HSC 2016 MX1 Marathon (archive) (1 Viewer)

calamebe · Nov 29, 2015

Re: HSC 2016 3U Marathon

InteGrand said:
$The first line of the working is incorrect. It is not true that if $f$ and $g$ are inverse functions, then $f^\prime (x)\cdot g^\prime (x) = 1$ (which is what you implied in your answer to (a)). What (a) states is that if $f$ is an invertible differentiable function with inverse $g$, and $y = f(x) \Longleftrightarrow x = g(y)$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}$, provided the denominator here is not 0. That is, if $b= f(a)$ (so $a = g(b)$), then $f^\prime (a) = \frac{1}{g^\prime (b)}$ provided $g^\prime (b)\neq 0$, where $f^\prime$ is the derivative of the function $f$ and similarly for $g^\prime$.$

Thanks, I understand it now.

InteGrand · Nov 29, 2015

Re: HSC 2016 3U Marathon

$So to start that question about the arcsin derivative, we should let $y =\sin^{-1}x, x = \sin y$ \Big{(}so $x \in \left[-1,1\right]$ and $y \in \left[ -\frac{\pi}{2},\frac{\pi}{2}\right]$\Big{)}. Then we need to prove that $\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{\sqrt{1-x^2}}$.$

$We have$

$\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}&= \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}, \frac{\mathrm{d}x}{\mathrm{d}y}\neq 0 \quad (\text{inverse function theorem}) \\ &= \frac{1}{\cos y},\cos y \neq 0\Longleftrightarrow y\neq \pm \frac{\pi}{2}\Longleftrightarrow x\neq \pm 1 \\ &= \frac{1}{\sqrt{1-\sin^2 y}}, y\neq \pm \frac{\pi}{2} \quad \Big{(}\text{as }\cos y > 0 \text{ for our range of }y,\text{ so } \cos y = +\sqrt{1-\sin^2 y}\Big{)} \\ &= \frac{1}{\sqrt{1-x^2}}, x\neq \pm 1 \quad (\text{as } x=\sin y). \end{align*}$

Drsoccerball · Nov 29, 2015

Re: HSC 2016 3U Marathon

$$ Use the first principle of differentiation to evaluate $ \frac{d}{dx}{sin^{-1}x}$

leehuan · Nov 29, 2015

Re: HSC 2016 3U Marathon

Part a of my question

The geometrical significance was that given the graphs of two mutually inverse functions, the gradients of the tangents are reciprocals of each other.

d(f-1(x))/dx = 1/f'(x)
---------

Drsoccerball said:
$$ Use the first principle of differentiation to evaluate $ \frac{d}{dx}{sin^{-1}x}$

Surely there is another way rather than a compound angle formula for inverse sine right? Cause that would require a derivation beforehand

leehuan · Nov 29, 2015

Re: HSC 2016 3U Marathon

Side note:

Whilst I don't expect anyone else to, I try to make the questions in the 2U and 3U forum manageable without a hassle in guessing what to do. A 4U student should be capable of adapting at almost any situation, however 2U is just to do some calculus and 3U should be given implicit hints when the question is too difficult by itself.

e.g. What method to do in a binomial coefficient proof is always hinted by what the things you're given look like.

InteGrand · Nov 29, 2015

Re: HSC 2016 3U Marathon

leehuan said:
d(f-1(x))/dx = 1/f'(x)

$This is not the case; rather, it is $f ^\prime (a) = \frac{1}{ \left( f^{-1} \right) ^\prime \left( b \right) }=\frac{1}{\left(f^{-1} \right) ^{\prime} \left(f(a)\right)}$, where $(a,b)$ is a point on the graph of $y=f(x)$, so $(b,a)$ lies on the graph of $y=f^{-1} (x)$ (i.e. $b=f(a)$ and $a = f^{-1}(b)$).$

$So in other words, given the graphs of $y=f(x)$ and $y=f^{-1} (x)$, the slopes of the tangents at \textsl{``corresponding points''} are reciprocals, provided neither of them is 0 (this is the geometrical significance you were after). For example, the function $f:\mathbb{R}\to \mathbb{R}^+,f(x) = \mathrm{e}^x$ has inverse $f^{-1} : \mathbb{R}^+ \to \mathbb{R}, f^{-1} (x) = \ln x$. A point on the graph of $y=f(x)$ is $P\left(1, \mathrm{e}\right)$, and so $Q \left( \mathrm{e},1\right)$ is a point on the graph of $y=f^{-1} (x)$. What the inverse function theorem then tells us is that the slope of the tangent to $y=f(x)$ at the point $P$ is the reciprocal of the slope of the tangent to $y=f^{-1} (x)$ at the point $Q$, since these points are ``corresponding points'' on each other's graphs (and neither slope is 0).$

