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Re: 2012 HSC MX2 Marathon your conics question is not entirely wrong, the m<0 condition restricts point of contact to quad1 only, so if you lift that conditon the question is fine

Re: 2012 HSC MX2 Marathon you're not the only one, it does that for all, however, how do robots gain weight by means of human food? I thought eating metal would make you gain weight

Re: 2012 HSC MX2 Marathon on thursday ill see what i can do :P

Re: 2012 HSC MX2 Marathon if you get 100 then it definitely wont go down

Re: 2012 HSC MX2 Marathon don't get anything wrong, k done EDIT: did you mean chippies instead of tippies, cause i can do those for you

Re: 2012 HSC MX2 Marathon come on so boring, (1) 2012 divisbe by 4 so 1, (2) collasping sum so boring, (3) AP with powers once you multiply them all. I hope schools can take these and make them harder. Combining these or putting in terms on n is a much better example as they would have...

Re: 2012 HSC MX2 Marathon yes, i simplied expression in terms of sin and cos, and got it down to ln|1/(1-sind)| so yes it diverges

Re: 2012 HSC MX2 Marathon This does seem true, but e ing both Sides then applying l hopitals rule Will show this is actually convergent As carrot said

Re: 2012 HSC MX2 Marathon f(x) represents the y values and 1/f(x) means 1 over every y value. So you use limits as f(x) appraoches 0 and infinity and note when f(x) equals 0 there becomes vertical asymptotes. Also the turning points on f(x) stay the same, only the y value changes and the...

Re: 2012 HSC MX2 Marathon derivative is taken to apply double root method for solving to find x value of point of contact.

Re: 2012 HSC MX2 Marathon Spirals question is flawed for the m condition The way it is worded

Re: 2012 HSC MX2 Marathon The equation of the ellipse is the top semi-ellipse Covering the first two quadrants. Now when you sub Y=mx+c into semi ellipse and form a quadratic in x The y int of the tangent must be positive otherwise It can't be a tangent to the top semi ellipse so c>0, now As...

Re: 2012 HSC MX2 Marathon The skill to using the reverse chain rule is to be very good at recognising powers of functions and their derivatives.

Re: 2012 HSC MX2 Marathon The reverse power rule is: \int f'(x)[f(x)]^{n} dx = \frac{f(x)^{n+1}}{n+1} + C

Re: 2012 HSC MX2 Marathon It was done using the reverse power Rule, the best tool in 4u integration, Substitution is hardly ever needed with It

Re: 2012 HSC MX2 Marathon pretty sure we will encounter this in complex analysis

Re: 2012 HSC MX1 Marathon This is what Drongoski's diagram would look like:

Re: 2012 HSC MX2 Marathon yes for ii) you first have to prove it is a rhombus, which can be done by taking the arg of the info given which shows the angle between the two diagonals is 90 hence it is a rhombus

IMO it is a true skill in maths to be able to just take one side (LHS or RHS) and turn it into the other using allowed manipulations and assumptions in the case of induction. As for how your school marks it well looking back at previous tests in my school, before tests are done solutions are...

Re: 2012 HSC MX2 Marathon ive seen that question in some 2011 trial, can't remember which though, fairly easy. I'll outline the method so you can try it yourself: (i) way obvious, if you cant do that by now 4u isn't for you. (ii) using the info they gave you for z1 and z2 make z1 the subject...