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March 28, 2024, 07:58:47 pm

Author Topic: VCE Methods Question Thread!  (Read 4802182 times)  Share 

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S_R_K

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Re: VCE Methods Question Thread!
« Reply #17880 on: April 22, 2019, 10:31:59 am »
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How do you get from \(\left(x-2\right)^2+\left(y+3\right)^2=27\) to the diameter?
root 27 times 2?

I'm not quite sure if you're going off track a bit here, so it might be worth clarifying that a diameter of a circle refers to a line or line segment that passes through the centre of a circle and intersects the circle twice; the length of a diameter is the length of this line segment.

The question asks you for the equation of the diameter, so you are not required to find its length. (And it's also worth noting that knowing the length of the diameter tells you nothing about the equation of a diameter passing through the origin).

Getting back to how to answer the question: if you know that the diameter passes through the centre of the circle and passes through the origin, then you have two points on the line. How do you find the equation of a line given two points?

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17881 on: April 22, 2019, 10:58:44 am »
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I'm not quite sure if you're going off track a bit here, so it might be worth clarifying that a diameter of a circle refers to a line or line segment that passes through the centre of a circle and intersects the circle twice; the length of a diameter is the length of this line segment.

The question asks you for the equation of the diameter, so you are not required to find its length. (And it's also worth noting that knowing the length of the diameter tells you nothing about the equation of a diameter passing through the origin).

Getting back to how to answer the question: if you know that the diameter passes through the centre of the circle and passes through the origin, then you have two points on the line. How do you find the equation of a line given two points?
Well that's the reason why I asked the question. I have no idea, that's why I looking for a step by step answer that will provide me with how to do it.

schoolstudent115

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Re: VCE Methods Question Thread!
« Reply #17882 on: April 22, 2019, 11:14:33 am »
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Well that's the reason why I asked the question. I have no idea, that's why I looking for a step by step answer that will provide me with how to do it.

So the equation for the circle is:

So the centre is at .
The origin is of course

You are told that the line must pass through the origin, and be the diameter of the circle. The diameter of any circle passes through the centre of the circle. (Imagine the radius being half a diameter, and there you go, it mast pass the centre.). So now we know the coordinates of 2 of the points this line passes through:

The line is in the form , since it passes through the origin: , what is m?

Use the coordinates for the centre:

Therefore
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17883 on: April 22, 2019, 11:17:29 am »
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So the equation for the circle is:

So the centre is at .
The origin is of course

You are told that the line must pass through the origin, and be the diameter of the circle. The diameter of any circle passes through the centre of the circle. (Imagine the radius being half a diameter, and there you go, it mast pass the centre.). So now we know the coordinates of 2 of the points this line passes through:

The line is in the form , since it passes through the origin: , what is m?

Use the coordinates for the centre:

Therefore

Ohhhhhhh l seee
Thanks a lot!!!

Omarrr_2163

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Re: VCE Methods Question Thread!
« Reply #17884 on: April 22, 2019, 02:10:23 pm »
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I need help with 2017 Exam 1 Q7) Bii) and C)
2018: Business Management {42} ~ Further Mathematics {41}
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17885 on: April 22, 2019, 02:22:48 pm »
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I need help with 2017 Exam 1 Q7) Bii) and C)

Just putting the answers to the previous parts for clarity.

a.\(\quad\ \ \text{range}(f)=[1,\,\infty)\)

b.i.\(\quad c=-3\)

b.ii.\(\ \ \ \mathbf{Hint:}\ \ (-\infty,\, -3]\overset{g}{\longrightarrow}[0,\,\infty)\overset{f}{\longrightarrow}[1,\, \infty)\)

c.\(\quad\ \ \mathbf{Hint:}\ \ \mathbb{R}\overset{h}{\longrightarrow}[3,\,\infty)\overset{f}{\longrightarrow}\ ??\)
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DBA-144

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Re: VCE Methods Question Thread!
« Reply #17886 on: April 24, 2019, 03:42:34 pm »
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WHen do we need to use set notation and when do we use interval notation? eg. for domain, range, values for which the derivative is zero/positive/negative?

A

Strictly speaking, what would {x:f(x)>0} = {x: x>2} U {x: x<-1} be? I suspect it is interval notation, but would the union make it a combination of interval and set? Why is this notation wrong?

Can someone please provide an example of what the 'correct' notation is, please? If possible, can you please provide some reasons as to why a specific notation may be incorrect in some cases but not others?

Thanks :D
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17887 on: April 24, 2019, 04:01:28 pm »
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WHen do we need to use set notation and when do we use interval notation? eg. for domain, range, values for which the derivative is zero/positive/negative?

A

Strictly speaking, what would {x:f(x)>0} = {x: x>2} U {x: x<-1} be? I suspect it is interval notation, but would the union make it a combination of interval and set? Why is this notation wrong?

Can someone please provide an example of what the 'correct' notation is, please? If possible, can you please provide some reasons as to why a specific notation may be incorrect in some cases but not others?

Thanks :D

Interval notation is just a short-hand notation for writing subsets of the real numbers. For example, instead of writing  \(\{x\in\mathbb{R}\mid a\leq x<b\}\),  you can equivalently write  \([a,\,b)\).  They mean identically the same things.

You can write whatever notation you want, so long as it is mathematically correct, and is clear to the reader.

