HSC 2016 Maths Marathon (archive) (2 Viewers)

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

Yeah but isn't that only if you assumed d/dx In(-x) = -1/x?

EDIT: So how does the marathon actually work, is there a process or do we (I) just wait until some kind samaritan posts up new questions
(subtly avoiding integrands averages questions because they make me lost lel)

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Yeah but isn't that only if you assumed d/dx In(-x) = -1/x?

EDIT: So how does the marathon actually work, is there a process or do we (I) just wait until some kind samaritan posts up new questions
(subtly avoiding integrands averages questions because they make me lost lel)

$\noindent In fact, $\frac{\mathrm{d}}{\mathrm{d}x} \left(\ln (-x)\right) =\frac{1}{x}$. More generally, if $k$ is a non-zero constant, then $\frac{\mathrm{d}}{\mathrm{d}x}\left( \ln (kx)\right) =\frac{1}{x}$.$

(Oh, by the way, the function is actually ln (lowercase L), not In. The l stands for 'logarithm' and the n stands for 'natural', because ln is the natural log (i.e. base e). So this means it's not pronounced "In" either.

)

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Regarding how the marathon is meant to work, I think we're meant to wait for the last unanswered Q., to be answered, then the person who answered that posts a new Q. Here's a new question if you don't want to do the ones about averages:

$\noindent Let $n$ be a positive integer. Find $\int \sqrt[n]{\mathrm{e}^{x}}\text{ d}x$.$

leehuan · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Yeah but isn't that only if you assumed d/dx In(-x) = -1/x?

EDIT: So how does the marathon actually work, is there a process or do we (I) just wait until some kind samaritan posts up new questions
(subtly avoiding integrands averages questions because they make me lost lel)

Lol that's not the problem. You can't log a negative number without complex numbers.

To which however fortunately with that knowledge we can still use the result. Because ln(-x) = ln(x) + i pi

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Let $n$ be a positive integer. Find $\int \sqrt[n]{\mathrm{e}^{x}}\text{ d}x$.$

$=\int { \sqrt [ n ]{ { e }^{ x } } } \\ =\int { ({ e }^{ x } } )^{ \frac { 1 }{ n } }\\ =\int { { e }^{ x\frac { 1 }{ n } } } \\ =\left[ ne^{ \frac { x }{ n } } \right] +C\\ =ne^{ \frac { x }{ n } }+C\\ =n\sqrt [ n ]{ e^{ x } } +C$

New Question
$Evaluate\quad \int _{ 1 }^{ 10 }{ \log _{ 10 }{ x } } dx$

So I tried to make a new question; until I couldn't get the answer to my own question, so I made an easier one so I could ask about my original question lel
$Find\quad \frac { dy }{ dx } \quad if\quad y={ \cos { x } }^{ \sin { { x }^{ \tan { x } } } }$

Is that the same as $y={ \cos { x } }^{ \sin { x.\tan { x } } }$ ?
I get different answers for the derivatives in Wolfram Alpha

Regardless I get a different answer from Wolfram Alpha

My working was as follows
$If\quad y=\cos { { x }^{ \sin { x } .\tan { x } } } \\ Let\quad u=\sin { x } .\tan { x } \\ Hence,\quad y=(\cos { x } )^{ u }\\ \frac { du }{ dx } =\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } \\ \frac { dy }{ du } =(\cos { x } )^{ u }\ln { \sin { x } } \\ \frac { dy }{ dx } =\frac { du }{ dx } .\frac { dy }{ du } \\ \frac { dy }{ dx } =((\cos { x } )^{ u }\ln { \sin { x } } )(\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } )\\ \frac { dy }{ dx } =((\cos { x } )^{ \sin { x } .\tan { x } }\ln { \sin { x } } )(\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } )$

