Print Page - Mathematics Extension 1 Challenge Marathon

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: RuiAce on February 16, 2016, 06:34:02 pm

Title: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 16, 2016, 06:34:02 pm

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.

$\text{There exists a very famous result that } \\ {n}^{7}-n \\ \text{is indeed divisible by 42.} \\ \text{Use AT LEAST one application of mathematical induction to verify this conjecture.}$

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. :)

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: KoA on February 16, 2016, 07:49:20 pm

Quote from: RuiAce on February 16, 2016, 06:34:02 pm

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.

$\text{There exists a very famous result that } \\ {n}^{7}-n \\ \text{is indeed divisible by 42.} \\ \text{Use AT LEAST one application of mathematical induction to verify this conjecture.}$

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

$\text{First, we begin by factoring the polynomial } n^7 - n = n(n-1)(n+1)(n^2+n+1)(n^2-n+1)\\\text{Note that }42 = 2\times3\times7\text{, so we just have to show that is divisible by all three of those numbers.}\\\text{Trivially, one of }n, n-1\text{ will be a multiple of two. Similarly, one of }n, n-1, n+1 \text{ is a multiple of three.}\\\text{So we just need to show the polynomial is divisible by 7 and our proof will be complete. }\\\text{The base case of } n= 0 \text{ is trivially true. Assume the result is true for }n = k \,\,(\text{i.e. } k^7 - k = 7m \text{ for some integer }m)\\ \text{ Then, for } n = k + 1\\(k+1)^7 - (k+1) = k^7 - k + 7k^6 + 21k^5 + 35k^4 + 21k^3 + 7k^2 + 7k = 7(m + k^6+3k^5+5k^4+3k^3+k^2)\\\text{Hence, by the principle of Mathematical induction, the result is true.}$

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.

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 16, 2016, 08:10:52 pm

Quote from: KoA on February 16, 2016, 07:49:20 pm

$\text{First, we begin by factoring the polynomial } n^7 - n = n(n-1)(n+1)(n^2+n+1)(n^2-n+1)\\\text{Note that }42 = 2\times3\times7\text{, so we just have to show that is divisible by all three of those numbers.}\\\text{Trivially, one of }n, n-1\text{ will be a multiple of two. Similarly, one of }n, n-1, n+1 \text{ is a multiple of three.}\\\text{So we just need to show the polynomial is divisible by 7 and our proof will be complete. }\\\text{The base case of } n= 0 \text{ is trivially true. Assume the result is true for }n = k \,\,(\text{i.e. } k^7 - k = 7m \text{ for some integer }m)\\ \text{ Then, for } n = k + 1\\(k+1)^7 - (k+1) = k^7 - k + 7k^6 + 21k^5 + 35k^4 + 21k^3 + 7k^2 + 7k = 7(m + k^6+3k^5+5k^4+3k^3+k^2)\\\text{Hence, by the principle of Mathematical induction, the result is true.}$

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:
$\int_{0}^{1}{\frac{\ln{\left(1+x\right)}}{1+{x}^{2}}dx}$
$\text{with the aid of the substitution }x=\frac{1-u}{1+u}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Happy Physics Land on February 16, 2016, 09:28:36 pm

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

(http://i.imgur.com/JFbMoyi.jpg)

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: jamonwindeyer on February 16, 2016, 09:39:43 pm

Quote from: Happy Physics Land on February 16, 2016, 09:28:36 pm

At least this is easier than the stuff you gave me last time...
My reasoning in the end was probably not strong enough

(http://i.imgur.com/JFbMoyi.jpg)

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:

$\int^1_0\frac{\ln{(1+x)}}{1+x^2}dx = \int^1_0\frac{\ln{(1+u)}}{1+u^2}du$

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

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Happy Physics Land on February 16, 2016, 09:43:06 pm

Quote from: jamonwindeyer on February 16, 2016, 09:39:43 pm

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:

