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The coordinate for the point T is \left(\alpha,\,a(\alpha - h)^2 + k\right) as the solutions posted by @anonymoushehe state. The coordinate for the point P is \left(h,\,k - a(\alpha - h)^2\right), matching the solution @eternallyboreduser gave. The coordinate for the vertex V is...

Note, also, the working on the second page can be substantially simplified, in my opinion. Starting from 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... < 1 + \lim_{k\to\infty}{\sum_{n = 2}^k{\ln{\left(\cfrac{1}{1-\frac{1}{n^2}}\right)}}} \qquad \text{. . . (*)} and then simplifying...

Coverage in the news suggests their actions may have gone beyond just getting together and posing for a photo... From The Guardian Australia: NSW Premier warns 'pathetic' neo-Nazia they will be exposed after attempted rally in Sydney park The story identified the park, which is on Sydney's...

I have no idea where they are from... but that photo was taken on Sydney's North Shore.

The fact that the photo was taken on the North Shore doesn't mean that is where these individuals live. Because Australia is an open and free society people are allowed to express unpopular / controversial views. So long as that doesn't extend to conduct that is threatening or causes harm to...

The first line on the second page, where the k is introduced, is flawed. It states that 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... < 1 + \sum_{n = 2}^k{\ln{\left(\cfrac{1}{1-\frac{1}{n^2}}\right)}} which may or may not be true, depending on the value of k: \begin{align*}...

Then out equal signs at the LHS of each ratio, showing that the ratio is equal to the line before... or use fractions. And, the distance to the axis is not \alpha - h, it is |\alpha - h|. You have drawn a diagram is which \alpha > h but nothing in the question requires it, so you need the...

19b goes wrong at the end. You have shown that PV = a(\alpha - h)^2 and the distance from A to the axis of symmetry is D = |\alpha - h|. You are required to show that PV \propto D^2 \qquad \qquad \text{or equivalently, that} \qquad \qquad \frac{PV}{D^2}\ \text{is constant} but you have...

For every permutation in which the E comes before the I, swapping these letters produces a corresponding permutation where the I comes before the E. It is impossible for one of them not to precede the other (as they can't both be in the same position), so the set of all possible permutations...

Note that a purely geometric approach would provide the graph, with the points z = -1 and z = i excluded, and is quicker: \begin{align*} \text{If $\frac{z -i}{z + 1}$ is purely imaginary, then} \quad \frac{z -i}{z + 1} &= ki \qquad \text{for some $k \in \mathbb{R}$, where $k \neq 0$} \\...

And z = i is also impossible, as it makes \frac{z - i}{z+1} = \frac{i - i}{i+1} = \frac{0}{i+1} = 0 which is not purely imaginary - so two open circles, at opposite ends of a diameter.

Having got to LHS > 3k3, you could then consider the graph of y = 3x3 - (x + 1)3 = 2x3 - 3x2 - 3x + 1, which is a cubic with both of its stationary points below the x-axis and a single x-intercept (a little above x = 2) and so y > 0 for all x > 3, which means LHS > RHS.

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\begin{align*} \text{LHS} &= 3^{k+1} \\ &= 3^1 \times 3^k \\ &> 3 \times k^3 \qquad \text{using the induction hypothesis (**)} \\ &= 2k^3 + k^3 \\ &> 2k^3 + 3k^2 \qquad \text{as $k^3 = k \times k^2 > 3k^2$ as $k > 3$} \\ &= k^3 + 3k^2 + k^3 \\ &> k^3 + 3k^2 + (3k + 1) \qquad \text{as $k^3 >...

True, and differentiation induction is one way to present something a bit different, and possibly in a form that wasn't anticipated (like with factorials, here).

It could be presented as an induction question, too. As in: \text{Consider the equation $y = \frac{1}{\sqrt{x}}$. Using mathematical induction, show that $\frac{d^ny}{dx^n} = \frac{(-1)^n(2n)!}{2^{2n}n!}x^{-\left(n + \frac{1}{2}\right)}$. Though induction isn't actually necessary, unless the...

Or, alternatively: \begin{align*} \tan{\left(\alpha - \beta\right)} &= \frac{\sin{\left(\alpha - \beta\right)}}{\cos{\left(\alpha - \beta\right)}} \\ &= \frac{\sin{\alpha}\cos{\beta} - \cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}} \\ \\ \text{Put $\alpha = x$ and...

Looking at the page you mentioned, https://en.wikipedia.org/wiki/List_of_trigonometric_identities, it goes way waaaaaay beyond the formulae covered in the HSC courses (across all levels) and, as illustrated above, many of the results can be derived using other results stated. Understanding how...

By angle sum identities, I think @liamkk112 is referring to results like: \begin{align*} \sin{\left(\frac{\pi}{2} - x\right)} &= \cos{x} \\ \sin{\left(\pi - x\right)} &= \sin{x} \\ \sin{\left(x + \pi\right)} &= -\sin{x} \\ \sin{\left(2\pi - x\right)} &= -\sin{x} \\ \\ \tan{\left(\frac{\pi}{2} -...

Permutations and Combinations is an MX1 topic in discrete maths