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March 28, 2024, 09:02:58 pm

Author Topic: Mathematics Extension 1 Challenge Marathon  (Read 26607 times)  Share 

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RuiAce

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Mathematics Extension 1 Challenge Marathon
« on: February 16, 2016, 06:34:02 pm »
+1
Here, I will occasionally post challenge questions for the daring 3U/4U student to attempt.

Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.

I invite everyone to also post their own questions at their own discretion, and for anyone who has completed 3U maths or equivalent to also answer.



Spoiler
Required knowledge: 2U Preliminary Basic Arithmetic and Algebra, 3U HSC Induction

Aside: An Extension 2 marathon will come in time after more topics have been taught. :)

KoA

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #1 on: February 16, 2016, 07:49:20 pm »
+2
Here, I will occasionally post challenge questions for the daring 3U/4U student to attempt.

Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.

I invite everyone to also post their own questions at their own discretion, and for anyone who has completed 3U maths or equivalent to also answer.



Spoiler
Required knowledge: 2U Preliminary Basic Arithmetic and Algebra, 3U HSC Induction

Aside: An Extension 2 marathon will come in time after more topics have been taught. :)

EDIT: Correct Solution in the Spoiler Below
Spoiler

I'd just like to say that in case anyone's interested, a somewhat easier (and possibly a bit more elegant?) proof would be to note that if the statement is true for P(k) then it is also true for P(k+7). Hence, if you can prove P(0) , P(1), ... P(6) all are multiples of 7, then you've proven the statement for all integers. Also, if anyone's familiar with modular arithmetic, once you're working mod 7, this problem is trivial by Fermat's Little Theorem.
« Last Edit: May 05, 2016, 01:48:43 pm by jamonwindeyer »
2015 HSC
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2016 HSC
Advanced English, Drama, Maths 3U, Maths 4U, Music 2, Extension Music, Chemistry, Software Design & Development

RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #2 on: February 16, 2016, 08:10:52 pm »
0


I'd just like to say that in case anyone's interested, a somewhat easier (and possibly a bit more elegant?) proof would be to note that if the statement is true for P(k) then it is also true for P(k+7). Hence, if you can prove P(0) , P(1), ... P(6) all are multiples of 7, then you've proven the statement for all integers. Also, if anyone's familiar with modular arithmetic, once you're working mod 7, this problem is trivial by Fermat's Little Theorem.

Spot on.

And yes, to those interested in extracurricular maths the most elegant approach is completed with an application of Fermat's Little Theorem.
____________________________________

NEXT QUESTION:


Happy Physics Land

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #3 on: February 16, 2016, 09:28:36 pm »
+2
At least this is easier than the stuff you gave me last time...
My reasoning in the end was probably not strong enough

EDIT: Correct solution in the spoiler below.

Spoiler

« Last Edit: May 05, 2016, 01:50:08 pm by jamonwindeyer »
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jamonwindeyer

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #4 on: February 16, 2016, 09:39:43 pm »
+2
At least this is easier than the stuff you gave me last time...
My reasoning in the end was probably not strong enough

(Image removed from quote.)

Spot on HPL. I think your reasoning was solid at the end there, though you can just say this, it is a mathematically correct statement after all:



Might fix the awkward-ness that you are worried about of your phrasing at the end there :)

Happy Physics Land

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #5 on: February 16, 2016, 09:43:06 pm »
0
Spot on HPL. I think your reasoning was solid at the end there, though you can just say this, it is a mathematically correct statement after all:



Might fix the awkward-ness that you are worried about of your phrasing at the end there :)

Thank you very much Jamon ^^
Mathematics: 96
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Maths Extension 1: 97
English Advanced: 92
Physics: 95
Chemistry: 92
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RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #6 on: February 16, 2016, 09:54:24 pm »
+1
Spot on HPL. I think your reasoning was solid at the end there, though you can just say this, it is a mathematically correct statement after all:



Might fix the awkward-ness that you are worried about of your phrasing at the end there :)
Believe it or not, the proper justification for that is simply to say that x and u are 'dummy variables' used in evaluation of a definite integral.