$(The notation $f:X\to Y, f(x) = \ldots$ just means that $f$ is a function with domain $X$ and codomain $Y$, and the rule for $f(x)$ is specified with the $\ldots$.)$

leehuan · Nov 29, 2015

Re: HSC 2016 3U Marathon

I was just thinking about the validity of that statement before I looked in this thread.

kawaiipotato · Nov 29, 2015

Re: HSC 2016 3U Marathon

$Question$

$$ If $ \log _{e}s $ and $ \log_{e}r $ are roots to the equation $ 4x^2 - 2x + 5 = 0$

$$ Find $ \log_{r}s + \log_{s}r$

RachelGreen · Nov 30, 2015

Re: HSC 2016 3U Marathon

I loved this question

Paradoxica · Dec 6, 2015

Re: HSC 2016 3U Marathon

$\noindent Prove that there cannot exist an equilateral triangle with lattice point vertices on the cartesian plane.$

leehuan · Dec 6, 2015

Re: HSC 2016 3U Marathon

You really should tone down the question difficulty a bit when you're outside the 4U marathons. I don't think any of the 3U kids know what that actually is.

InteGrand · Dec 6, 2015

Re: HSC 2016 3U Marathon

$\noindent For the benefit of those who don't know, \textsl{lattice points} means points with integer coordinates. E.g. points like $(1, 0), (-5, 3)$, etc.$

$\noindent So you should have a picture like this in mind when thinking of lattice points:$
http://mathworld.wolfram.com/images/eps-gif/LatticePoints_1000.gif.

$\noindent So the goal is basically to show that on such a grid, we cannot have an equilateral triangle.$

Paradoxica · Dec 6, 2015

Re: HSC 2016 3U Marathon

leehuan said:
You really should tone down the question difficulty a bit when you're outside the 4U marathons. I don't think any of the 3U kids know what that actually is.

The question is intrinsically lacking in difficulty. What it does contain is terminology, which is severely neglected in the current curriculum.

leehuan · Dec 6, 2015

Re: HSC 2016 3U Marathon

I could tell the difficulty was absolutely fine. These are just average 12th grade students though; there are hardly any places in the world where such terminology is required.

InteGrand · Dec 6, 2015

Re: HSC 2016 3U Marathon

leehuan said:
I could tell the difficulty was absolutely fine. These are just average 12th grade students though; there are hardly any places in the world where such terminology is required.

"Lattice points" is a standard mathematical term. But I agree that if this question came up in the HSC (unlikely in itself these days), instead of saying "lattice points", they would probably explain it (by saying "integer coordinates" or something), and maybe even provide a diagram like the one I linked above.

Paradoxica · Dec 6, 2015

Re: HSC 2016 3U Marathon

leehuan said:
I could tell the difficulty was absolutely fine. These are just average 12th grade students though; there are hardly any places in the world where such terminology is required.

Mathematicians are nitpicky.

leehuan · Dec 6, 2015

Re: HSC 2016 3U Marathon

When I meant hardly any places, I meant any places in the world in their 11th/12th/13th grade curriculum.

Part of why I even want to do maths at uni is to extend above these simplistic questions. But without the explanations, no assumptions should be made that a person who's basically just leaving behind childhood is capable of interpreting language themselves, or to research it lol

Paradoxica · Dec 6, 2015

Re: HSC 2016 3U Marathon

leehuan said:
When I meant hardly any places, I meant any places in the world in their 11th/12th/13th grade curriculum.

Part of why I even want to do maths at uni is to extend above these simplistic questions. But without the explanations, no assumptions should be made that a person who's basically just leaving behind childhood is capable of interpreting language themselves, or to research it lol

There needs to be a single quick topic in the current one labelled : Terminology

davidgoes4wce · Dec 7, 2015

Re: HSC 2016 3U Marathon

kawaiipotato said:
$Question$

$$ If $ \log _{e}s $ and $ \log_{e}r $ are roots to the equation $ 4x^2 - 2x + 5 = 0$

$$ Find $ \log_{r}s + \log_{s}r$

Not sure what has happened but I have tried accessing your link but can't read it from 2 computers (a uni) and my mac book (which I use regularly)

InteGrand · Dec 7, 2015

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Not sure what has happened but I have tried accessing your link but can't read it from 2 computers (a uni) and my mac book (which I use regularly)

I think LaTeX is just messing up on this site, it happens once in a while.

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