VCE Methods Question Thread!

[snowisawesome]

Well it really depends on how comfortable you feel with the course. If you are a good maths student, you probably won't have to work that hard to get 35+, however some people who aren't as into maths might struggle. The textbook is there to give you a good foundation so you understand the underlying concepts, which is essential if you want to extend and apply these principals in harder exam-style questions. Once you fully grasp these concepts and have reinforced them in your mind, there is no use in continuing to do basic questions.

So once you have fully grasped and understood the concepts, you can skip all the easy/basic questions and just focus a lot on the harder application style questions?

[Quantum44]

So once you have fully grasped and understood the concepts, you can skip all the easy/basic questions and just focus a lot on the harder application style questions?

Yes that is what I was saying, although sometimes it can be good to go over basic questions as revision if you haven't covered a certain topic in a long time.

[snowisawesome]

Can someone please explain why the y value when x = 2pi, which has a y value of ~1.4 is not halfway between the y value when x = (3pi/2), which has a y value of 5, and when x = (5pi/2), which has a y value of 0

Also, when studying for a methods sac in year 12, is doing the textbook questions+ the chapter review questions+checkpoints+past sacs the best way to revise?

[VanillaRice]

Can someone please explain why the y value when x = 2pi, which has a y value of ~1.4 is not halfway between the y value when x = (3pi/2), which has a y value of 5, and when x = (5pi/2), which has a y value of 0

What you've suggested assumes a linear relationship - trigonometric functions are not linear. Have a look at other non-linear functions such as quadratics, cubics, exponentials, etc. and you will see something similar.

Also, when studying for a methods sac in year 12, is doing the textbook questions+ the chapter review questions+checkpoints+past sacs the best way to revise?

In my opinion, exposing yourself to and attempting as many different types of questions and practise as possible is a good way to revise (so, yes). I would begin with textbook questions to solidify your knowledge (to help you attempt extended response-style questions), and identify any mistakes you may make repeatedly.

Hope this helps

[Sine]

Can someone please explain why the y value when x = 2pi, which has a y value of ~1.4 is not halfway between the y value when x = (3pi/2), which has a y value of 5, and when x = (5pi/2), which has a y value of 0

Also, when studying for a methods sac in year 12, is doing the textbook questions+ the chapter review questions+checkpoints+past sacs the best way to revise?

You cant assume the mid value of two x values will result in the mid value of the y values since the graph is not linear. That is - for your assumption to work the gradient over an interval must be constant

Revising for sacs i would do chapter reviews since the exercise questions are just the bare minimums.

EDIT: beaten by vanillarice

[snowisawesome]

f(x) = x^2 - 4, domain = R
g(x) = 1/(x-2), domain = (2, infinity)

restrict the domain of f(x) to obtain a function f1(x) such that g(f1(x)) exists

Could someone please explain how to do this?

Stupid question but if f(x) = x^3, then is f(x-y) = x^3 - y

[VanillaRice]

f(x) = x^2 - 4, domain = R
g(x) = 1/(x-2), domain = (2, infinity)

restrict the domain of f(x) to obtain a function f1(x) such that g(f1(x)) exists

Could someone please explain how to do this?

I'll try get you started.
Remember that for the composite function g[f1(x)] to exist, the range of f1 must be a subset of the domain of g.

1) What does the range of f1 have to be to be a subset of the domain of g? Since the question doesn't state otherwise, any subset will work. Hint: for this question, remember the difference between closed and open brackets
2) How do we restrict the domain of f1 to achieve the range you just determined? Hint: draw a quick sketch of the graph

Hope this helps. Post if you get stuck

EDIT: See Rui's solution below. I would highly recommend attempting the question yourself first using the steps above though

[RuiAce]

f(x) = x^2 - 4, domain = R
g(x) = 1/(x-2), domain = (2, infinity)

restrict the domain of f(x) to obtain a function f1(x) such that g(f1(x)) exists

Could someone please explain how to do this?

Enclosed in spoiler because a reply was posted 36 seconds prior.

(Or a smaller subset of that interval.) The idea is that if \( x\notin (2,\infty) \), then \( g(x) \) is not defined. In a similar way, if \( f(x)\notin (2,\infty)\), then \(g\left[f^{-1}(x)\right] \) is not defined. The easiest way to make sure \(f^{-1}(x)\in (2,\infty) \) is true is to simply make the range of \(f^{-1}\) exactly that.

Regarding the edit:

[snowisawesome]

Just confirming that (x-y)^3 is not the same as x^3 - y^3 right?

And also, (x^3)/(y^3) = x^3 - y^3 right?
Also, is e^(x-y) = e^x - e^y

[Opengangs]

Just confirming that (x-y)^3 is not the same as x^3 - y^3 right?

And also, (x^3)/(y^3) = x^3 - y^3 right?
Also, is e^(x-y) = e^x - e^y

[Lear]

1. Just as (a-b)^2 is NOT a^2-b^2, (x-y)^3 is not x^3-y^3. I would recommend expanding it yourself to see how it works - (x-y)(x-y)(x-y)

2. I think you might be confusing this with the law where x^a/x^b= x^a-b
Note: the bases must be the same and only the exponentials are taken away. Therefore (x^3)/(y^3) is NOT = x^3-y^3

3. again, this is NOT true due to index laws.

I would recommend trying some substitutions yourself and seeing if the two sides equal each other. This was in a test situation you have a way to determine if you are applying your index laws correctly.

[snowisawesome]

edit: Is (1/x)+(1/y) = 1/(x+y)

So what would (x^3)/(y^3) simplify to since they have different bases of x and y?

MOD EDIT: Merge//Sine

[Lear]

I would highly recommend you you try adding them yourself first and see if you can reach the result. This way you will remember in the long term. However, in this case, no it is not.

[snowisawesome]

I would highly recommend you you try adding them yourself first and see if you can reach the result. This way you will remember in the long term. However, in this case, no it is not.

Am I right in saying that (x^3)/(y^3) will be left as it is as x and y are different bases?

Also, would I be right in saying that (2^x)^2 is the same as (2^2)^x?

[Opengangs]

Am I right in saying that (x^3)/(y^3) will be left as it is as x and y are different bases?

Also, would I be right in saying that (2^x)^2 is the same as (2^2)^x?

Yep to both questions