HSC 2015 MX1 Marathon (archive) (1 Viewer)

InteGrand · Sep 28, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I understand that by using first principles, the binomial theorem would come into play.

I have not seen how it is done by induction but would be interested to see.

A proof by induction was asked in a past HSC 3U paper (required use of the product rule too). See Question 7(a) on Page 10 of the 2009 HSC 3U paper: http://www.boardofstudies.nsw.edu.a...doc/2009-hsc-exam-mathematics-extension-1.pdf.

rand_althor · Sep 28, 2015

Re: HSC 2015 3U Marathon

Is this correct for induction?

$\begin{align*}\textup{RTP }\frac{\mathrm{d}}{\mathrm{d}x}x^{n}&=nx^{n-1} \\\textup{1. Show true for }n&=1 \\f(x)&=x^1=x \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x \\ &=\frac{f(x+h)-f(x)}{h}\\&=\frac{x+h-x}{h} \\&=\frac{h}{h} \\&=1 \\\textup{RHS}&=1\times x^{1-1}=1\times x^0 =1=\textup{LHS} \\\textup{Therefore, true for }n&=1. \\\end{align*}$

$\begin{align*}\textup{2. Assume true for }n&=k \\\frac{\mathrm{d}}{\mathrm{d}x}x^k&=kx^{k-1} \\\end{align*}$

$\begin{align*}\textup{3. Prove true for }n&=k+1 \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1} \\&=\frac{\mathrm{d}}{\mathrm{d}x}x^{k}\cdot x \\&=\left (\frac{\mathrm{d}}{\mathrm{d}x}x^{k} \right )\cdot x+\left (\frac{\mathrm{d}}{\mathrm{d}x}x \right )\cdot x^k \\&=kx^{k-1} \cdot x + 1\cdot x^k \\&=kx^k+x^k \\&=(k+1)x^k \\\textup{RHS}&=(k+1)x^{k+1-1}=(k+1)x^{k}=\textup{LHS} \\.\end{align*}$

leehuan · Sep 29, 2015

Re: HSC 2015 3U Marathon

Yes it is, cause I had to do the induction proof before.

leehuan · Sep 29, 2015

Re: HSC 2015 3U Marathon

$\\ $In the 2U marathon, it has been asked what the probability of was to find a royal flush in a standard poker game. \\ 1. What is the probability of a straight flush in a standard 5 card poker game? (Given that a straight flush is a sequence of five cards in a row, all of the same suit, without counting the royal flush). \\ 2. What is the probability of a four-of-a-kind? ('kind' refers to the rank of the card). \\ Explain all logic and reasoning.$

You may choose to use combinatoric notation in the 3U marathon.

braintic · Sep 29, 2015

Re: HSC 2015 3U Marathon

leehuan said:
$\\ $In the 2U marathon, it has been asked what the probability of was to find a royal flush in a standard poker game. \\ 1. What is the probability of a straight flush in a standard 5 card poker game? (Given that a straight flush is a sequence of five cards in a row, all of the same suit, and the royal flush is NOT a straight flush). \\ 2. What is the probability of a four-of-a-kind? ('kind' refers to the rank of the card). \\ Explain all logic and reasoning.$

You may choose to use combinatoric notation in the 3U marathon.

The royal flush IS a straight flush. It is an ace-high straight flush.
Only pub gambling machines treat it as being distinct from a straight flush.

braintic · Sep 29, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Is this correct for induction?

$\begin{align*}\textup{RTP }\frac{\mathrm{d}}{\mathrm{d}x}x^{n}&=nx^{n-1} \\\textup{1. Show true for }n&=1 \\f(x)&=x^1=x \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x \\ &=\frac{f(x+h)-f(x)}{h}\\&=\frac{x+h-x}{h} \\&=\frac{h}{h} \\&=1 \\\textup{RHS}&=1\times x^{1-1}=1\times x^0 =1=\textup{LHS} \\\textup{Therefore, true for }n&=1. \\\end{align*}$

$\begin{align*}\textup{2. Assume true for }n&=k \\\frac{\mathrm{d}}{\mathrm{d}x}x^k&=kx^{k-1} \\\end{align*}$

$\begin{align*}\textup{3. Prove true for }n&=k+1 \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1} \\&=\frac{\mathrm{d}}{\mathrm{d}x}x^{k}\cdot x \\&=\left (\frac{\mathrm{d}}{\mathrm{d}x}x^{k} \right )\cdot x+\left (\frac{\mathrm{d}}{\mathrm{d}x}x \right )\cdot x^k \\&=kx^{k-1} \cdot x + 1\cdot x^k \\&=kx^k+x^k \\&=(k+1)x^k \\\textup{RHS}&=(k+1)x^{k+1-1}=(k+1)x^{k}=\textup{LHS} \\.\end{align*}$

It is correct. But you don't need a first principles proof for the n=1 case - it is the gradient of a straight line.

davidgoes4wce · Sep 30, 2015

Re: HSC 2015 3U Marathon

A major supplier of mobile devices supplies a range which come with 8 choices of casing color and 15 choices of cover designs. The devices can be paired such that any of the 8 color choices goes with any of the 15 cover designs, for a distinctive look.

