VCE Methods Question Thread!

[~T]

question 3a, 2008 VCAA exam 2, when it asks for the time, why do they round up? Shouldn't they round the time down? Cuz if its more than the exact time, he would die? thanks

yes, they should. I think I was debating this with someone last time, one of the inconsistencies of VCAA...

Yeah, that was me. I have come to a kind of conclusion in that it's the specific wording. Because it says "find the answer" first [thus an exact solution which was 191.67 or whatever] and THEN it states "correct to the nearest minute", we must round our previous answer to the nearest minute. Thus, upwards.

If it had have stated "find the maximum integer number of minutes" or something similar, then we would round down. I'm still not sure if I agree with VCAA, but the above is my interpretation of why they chose to round as they did.

Perhaps, if there is something similar this year, qualify your answer with something along the lines of:

191.67 minutes. To "the nearest minute" this value is 192 minutes. However, should he wish to survive he needs to take it before 191.67 minutes.




EDIT: Not worth a new post, but I just realised I've been using the incorrect normal distribution notation of instead of for the entire past year.
'Tis a bit unsettling two days before the exam

[Sanguinne]

can someone explain in detail how to do q4f in vcaa 2010 exam 2

[Lejn]

Okay so as previously known, the x coords of the stationary points are:
x=ab+1/4a or 1/a

Since we are told the x coordinates of one of the stationary points is 1, then we know ab+1/4a or 1/a = 1. We can choose either one, as the other coordinate will remain the same as well. For the sake of ease, we will equate 1/a and 1
so 1/a=1 therefore a=1

The other x coordinate of the stationary point is the subject, p, which now must = ab+1/4a, since we already used the other one.
Subbing in a=1, we have p=b+1/4 (1)

We also know that the stationary point is at (p,p). Since this is the case, f(p)=p (2) (remember that a in this circumstance = 1).

We now have two equations in terms of p and b [(1) and (2)], and we can solve simultaneously.

[Sanguinne]

thanks lejn i understand

[Zealous]

He receives "calls" on each day. Is this in relation to it being conditional, or how we interpret how many calls a day he can receive?So whenever a question usually has a statement that is relating to the first part of the question, we should try and see if its a conditional question or not?

I looked at it something like this.
Because we have been told he received calls on both days, we therefore eliminate the probability that he gets zero calls on both days (if he received calls each day, how could x=0?), and we can treat this as a condition, where x>0 - therefore conditional probability. <- I might have just stated the obvious though haha

I stuffed this up though, I forgot to square 0.8 =(

EDIT: Not worth a new post, but I just realised I've been using the incorrect normal distribution notation of instead of for the entire past year.
'Tis a bit unsettling two days before the exam

Ahaha! I checked that up from multiple sources to make sure if standard deviation was squared or not in the notation before my probability SAC. That worried me also =p

[Alwin]

can someone explain in detail how to do q4f in vcaa 2010 exam 2

Okay, so assuming that you got parts a-e, this is how I would approach part f:

From part b, we know that stationary points at:


Now, one of these solutions has to be x=1, and the other is x=p yeah

So either:




But if we look closely, we can't actually solve (easily) the second equation for a, b, p coz we can't pin anything down (they'll be one free variable). So, we chose the first option.




Then put this value back into the equation f(p)=p and solve

Good luck with it!

@Tim...blahhh I agree, if you explain your rounding in an ambiguous situation I think they'll accept it. It might have happened once where they accepted two values, but not sure.

EDIT: beaten.

[Zealous]

Okay, so assuming that you got parts a-e, this is how I would approach part f:
.........

Okay so as previously known, the x coords of the stationary points are:
x=ab+1/4a or 1/a
..............

^^^ That method is so much better than what I tried to do when I did this paper a few weeks ago (although I still got the answer... I completely started from scratch as if I started q4 again and used so much space working it out, it wasn't aesthetically pleasing.

Sometimes I completely forget that VCAA is actually nice in some cases, and previous answers often help and lead into harder question.

Perhaps, if there is something similar this year, qualify your answer with something along the lines of:

191.67 minutes. To "the nearest minute" this value is 192 minutes. However, should he wish to survive he needs to take it before 191.67 minutes.

Hmm I may do that... eh I was a little annoyed when I had to mark that wrong on my practice paper. So used to VCAA actually rounding in ways that make sense (especially in further with business math and sequences).

[ahat]

should he wish to survive

Haha, I love how cold and precise this analysis is, "should he wish to survive" hahahaha 

[sasa]

Does anyone here have access to kilbaha papers? I can't be bothered writing out the stem 0_0

[SocialRhubarb]

I've got 2012 in front of me and 2013 ... somewhere ...

[sasa]

I'm correcting stuff from 2013...it doesn't make sense

[Damoz.G]

I've got the 2010, 2011 and 2012 Kilbaha Trials, but not the 2013 ones. Sorry.

[SocialRhubarb]

Okay, I've got the Kilbaha 2013 trial methods exams here.

[sasa]

Have you done it and checked it?

[SocialRhubarb]

I've done it, I haven't marked it.