VCE Methods Question Thread!

[IndefatigableLover]

the equation of the graph is y=x^2 -1
  (m,n)
n=m^2 -1
so (n-1)^2 = ( m^2 -1 -1)^2 =(m^2 -2)^2

URGH it was just right in front of me (need to make sure I bring a highlighter into the exam now )
Thanks for that though RKTR

[chocolatechipcookies]

so basically I was asked to differentiate this in terms of X

but it's just log no log e ??

[keltingmeith]

so basically I was asked to differentiate this in terms of X

but it's just log no log e ??

In cases like this, normally you assume it's base 10 - but I think they mean base e.

Don't worry, VCAA will always specify the number base.

[speedy]

I think he's referring to Question 4c in 2012? Seems like it's that question imo..

Yeah that's what I thought - I understood it though. Maybe because I've done something similar to it before :s

[Brunette15]

How does everyone usually approach these kind of questions? Does anybody have any tips as to how to know you have found all the values within the required domain. I always end up finding extra unnecessary values to make sure but this is time consuming...

[IndefatigableLover]

Yeah that's what I thought - I understood it though. Maybe because I've done something similar to it before :s

Haha lucky, lost 2 marks on that question since I forgot to restrict the sample space (but I guess I'm lucky in that majority of my mistakes happen to be 'knowledge errors' rather than silly mistakes :S)

[speedy]

Haha lucky, lost 2 marks on that question since I forgot to restrict the sample space (but I guess I'm lucky in that majority of my mistakes happen to be 'knowledge errors' rather than silly mistakes :S)

Ahh... I make so many silly mistakes LOL. (along with knowledge errors aha)

[IndefatigableLover]

How does everyone usually approach these kind of questions? Does anybody have any tips as to how to know you have found all the values within the required domain. I always end up finding extra unnecessary values to make sure but this is time consuming...

I would have done it the same way as you Brunette15 except I would have kept my new domain as:

rather than simplify it down to 'x'.

And also I would write out the quadrants I'll be taking my values from (like if it's positive, then I'll write Q1 & Q2 to know I'll be getting those values from the unit circle). Now you know where to get your values from (just keep practising and all the unit circle values will get to you)!

[speedy]

This should be simple, but could someone explain question 4e) from 2012 VCAA Exam 2?

I let the dV/dt that was found in part cii) equal to 0.2 and then subbed in height and solved for time, but that is wrong.

They seem to be just dividing volume by dV/dt, to find time. Actually, now that I think about it, that makes sense generally with rates.

If you have a cup with volume 2L, and you're pouring in water at 0.2 L/s; then to find how long it takes you just go 2/0.2...

So now I guess my question is, why didn't my original way work?

[skeletalclown]

Hi, I'm having a fair bit of trouble understanding VCAA 2013 Exam 2 question 1fi and 1fii (Trigg the gardener and the dilations), and would really really appreciate an explanation, thanks guys.

[kawfee]

^^ Yeah me too

Posted the q up today, just in case it was missed ,

thank you!

Methods [3/4] Question Thread!

[rosieee]

Could someone please help me with question 2c of the 2012 VCAA methods exam 2?!
I've looked at worked solutions and the examiners report but I still don't understand how to manipulate the answer into the form that they're looking for!
Help would be greatly appreciated!!

[kawfee]

Another Q,

exam 2012 vcaa exam 1

question 10 ii

i dont understand the working out :/

can we treat inequalities the same way as we do if we have '=' sign? hope this makes sense, but regardless, please explain

[keltingmeith]

can we treat inequalities the same way as we do if we have '=' sign? hope this makes sense, but regardless, please explain

Yes, that's how you're meant to do it - but, if you multiply/divide by a negative number, you need to swap it. So, a greater than becomes a less than and vice-versa.

[Yacoubb]

Kudos to all those who did Methods last year! That exam was just....horrendous! Exam 2...I think I would have broken down. Seriously.