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November 08, 2025, 03:25:38 pm

Author Topic: VCE Methods Question Thread!  (Read 5782936 times)  Share 

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Lasercookie

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Re: VCE Methods Question Thread!
« Reply #840 on: August 22, 2012, 06:47:07 pm »
+2
1. When it comes to the constant 'c' in Integrated functions, if we had to divide both sides by two or something like that, would the 'c' remain as 'c' (which my teacher says is the answer) or is it, 2c?
So what does that constant 'c' represent? We add in that constant of integration to represent the multitude of different functions that have that same derivative (the family of curves).

Let's take the example of (chosen an example like this for a particular reason which we'll see later)











But is just a constant, so we can reduce the answer to the indefinite integral to:



Elaborating a bit from that, if we took the function:
and differentiated it, we would get

(using chain rule)

To summarise:

If we had , we'd get the same derivative. The same goes for . Or if c was equal to 123, we get that same derivative for  and

2. √(x^2 ) does that equal, x or |x|?

over all real numbers, but for positive real numbers, than is true.

We can just sub in two numbers to see this I guess,
and and

and and

generalkorn12

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Re: VCE Methods Question Thread!
« Reply #841 on: August 23, 2012, 05:28:54 pm »
0
I see now! Thanks Laseredd.

There's just another thing that's been bugging me, in Transition Matrices, to look for the Steady State Matrix, how do we know when to use
T^n x So (where n is a really big number) or the Formula b/a+b?

Also, this is something my teacher couldn't explain, but if we were integrating and the 'with respect to' was changed from 'dx' to 'd(4x)', what does that exactly entail?
« Last Edit: August 23, 2012, 05:41:21 pm by generalkorn12 »

TrueTears

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Re: VCE Methods Question Thread!
« Reply #842 on: August 23, 2012, 07:17:24 pm »
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Also, this is something my teacher couldn't explain, but if we were integrating and the 'with respect to' was changed from 'dx' to 'd(4x)', what does that exactly entail?
As far as I know, that notation isn't defined.
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generalkorn12

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Re: VCE Methods Question Thread!
« Reply #843 on: August 23, 2012, 07:37:59 pm »
0
Here's the question, in question  ;)

Jenny_2108

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Re: VCE Methods Question Thread!
« Reply #844 on: August 23, 2012, 07:52:07 pm »
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Here's the question, in question  ;)

let 4x=u then antidiff

kensan

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Re: VCE Methods Question Thread!
« Reply #845 on: August 23, 2012, 08:13:08 pm »
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Probability question here, not sure what method to use.
Ben selects a card at random from a standard 52 deck of cards. He replaces it and randomly selects another, and continues this way. Find the minimum number of cards he needs to select so that it is more likely than not that he will select at least one ace.
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brightsky

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Re: VCE Methods Question Thread!
« Reply #846 on: August 23, 2012, 09:51:05 pm »
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pr of getting at least one ace from n draws = 1-ncr(n,0)*(1/52)^0*(51/52)^n
pr of not getting at least one ace from n draws = ncr(n,0*(1/52)^0*(51/52)^n
we require min n such that 1-ncr(n,0)*(1/52)^0*(51/52)^n > ncr(n,0*(1/52)^0*(51/52)^n
plug that into the calc and see wha it gives you. use nsolve if solve does not yield any meaningful solutions.
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TrueTears

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Re: VCE Methods Question Thread!
« Reply #847 on: August 23, 2012, 10:57:07 pm »
+1
Here's the question, in question  ;)
Either very sloppy notation or a typo lol, pretty sure you should just treat it as dx
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generalkorn12

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Re: VCE Methods Question Thread!
« Reply #848 on: August 29, 2012, 04:37:10 pm »
0
Another thing, for composite functions, is there a quick method to determine the domain of f(g(x)) quickly, without having to draw tables of each individual function and then determine what parts to restrict?

My teacher indicated to me that there's really no short-cut for composite functions, and I'm noticing on a lot of the practice exams i spend about 5 minutes determining the domain of f(g(x)) for a multi-choice!

WhoTookMyUsername

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Re: VCE Methods Question Thread!
« Reply #849 on: August 29, 2012, 04:44:40 pm »
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Here's the question, in question  ;)
Either very sloppy notation or a typo lol, pretty sure you should just treat it as dx
so is it just "technically" incorrect?
cos if you go let u = 4x
... it works ?

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Re: VCE Methods Question Thread!
« Reply #850 on: September 02, 2012, 10:59:51 am »
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How come when i type -pi/6= 2pi-(pi/6) on the CAS it says false? I know when i calculate the numerical value its -0.5236=5.7596 which is why it said false (stupid me, i just answered my first question...) but what im so confused about it that the symmetry properties say that sin(2pi-theta)=-sin(theta) , so the property says they should equal but their numerical value does not. They are both in the 4th quadrant but they aren't the same???

This issue came up when i was trying to solve for a general solution question and wasn't sure whether to start of with -pi/6 or 11pi/6.

Any help please???
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paulsterio

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Re: VCE Methods Question Thread!
« Reply #851 on: September 02, 2012, 11:04:45 am »
+1
The value of the sine is equal, but not the actual angle themselves.

i.e. sin(11pi/6) = -sin(pi/6), but 11pi/6 does not equal -pi/6

The sine function is a many to one function.

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Re: VCE Methods Question Thread!
« Reply #852 on: September 03, 2012, 08:07:28 pm »
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Gaah, these probability questions keep tripping me up:

I'm sure someone else knows how to solve this! Thanks in advance :D
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Jenny_2108

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Re: VCE Methods Question Thread!
« Reply #853 on: September 03, 2012, 08:25:41 pm »
+1
The sum value of two card can be: 10,15,20

You calculate their probabilities

Pr(sum $10)=

Pr(sum $15)=

Pr(sume $20)=



The player should pay $14

barydos

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Re: VCE Methods Question Thread!
« Reply #854 on: September 03, 2012, 08:35:00 pm »
0
The sum value of two card can be: 10,15,20

You calculate their probabilities

Pr(sum $10)=

Pr(sum $15)=

Pr(sume $20)=



The player should pay $14


Thank you very much! Much appreciated :)
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