Recent content by Aykay

Yeah, reckon I screwed up Eng Adv. Then I checked the HSC mark and which percentile it equates to. Then I realised I'm fine, considering my internal rank gives me an assessment mark of about 91-2. :)

Of course. OP, whenever you have a function of a function, the chain rule must be used. I just did it more systematically so you'd understand.

\frac{d}{dx}(ln(\frac{x}{x^{2}+1})) \\ \\ \text{When differentiating logs, it's just} \frac{d}{dx}ln(f(x)) = \frac{f`(x)}{f(x)} \\ \text{So first we differentiate} \frac{x}{x^{2} + 1}, \text{ using the quotient rule, which gives us:} \frac{1-x^{2}}{({1+x^{2})^{2}}{}} \\ \\ \text{And then we...

The focus is (0,a). So find the line which foes through Q and (0,a) and that's your focal chord.

A guy from our school has a pretty good chance at getting first this year. Hopefully we can top 2 years in a row. :p

They don't. Refer to the above post.

You could've said a lot for Q28, as long as you had more than 1 or 2 policies, it's fine to leave others out.

It was just 1 mark, don't beat yourself up over it.

1: 14/15 2: 14/15 3: 13/15

Easy exam.

Use the identity: tan(2x) = 2tanx / [1 - tan^2(x)] to simplify the expression and it is much easier. You should get 1/2 - tan^2(x)/2 after simplifying which is easy to integrate.

Okay, thanks for that!

You can definitely use those abbreviations.

Hey guys, How do I do this integral? \int e^xln(x) What I do is: \int e^xln(x)dx\\ \\ =e^xln(x) - \int e^xx^{-1}dx \\ \\ = e^xln(x) - \left [ e^xx^{-1} + \int e^xx^{-2}dx\right ] \\ \\ = e^xln(x)- \left [ e^xx^{-1} + e^xx^{-2} + 2\int e^xx^{-3}dx \right ] \\ \\ = e^xln(x)- \left [ e^xx^{-1}...

The question 6 in that paper wasn't even a very hard question - in fact it wasn't even the last question of the 1993 (i think) paper, it was Q7. I'm thinking like the top 25% of the state could get the last questions in this.