$\int^1_0\frac{\ln{(1+x)}}{1+x^2}dx = \int^1_0\frac{\ln{(1+u)}}{1+u^2}du$

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

Thank you very much Jamon ^^

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 16, 2016, 09:54:24 pm

Quote from: jamonwindeyer on February 16, 2016, 09:39:43 pm

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:

$\int^1_0\frac{\ln{(1+x)}}{1+x^2}dx = \int^1_0\frac{\ln{(1+u)}}{1+u^2}du$

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
$\text{Simplify fully } \\8\cot{8x}+4\tan{4x}+2\tan{2x}+\tan{x}-\cot{x}$

Spoiler

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

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: jamonwindeyer on February 16, 2016, 10:53:30 pm

Quote from: RuiAce on February 16, 2016, 09:54:24 pm

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
$\text{Simplify fully } \\8\cot{8x}+4\tan{4x}+2\tan{2x}+\tan{x}-\cot{x}$

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

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Happy Physics Land on February 17, 2016, 05:59:33 pm

Quote from: RuiAce on February 16, 2016, 09:54:24 pm

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
$\text{Simplify fully } \\8\cot{8x}+4\tan{4x}+2\tan{2x}+\tan{x}-\cot{x}$

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

(http://i.imgur.com/l4E0Yz4.jpg)

I did it only to satisfy your eagerness jamon :D

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 17, 2016, 08:17:36 pm

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.

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 28, 2016, 07:27:06 pm

The above question is probably easier than this one.

$\text{Prove the statement of the binomial theorem using mathematical induction, that is}\\ {\left(1+x\right)}^{n}=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}{x}^{2}+\dots+\binom{n}{n}{x}^{n}$

Spoiler

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

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Happy Physics Land on February 28, 2016, 07:31:49 pm

Quote from: RuiAce on February 28, 2016, 07:27:06 pm

The above question is probably easier than this one.

$\text{Prove the statement of the binomial theorem using mathematical induction, that is}\\ {\left(1+x\right)}^{n}=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}{x}^{2}+\dots+\binom{n}{n}{x}^{n}$

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

This one is gonna be a tedious as f proof

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 28, 2016, 10:50:35 pm

Quote from: Happy Physics Land on February 28, 2016, 07:31:49 pm

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.

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on March 18, 2016, 08:56:44 am

$\text{Redefine }y=\sec^{-1}{x}\text{ and hence show that} \\ \frac{d}{dx}\sec^{-1}{x}=\frac{1}{\sqrt{{x}^{4}-{x}^{2}}}$

Required knowledge

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

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on April 18, 2016, 08:40:45 am

'Infinitely' times more easier than what meets the eye

$\lim_{x\rightarrow\infty}{\frac{8{x}^{8}+9{x}^{5}+951{x}^{2}-{e}^{-{x}^{1000}}+351-4\cos{\left(210x-\frac{93\pi}{7}\right)}}{21\tan^{-1}{\left(3x\right)}-\cot{\frac{\pi}{4}}-\ln{200}+{x}^{8}-9{e}^{-2141x}+100!{x}^{\pi}}}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: birdwing341 on April 30, 2016, 02:17:36 pm

Quote from: RuiAce on March 18, 2016, 08:56:44 am

$\text{Redefine }y=\sec^{-1}{x}\text{ and hence show that} \\ \frac{d}{dx}\sec^{-1}{x}=\frac{1}{\sqrt{{x}^{4}-{x}^{3}}}$

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

Are you sure the last part of the root is x^3? My answer is with an x^2.

Is this right?

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: jamonwindeyer on May 01, 2016, 12:12:57 am

Quote from: birdwing341 on April 30, 2016, 02:17:36 pm

Are you sure the last part of the root is x^3? My answer is with an x^2.

Is this right?