All the notation you provided in your post above was correct. Could you perhaps expand on your question about incorrect notation so I can see where the confusion is?
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DBA-144

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Re: VCE Methods Question Thread!
« Reply #17888 on: April 24, 2019, 04:23:36 pm »
0
Interval notation is just a short-hand notation for writing subsets of the real numbers. For example, instead of writing  \(\{x\in\mathbb{R}\mid a\leq x<b\}\),  you can equivalently write  \([a,\,b)\).  They mean identically the same things.

You can write whatever notation you want, so long as it is mathematically correct, and is clear to the reader.

All the notation you provided in your post above was correct. Could you perhaps expand on your question about incorrect notation so I can see where the confusion is?


What I am asking is, regarding the notation I used above for f(x)>0, could you please confirm that is correct? IS this set or itnerval notation?
I was doing some questions and the answers said that set notation was required, but I can't see why we cannot use interval notation(like I believe I have). Could you please provide an example of the 'correct' notation that you would use please?

And why have the answers said that I am wrong in using interval notation? Why is using my notation wrong for this question is the main issue that I have.

Sorry for the confusion. If my question does not make sense, please let me know so I can clarify.

Thanks for your help! :)
PM me for Methods (raw 46) and Chemistry (raw 48) resources (notes, practice SACs, etc.)

I also offer tutoring for these subjects, units 1-4 :)

schoolstudent115

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Re: VCE Methods Question Thread!
« Reply #17889 on: April 24, 2019, 04:57:56 pm »
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I need help with 2017 Exam 1 Q7) Bii) and C)
Here is the step by step explanation to solve the questions:

Spoiler

Given that
b.ii. Find

Notice that the  , hence , as solved in part a

Spoiler
c. Find the range of ,

First, solve for the range of : notice that for all , therefore , hence

Notice that , and the lower bound of the range of h is 3.

Since is an increasing function, the minimum of the range of is simply . Of course as  , it can be concluded that the range of
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17890 on: April 24, 2019, 05:27:52 pm »
0
What I am asking is, regarding the notation I used above for f(x)>0, could you please confirm that is correct? IS this set or itnerval notation?
I was doing some questions and the answers said that set notation was required, but I can't see why we cannot use interval notation(like I believe I have). Could you please provide an example of the 'correct' notation that you would use please?

And why have the answers said that I am wrong in using interval notation? Why is using my notation wrong for this question is the main issue that I have.

Sorry for the confusion. If my question does not make sense, please let me know so I can clarify.

Thanks for your help! :)

The notation  \(\{x:f(x)>0\}\)  is perfectly fine and reads:  "the set of all values \(x\) such that \(f(x)\) is greater than \(0\)".

The notation you provided,  \(\{x:x>2\}\cup \{x:x<-1\}\),  is not 'interval notation'. The corresponding 'interval notation' would be  \((-\infty,\,-1)\cup(2,\,\infty)\).

If a question specifies a required form for your answer, you must adhere to the instructions and do so. Otherwise, use any notation you want. For example, all the following are equivalent:\[(-2,\,-1)\cup [0,\,\infty)\qquad \{x\in\mathbb{R}\mid -2<x<-1\}\cup \{x\in\mathbb{R}\mid x\geq 0\}\qquad \mathbb{R}^+\cup (-2,\, -1)\cup\{0\}\] For this particular set of numbers, I prefer the first one because I believe it's the easiest to understand, but feel free to use whatever you want if the question allows you to.
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JR_StudyEd

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Re: VCE Methods Question Thread!
« Reply #17891 on: April 24, 2019, 07:37:44 pm »
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Any tips for making Methods more fun (or less boring/monotonous/painful) to study? Just in general, not referring to any topic.
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S_R_K

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Re: VCE Methods Question Thread!
« Reply #17892 on: April 25, 2019, 09:54:17 am »
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Any tips for making Methods more fun (or less boring/monotonous/painful) to study? Just in general, not referring to any topic.

Methods can be an extremely dull subject if it is approached as a list of basic skills / knowledge to memorise or rote learn (not suggesting this is what you are doing, of course, it sounds like you want to avoid this). The strengths of the subject lie in the many opportunities for connecting the various concepts (which is often tested in tougher exam questions). So if possible, try to minimise the amount of routine textbook questions you are doing, and prioritise exam qs and qs from good quality commercial papers. The course will feel more rewarding if you are spending your time doing questions that require more application and understanding.

f0od

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Re: VCE Methods Question Thread!
« Reply #17893 on: April 26, 2019, 09:52:30 pm »
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Is it possible to find the y asymptote through the division of ordinates, or is it only possible to find the x asymptote using this method?

Also: quick question: has anyone done anything related to the 'investigation of graphs of the form y = k + kx/x , k > 0? If so, would I be able to ask you a few questions?

Thanks! :)
« Last Edit: April 26, 2019, 10:14:55 pm by f0od »
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17894 on: April 26, 2019, 11:14:58 pm »
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Is it possible to find the y asymptote through the division of ordinates, or is it only possible to find the x asymptote using this method?

This seems like a question for the Spesh Questions Thread. There are no functions in the course that would require you to use 'division of ordinates' to find horizontal asymptotes.

Also: quick question: has anyone done anything related to the 'investigation of graphs of the form y = k + kx/x , k > 0? If so, would I be able to ask you a few questions?
I think you need to include some brackets in your expression. Your inline typeset currently reads:  \(y=k+\dfrac{kx}{x}\),  which can obviously be simplified to just  \(y=2k\), which is a boring horizontal line with a hole at  \(x=0\).  Did you mean:  \(y=\dfrac{k+kx}{x}\) ? In either case, feel free to ask any questions  :)
« Last Edit: April 26, 2019, 11:19:22 pm by AlphaZero »
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