I honestly don't recall learning the inverse function theorem until first year uni to be fair (and basically never used it from 2nd year and later) , so I doubt the majority of the high-school students would even consider it.
Please share some more rants pls this is hilarious.
Wow this is definitely harder than the previous years. I see that Carrotsticks having extra assistance from former BOS members (who are also math whizzes :P) definitely revoluntionised the content.
However, with the new syllabus being implemented, I wonder what mathematical theorem do you plan...
You can't compare the BOS trials with the HSC. It's just at a completely different level since it's specifically aimed to challenge the students who have mastered virtually every single topics, i.e people who have the potential to easily achieve 95+ HSC. Even if you did do badly, it wouldn't...
When you're finding the rate, or how fast or slow something is, you use the derivative expression.
so for c)
you find dS/dt and then sub in t = 5 and your value of k from part b), that's your rate.
d)sub S = 4 (as that's the amount that's left undissolved ) and your k from b)
Then you solve...
The problem with checking your answers over, is that sometimes you would think it's right in the back of your mind so it could blind you from seeing the mistake.
What I would do is that I would read the question thoroughly a few times (even though it's something I'm familiar with), and then do...
I'm not sure what you mean by "New" HSC Math courses (you mean progressing to senior or a complete revamp in the content?)
Anyhow, all I can say is that you general math is basically learning applied "real life" math like finding areas and doing some basic finance stuff you might encounter in...
My parents and relatives would talk about how my mum and dad did really well in math back at school in China. I'm not sure if it's legit or not haha but I was able to gasp at most concepts a bit better than the average student so I guess it's a factor.
That's not even true since you're squaring a number less than 1
0<ln(2)<ln(e) = 1 \\ \\ ln(2) > (ln(2))^2
2ln(2) > (ln(2))^2 \Rightarrow 2 > ln(2)
At the moment I still can't figure the lower bound either.
In order to find the horizontal asymptote, you need to take the limit as the x-value approaches infinity. You also need use this concept.
lim_{x \rightarrow 0} \frac{sinx}{x} =lim_{x \rightarrow \infty} \frac{sin(\frac{1}{x})}{\frac{1}{x}} = 1
lim_{x \rightarrow \infty} \frac {x^2...
Note that is says distinct points so that already implies p \neq q
You can easily see that's true if you simply observe the equation mQRmPR=-1. which leads to (p+1)(q+1) = -4 and you'll understand that p and q can't be -1 as you'll get zero.
However also observe that both p+1 and q+1 can't...