I think (pending the thumbs up from Rui, I hadn't tried this question before you mentioned the possible mistake) that you are indeed correct!! Great solution, well set out, great job! ;D

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on May 01, 2016, 03:55:46 pm

I don't know where I got x³ from either, sorry

Edit: Original post fixed! :)

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Empathy on May 05, 2016, 12:26:03 pm

Quote from: RuiAce on April 18, 2016, 08:40:45 am

'Infinitely' times more easier than what meets the eye

$\lim_{x\rightarrow\infty}{\frac{8{x}^{8}+9{x}^{5}+951{x}^{2}-{e}^{-{x}^{1000}}+351-4\cos{\left(210x-\frac{93\pi}{7}\right)}}{21\tan^{-1}{\left(3x\right)}-\cot{\frac{\pi}{4}}-\ln{200}+{x}^{8}-9{e}^{-2141x}+100!{x}^{\pi}}}$

Answer/method in spoiler

Spoiler

Is the answer 8?
arctan(infinity)=pi/2.
e^(-infinity)=0
cos(infinity) oscillates between 1, -1 but this is actually irrelevant.

After removing/subbing these in (aside from the cos one), divide the top and bottom by x^8 as usual with lim questions. The reason the cos was irrelevant is that its basically (1/x^7)*((cosx)/x) which is 0/x^7. After cancelling out all the non whole numbers, I got 8/1=8

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: jamonwindeyer on May 05, 2016, 01:47:58 pm

Quote from: Empathy on May 05, 2016, 12:26:03 pm

Answer/method in spoiler
Spoiler
Is the answer 8?
arctan(infinity)=pi/2.
e^(-infinity)=0
cos(infinity) oscillates between 1, -1 but this is actually irrelevant.

After removing/subbing these in (aside from the cos one), divide the top and bottom by x^8 as usual with lim questions. The reason the cos was irrelevant is that its basically (1/x^7)*((cosx)/x) which is 0/x^7. After cancelling out all the non whole numbers, I got 8/1=8

Good thinking putting the solution in a spoiler, I'm going to go back and put other solutions in a spoiler as well ;D but yes, pending the Rui tick of approval, that is correct!! Nicely done! ;D

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Empathy on May 05, 2016, 02:57:06 pm

Quote from: jamonwindeyer on May 05, 2016, 01:47:58 pm

Good thinking putting the solution in a spoiler, I'm going to go back and put other solutions in a spoiler as well ;D but yes, pending the Rui tick of approval, that is correct!! Nicely done! ;D

One of my friends just pointed out to me that i wasted alot of time on that, I could have just skipped everything up to the last step and solved it in 1-2 lines... lol

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on June 22, 2016, 01:51:27 pm

Didn't even realise that someone actually attempted one of these questions.
$\text{The answer is indeed 8.}\\ \text{Whilst formally speaking a good answer would have divided the numerator and denominator by }x^8\\ \text{It is really the fact that the leading terms of the polynomial dictate the value of the limit}$
$\text{This arises because for large x, polynomials always dominate over the sine, cosine functions, }\\ \text{the (RELEVANT) inverse trigonometric functions and constants}\\ \text{and for this question we also use }\lim_{x \to \infty}e^{-ax}=0 \quad \forall a>0$

$\text{At the Extension 1 level I am more than happy for one to assume:} \\ \lim_{x \to \infty}\frac{\cos{x}}{x^n}=0 \\ \lim_{x\to \infty}\frac{\ln{x}}{x^n}=0\\ \lim_{x\to \infty}{\frac{e^{-ax}}{x^n}}=0\\ \lim_{x \to \infty}{\frac{\arctan{x}}{x^n}}=0\\ \text{just by analysing what dominates over what}$
For anyone's interest, the formal way to evaluate them is:

1. By considering -1≤cos(x)≤1, apply the squeeze theorem by choosing an appropriate function to multiply every side with
2. Use one application of L'Hopital's rule
3. (DOABLE) Split the limit into the limit of arctan(x) and the limit of x^-n which was included in the given answer
4. (DOABLE) This is just 0 * 1/inf

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: conic curve on July 10, 2016, 10:27:43 am

