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Note, there is no curve with x,\ y \in \mathbb{R} if c = -2 because \begin{align*} x^2 + cxy + y^2 + 1 &= 0 \\ \text{Put $c = -2$:} \qquad x^2 + -2xy + y^2 + 1 &= 0 \\ (x - y)^2 + 1 &= 0 \\ (x - y)^2 &= -1 = i^2 \\ x - y &= \pm i \end{align*}

The working provided appears correct, but is flawed because there is only one curve (and one value of c) which actually passes through (1, 1). It is true that all curves have the same gradient because the total number of curves in the 'family' passing through the point (1, 1) is 1.

I agree, but there are surprisingly strong commonalities in views that don't really withstand rational analysis. For example (in social policy), there are plenty of Americans who strongly oppose access to abortion on sanctity of life grounds but also endorse capital punishment - two positions...

Yes, economic left/right and social left/right are regularly poorly correlated

On this point, I would encourage not only working on topics that you find challenging, but also seeking out questions that cross multiple topics. These can often involve different approaches and new ways of looking at questions. They also help you to develop insight into the links between...

I agree that understanding fundamentals is of great value, and looking into the "why"s. @liamkk112, your understanding of physics can explain the propane v. cyclopropane without knowing the chemistry. Many chemical reactions occur for the same reason that a ball at the top of a hill will roll...

It sounds like a similar situation as Ben Roberts-Smith, claiming defamation when you know there is at least significant truth in what was said and yet claiming it is all false. It's a legal strategy that is very high risk, especially as the standard of proof in a civil trial is "on the balance...

To work as a psychologist needs four years of training in Australia, but BA / BSc pass degrees are three years, so getting into honours is much more competitive. Having an Honours degree will be helpful for future study in the US as it may allow access direct to PhD, whereas without it you...

I agree with most of what @ZakaryJayNicholls is saying about the ordering of topics being questionable in some subjects. In chemistry, for example, the way content in Module 1 (Year 11) is typically examined requires content from Module 7 (of the 8 Modules), not encountered until mid-Year 12 at...

Carrotsss is right, it is rankings not marks that you need to focus on. My trial result in one subject was about twenty marks below my HSC result and moderated assessment because the exam was so difficult / challenging. Also, for rankings, look at how many students there are who are really...

Once you've got calculus of exponentials, you can explore ideas like: x^0 = 1 0^x = 0 \text{So, $0^0 = 0$ or $1$ or $\frac{1}{2}$ (on average)?} Or, what happens to a function like x.2^x as x becomes increasingly negative, or as x\ln{x} gets close to zero...

Pretty much, though this uses the reasoning mentioned above and would actually be at Advanced level. This is a result that appears at the start of plenty of MX2 questions, and which can be done without calculus... so I think it is useful to know. It can also be extended into other problems...

Another question like this would be: \text{Use calculus to prove that, for any $a \in \mathbb{R}^+$, $a + \frac{1}{a} \geqslant 2$.}

Yes, applying the corrections that have been noted above by @scaryshark09, the working should be: \begin{align*} \text{LHS} &= x\left(x^{k+2}\right) + (x + 1)^2(x + 1)^{2k + 1} \\ &= x\left(x^{k+2}\right) + \left(x^2 + 2x + 1\right)(x + 1)^{2k + 1} \\ &= x\left(x^{k+2}\right) + \left(x^2 + x +...

Don't underestimate the difficulty of achieving a WAM of 75, @jeonghannie. When there were guidelines applying to grades, the guideline was that grades of distinction (D) and high distinction (HD) would typically be achieved by about 15% of the students who pass the unit of study. Marks above...

Yes, as I said, it's a form of reasoning that you might not be very familiar with yet, but you will use it from time to time going forwards :) And it is needed for most of the questions you posted, FYI.

I was showing not only that x = c/2 is a stationary point, but also that there are no other stationary points. Draw a set of coordinate axes. Mark on them the max at \left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right). Now draw a continuous curve that has that point as a maximum and has no...

I have seen this exact result in an MX2 assignment, though it was meant to be done by induction - this is actually a nicer approach. A question like this is certainly possible in an assessment task for a school with a strong candidature. Remember that exams are meant to facilitate ranking the...

This is for the case where 0 < k < 1, in which case we have shown that there is a maximum turning point at coordinates \left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right) There are no other stationary points on the curve as the derivative \frac{dy}{dx} = k\left[x^{k-1} - (c -...

@eternallyboreduser, does this solution make sense for the question you were looking at? The part that I have left out is very similar to the solution provided.