HSC 2016 Maths Marathon (archive) (1 Viewer)

InteGrand · Apr 24, 2016

Re: HSC 2016 2U Marathon

si2136 said:
Yes ik

But people don't write the proof down do they

No. Only Russell and Whitehead did probably.

Green Yoda · Apr 24, 2016

Re: HSC 2016 2U Marathon

2+2=4
http://www.cs.yale.edu/homes/as2446/224.pdf

KingOfActing · Apr 24, 2016

Re: HSC 2016 2U Marathon

2+2 = 5

ml125 · Apr 24, 2016

Re: HSC 2016 2U Marathon

1+1=window

porcupinetree · Apr 24, 2016

Re: HSC 2016 2U Marathon

KingOfActing said:
2+2 = 5

Paradoxica · Apr 27, 2016

Re: HSC 2016 2U Marathon

$$\noindent a) Sketch a horizontally truncated, vertically dilated, graph of $y=\frac{1}{x}$ for $2^n -2 \leq x \leq 2^{n+1} $ where $n$ is a positive integer.$ \\\\ $b) Use each of the integer $x$-values to construct 1-unit wide rectangles above and below each 1-unit interval of the graph.$ \\\\ $c) Explain, with reference to the graph, why $\\ \frac{1}{2^n -1} + \frac{1}{2^n} + \frac{1}{2^n +1} + \cdots + \frac{1}{2^{n+1}-2}+\frac{1}{2^{n+1}-1} < \int_{2^n-2}^{2^{n+1}-1} \frac{\text{d}x}{x} \\\\ $d) Similarly, show that $\\ \int_{2^n-1}^{2^{n+1}} \frac{\text{d}x}{x} < \frac{1}{2^n -1} + \frac{1}{2^n} + \frac{1}{2^n +1} + \cdots + \frac{1}{2^{n+1}-2}+\frac{1}{2^{n+1}-1} \\\\ $e) Deduce that $\\ \log{\frac{2^{n+1}}{2^n -1}} < \frac{1}{2^n -1} + \frac{1}{2^n} + \frac{1}{2^n +1} + \cdots + \frac{1}{2^{n+1}-2}+\frac{1}{2^{n+1}-1} < \log{\frac{2^{n+1}-1}{2^n-2}}$

$\noindent f) Hence, find the limit of $ \frac{1}{2^n -1} + \frac{1}{2^n} + \frac{1}{2^n +1} + \cdots + \frac{1}{2^{n+1}-2}+\frac{1}{2^{n+1}-1}$ as $n$ goes to positive infinity.$

This problem may have incorrect borders for the integrals, if that happens to be the case, let me know and I will fix it, as I have only checked the values three times so far.

InteGrand · May 1, 2016

Re: HSC 2016 2U Marathon

$\noindent Find $\frac{\mathrm{d}}{\mathrm{d}x}\big{(}\ln |x|\big{)}$.$

Nailgun · May 1, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Find $\frac{\mathrm{d}}{\mathrm{d}x}\big{(}\ln |x|\big{)}$.$

$\noindent As ln$x$ is undefined when $x$ is negative, the expression is equivalent to $\frac{\mathrm{d}}{\mathrm{d}x}\big{(}\ln x\big{)}$

$Therefore $\frac { d }{ dx } \ln { |x| } =\frac { 1 }{ x } $$

$I think.$

InteGrand · May 1, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$\noindent As ln$x$ is undefined when $x$ is negative, the expression is equivalent to $\frac{\mathrm{d}}{\mathrm{d}x}\big{(}\ln x\big{)}$

$Therefore $\frac { d }{ dx } \ln { |x| } =\frac { 1 }{ x } $$

$I think.$

$\noindent Your final answer is correct, but the reasoning is a bit faulty. The function $\ln |x|$ is defined when $x$ is negative. Its graph is here:$

https://graphsketch.com/?eqn1_color..._lines=1&line_width=4&image_w=850&image_h=525

$\noindent (I.e. flip the normal log graph about the vertical axis, and keep the original graph too.)$

$\noindent In other words, the function is $f(x) = \ln x$ for $x>0$ and $f(x) = \ln (-x)$ for $x<0$ (for $x=0$, $f(x)$ is undefined). This is due to the definition of the absolute value function.$

So using this knowledge, here's the answer in a spoiler (so that you can have a go first at the reasoning if you wish).

