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They were standard integrals on the old syllabus, but that's going back a way - see page 16 of the 2001 paper accessible from https://educationstandards.nsw.edu.au/wps/portal/nesa/resources-archive/hsc-exam-papers-archive/mathematics/mathematics-2001-hsc-exam-pack-archive, for example

Since element B has the highest first IE amongst the four consecutive elements, it must be a noble gas. Thus: element A is a halogen (group 17) and forms an A- anion element C is in group 1 and forms a C+ cation element D is in group 2 and forms a D2+ cation Forming a D3+ cation requires much...

You can also find the boundary values by solving \begin{align*} |3x - 2| &= |x + 4| \\ \\ \textbf{Case 1:} \qquad 3x - 2 = x + 4 \qquad \implies \qquad 2x &= 6 \qquad \implies \qquad x = 3 \\ \\ \textbf{Case 2:} \qquad 3x - 2 = -(x + 4) \qquad \implies \qquad 4x &= -2 \qquad \implies \qquad x =...

Define a height axis, y, along the central axis of the cylinder, taking y = 0 at the top and placing the light source at y = h above the cylinder. When h > 0, the light casts a shadow of radius i cm from the axis. We are told that h is increasing at dh/dt = 3 cm/s, and r decreases as this...

Right-angle triangle trigonometry should allow you to show that \text{LHS} = \sin{\frac{\theta}{2}} + \frac{1}{2}\cos{\frac{\theta}{2}} = \frac{b + h}{2x} Applying the AM-GM inequality leads to \text{LHS} \geq \sqrt{\frac{bh}{x^2}} Now, using the fact that \text{Area of } \Delta =...

The cover is a modified version of that used by the Independent Schools papers, as is the table on page 2 for MCQ papers... but it's not the 2020 Independent paper. The way things like square roots appear makes it look like it's not from a professional source. The repeating of the diagram in...

Some excellent points. From a cognition perspective, there is a significant difference depending on how you are with writing on a computer and creating illustrations, etc. If you are typing up notes that are largely already written, then the cognitive demand of composing prose is relatively...

I think the observation implicit in these questions / comments is important... why is this worth 4 marks? Is it just a mistake, or an indication that something is being missed? Start by considering the term "valency"... it is more than just the charges in @wizzkids' table, which tells us about...

The problem can be approached as an area between two curves, just relative to the y-axis. The curves that bound the region are x=e^{2 - y}, which is the "top" curve, and x = 1, which is the "bottom" curve. The area, against the y-axis, is then \begin{align*} \text{Shaded Area}\ &= \int_0^2...

\text{Since} \qquad \tan^{-1}{x} + \tan^{-1}{\left(\frac{1}{x}\right)} = \frac{\pi}{2} \qquad \text{for all $x > 0$} \text{And} \qquad \tan^{-1}{x} + \tan^{-1}{\left(\frac{1}{x}\right)} = -\frac{\pi}{2} \qquad \text{for all $x < 0$} \text{Your answer and the desired answer match but with...

Since the area is symmetric about the x-axis, its area can be found by doubling the integral of the function above the x-axis: \begin{align*} \text{The area of the property is given by } 2\int_0^1 y\,dx &= 2\int_0^1 \sqrt{9x}\,dx \qquad \text{as $y^2 = 9x \implies y = \sqrt{9x}$ above the...

Definitely positively certainly B and not D.

Question 1: \text{$y = \sin^{-1}{x}$ takes ratios between $-1 \le x \le 1$ and outputs angles between $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.} \text{However, you are using this output as an input into a log function, which can only accept positive values. Hence, you need $\sin^{-1}{x} > 0$.}...

Sure... \begin{align*} \text{Going from Line 1:} \qquad \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ \text{to Line 2:} \qquad \qquad 4!(n - 5)(n - 4) - 6! &= 0 \end{align*} \\ \text{was done by either dividing both sides of the equation by $\frac{n!}{(n -4)!6!4!}$, or...

Your algebra is correct, you got to \begin{align*} \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ 4!(n - 5)(n - 4) - 6! &= 0 \qquad \text{as $\frac{n!}{(n -4)!6!4!} \neq 0$} \\ (n - 5)(n - 4) &= \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} \\ (n - 5)(n - 4) & = 30 \\ n &=...

Each row of Pascal's Triangle has terms that are distributed symmetrically. In binomial coefficient form, this symmetry property can be expressed as: \binom{n}{r} = \binom{n}{n-r} \begin{align*} \text{Applying the symmetry property to this problem yields:} \qquad \binom{n}{4} &= \binom{n}{n -...

Past papers

@wizzkids is correct that some metal hydroxide increase their solubility at high pH - aluminium hydroxide, for example - which the HSC covers in the form of an example of amphotericism. It pops up in modules 6 and 8, though is generally covered very briefly. Aluminium hydroxide is amphoteric...

Yes, they do. Hydroxide ions will react with hydronium / hydrogen ions and with unionised propanoic acid molecules, in each case producing water and (in the latter case) the propanoate anion. Which they react with principally will depend on relative concentrations as well as collision...

(a) is bounded above but not below as its a negative cubic, and so the region above y > 1 will be x \in (-\infty,\ a) or x \in (-\infty,\ a)\cup(b,\ c). (b) must be bounded below as \ln{x} \in \mathbb{R} \implies x > 0. It's also bounded above as x\ln{(x + 1) - (x + 1)\ln{x} < 0\ \forall x >...