HSC 2016 Maths Marathon (archive) (3 Viewers)

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:

HOW.

I never thought about this at all.

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
HOW.

I never thought about this at all.

what do you mean?

lol think about it, its pretty apparent when you play with the formulas

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
what do you mean?

lol think about it, its pretty apparent when you play with the formulas

d/dx x^x
= d/dx e^(x.ln(x))
= (ln(x)+1)e^(x.ln(x))
= x^x(ln(x)+1)

Uh no I don't see how by adding the power and exponential rule mistake method works

Paradoxica said:
$\noindent Power Rule: $ x \cdot x^{x-1} = x^x $ - FALSE$\\ $Exponential Rule: $ x^x \log{x} $ - FALSE$ \\ $Sum of these: $(1+\log{x})x^x $ - TRUE$

Edit#2: No idea how that x creeped it's way in

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
d/dx x^x
= d/dx e^(x.ln(x))
= (x.ln(x)+1)e^(x.ln(x))
= x^x

Uh no I don't see how by adding the power and exponential rule mistake method works

uhhm.....you made a mistake lol

InteGrand · Apr 10, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
uhhm.....you made a mistake lol

The emboldened part is correct, it's just he forgot to include the (x.ln(x)+1) factor in the next line.

Edit: oh wait, saw he had an unnecessary x. Should just have been (ln(x)+1) rather than (x.ln(x)+1)

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
The emboldened part is correct, it's just he forgot to include the (x.ln(x)+1) factor in the next line.

Edit: oh wait, saw he had an unnecessary x. Should just have been (ln(x)+1) rather than (x.ln(x)+1)

i had no idea if he made a mistake lel
i just bolded a random line so it would confuse him

its all paradoxica's fault
he put me up to it

*runs away*

InteGrand · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Edit#2: No idea how that x creeped it's way in

Which x are you referring to?

InteGrand · Apr 10, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
i had no idea if he made a mistake lel
i just bolded a random line so it would confuse him

its all paradoxica's fault
he put me up to it

*runs away*

Haha – you actually ended up getting the right line. And why would Paradoxica put you up to this?

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Which x are you referring to?

The one I made a mistake in

I'm just mindblown at Para's statement.

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
The one I made a mistake in

I'm just mindblown at Para's statement.

yeha it was just a coincidence lel

you were confused

we aimed to confuse you further

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
yeha it was just a coincidence lel

you were confused

we aimed to confuse you further

Nah you were just being a dork

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Nah you were just being a dork

leehuan said:
HOW.

I never thought about this at all.

turntaker · Apr 10, 2016

Re: HSC 2016 2U Marathon

InteGrand · Apr 10, 2016

Re: HSC 2016 2U Marathon

turntaker said:

= e^x + C.

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

$$Suppose $x={e}^{t}$. $\textbf{Verify} $ that the following differential equation is true for all functions $y.$

$\\ \frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=\frac{1}{{x}^{2}}\left(\frac{{\text{d}}^{2}y}{\text{d}{t}^{2}}-\frac{\text{d}y}{\text{d}t}\right)$

Terminology is a bit out of the course. But this basically means that y is some arbitrary function of x. I.e. y=f(x)

Paradoxica · Apr 10, 2016

Re: HSC 2016 2U Marathon

turntaker said:

$xe^x + \mathcal{C}$

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
$xe^x + C$

I was trying to find the troll when I realised oh wait wow you troll differentiated

InteGrand · Apr 12, 2016

Re: HSC 2016 2U Marathon

$\noindent Let $f$ be a function with domain $[0,1]$, i.e. $\left \{ t \in \mathbb{R} : 0 \leq t \leq 1 \right \}$ (and codomain $\mathbb{R}$). Let $a,b$ be real numbers with $a<b$. Let $f$ be defined by $f(t) = a\left(1-t)+bt$ for $t\in \left[0,1\right]$. Find the range of $f$.$

parad0xica · Apr 13, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Let $f$ be a function with domain $[0,1]$, i.e. $\left \{ t \in \mathbb{R} : 0 \leq t \leq 1 \right \}$ (and codomain $\mathbb{R}$). Let $a,b$ be real numbers with $a<b$. Let $f$ be defined by $f(t) = a\left(1-t)+bt$ for $t\in \left[0,1\right]$. Find the range of $f$.$

2U HS students don't learn this dense notation so it's going to take a while for them to comprehend it unless they were already used to it in the past posts...

InteGrand · Apr 13, 2016

Re: HSC 2016 2U Marathon

parad0xica said:
2U HS students don't learn this dense notation so it's going to take a while for them to comprehend it unless they were already used to it in the past posts...

Rewritten in more 2U style:

$\noindent Let $f(t)$ be a function with domain $0\leq t \leq 1$. Let $a<b$ be fixed real numbers. Let $f(t)$ be defined as $f(t) = a(1-t) + bt$ for $0\leq t \leq 1$. Find the range of $f(t)$.$

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