Just in case anyone's interested feel free to try out this question

1. a. SKetch f(x)=(e^x)-4, showing clearly the points of intersection with the axes and the equations of any asymptotes
b. On the same diagram, sketch the graph of the inverse function f^-1(x), showing clearly any important features
c. Explain why the x-coordinate of any points of the intersection y=f(x) and y=f^-1(x) satisfies e^x - x - 4=0
d. Show the equation e^x - x - 4 =0 between x=0 and x=2 and use the method of 'halving by intervals' to find this root correct to the nearest whole number

Also anyone here wishing to try a few challenge circle geometry questions feel free to

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: relativity1 on July 12, 2016, 08:23:42 pm

for c) Would the answer be becuase the inverse is a reflection about the line y=x so the point of intersection is where x=y ie e^x-4=x therefore e^x-x-4=0?

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on July 12, 2016, 09:00:15 pm

Quote from: relativity1 on July 12, 2016, 08:23:42 pm

for c) Would the answer be becuase the inverse is a reflection about the line y=x so the point of intersection is where x=y ie e^x-4=x therefore e^x-x-4=0?

Yes.

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Paradoxica on October 05, 2016, 06:32:45 pm

The acute triangle ABC has sides a>b>c, where A is opposite a and likewise for B and C.

A square is inscribed with vertices on the edges of the triangle.

It is given that only three such squares exist, and that one side of each square is concurrent with the side of a triangle.

Suppose the side length of the squares are p,q,r (with any correspondence you want), and the altitudes of the triangle are a',b',c', which connect the vertices A,B,C with the sides a,b,c respectively.

i) Find expressions for each of the three squares in terms of all the variables described above.

ii) Determine which of the three squares has the largest area.

iii) Hence, or otherwise, find the smallest possible area of any such triangle ABC which encloses a square of unit area.

Title: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on June 08, 2017, 10:16:21 am

$\text{Let }a\text{ be a fixed real number and let }n\text{ be a positive integer.}$
$\text{The trapezoidal rule asserts that given }n\text{ sub-intervals, we have the approximation}\\ \int_{a}^{a+n} f(x) \, dx \approx \frac{b-a}{2n}[f(a) + 2f(a+1) + 2f(a+2) + \dots + 2f(a+n-1) + f(a+n)]\\ \text{where in our case, }b=a+n$
$\text{Use mathematical induction to prove that this result is valid}\\ \text{for any number of subintervals, i.e. for all possible values of }n$
$\textit{Extension: }\text{Does the above formula give us an approximation for }\int_a^bf(x)\,dx\\ \text{where }a\text{ and }b\text{ are }\textit{both}\text{ fixed?}\\ \text{This extension question goes beyond the scope of the course}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on July 21, 2017, 07:34:53 pm

$\text{In a 5-card poker hand}\\ \text{What is the probability of obtaining a four-of-a-kind?}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on July 22, 2017, 01:57:56 pm

$\text{1. Interpret geometrically the result }\frac{d}{dx}mx = m$
$\text{2. Prove, by way of mathematical induction,}\\ \frac{d^n}{dx^n}x^n = n!\\ \text{(without assuming the power rule)}$
$\text{3. Explain, or mathematically show, why given a polynomial }P(x)\text{ of degree }n\\ \text{the }(n+1)\text{-th derivative of }P(x)\text{ is always}\\ \text{the zero polynomial, i.e. }P^{(n+1)}(x) \equiv 0$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on August 14, 2017, 06:10:43 pm

$\text{1. Verify that }\ln x = \int_1^x \frac{dt}{t}$
$\text{2. Use this definition to prove that}\\ \ln A + \ln B = \ln AB$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on September 23, 2017, 08:39:54 pm