Halp pls

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$=\int { \sqrt [ n ]{ { e }^{ x } } } \\ =\int { ({ e }^{ x } } )^{ \frac { 1 }{ n } }\\ =\int { { e }^{ x\frac { 1 }{ n } } } \\ =\left[ ne^{ \frac { x }{ n } } \right] +C\\ =ne^{ \frac { x }{ n } }+C\\ =n\sqrt [ n ]{ e^{ x } } +C$

New Question
$Evaluate\quad \int _{ 1 }^{ 10 }{ \log _{ 10 }{ x } } dx$

So I tried to make a new question; until I couldn't get the answer to my own question, so I made an easier one so I could ask about my original question lel
$Find\quad \frac { dy }{ dx } \quad if\quad y={ \cos { x } }^{ \sin { { x }^{ \tan { x } } } }$

Is that the same as $y={ \cos { x } }^{ \sin { x.\tan { x } } }$ ?
I get different answers for the derivatives in Wolfram Alpha

Regardless I get a different answer from Wolfram Alpha

My working was as follows
$If\quad y=\cos { { x }^{ \sin { x } .\tan { x } } } \\ Let\quad u=\sin { x } .\tan { x } \\ Hence,\quad y=(\cos { x } )^{ u }\\ \frac { du }{ dx } =\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } \\ \frac { dy }{ du } =(\cos { x } )^{ u }\ln { \sin { x } } \\ \frac { dy }{ dx } =\frac { du }{ dx } .\frac { dy }{ du } \\ \frac { dy }{ dx } =((\cos { x } )^{ u }\ln { \sin { x } } )(\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } )\\ \frac { dy }{ dx } =((\cos { x } )^{ \sin { x } .\tan { x } }\ln { \sin { x } } )(\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } )$

Halp pls

$\noindent Nested powers are done from the top down. So something like $a^{{b}^{c}}$ means $a^{\left(b^{c}\right)}$. So this is distinct from $a^{bc}$; instead, $a^{bc}$ refers to $\left(a^{b}\right)^{c}$, which is different.$

Paradoxica · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$=\int { \sqrt [ n ]{ { e }^{ x } } } \\ =\int { ({ e }^{ x } } )^{ \frac { 1 }{ n } }\\ =\int { { e }^{ x\frac { 1 }{ n } } } \\ =\left[ ne^{ \frac { x }{ n } } \right] +C\\ =ne^{ \frac { x }{ n } }+C\\ =n\sqrt [ n ]{ e^{ x } } +C$

New Question
$Evaluate\quad \int _{ 1 }^{ 10 }{ \log _{ 10 }{ x } } dx$

So I tried to make a new question; until I couldn't get the answer to my own question, so I made an easier one so I could ask about my original question lel
$Find\quad \frac { dy }{ dx } \quad if\quad y={ \cos { x } }^{ \sin { { x }^{ \tan { x } } } }$

Is that the same as $y={ \cos { x } }^{ \sin { x.\tan { x } } }$ ?
I get different answers for the derivatives in Wolfram Alpha

Regardless I get a different answer from Wolfram Alpha

My working was as follows
$If\quad y=\cos { { x }^{ \sin { x } .\tan { x } } } \\ Let\quad u=\sin { x } .\tan { x } \\ Hence,\quad y=(\cos { x } )^{ u }\\ \frac { du }{ dx } =\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } \\ \frac { dy }{ du } =(\cos { x } )^{ u }\ln { \sin { x } } \\ \frac { dy }{ dx } =\frac { du }{ dx } .\frac { dy }{ du } \\ \frac { dy }{ dx } =((\cos { x } )^{ u }\ln { \sin { x } } )(\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } )\\ \frac { dy }{ dx } =((\cos { x } )^{ \sin { x } .\tan { x } }\ln { \sin { x } } )(\sin { x } .\sec ^{ 2 }{ x } +\cos { x } .\tan { x } )$

Halp pls

Exponentiation is not associative. Without brackets, the exponential tower is always read top down, by international convention.

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

Oh right, that does makes sense lol, thanks.
Regarding the working out though?

