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Question 10 of the 2020 Advance Maths HSC was a challenging question. There are (at least) two approaches that will work here. One is to use theory and differential calculus, which is quick but sophisticated. The other is to take a more straight-forward but longer approach. Starting with the...

It would also be sufficient to take the given value for a and show that P(a) = P'(a) = 0.

This result can be proved by summing APs without induction, considering the cases where n is odd and even separately: \begin{align*} 1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{n - 1}n^2 &= 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ... + (n-1)^2 - n^2 \qquad \text{supposing $n$ is even} \\ &= (1 + 2)(1 - 2)...

By the way, the @blob063540 approach is more rigorously re-written as: \begin{align*} \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx &= \int \sqrt{\tan{x}} + \frac{1}{\sqrt{\tan{x}}}\ dx \\ &= \int \frac{\left(\sqrt{\tan{x}}\right)^2 + 1}{\sqrt{\tan{x}}}\ dx \\ &= \int \frac{\left|\tan{x}\right| +...

Thanks for that information. I can't say that I am surprised that IUPAC decided to declare a new practice that differs from one already in use. It doesn't mean that many will follow it, however. It can see the sense in not using a single term (sol) with different meanings, but that doesn't...

Yes, another approach is to express the integral as piece-wise defined: \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx = \sqrt{2}\left[\sin^{-1}{(\sin{x} - \cos{x})}\right] + C \qquad \text{provided $x$ is an angle in the first quadrant} \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ dx =...

Prove that the value is a=\frac{pq-r}{2\left(q - p^2\right) means that you need to find the value and, in so doing, establish that it can be no other value. So, if you used the approach I noted above with \alpha = -p \pm \sqrt{p^2 - q}, you would need to establish which of these two possible...

@dan964, thanks so much for THSC. Just FYI, there is a link for a Fort St advanced paper mixed in with the Barker MX1 papers, and there is a duplicate entry for the Hurlstone 2022 MX1 paper.

Start question 26 by letting the three roots be \alpha,\ \alpha, \text{ and } \beta \text{ where } \beta \neq \alpha. Root theory (sum of roots, product of roots, etc) can then be used to get a = \alpha for part (a) and then use the three equations to eliminate \alpha and \beta to get the...

The only thing I see wrong is that the question sought the reciprocal of the ratio written here, i.e.: \frac{T_{r + 1}}{T_r} = \frac{(n - r + 1)b}{ar}

The problem with the approach from @blob063540 arises from the initial transformation of the integrand \sqrt{\tan{x}} + \frac{1}{\sqrt{\tan{x}}} = \sqrt{\frac{\sin{x}}{\cos{x}}} + \sqrt{\frac{\cos{x}}{\sin{x}}} = \frac{\sqrt{\sin^2{x}} + \sqrt{\cos^2{x}}}{\sqrt{\sin{x}\cos{x}}} = \frac{\sin{x}...

Yes

Or are the rare and high-achieving Standard-to-MX2 upgrader... :P

Extension 2 only exists in year 12, all MX2 students were MX1 students in year 11.

This is making use of the same sort of approach as I used for factorising, but applied to greatly simplify the solution. Excellent!

Your f(x) is missing a factor of 1 over the square root of 2. That is, it should be: f(x) = \sqrt{2}\tan^{-1}{\left[\frac{1}{\sqrt{2}}\left(\sqrt{\tan{x}} - \frac{1}{\sqrt{\tan{x}}}\right)\right]} = \sqrt{2}\tan^{-1}{\left(\frac{\tan{x} - 1}{\sqrt{2\tan{x}}}\right)} It is the same result as I...

\int \sqrt{\tan{x}}\ dx = \int u \times \frac{2u\,du}{u^4 + 1} = \int \frac{2u^2}{u^4 + 1}\ \!du \qquad \text{using the substitution $u = \sqrt{\tan{x}}$} The partial fractions result will now yield both inverse tan and log terms. Alternatively, focus on \int \sqrt{\cot{x}}\ dx = \int...

Are there THSC updates beyond Adv Maths / MX1 / MX2 Trials? Are there more coming? Please let there be more :)

I typed it in the box for posting replies. Note that the original solution posted: \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ \!dx = \sqrt{2}\bigg[\sin^{-1}{\left(\sin{x}-\cos{x}\right)}\bigg] + C does works on some domains - for example, on 0 < x < \frac{\pi}{2}. However, for the domain \pi < x...

\text{Let } I = \int \sqrt{\tan{x}} + \sqrt{\cot{x}}\ \!dx = \int \sqrt{\tan{x}} + \frac{1}{\sqrt{\tan{x}}}\ \!dx = \int \frac{\tan{x} + 1}{\sqrt{\tan{x}}}\ \!dx \begin{align*} \text{Let} \qquad u &= \sqrt{\tan{x}} = \(\tan{x})^{\frac{1}{2}} \\ du &= \frac{1}{2} \times (\tan{x})^{-\frac{1}{2}}...