$\text{Suppose that }f_1(x), f_2(x), \dots \text{ are differentiable functions.}$
$\text{1. Prove the product rule from first principles, i.e. prove that}\\ \frac{d}{dx} \bigg(f_1^\prime(x) f_2^\prime(x)\bigg) = f_1^\prime(x)f_2(x) + f_1(x) f_2^\prime(x)$
$\text{2. Prove by mathematical induction the generalised product rule, i.e.}\\ \frac{d}{dx} \bigg(f_1(x) f_2(x) \dots f_n(x)\bigg) = \big[f_1^\prime(x) f_2(x)\dots f_n(x)\big] +\big[f_1(x) f_2^\prime(x) \dots f_n(x) \big] + \dots + \big[f_1(x) f_2(x) \dots f_n^\prime(x)\big]$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on December 03, 2017, 08:09:48 pm

$\text{Sketch, showing intercepts, asymptotes and stationary points,}\\ y=\frac{x^4-2}{x^2-1}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on December 23, 2017, 12:40:04 am

$\text{Suppose }\triangle ABC\text{ is equilateral.}$
$\text{Prove that any altitude of }\triangle ABC\\ \text{must pass through the centre of the circle}\\ \text{formed by the points }A,B\text{ and }C.$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on January 01, 2018, 12:21:06 pm

$\text{New year's special - 3U}\\ \text{Prove that }\sum_{k=0}^n \binom{2018}{k}=2^{2017}+\frac{2018!}{2(1009!)^2}\\ \text{without any calculator or computer assistance}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on January 01, 2018, 01:58:40 pm

Quote from: RuiAce on January 01, 2018, 12:21:06 pm

$\text{New year's special - 3U}\\ \text{Prove that }\sum_{k=0}^n \binom{2018}{k}=2^{2017}+\frac{2018!}{2(1009!)^2}\\ \text{without any calculator or computer assistance}$

Possible soln

$(1+x)^{2018} = \binom{2018}{0} + \binom{2018}{1}x + ... + \binom{2018}{2018}x^{2018}$
$\begin{align*}2^{2018} &= \binom{2018}{0} + \binom{2018}{1} + ... + \binom{2018}{2018} \\<br />&= 2\left(\binom{2018}{0} + \binom{2018}{1} + \binom{2018}{1008}\right) + \binom{2018}{1009}\end{align*}$
$\begin{align*}2^{2018} + \binom{2018}{1009} &= 2\left(\binom{2018}{0} + \binom{2018}{1} + \binom{2018}{1008} + \binom{2018}{1009}\right) \\<br />&= 2\sum_{k=0}^{1009}\binom{2018}{k} \end{align*}$
$2\sum_{k=0}^{1009}\binom{2018}{k} = 2^{2018} + \frac{2018!}{1009! \cdot \left(2018 - 1009\right)!}$
$\begin{align*}\therefore \sum_{k=0}^{1009}\binom{2018}{k} &= \frac{1}{2}\left(2^{2018} + \frac{2018!}{(1009!)^2}\right) \\<br />&= 2^{2017}+\frac{2018!}{2(1009!)^2}\end{align*}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on January 14, 2018, 05:20:26 pm

$\text{Prove that }\binom{n}{k}+\binom{n}{k+1} = \binom{n+1}{k+1}\\ \text{using only combinatorics.}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on January 14, 2018, 05:45:26 pm

Quote from: RuiAce on January 14, 2018, 05:20:26 pm

$\text{Prove that }\binom{n}{k}+\binom{n}{k+1} = \binom{n+1}{k+1}\\ \text{using only combinatorics.}$