This is what Wolfram Alpha outputs http://www.wolframalpha.com/input/?i=dy%2Fdx+If+y%3Dcosx%5E%28sinx.tanx%29

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Oh right, that does makes sense lol, thanks.
Regarding the working out though?

This is what Wolfram Alpha outputs http://www.wolframalpha.com/input/?i=dy%2Fdx+If+y%3Dcosx%5E%28sinx.tanx%29

$\noindent We can't do the working out quite how you did it unfortunately, because there's a variable in both the base and the exponent. When we let $u$ equal something, it should get rid of all $x$'s from the expression. Like how if we had $y=\ln (\cos x)$, to find $y^\prime$, we'd let $u=\cos x$, so $y=\ln u$, and there's no $x$ here now. You would need to replace your $\cos x$ in the base in terms of $u$ too to make the method valid, but this still wouldn't simplify things, since the base would still be ugly (extremely ugly in fact). For this one, it really helps to use logarithmic differentiation.$

$\noindent Since $y=\cos x ^{\sin x \tan x}$, we have $\ln y = \sin x \cdot \tan x \cdot \ln \left(\cos x \right)$. From here, we'd differentiate both sides to get $\frac{y^\prime}{y}$ on the L.H.S. and differentiate the R.H.S. using the extended product rule for three functions, which states that $(uvw)^\prime = u^\prime vw + uv^\prime w + uvw^\prime$, where $u,v,w$ are differentiable functions of $x$. For us, $u=\sin x$, $v=\tan x$ and $w= \ln \left(\cos x \right)$, so $u^\prime = \cos x$, $v^\prime = \sec^2 x$, and $w^\prime = \frac{\frac{\mathrm{d}}{\mathrm{d}x}\left(\cos x\right)}{\cos x}=\frac{-\sin x}{\cos x} = -\tan x$. So we have$

$\begin{align*}\frac{y^\prime}{y}&= \cos x \tan x \ln \left(\cos x \right) + \sin x \sec^2 x \ln \left(\cos x \right) + \sin x \tan x \cdot \left(-\tan x \right).\end{align*}$

$\noindent We would then simplify this by cancelling cosines where possible etc., and then multiply through by $y$.$

Drsoccerball · Jan 17, 2016

Re: HSC 2016 2U Marathon

$New Question$

$$ (i) Find : $ \frac{d}{dx}{\frac{f(x)}{g(x)}}$

$$ (ii) Hence or otherwise, solve : $ \int{\frac{\log{x} \cos{x} - \sin{x} \cdot \frac{1}{x}}{(\log{x})^2}}dx$

Drsoccerball · Jan 19, 2016

Re: HSC 2016 2U Marathon

$$ Using the fact that $ \int{f'(x)}dx=f(x) ($ So that new comers who don't know integration can attempt this. $)$

$$ Hence, or otherwise $ \int{x^{2x}(1+\ln{x})}dx$

leehuan · Jan 20, 2016

Re: HSC 2016 2U Marathon

I don't think a 2U kid will easily figure out that f(x) = 1/2 x^(2x) lol

__________
BUMP: Perfectly manageable 2U question

Drsoccerball said:
$New Question$

$$ (i) Find : $ \frac{d}{dx}{\frac{f(x)}{g(x)}}$

$$ (ii) Hence or otherwise, solve : $ \int{\frac{\log{x} \cos{x} - \sin{x} \cdot \frac{1}{x}}{(\log{x})^2}}dx$

Drsoccerball · Jan 20, 2016

Re: HSC 2016 2U Marathon

leehuan said:
I don't think a 2U kid will easily figure out that f(x) = 1/2 x^(2x) lol

__________
BUMP: Perfectly manageable 2U question

I think a 2 unit student will be more likely to get it than a 3U student as 3U students are introduced to substitution including trig etc... while a 2 unit student only knows one way of integration which is the one used in that question.