$\noindent For $x>0$, we find $f^\prime (x) = \frac{1}{x}$ and for $x<0$ we find $f^{\prime} (x) = \frac{-1}{-x}=\frac{1}{x}$ (using rules for differentiating $\ln x$ and $\ln (-x)$. Hence $f^\prime (x) = \frac{1}{x}$ for all $x$ in the domain of $f$.$

loje · May 2, 2016

Re: HSC 2016 2U Marathon

This should be an easy one but for some reason it's messing my head really badly lol.

A coin is tossed three times. A is the event "at least two tails"; B is the event "three heads or three tails"; C is the event "at least one tail". Which of the following are independent?

a) A and B (ans: indepedent)
b) A and C (ans: dependent)
c) B and C (ans: dependent)

Why aren't they all independent?

Cheers

leehuan · May 2, 2016

Re: HSC 2016 2U Marathon

Getting at least 2 tails is dependent on getting at least 1 tail first

loje · May 2, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Getting at least 2 tails is dependent on getting at least 1 tail first

Ok. But whats the difference between part a and part b that makes it indepdent and depedent, respectively? Thanks

InteGrand · May 2, 2016

Re: HSC 2016 2U Marathon

$\noindent The definition of two events $\mathcal{A}$ and $\mathcal{B}$ being independent is that $\Pr \left(\mathcal{A} \text{ and }\mathcal{B}\right) =\Pr \left(\mathcal{A} \right)\Pr \left(\mathcal{B}\right)$. If this is not the case, then the events $\mathcal{A}$ and $\mathcal{B}$ are said to be dependent (i.e. not independent). So we can use this definition to obtain the right answers.$

InteGrand · May 2, 2016

Re: HSC 2016 2U Marathon

Intuitively if two events A and B are independent, it means that knowledge about whether A happened doesn't affect the probability of B happening.

For example, say someone tosses a coin and rolls a die. Let A be the even the coin landed Heads and B be the event that the die landed 6.

If A and B are independent events, it means that if we know A happened (coin landed Heads), this gives us no further or less confidence in B having happened given A happened (6 being rolled too given a Heads). So if someone tossed the coin and rolled the die (without you observing them), and they told you that A occurred (Heads), it doesn't make it any more or less likely that B happened (that the die landed 6).

If A and B are dependent events though, intuitively it means that knowledge about whether A occurred alters the probability that B occurred given this knowledge.

For example, say two players Alice and Bob are playing a game of tennis. Say A is the event that Alice was sick on the day and B is the event that Alice wins.

Intuitively speaking, if we know A occurred (Alice was sick on the day), it affects the chance that B occurred given A (that Alice won the game given we know she was sick).

leehuan · May 10, 2016

Re: HSC 2016 2U Marathon

Instructive question:

$For positive $a,b$, find further conditions on $a, b$ such that$\\ \frac{a+x}{b+x}>\frac{a}{b}\\ $where $x$ is positive, or negative.$

davidgoes4wce · Jun 19, 2016

Re: HSC 2016 2U Marathon

$Geometric Series$

$Find the value of the common ratio if (a) the first term is 11, sum of the first 12 terms is 2,922,920$

$Ans r=3$

davidgoes4wce · Jun 19, 2016

Re: HSC 2016 2U Marathon

$S_n=\frac{a(r^n-1)}{r-1}$

We are supposed to use that formula

$2922920=\frac{11(r^{12}-1)}{r-1}$

That's where I got up to

leehuan · Jun 19, 2016

Re: HSC 2016 2U Marathon

I fail to see how 2U methods by themselves are sufficient to deduce that

$\begin{align*}2922920&=\frac{11\left(r^{12}-1\right)}{r-1}\\ 265720&=1+r+r^2+\dots+r^{11}\end{align*}$

davidgoes4wce · Jun 19, 2016

Re: HSC 2016 2U Marathon

OK I got this question from a Cambridge University Maths SL book (which I consider to be on par with ) Maths 2U. I think I might have to create an own thread for IB Maths.

davidgoes4wce · Jun 19, 2016

Re: HSC 2016 2U Marathon

I will also admit this Cambridge textbook is just like the HSC version (i.e the Pender Sadler type), its Cambridge for a reason.

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