VCE Methods Question Thread!

[IndefatigableLover]

I see what you mean but I sketched both graphs and a translation of 1 unit left without a reflection already brings in to the same shape as the other graph

Well remember that the question is asking for a sequence of two transformations so if you only did one (which is still the same graph with that translation you stated), then you're still wrong since you haven't answered the question

[psyxwar]

if part of a graph is under the x-axis do we need to put a -ve sign in front of the integral? In one vcaa exam they didn't so it confuses me (VCAA 2007 Q3bii)

Yeah you have to, but you can also absorb it into the integral because of the following property.

[shadows]

I see what you mean but I sketched both graphs and a translation of 1 unit left without a reflection already brings in to the same shape as the other graph

Not quite sure what you're trying to ask?
But in this question, you are reflecting the already translated  function. Try think about it visually. If there was a mirror in the middle (and you were looking at it from the top view) that reflected everything for the left. The more left it is pushed, once reflected, it will be more right.

[keltingmeith]

Thanks!! You guys are really helping But I guess I'll just have to memorise the simple triangles and basics such as sin(pi/2)=1 etc...
Any other essentials that I should rote for exam 1? 

Quadratic equation.

[faredcarsking123]

Well remember that the question is asking for a sequence of two transformations so if you only did one (which is still the same graph with that translation you stated), then you're still wrong since you haven't answered the question

aahh yes

[IndefatigableLover]

Yeah you have to, but you can also absorb it into the integral because of the following property.

Alternatively you could just put a modulus sign over the entire integral too though I'd recommend following psyxwar's technique!

[faredcarsking123]

What about 5fii in 2010 exam 2?
I've got no clue about that one.

Thanks in advance again and thanks for all the help!

[Reus]

For exam 1 2012 q2:

I got instead of
This came from ... where did I go wrong?

[shadows]

Alternatively you could just put a modulus sign over the entire integral too though I'd recommend following psyxwar's technique!

Definitely recommend just taking the integral of the modded function (not the mod of the integral) in Exam 2 when you have you're cas. It's a lot faster method of finding the area (not simply just evaluating the integral) under the curve.

Split it into parts in Exam 1.  Modding the function in exam 1, and having to solve it by hand takes way too long.

[keltingmeith]

For exam 1 2012 q2:

I got instead of
This came from ... where did I go wrong?

Man is that some messy LaTeX... Hopefully you can use my edits to help yourself out.

Also, you flipped the -4 up for literally no reason - it doesn't have a negative power, so it stays on the bottom.

[Valyria]

I actually answered question a few pages back Valyria!

Aha, I replied soon after;

I see where you are coming from but I'm still not sure to why we halve 8/3. I understand he has to make a smooth trip without any interruptions but why would this cause 8/3 to be divided by 2?

[Reus]

Man is that some messy LaTeX... Hopefully you can use my edits to help yourself out.

Also, you flipped the -4 up for literally no reason - it doesn't have a negative power, so it stays on the bottom.

Hahaha kinda new to it! And I see makes sense now thanks.

[Rishi97]

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is:
A) 8
B) 9
C) 10
D) 11
E) 12

probabaility is gross..any help appreciated :p

[Jason12]

how to approach questions where it says things like f(u)f(v) = f(u) + f(v) - f(u+v)? I usually sub u and v in and make it equal to each other and the calc says its true but that doesn't show anything.

[vcestudent123]

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is:
A) 8
B) 9
C) 10
D) 11
E) 12

probabaility is gross..any help appreciated :p

Since to get heads is 0.5
You do solve ((0.5)^n = 0.0005)

= 10.965...
so = 11, D

You could also use logs and do it manually, but I think that is more a calculator question anyway.