leehuan · Jan 20, 2016

HSC 2016 2U Marathon

Both 2u and 3u can get easily swayed just by not knowing to use the same trick for 2^x to differentiate x^x

Consequently they wonder where did a log even come from

Sent from my iPhone using Tapatalk

Nailgun · Jan 20, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$New Question$

$$ (i) Find : $ \frac{d}{dx}{\frac{f(x)}{g(x)}}$

$$ (ii) Hence or otherwise, solve : $ \int{\frac{\log{x} \cos{x} - \sin{x} \cdot \frac{1}{x}}{(\log{x})^2}}dx$

For i
$\frac { d }{ dx } \frac { f(x) }{ g(x) } =\frac { d }{ dx } \frac { 1 }{ g(x) } .f(x)\\ \\ \\ Let\quad y'=\frac { d }{ dx } \frac { f(x) }{ g(x) } \\ \\ \\ y'=vu'+uv'\\ \\ \\ Let\quad v=(g(x))^{ -1 }\quad and\quad u=f(x)\\ \\ \\ y'=f(x).-(g(x))^{ -2 }.g'(x)+f'(x).(g(x))^{ -1 }\\ \\ \\ y'=\frac { -f(x).g'(x) }{ (g(x))^{ 2 } } +\frac { f'(x) }{ g(x) } \\ \\ \\ y'=\frac { f'(x).g(x)-f(x).g'(x) }{ (g(x))^{ 2 } } \\ \\ \\ \therefore \frac { d }{ dx } \frac { f(x) }{ g(x) } =\frac { f'(x).g(x)-f(x).g'(x) }{ (g(x))^{ 2 } }$

Solid quotient rule derivation lel

For ii
$\int { \frac { \log { x } \cos { x } -\sin { x } \cdot \frac { 1 }{ x } }{ (\log { x } )^{ 2 } } } dx\\ \\ \\ Let\quad g(x)=\log { x } \quad and\quad f(x)=\sin { x } \\ \\ \\ Hence\quad g'(x)=\frac { 1 }{ x } \quad and\quad f'(x)=\cos { x } \\ \\ \\ So,\quad \int { \frac { f'(x).g(x)-f(x).g'(x) }{ (g(x))^{ 2 } } } dx\\ \\ \\ From\quad previous\quad result\quad \frac { d }{ dx } \frac { f(x) }{ g(x) } =\frac { f'(x).g(x)-f(x).g'(x) }{ (g(x))^{ 2 } } \\ \\ \\ \therefore \int { \frac { f'(x).g(x)-f(x).g'(x) }{ (g(x))^{ 2 } } } dx=\frac { f(x) }{ g(x) } +C\\ \\ \\ \therefore \int { \frac { f'(x).g(x)-f(x).g'(x) }{ (g(x))^{ 2 } } } dx=\frac { \sin { x } }{ \log { x } } +C$

Drsoccerball · Jan 20, 2016

Re: HSC 2016 2U Marathon

Correct. But for part (i) I wasnt looking for a derivation I just wanted you to give me the value which could've been done in 2 lines with the quotient rule.

Nailgun · Jan 20, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Correct. But for part (i) I wasnt looking for a derivation I just wanted you to give me the value which could've been done in 2 lines with the quotient rule.

Oh lol now that I look at it, I have no idea why I used the product rule lel
oh well

Drsoccerball · Jan 21, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$$ Using the fact that $ \int{f'(x)}dx=f(x) ($ So that new comers who don't know integration can attempt this. $)$

$$ Hence, or otherwise $ \int{x^{2x}(1+\ln{x})}dx$

Bumpppp

leehuan · Jan 24, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Bumpppp

Just put another question. It's also something scary to encounter and pretty much never even given a slightest mention in the 2U course; x^x

Drsoccerball · Jan 24, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Just put another question. It's also something scary to encounter and pretty much never even given a slightest mention in the 2U course; x^x

It was in one of my school's exams though? And that year someone state ranked it

HSC 2016 Maths Marathon (archive) (2 Viewers)

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