Possible soln

$\text{Revisiting the combinatoric formula, note: } \\ \binom{n}{k} = \frac{n!}{k!(n - k)!}$
$\text{Using this formula, we can rewrite the left hand side as: } \\ \frac{n!}{k!(n - k)!} + \frac{n!}{(k + 1)![n - (k + 1)]!}$
$\text{If we compare the denominators of both fractions, we notice a few things: } \\ (k+1)! = (k+1)\cdot k! \text{ (by definition)} \\ \text{Using the same argument: } (n - k)! = (n - k) \cdot (n - k - 1)! = (n - k) \cdot (n - (k + 1))!$
$\text{Combining the two fractions together, we get: } \\ \begin{align*}LHS &= \frac{(k+1)n! + (n - k)n!}{(k+1)! \cdot (n - k)!} \\ &= \frac{n!\left(k+1 + n - k\right)}{(k+1)! \cdot (n - k)!} \\ &= \frac{n!(n + 1)}{(k+1)! \cdot (n - k)!} \\ &= \frac{(n+1)!}{(k+1)! \cdot (n - k + 1 - 1)!} \\ &= \frac{(n+1)!}{(k+1)![(n+1) - (k + 1)]!} \\ &= \binom{n+1}{k+1} \text{ (by definition)} \end{align*}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on January 14, 2018, 05:55:17 pm

Quote from: Opengangs on January 14, 2018, 05:45:26 pm

Possible soln
$\text{Revisiting the combinatoric formula, note: } \\ \binom{n}{k} = \frac{n!}{k!(n - k)!}$
$\text{Using this formula, we can rewrite the left hand side as: } \\ \frac{n!}{k!(n - k)!} + \frac{n!}{(k + 1)![n - (k + 1)]!}$
$\text{If we compare the denominators of both fractions, we notice a few things: } \\ (k+1)! = (k+1)\cdot k! \text{ (by definition)} \\ \text{Using the same argument: } (n - k)! = (n - k) \cdot (n - k - 1)! = (n - k) \cdot (n - (k + 1))!$
$\text{Combining the two fractions together, we get: } \\ \begin{align*}LHS &= \frac{(k+1)n! + (n - k)n!}{(k+1)! \cdot (n - k)!} \\ &= \frac{n!\left(k+1 + n - k\right)}{(k+1)! \cdot (n - k)!} \\ &= \frac{n!(n + 1)}{(k+1)! \cdot (n - k)!} \\ &= \frac{(n+1)!}{(k+1)! \cdot (n - k + 1 - 1)!} \\ &= \frac{(n+1)!}{(k+1)![(n+1) - (k + 1)]!} \\ &= \binom{n+1}{k+1} \text{ (by definition)} \end{align*}$

I guess I wasn't clear enough, sorry :(

That was the "algebraic" proof I had in mind. I was more of looking for one based off actual counting and enumeration.

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on January 14, 2018, 06:09:44 pm

oh my bad :-[
Didn't read it properly aha

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on January 29, 2018, 09:54:19 am

$\text{It is known that the exponential graph } y = a^x \text{ and the line } y = x \text{ intersect at one point.} \\ \text{Find the value of a for this to be true.}$

Assumed knowledge

Calculus, Basic algebra, Exponential and Logarithms.

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on February 15, 2018, 11:48:44 pm

$\text{Prove by mathematical induction:}\\ \frac{d^n}{dx^n} [f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} \left[\frac{d^{n-k}}{dx^{n-k}}f(x)\right] \left[\frac{d^k}{dx^k}g(x) \right]$
You are welcome to use the notation $ f^{(n)}(x) $ for the $n$-th derivative of $f$ with respect to $x$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on April 14, 2018, 10:40:37 pm

$\text{Let }x_1, \dots, x_k\text{ be non-negative integers.}$
$\text{Prove that the number of distinct solutions to the equation}\\ x_1+\dots + x_k = n \\ \text{is equal to } \binom{n+k-1}{k-1}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on April 23, 2018, 11:28:11 am

$\text{Source: MIT Integration Bee 2015}$ $\text{Note: this can be done } \textbf{without} \text{ any 3U techniques.}$ $\text{Sometimes, the most elegant solution is in its simplicity.}$
$\text{Evaluate: } \int_{0}^{\frac{\pi}{2}} \frac{\,dx}{\cot x + \tan x}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on April 23, 2018, 11:47:00 am

