Recent content by Luukas.2

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 &=...