Search results

Just conclude every essay with "therefore there is no god".

Implicit differentiation is definitely allowable. From the official marker's comments to Q3c of the 2004 HSC: "Some candidates successfully used implicit differentiation methods".

Well, I'm a teacher of more than 25 years, and although I have never marked the HSC myself, I regularly talk to teachers who have. Also, I didn't say it applied to every question, probably even a minority of questions. The markers generally decide beforehand what will be accepted and what will...

Re: HSC 2015 3U Marathon Solution: Edit: I'm struggling to come up with an alternate combinatorics-based solution for this question. Can anyone come up with one?

Then assuming you also offered a non-crossed-out solution, that is non-standard marking.

How many HSC markers do you know? They WILL often give full marks for a calculation error (not just a transcription error), provided the provided it is ONLY a calculation error, and provided the error doesn't make the question easier.

Re: HSC 2015 3U Marathon The proof comes out much more nicely when you use an alternative formula for the derivative:

Pick one person (say 'person A') to be in one group. Each of the other 7 people have 2 choices - they are in A's group or they are in the other group. That's 2^7 possibilities. Then subtract the case where they all end up in A's group.

Re: HSC 2015 3U Marathon How about: Assuming k! > 2^k (and k>=4) (k+1)! = (k+1).k! > 2.2^k = 2^(k+1) [BTW, "n=k+1 is true" doesn't mean too much. Not that the examiners are going to care, but it should be "true for n=k+1 (when true for n=k)" ]

I agree with your answer. But the working is much simpler: 7C3 / (2^7 - 1)

Re: HSC 2015 4U Marathon Those 'rules' .... they are all those things you've derived in doing all those Conics questions.

Re: HSC 2015 4U Marathon I couldn't find what this was referring to. Would you explain what you mean in the non-probabilistic sense, perhaps with an example or two.

If the sum was 3^1 + 3^2 + 3^3 + ... + 3^9 + 3^10 + 3^11 + ... + 3^20, then there would be 20 terms (powers 1 to 20). But if the sum was 3^10 + 3^11 + ... + 3^20, the first 9 terms are missing, so there are (20-9) terms. If the sum was 3^1 + 3^2 + 3^3 + ... + 3^(n-1) + 3^n + 3^(n+1) + ... +...

A good way to learn about functions like this is to download Geogebra or some other graphing program and just experiment. Try sketching y=x^2, y=x^3, y=x^4, etc on the same axes. Then start again and sketch y=x^7+1, y=x^7 - 2, y=(x+1)^7, y=(x-2)^7. It will sink in much more easily if you...

There is no perms and combs required here. For the first question, draw up a grid as you would for rolling 2 dice. But instead of labelling the rows and columns 1 to 6, label them with the outcomes given in Ambility's post. That is, a 5 by 5 grid. The outcomes across the top are Max's; the...

Re: HSC 2015 4U Marathon Imaginary roots can provide only so much enjoyment.

Re: HSC 2015 4U Marathon I can think of good reasons. But if those reasons are correct, I sure hope he goes to a co-ed school.

Re: HSC 2015 4U Marathon Not important?? Tell your teacher they are an idiot.

Re: HSC 2015 4U Marathon - Advanced Level Have you at the solutions on BOSTES yet?

Re: HSC 2015 4U Marathon You haven't done 3U Parameters yet?