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Suggest checking out this flow chart https://www.boredofstudies.org/threads/flowchart-of-how-hsc-marks-and-atar-are-determined.390261/

A corollary you need first is that for any complex number z (something you can easily prove): \text{Re}(z) \leq |z| Notice that since the terms of the sum are conjugates of each other: z\bar{w} + \bar{z}w = 2\text{Re}(z\bar{w}) \leq 2|z\bar{w}| = 2|zw|

Only 4 more days to go! :skip: Thanks to everyone who has registered so far! I would like to provide an important update on the registration availability. For Mathematics Extension 1, we are almost at full capacity. I will continue to take registrations for this subject up until 11:59pm...

Easiest way is to just substitute each given root into the LHS and show it equals zero.

It’s hard to make any generalisation on “reliability” when the level of uncertainty is quite high for something the users want good precision for. At the end of the day they’re just using mathematical models based on very limited data points available to the public. In that regard, none of the...

Bump! Still got spots left.

Yes. This can happen. In fact it has happened to the founder of this website iirc…

Want some valuable exposure to harder questions for the HSC exam? Want to do a practice paper under simulated exam conditions in preparation for your HSC? Know anyone at school who is a gun at HSC maths and/or HSC chemistry? It is that time of year again! This year, Bored of Studies will be...

Coming soon…:spin:

If you have any trial papers on hand, please upload them in our resources section so the rest of the student community can access them not just for this year but for future years as well

You utilise the symmetry properties of the normal distribution. P(Z < -1.12) = P(Z > 1.12) = 1 - P(Z < 1.12)

A bit beyond syllabus but consider a parametric polar form where both r and theta are variables: x = r\cos\theta y = r\sin\theta This reduces the condition to r^2 \sin 4\theta = 4 Your objective is to find the minimum value of r^2. Since the RHS is fixed, we minimise r^2 by maximising \sin...

I am proud to say that it has been 20 YEARS since Bored of Studies was established! :cake: For a bit of history, Bored of Studies was established by four ex-students who did their HSC in 2001 and found a lack of resources for the new syllabus at the time, which inspired them to create the Bored...

It's not as exact as that. It's more like if there is a large gap between 1st and 2nd and a small gap between 2nd and 3rd, then there will still be a large gap between 1st and 2nd and a small gap between 2nd and 3rd after moderation.

Start with the general displacement equation of the particle (and differentiate to get its velocity). Substitute all the information provided to solve for the value of n which is what you need to work out the period.

Generally speaking, if you have a division of polynomials and the polynomial in the denominator can be fully factorised into linear and/or irreducible quadratic factors (i.e the quadratic doesn’t have real solutions) then partial fractions would the first thing you consider trying. Anything else...

Making x the subject gives x = 0.5*(u+1) so dx = 0.5*du. Substitute for x, dx and change the limits accordingly and it should reduce to a simpler integral.

No, you’re not expected to recall integral results outside your formula sheet without working. Just familiarise yourself with them.

Assuming k is a positive integer, another rather crude way to approach it is to realise that \dfrac{k}{(2k+1)(2k+2)}=\dfrac{1}{2}\times\dfrac{k}{(2k+1)(k+1)} Notice that \dfrac{k}{(2k+1)(k+1)} < 1 because each factor in the denominator is clearly larger than the numerator, hence the whole...

It's not too different to training in sport or gaming. The principle is to familiarise yourself with learning a new skill/concept/technique/strategy that is initially brand new to you. With enough practice, over time you will gain enough understanding and confidence so that when you are tested...