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March 29, 2024, 07:55:53 am

Author Topic: VCE Methods Question Thread!  (Read 4802774 times)  Share 

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a weaponized ikea chair

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Re: VCE Methods Question Thread!
« Reply #19095 on: February 25, 2021, 10:07:27 pm »
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where's everyone up to in class?

We're in chapter 4.

pans

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Re: VCE Methods Question Thread!
« Reply #19096 on: February 25, 2021, 10:10:39 pm »
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where's everyone up to in class?

We're in chapter 4.

same. I know a lot of people r doing vectors tho

Corey King

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Re: VCE Methods Question Thread!
« Reply #19097 on: February 26, 2021, 11:12:06 am »
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We are finishing chapter 3 still.

Ruchir

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Re: VCE Methods Question Thread!
« Reply #19098 on: February 28, 2021, 08:08:20 pm »
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Hello, I need help with this question
Answer for 6b is 430000, can anyone explain me how and why that is correct?

parieeelol

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Re: VCE Methods Question Thread!
« Reply #19099 on: March 06, 2021, 09:52:50 pm »
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Just a really standard question about set notation/range -

For this picture, my answer for range was (0, ∞) or just R+
Could anyone explain why the book's answer uses the union symbol? Does this mean that the range is only either positive or zero?

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19100 on: March 06, 2021, 10:28:15 pm »
+2
The set union means that it includes both sets. Any two sets \(A, B\) defined by the relation \(A \cup B\) will include all items in set \(A\), set \(B\) and items in both sets. In our case, it means that the range of the function includes positive reals and zero (since no real number lies in both the positive reals and the set including only zero). It's effectively the same as the range \([0, \infty)\) but not the same as \(\mathbb{R}^+\) as the latter does not include zero (but the graph clearly does).

Hope this clears things up :)
« Last Edit: March 08, 2021, 11:38:47 pm by fun_jirachi »
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Ruchir

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Re: VCE Methods Question Thread!
« Reply #19101 on: March 07, 2021, 09:28:19 am »
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Can anyone explain me Q7a
Why is the answer 0<k<1/4
And not just k<1/4

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19102 on: March 07, 2021, 10:28:15 am »
+1
Notice that \(\sqrt{x}-1\) is a one-to-one function ie. any horizontal line will meet the curve only once. Try drawing it out on desmos or geogebra (as well as lines with negative gradient, which you are including by using k < 1/4 as opposed to 0 < k < 1/4).
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Re: VCE Methods Question Thread!
« Reply #19103 on: March 07, 2021, 12:40:13 pm »
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where's everyone up to in class?

We're in chapter 4.
We're still on chapter 3I, probably because we have a pretty weak maths cohort overall
2021 - Further
2022 - Eng | Physics | Methods | Specialist

Ruchir

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Re: VCE Methods Question Thread!
« Reply #19104 on: March 07, 2021, 04:59:30 pm »
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Notice that \(\sqrt{x}-1\) is a one-to-one function ie. any horizontal line will meet the curve only once. Try drawing it out on desmos or geogebra (as well as lines with negative gradient, which you are including by using k < 1/4 as opposed to 0 < k < 1/4).

Thanks a lot bro I understand now.
But can you find this out using algebra or do you just have to graph it and check
Cause if you do it using algebra you get my answer

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19105 on: March 07, 2021, 06:37:09 pm »
+2
I assume you did something like this?


We want two solutions, so the discriminant \(\Delta = b^2 - 4ac > 0\) - so we now have:


The thing is by doing all this, we're actually also adding a bunch of solutions in that first step: where the line \(kx\) cuts the negative root \(-\sqrt{x}-1\). Squaring both sides is dangerous in this way (in the same way taking a particular root removes some solutions). If it helps you should definitely do it graph it. Intuitively you should be thinking about this already (especially when you add 'solutions' like this) ie. are there any values that don't work in my proposed solution?

There are also other ways to verify this. Subbing certain values of \(k\) into \(k^2x^2+ (2k-1)x +1 = 0\) eg. -1 will yield you the equation \(x^2 -3x + 1 = 0\). In this example, there are two solutions for \(x\), but you'll find that subbing them into \(kx\) and \(\sqrt{x}-1\) will give you three different points (one in common, one unique to each equation).




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Corey King

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Re: VCE Methods Question Thread!
« Reply #19106 on: March 08, 2021, 09:27:03 pm »
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The set union means that it includes both sets. Any two sets \(A, B\) defined by the relation \(A \cup B\) will include all items in set \(A\), set \(B\) and items in both sets. In our case, it means that the range of the function includes positive reals and zero (since no real number lies in both the positive reals and the set including only zero). It's effectively the same as the range \((0, \infty)\) but not the same as \(\mathbb{R}^+\) as the latter does not include zero (but the graph clearly does).

Hope this clears things up :)

Hey Jirachi,
Since there is no dot on the 0 for the graph, does this not mean that 0 is not included but the graph approaches zero?

Therefor it would be just R+?

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19107 on: March 08, 2021, 11:42:50 pm »
+1
Decent question!

A dot usually implies the inclusion of a point not already on the curve - note that for \(f(x) = \sqrt{2(x+3)}, f(-3)\) is well defined ie. its corresponding y-value is clearly included in the range. It's not seen on most graph sketches but is more commonly seen when denoting closed and open intervals (which is probably where this confusion arises), or when marking out discontinuities on a graph. It's clear enough that the graph does include zero and that the endpoint is well-defined so there's no need for a dot.

That being said I did make a mistake in my post (re: 'the same as the range \((0, \infty)\)') which has now been changed to the correct 'the same as the range \([0, \infty)\)'.
« Last Edit: March 09, 2021, 01:13:50 pm by fun_jirachi »
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jasmine24

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Re: VCE Methods Question Thread!
« Reply #19108 on: March 13, 2021, 03:43:23 pm »
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I know this is a really stupid question but does anyone know how to simplify the first equation to get the second equation?😅

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19109 on: March 13, 2021, 04:38:11 pm »
+3
Hi :D

The answer is wrong, so it makes total sense that this doesn't make sense! What's in the spoiler is correct :D

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Hope this helps!
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