But yes the solution from HPL was spot on.
_____________________
NEXT QUESTION


Spoiler
Required knowledge: Preliminary 2U Trigonometric Ratios, Preliminary 3U Trigonometric Ratios
« Last Edit: February 16, 2016, 09:56:17 pm by RuiAce »

jamonwindeyer

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #7 on: February 16, 2016, 10:53:30 pm »
+2
Believe it or not, the proper justification for that is simply to say that x and u are 'dummy variables' used in evaluation of a definite integral.

But yes the solution from HPL was spot on.
_____________________
NEXT QUESTION


Spoiler
Required knowledge: Preliminary 2U Trigonometric Ratios, Preliminary 3U Trigonometric Ratios

Indeed. I usually use an equation for similar proofs at uni but I have seen others use the "dummy variable" explanation. For those interested, questions like this (where the original integral appears again in the result) are formalised using Reduction Formulae at higher levels of mathematical study. I believe this is similar to the Recurrence Relations studied at the Extension 2 Level.

I love that we have a Trig Proof. Eager to see someone solve this one!  ;D

Happy Physics Land

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #8 on: February 17, 2016, 05:59:33 pm »
+1
Believe it or not, the proper justification for that is simply to say that x and u are 'dummy variables' used in evaluation of a definite integral.

But yes the solution from HPL was spot on.
_____________________
NEXT QUESTION


Spoiler
Required knowledge: Preliminary 2U Trigonometric Ratios, Preliminary 3U Trigonometric Ratios

Stuff you rui, I spent 15 mins to see what the magic will turn out to be and the answer is SPOILERS  :-X.

EDIT: Correct solution in spoiler below.
Spoiler


I did it only to satisfy your eagerness jamon :D
« Last Edit: May 05, 2016, 01:51:06 pm by jamonwindeyer »
Mathematics: 96
Maths Extension 2: 93
Maths Extension 1: 97
English Advanced: 92
Physics: 95
Chemistry: 92
Engineering Studies: 90
Studies of Religion I: 98

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RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #9 on: February 17, 2016, 08:17:36 pm »
0
Love you too.

The most elegant solution is essentially the above. The fastest solution, however, was to work on both sides. Note that 8cot(8x) + 4tan(4x) could also be decomposed to something that will allow the final result to be in the form of (a-a)=0

NEXT QUESTION:

TBD - I have a series of questions planned for the next MX1 Marathon question. Until that gets posted, reminder that at any time someone may post their own challenge.

RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #10 on: February 28, 2016, 07:27:06 pm »
0
The above question is probably easier than this one.



Spoiler
Required knowledge: HSC 3U Induction, HSC 3U Binomial Theorem

Happy Physics Land

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #11 on: February 28, 2016, 07:31:49 pm »
0
The above question is probably easier than this one.



Spoiler
Required knowledge: HSC 3U Induction, HSC 3U Binomial Theorem

This one is gonna be a tedious as f proof
Mathematics: 96
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Maths Extension 1: 97
English Advanced: 92
Physics: 95
Chemistry: 92
Engineering Studies: 90
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RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #12 on: February 28, 2016, 10:50:35 pm »
0
This one is gonna be a tedious as f proof

You'd be surprised. It's actually quite easy if you know what you're doing.

RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #13 on: March 18, 2016, 08:56:44 am »
0


Required knowledge
Preliminary 2U Trigonometric Ratios, Preliminary 2U Introduction to Calculus, HSC 3U Inverse Functions and the Inverse Trigonometric Functions

« Last Edit: May 05, 2016, 01:51:34 pm by jamonwindeyer »

RuiAce

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Re: Mathematics Extension 1 Challenge Marathon
« Reply #14 on: April 18, 2016, 08:40:45 am »
+1
'Infinitely' times more easier than what meets the eye