducing the nCk is to quantify the fact we could've chosen any k of the 3 letters. In P(1), we could've chosen a, b or c to be our letter. (And indeed, 3Ck=3 possible letters.)

And now for the question





_________________________________

_________________________________

If a 4U student has not seen that Greek letter before, that is rho.
Explanation
Going back to the case n=3

If the words satisfied k=1 (e.g. aaa), then ρ(X) = 2  (Note: 1+2=3)
If the words satisfied k=2 (e.g. aab), then ρ(X) = 1  (Note: 2+1=3)
If the words satisfied k=3 (e.g. abc), then ρ(X) = 0  (Note: 3+0=3)

This can be seen by just realising how man letters we did not use, in each case.

The important thing to realise is that for each value of k, ρ(X) differs. As a matter of fact, ρ(X) always goes down by 1.

In fact, ρ(X) = 3 - k



_________________________________


I don't fully trust what I say from here due to how P(k) has been defined.



Paradoxica

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #24 on: March 16, 2017, 10:23:44 pm »
0
Here's a question for people to try:

Suppose we have the hyperbola y = 1/x defined over the positive real numbers (first quadrant). Let P(p, 1/p) and Q(q, 1/q) be two arbitrarily fixed points along the curve, with p < q. Define M as the midpoint of the chord PQ. The line segment OM intersects the hyperbola at R(r, 1/r), where O is the origin (0,0).

Without expressing the coordinates of R in terms of p and q, i.e. without deriving the equation of the line OM, prove that the tangent to the hyperbola at R is parallel to the chord PQ.

Mod edit: Altered the language to make it a bit "easier" to comprehend with respect to the HSC 4U course

Consider an arbitrary chord AB parallel to PQ using the following parameters: A(p/r,r/p), B(qr,1/qr) where r is a positive scaling factor. Construct tangents at A and B and denote the point of intersection I.

With some algebra, it is verified that OI and OM share the same gradient. Thus, I always lies on OM for any and every positive value of r.

Taking the limit as , the chord becomes a tangent.

The tangent at point P is unique to P, so conversely, the point I degenerates and becomes P.


wu345

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #25 on: April 06, 2017, 07:33:58 pm »
0
Let be a root of the equation , where

The root locus of a complex quadratic is the set of all points on the Argand Diagram that could be a root of the quadratic.
Sketch the root locus of
« Last Edit: April 07, 2017, 05:07:50 pm by wu345 »

wu345

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #26 on: April 06, 2017, 07:54:23 pm »
0
Define .
i) Show that
ii) Show that
iii) Hence show that and find a similar expression for
iv) Hence show that
« Last Edit: April 06, 2017, 07:59:30 pm by wu345 »

RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #27 on: April 07, 2017, 04:49:14 pm »
+3
Let be a root of the equation , where

The root locus of a complex quadratic is the set of all points on the Argand Diagram that could be a root of the quadratic.
Sketch the root locus of



________________________________________


Note that through these computations, the points (1,0) and (-1,0) are technically excluded. Coincidentally, the other case(s) brings it back.
________________________________________




________________________________________


Note that the above line was an abuse of notation. Infinity is not a number.

Proof of the limit used on g(p)

(Brief) Explanation as to why the functions decrease/increase from -1 respectively



In a similar way, the other case of \( p\le -2\) shows that \( y=0\) is a part of the locus for all \( x > 0\).

________________________________________

wu345

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #28 on: April 07, 2017, 07:22:23 pm »
0
Correct! Thanks for pointing out my flaw too, fixed it now. The intended shorter method however, was to simply sub the root into the quadratic and equate real and imaginary parts

RuiAce

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Re: Mathematics Extension 2 Challenge Marathon
« Reply #29 on: June 28, 2017, 05:13:03 pm »
+4
A classic complex numbers question I just did a few minutes ago.

Hint
You only have to consider when \(|\alpha|=1\). After you prove it holds for \( |\alpha|=1\), the case \( |\alpha| < 1 \) falls out pretty easily from it
Spoiler
You need to know circle geometry as well as have a substantial understanding of complex number tricks.
« Last Edit: June 28, 2017, 05:18:17 pm by RuiAce »