Quote from: Opengangs on April 23, 2018, 11:28:11 am

$\text{Source: MIT Integration Bee 2015}$ $\text{Note: this can be done } \textbf{without} \text{ any 3U techniques.}$ $\text{Sometimes, the most elegant solution is in its simplicity.}$
$\text{Evaluate: } \int_{0}^{\frac{\pi}{2}} \frac{\,dx}{\cot x + \tan x}$

lol

\[ \int_0^{\pi/2} \frac{dx}{\frac{\cos x}{\sin x}+ \frac{\sin x}{\cos x}} = \int_0^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + \sin^2 x}\,dx = \int_0^{\pi/2} \frac12 \sin 2x\,dx \]

Too lazy to figure out a 2U-only way though

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: RuiAce on September 21, 2018, 05:00:00 pm

$\text{Simplify this expression to the best of your ability.}\\ \left(-\sin \theta + t\sin \frac\theta2 \sin \theta - \frac12 t \cos\frac\theta2\cos \theta\right)^2 + \left(\cos\theta -t \sin \frac\theta2 \cos \theta - \frac12 t\cos \frac\theta2 \sin\theta\right)^2 + \left( -\frac12 t\sin \frac\theta2 \right)^2$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on December 29, 2018, 02:14:35 am

\[ a) \text{ Show that } \\ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)\]
\[ \text{Let } p(x) = x^3 + 12x^2 - 2x + 1 \text{ have roots } \alpha, \beta, \gamma. \\ b) \text{ Explain why} \\ \alpha^3 = 2\alpha - 12\alpha^2 - 1.\]\[c) \text{ Hence, or otherwise, find } \alpha^3 + \beta^3 + \gamma^3.\]

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: david.wang28 on December 30, 2018, 11:41:07 pm

Quote from: Opengangs on December 29, 2018, 02:14:35 am

\[ a) \text{ Show that } \\ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)\]
\[ \text{Let } p(x) = x^3 + 12x^2 - 2x + 1 \text{ have roots } \alpha, \beta, \gamma. \\ b) \text{ Explain why} \\ \alpha^3 = 2\alpha - 12\alpha^2 - 1.\]\[c) \text{ Hence, or otherwise, find } \alpha^3 + \beta^3 + \gamma^3.\]

Are these the answers?

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on February 05, 2019, 02:45:56 pm

Not so much of a "challenge" but an interesting proof to a property you guys should be familiar with.
$\begin{align*}&a) \text{ Explain why } \int_{1}^{x} \frac{dt}{t} = \ln x. &\quad \textbf{(1 mark)} \\ &b) \text{ By rewriting } \int_{1}^{ab} \frac{dx}{x} \text{ as } \int_{1}^a \frac{dx}{x} + \int_a^{ab} \frac{dx}{x}, \\ &\text{show that } \ln(ab) = \ln a + \ln b, \text{ where } a, b > 1. &\quad \textbf{ (2 marks)}\end{align*}$

Title: Re: Mathematics Extension 1 Challenge Marathon
Post by: Opengangs on May 14, 2020, 10:55:24 pm

A fun (and not too hard) problem that ties into number theory a little :-))

Our goal here is to investigate $n^5 - n$ and show that, indeed it is divisible by 5. We will verify this is true using induction (that's for you to prove) at the end. But for now, let's explore this using some number theory.

Part 1: Show that $n^5 - n$ can be written as $n(n - 1)(n + 1)(n^2 + 1)$.
Part 2: If we choose $n = 5k$, $n = 5k + 1$ or $n = 5k + 4$, briefly explain why $n^5 - n$ is divisible by 5 for any integer $k$.
Part 3: Show that, if we choose $n = 5k + 2$ or $n = 5k + 3$, then $n^5 - n$ is also divisible by 5.
Part 4: Conclude that, regardless of our choice of $n$, $n^5 - n$ is always divisible by 5.
Part 5: Verify this statement using induction. That is, prove that $n^5 - n$ is divisible by 5 using mathematical induction.

ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: RuiAce on February 16, 2016, 06:34:02 pm