VCE Methods Question Thread!

[fun_jirachi]

I'm not Rui, but I'm pretty sure he's using his beloved GeoGebra.

EDIT: GeoGebra is essentially a graphing calculator + shape drawing thingo ie. good for constructions, circle geometry and whatnot. It's basically Desmos that's slightly tackier (depends on who you ask) and has way more features.

[aspiringantelope]

I'm not Rui, but I'm pretty sure he's using his beloved GeoGebra.

EDIT: GeoGebra is essentially a graphing calculator + shape drawing thingo ie. good for constructions, circle geometry and whatnot. It's basically Desmos that's slightly tackier (depends on who you ask) and has way more features.

Ok thanks so much!!

[WhompingWillow]

Hey guys, I need help with this question.

f : (−∞, 3] → R, f (x) = 3 − x and g : R → R, g(x) = x^2 − 1
a Show that f ◦ g is not defined.
b Define a restriction g∗ of g such that f ◦g∗ is defined and find f ◦g∗.

[Sine]

Hey need help with this question.

f : (−∞, 3] → R, f (x) = 3 − x and g : R → R, g(x) = x^2 − 1
a Show that f ◦ g is not defined.
b Define a restriction g∗ of g such that f ◦g∗ is defined and find f ◦g∗.

what is your current working out or which aspect are you unsure of?

For a I would start by finding f o g and testing out it's domain

[RuiAce]

Hey need help with this question.

f : (−∞, 3] → R, f (x) = 3 − x and g : R → R, g(x) = x^2 − 1
a Show that f ◦ g is not defined.
b Define a restriction g∗ of g such that f ◦g∗ is defined and find f ◦g∗.

___________________________________

So essentially we now need to solve \( -1 \leq x^2-1 \leq 3  \). But that "less than/equal to -1" is completely redundant because \(x^2 -1 \geq -1\) is always true, So really we're just solving \(x^2-1 \leq 3\).
\begin{align*}x^2-1 &\leq 3\\ x^2-4 &\leq 0\\ (x-2)(x+2)&\leq 0 \end{align*}
\[ \text{From a graph, the solution is }-2 \leq x\leq 2 \\ \text{so }g^*: [-2,2] \to \mathbb{R},\quad g^*(x) = x^2-1\text{ should do.}\]
From here, computing \( f (g^*(x))\) should be easy.

Subject to some minor computational errors.

[WhompingWillow]

___________________________________

So essentially we now need to solve \( -1 \leq x^2-1 \leq 3  \). But that "less than/equal to -1" is completely redundant because \(x^2 -1 \geq -1\) is always true, So really we're just solving \(x^2-1 \leq 3\).
\begin{align*}x^2-1 &\leq 3\\ x^2-4 &\leq 0\\ (x-2)(x+2)&\leq 0 \end{align*}
\[ \text{From a graph, the solution is }-2 \leq x\leq 2 \\ \text{so }g^*: [-2,2] \to \mathbb{R},\quad g^*(x) = x^2-1\text{ should do.}\]
From here, computing \( f (g^*(x))\) should be easy.

Subject to some minor computational errors.

Thanks again

[Aaron]

PSA:

* If you are going to post a question, try to show current understanding so you can get meaningful help, not just answers/working out so you can rote learn that particular question. As we know, maths requires application of knowledge to a broad range of questions. If you just post a question with zero effort (meaning no working out and just a copy/pasted question expecting a full blown answer), it not only shows a lack of respect to the people who volunteer their time to help you but it also doesn't address the main issue that you have with your understanding.

Examples of showing understanding

* A brief written explanation as to what you know about this topic
* The particular point where you are stuck / the point where you are doubtful (e.g. is it -1 or +1?)
* Your current working out that you have.

* If you respond to questions, please encourage them to show to you what they currently know or even start off with a prompting question (Sine has done this perfectly as above) before giving them the answer/full solution. If they "do not know" or "don't know where to start", do they have prior content knowledge or is there a topic that they can think back to, to assist with their problem? Prompt them to think for themselves.

Examples of prompting

* What is the first step to solving this problem?
* Think about concept x or concept y... how does this relate to this problem?
* Giving short hints that will prompt the user to go back and have a think before posting again.

Thank you for reading and I hope that you'll consider this to ensure that student learning benefits in a meaningful way. Obviously I have posted this a few times now so I won't be posting it anymore, I really hope people will think about this (on both ends - those asking questions and those answering questions).

[Lear]

Unfortunately the above will likely be ignored/not seen by users in the future.

I wish this could be shown permanently when people post questions on these math/science threads.

Perhaps a suggestion for the site? Even an enforceable rule?

[Yertle the Turtle]

As Lear says above, in the development of the new site, would it be possible to have the OP on all subject question threads, with updated rules, guidelines etc. showing at the top of each page? Because there are plenty of times where this would be of use, I think.

[aspiringantelope]

What fraction of the shaded area (triangle) is compared to the hexagon?
I have no idea and no current understanding so I hope someone will be able to give an answer and explanation.
https://imgur.com/iU80scx

[lzxnl]

Let the radius be r. A regular hexagon can be divided into six congruent equilateral triangles. Find the area of each of those in terms of r. Then, for your current question, you will need a circle theorem to find the angles of the triangle (hint: angle at the centre equals?). You should be able to find the height and base of the triangle in terms of r.

[aspiringantelope]

Let the radius be r. A regular hexagon can be divided into six congruent equilateral triangles. Find the area of each of those in terms of r. Then, for your current question, you will need a circle theorem to find the angles of the triangle (hint: angle at the centre equals?). You should be able to find the height and base of the triangle in terms of r.

Are you sure it is that complicated?
Can you divider the triangle into two and find it out that way?

[lzxnl]

Are you sure it is that complicated?
Can you divider the triangle into two and find it out that way?

You still need some way to compare the triangle to the hexagon. You also need some way of finding the angles in the triangle. The right circle theorem will solve this immediately. I'm sure you can do this in other ways too, but I don't think those are simpler.

[aspiringantelope]

You still need some way to compare the triangle to the hexagon. You also need some way of finding the angles in the triangle. The right circle theorem will solve this immediately. I'm sure you can do this in other ways too, but I don't think those are simpler.

Ok and what answer do you get from that way can I ask?

[redpanda83]

Ok and what answer do you get from that way can I ask?

Hey!
i remember doing this question in 2018 Australian Maths comp.
if i am not wrong the the fraction of the shaded area (triangle) is compared to the hexagon should be 1/3.
The way i did is, just found the area of 6 tiny triangles in hexagon with height of h/2
so, area of hexagon = 3sh/2
and area of shaded area triangle = sh/2
fraction of comparison = area of triangle(shade)/area of hexagon = 1/3