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$\noindent For (c), calling the function $f(x,y)$, it is easy to show that $(0,0)$ is a stationary point (I think they let you assume that even). Note $f(0,0) = 0$. Along the line $y=0$, we have $f(x,0) = x^{2}$. So in any neighbourhood of $(0,0)$, there exist points of the form...

$\noindent We have $f(x) = x\sqrt{3x-4} = x\left(3x-4\right)^{\frac{1}{2}}$. Recall that $\left(uv\right)'' = u''v + 2u'v' + uv''$ for twice-differentiable functions $u,v$. Note that $x'' = 0, x' = 1, v' = \frac{3}{2}\left(3x-4\right)^{-\frac{1}{2}}$ and $v'' = -...

$\noindent What is the function (need brackets, it's hard to tell what it is)? Also, is it saying to evaluate the derivative at a particular point? If so, where?$

Re: HSC 2016 2U Marathon Since you have the actual data, you can get a computer package to fit the data to that model, as well as produce the graphs.

It's 30+, so avoids the * ATAR. :p

Re: HSC 2018 4U Marathon Thread title currently says 2018 marathon. Did you mean 2017 (typo)?

Easy part as in Paper 1 (rather than Paper 2) is what he meant, I think.

$\noindent For the second one, recall that $f(x,y) = xy \left(1-(x+y)\right)$. The domain is a triangle with $x,y \geq 0$ and due to the given vertices, $x+y$ takes values between $0$ and $2$. Consider $f(x,y)$ on the family of lines $x+y = c$ (and only within the triangle) for $c$ between $0$...

Note for a), you just need to maximise and minimise x - x^2 for x in [0, 3] and maximise and minimise y^2 for y in [0,2] (the latter obviously has min. value 0 and max. value 4, no real calculations needed). Add the max. values to get f(x,y)'s max. and add the min. values to get the min. When...

Posted a year in advance. :p

You could try searching up interpretations/analyses of it online (sites like Wikipedia or others). But I guess it's getting a bit late now.

Oh yeah, forgot to mention there are sometimes shortcuts. Was going to edit it in with a sample answer but ended up being too lazy then to type up the answer (it would be fairly long) and left the post unedited.

You can do these fairly algorithmically, just tedious. The fact that the regions are closed and bounded (compact) and the functions are continuous tells us there is a maximum and a minimum. The functions are differentiable, so the extrema occur either at stationary points in the interior of the...

Incidentally, that's the lowercase Greek letter 'tau'.

$\noindent Yep. In general if we have $N = ng$ people and want to split them up into $g$ groups of $n$ people each, the no. of ways is: $\binom{N}{n}\times \binom{N-n}{n}\times \binom{N-2n}{n}\times \cdots \times \binom{2n}{n}\times \binom{n}{n}\bigg/ g!$. I.e. if dividing into \emph{equally...

In part (ii), the groups are indistinguishable, so we over-count by a factor of 3! if we just do (12C4)*(8C4). For part (i), the two groups are indistinguishable, so there's no over-counting. To see why there's overcounting in part (i), note that what we're doing with (12C4)*(8C4)*(4C4) is...

The coefficient of the (z + (p^2 + q^2)) part should just be 1. But yeah, set x = y = 0 to find the z-intercept.

$\noindent c) The sub-intervals used are $I_{n} = [n,n+1]$ (so spacing is $h=1$), for $n=50,51,52,\ldots$. Note $f''(x) = 6x^{-4}$. Since $f''$ is decreasing on each $I_{n}$, the maximum of $|f''(x)|$ on $I_{n}$, say $M_{n}$, occurs on the left endpoint of $I_{n}$, i.e. at $x=n$. So $M_{n} =...

The equation of a plane with normal n = (n1,n2,n3) and through the point a = (a,b,c) is just n•(x - a) = 0, i.e. n1*(x-a) + n2*(y-b) + n3*(z-c) = 0.

You can try testing whether it's right by checking whether the given point (4,2,1) lies on your plane. Subbing it in, we see the plane equation isn't satisfied, so the answer's not right. You did everything right up till the "Equation of tangent" part. The equation is just meant to be what...