A group of 6 friends all wish to buy the same device, however, they will each choose a different color and case configuration so that they can see at a glance which device belongs to which person. How many different configurations are there for the entire set of 6 devices (each one has an independent configuration and the order is important)

My thinking was somewhere along the lines of, the first selection has 8 choices in terms of casing colour and 15 choices of cover designs. After the initial person has selected his option, the second person has 1 less choice in terms of casing color and cover design.

davidgoes4wce · Sep 30, 2015

Re: HSC 2015 3U Marathon

Second part to this question,

b) A set of 5 team devices for train ticket inspectors needs to be purchased. The team devices need to have a common cover design. Given that the devices need to be able to be identified so that each team member knows which one is theirs, how many possible team sets can be configured?

davidgoes4wce · Oct 1, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Is this correct for induction?

$\begin{align*}\textup{RTP }\frac{\mathrm{d}}{\mathrm{d}x}x^{n}&=nx^{n-1} \\\textup{1. Show true for }n&=1 \\f(x)&=x^1=x \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x \\ &=\frac{f(x+h)-f(x)}{h}\\&=\frac{x+h-x}{h} \\&=\frac{h}{h} \\&=1 \\\textup{RHS}&=1\times x^{1-1}=1\times x^0 =1=\textup{LHS} \\\textup{Therefore, true for }n&=1. \\\end{align*}$

$\begin{align*}\textup{2. Assume true for }n&=k \\\frac{\mathrm{d}}{\mathrm{d}x}x^k&=kx^{k-1} \\\end{align*}$

$\begin{align*}\textup{3. Prove true for }n&=k+1 \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1} \\&=\frac{\mathrm{d}}{\mathrm{d}x}x^{k}\cdot x \\&=\left (\frac{\mathrm{d}}{\mathrm{d}x}x^{k} \right )\cdot x+\left (\frac{\mathrm{d}}{\mathrm{d}x}x \right )\cdot x^k \\&=kx^{k-1} \cdot x + 1\cdot x^k \\&=kx^k+x^k \\&=(k+1)x^k \\\textup{RHS}&=(k+1)x^{k+1-1}=(k+1)x^{k}=\textup{LHS} \\.\end{align*}$

What does 'RTP' mean again?

InteGrand · Oct 1, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
What does 'RTP' mean again?

"Required to prove"

davidgoes4wce · Oct 2, 2015

Re: HSC 2015 3U Marathon

Can you graph an inverse trig function, in the case consider y=sin^-1 (x) from, a range from [Pi/2, 3Pi/2]?with the same domain values [-1,1]

This is based on what I have read:

"A function that has an inverse has exactly one output (belonging to the range) for every input (belonging to the domain) "

Paradoxica · Oct 2, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Can you graph an inverse trig function, in the case consider y=sin^-1 (x) from, a range from [Pi/2, 3Pi/2]?with the same domain values [-1,1]

This is based on what I have read:

"A function that has an inverse has exactly one output (belonging to the range) for every input (belonging to the domain) "

Functions that are periodic can only have partial inverses. We restrict the range of $\sin(x)$ to $|x| \leq \frac{\pi}{2}$ so that it can have a defined inverse.

davidgoes4wce · Oct 3, 2015

Re: HSC 2015 3U Marathon

Just wondering if the step in the 3rd line is correct? I watched a video off Youtube (not sure what their credentials are) and compared it with a comment I recently heard from a lecturer, in that he mentions that when integrating 'the terms must be all the same'. (e.g. dx must be integrated with x, du must be integrated with u terms)

leehuan · Oct 3, 2015

Re: HSC 2015 3U Marathon

I would've gone straight from line 2 to line 4 if I were doing integration by substitution. But it's not hard to really just go from the first line to the last line. It's just integrating f'(x).(f(x))^n

rand_althor · Oct 3, 2015

Re: HSC 2015 3U Marathon

$$Prove by mathematical induction that $\sum_{r=n}^{2n-1}2r+1=3n^2$ for $n\geq1$.$

Paradoxica · Oct 3, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Prove by mathematical induction that $\sum_{r=n}^{2n-1}2r+1=3n^2$ for $n\geq1$.$

*true for all $n \geq 1$

davidgoes4wce · Oct 3, 2015

Re: HSC 2015 3U Marathon

leehuan said:
I would've gone straight from line 2 to line 4 if I were doing integration by substitution. But it's not hard to really just go from the first line to the last line. It's just integrating f'(x).(f(x))^n

It still doesn't quite answer my question, is that 3rd line an allowable step in the integration by substitution?

photastic · Oct 3, 2015

Re: HSC 2015 3U Marathon

rand_althor said:

Clarify if it's suppose to be r=1 or k=1?

rand_althor · Oct 3, 2015

Re: HSC 2015 3U Marathon

photastic said:
Clarify if it's suppose to be r=1 or k=1?

It was k=1, my bad. Is it just me or does it not work for n=1?

dan964 · Oct 3, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
It was k=1, my bad. Is it just me or does it not work for n=1?

does not.

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