VCE Methods Question Thread!

[keltingmeith]

Is it bad if I do the du/dx substituion method that we learn in 3&4 Specialist in the methods exam when antidifferentiating?

It's not bad - you shouldn't need to, but it's okay if you do that. If you're using mathematically correct techniques, you shouldn't lose marks - but it's highly recommended you stick to the techniques you learn in class.

[lzxnl]

Thanks EulerFan101
So this means that just having √16 equals to just 4 right?, not 4 AND -4

By definition, 'the' square root of y means the positive solution to x^2 = y

It's not bad - you shouldn't need to, but it's okay if you do that. If you're using mathematically correct techniques, you shouldn't lose marks - but it's highly recommended you stick to the techniques you learn in class.

To add to that, if you use a technique not in the course and you get the question wrong...bye bye marks...

[LiquidPaperz]

One student doing their homework can be modelled as Bernoulli. Since they are independent, we can then model this as Binomial with n=2 and p=0.9.

So, we work:



So, the probability of only one student doing their homework is 0.09.

You could also model this in other ways, this is just a nice result of independence.

Hey Euler, could you show me in Year 11 probability terms?

thanks

[GeniDoi]

Hey Euler, could you show me in Year 11 probability terms?

thanks

Bernoulli is essentially a way to quickly calculate lots of different branches in a of a very long tree diagram. This question doesn't require Bernoulli as its only 2 branches deep but if the question had, say, 10 students and asked for the probability of exactly 6 completing their homework, it would be completely impractical to draw out a tree diagram and add up the probabilities of all the branches that end up with 6 students.

Anyway, you want only one student to do their homework so:

Case 1: Student 1 does hw, student 2 doesn't:
0.9 * 0.1 = 0.09
Case 2: Student 1 doesn't do hw, student 2 does:
0.1 * 0.9 = 0.09

Not these two probabilities are not the same scenarios and must be added together:

0.09 + 0.09 = 0.18 as EulerFan101 calculated.

[keltingmeith]

Hey Euler, could you show me in Year 11 probability terms?

thanks

Alright - since they're independent,



Since each of the probability must add up to 0.9, then
So, the probability of one of them doing their homework (not both) is:



EDIT: Beaten, but since I answered differently, thought I'd keep my stuff there.

[Escobar]

Do we need to learn how to differentiate trigonometric functions to a higher degree?
eg
cos^3(2x)?
this isn't in the maths quest textbook but it is in the cambridge one.
thanks
I came across a question that needed it in MAV 2011 exam 1 (attached)

[keltingmeith]

Do we need to learn how to differentiate trigonometric functions to a higher degree?
eg
cos^3(2x)?
this isn't in the maths quest textbook but it is in the cambridge one.
thanks
I came across a question that needed it in MAV 2011 exam 1 (attached)

You know how to do it - just apply the chain rule. Remember:

[psyxwar]

By definition, 'the' square root of y means the positive solution to x^2 = y

To add to that, if you use a technique not in the course and you get the question wrong...bye bye marks...

Isn't this technically the principle square root?

In any case I think I'm gonna stick to u-sub for those transformation integral questions, so much easier.

[Yacoubb]

Do we need to know how to manually solve Binomial Distribution qs asking for probability?

[GeniDoi]

Do we need to learn how to differentiate trigonometric functions to a higher degree?
eg
cos^3(2x)?
this isn't in the maths quest textbook but it is in the cambridge one.
thanks
I came across a question that needed it in MAV 2011 exam 1 (attached)

This is not a very easy question to differentiate manually. Highly unlikely to be asked on a Cas-free and if it were it would be 3 if not 4 marks.

Nonetheless:

[speedy]

In any case I think I'm gonna stick to u-sub for those transformation integral questions, so much easier.

How would you do that? I literally just got stuck on one for multi-choice and would love an easier method. (I do spesh btw so I know how to u-sub)

[GeniDoi]

Do we need to know how to manually solve Binomial Distribution qs asking for probability?

Yes, there was a tricky question at the very end of one of the earlier VCAA exam 2's (2006 or 2007 I think) that involved having a probability of "p" and required you to expand a binomial for p, which ended up giving you a cubic function for f(p) = 0 that you would need to solve.

You couldn't do that in your cas, as binomPdf/binomCdf doesn't accept algebraic arguments.

[Yacoubb]

Yes, there was a tricky question at the very end of one of the earlier VCAA exam 2's (2006 or 2007 I think) that involved having a probability of "p" and required you to expand a binomial for p, which ended up giving you a cubic function for f(p) = 0 that you would need to solve.

You couldn't do that in your cas, as binomPdf/binomCdf doesn't accept algebraic arguments.

Could you perhaps give me an example/solution to solving it because my teacher hasn't taught this method!

[psyxwar]

How would you do that? I literally just got stuck on one for multi-choice and would love an easier method. (I do spesh btw so I know how to u-sub)

if you had f(2x) for example, you let u=2x and work from there.

This is not a very easy question to differentiate manually. Highly unlikely to be asked on a Cas-free and if it were it would be 3 if not 4 marks.

Nonetheless:

(Image removed from quote.)

this really isn't particularly difficult to do, it's just a bit lengthier than usual.

[psyxwar]

Could you perhaps give me an example/solution to solving it because my teacher hasn't taught this method!

all you need to know is that Pr(X=x) = nCr(n,x) * (p)^x * (1-p)^(n-x), where nCr(n,x) = n!/(n-r)!r!

There isn't much else to it