VCE Methods Question Thread!

[RuiAce]

... What other R could he mean? O.o

Mhm. Without some emphasis I just kept too far in context and interpreted it as range.

[LPadlan]

as= a + h, h + ah=1, For each of the following pairs of eqns, write s in terms of a only

[jamonwindeyer]

as= a + h, h + ah=1, For each of the following pairs of eqns, write s in terms of a only

This is just a slightly strange application of the Substitution Rule. We want to eliminate \(h\), so we find an expression equal to \(h\) in one equation and then substitute it into the other. Then, it's just some algebra to tidy it up (and then I put it into a single fraction, which I think looks the nicest) remember, you are always looking to define a variable you don't want in terms of the others, and then you put that new expression in place of your variable-reject

Let me know if any of those steps need clarification! Hope it helps

[LPadlan]

This is just a slightly strange application of the Substitution Rule. We want to eliminate \(h\), so we find an expression equal to \(h\) in one equation and then substitute it into the other. Then, it's just some algebra to tidy it up (and then I put it into a single fraction, which I think looks the nicest) remember, you are always looking to define a variable you don't want in terms of the others, and then you put that new expression in place of your variable-reject
Let me know if any of those steps need clarification! Hope it helps

Hello, thanks this has been very helpful. But how did you get s= 1 + 1/a+a2?

[MightyBeh]

He divided the expression \(as = a + h\) by \(a\), giving \(s = 1 + \frac{h}{a}\).

[LPadlan]

He divided the expression \(as = a + h\) by \(a\), giving \(s = 1 + \frac{h}{a}\).

Thanks a lot!

[One Step at a Time]

Hey everyone 

I've got 2 questions. The 1st one is related to the 1st attached pic and the 2nd question to the 2nd pic. (Hope the order isn't mixed up or my questions will definitely be quite confusing ahah)

Q1. (1st pic) A bit unsure as to what the notation used in 4a) and b) mean.
Does 4a) refer to the equation y=4, whilst 4b) refers to the equation x=2?

Q2. (2nd pic) A bit of a general question: what is the notation used in 9a), b) and c)?called? Is it set notation, or is that something else?
Also, for 9b), do I need to rearrange the equation 3y+4x=12 into f(x)= -4x/3 +4 when writing it in function notation? Or can I leave the equation as 3y=4x=12 for function notation?

Thanks heaps 
 

[RuiAce]

Hey everyone 

I've got 2 questions. The 1st one is related to the 1st attached pic and the 2nd question to the 2nd pic. (Hope the order isn't mixed up or my questions will definitely be quite confusing ahah)

Q1. (1st pic) A bit unsure as to what the notation used in 4a) and b) mean.
Does 4a) refer to the equation y=4, whilst 4b) refers to the equation x=2?

Q2. (2nd pic) A bit of a general question: what is the notation used in 9a), b) and c)?called? Is it set notation, or is that something else?
Also, for 9b), do I need to rearrange the equation 3y+4x=12 into f(x)= -4x/3 +4 when writing it in function notation? Or can I leave the equation as 3y=4x=12 for function notation?

Thanks heaps 

Sounds right for a). With b) however, take care that y is an integer; not necessarily a real number. So you don't have a smooth line. (If you tried plotting that one you'd get a ton of dots instead.)
__________________

If they say to define a function f:X->Y then you need to put f(x) in there.
However, what I'm not sure if they care about is if they let you leave it as 3f(x) + 4x = 12

Notation wise, I haven't seem a name strictly made for that, but the closest you can get would definitely be set notation.

[One Step at a Time]

Sounds right for a). With b) however, take care that y is an integer; not necessarily a real number. So you don't have a smooth line. (If you tried plotting that one you'd get a ton of dots instead.)
__________________

If they say to define a function f:X->Y then you need to put f(x) in there.
However, what I'm not sure if they care about is if they let you leave it as 3f(x) + 4x = 12

Notation wise, I haven't seem a name strictly made for that, but the closest you can get would definitely be set notation.

Thanks so much RuiAce for the speedy reply and explaining everything so clearly

[One Step at a Time]

Have some more pretty basic qs but I don't understand them ahaha would appreciate anyone's help. Thanks

1. (Question in 1st pic and answer in 2nd pic) Why is 0 included in the range of both functions?

2. (Question in 3rd pic and answer in 4th pic) Super confused. I thought 0 should've been included in the domain of both functions (like it was in the range in the previous question), but it's not 

[keltingmeith]

1. (Question in 1st pic and answer in 2nd pic) Why is 0 included in the range of both functions?

This is going to sound like a pretty poor excuse for an answer, so sorry about that. However, there's no reason that 0 shouldn't be included in either of them, so it is included in both. What makes you think it shouldn't be included?

2. (Question in 3rd pic and answer in 4th pic) Super confused. I thought 0 should've been included in the domain of both functions (like it was in the range in the previous question), but it's not 

So the difference here is that in this question, you want to form two functions that combine to give the original function. In that sense, it's kind of contradictory to have 0 included twice - remember, a function should only specify each point once. In the first one, they wouldn't "combine" to give the same function, because it wasn't even a function to begin with.


Having said that, both of these answers are fairly subjective, and VCAA will never ask them because they are so subjective. Whilst there's no reason to ignore 0 in the first one, there's also no reason to take it for both. Similarly with the second, there's no reason to not accept 0 in both functions. I wouldn't worry about these ones too much, as long as you get the general idea.

[One Step at a Time]

Thanks so much EulerFan101, appreciate it! Plus I'm relieved to hear that VCAA won't  chuck in a question like this (hopefully)

What makes you think it shouldn't be included?

When I was doing the question, it didn't make any sense to include it. But now that you've explained it, I've realised that it DOES in fact make sense since the range can "overlap", unlike the domain for a function 

[One Step at a Time]

Hey everyone,

Trying to finish the holiday homework and now I have some more basic questions ahah thanks for your help 

1. (1st pic) In c), why does the book suddenly switch from substituting a to substituting 1? Am I missing something... 

2. (2nd pic) For Q6), can we work out the range without drawing the graph? I resorted to drawing the graph for each equation to figure out the range, but I'm not sure if it's the quickest/ most effective way to do so.

3. (3rd pic) For 8c), does the fact that the discriminant < 0/ there are no X-int for x^2+3≥0 affect the implied domain? Or is it totally irrelevant to calculate the implied domain?

4. (Also 3rd pic) For 8h) , the answer for the implied domain= (-∞,2) ∪ [ 1,∞], but I only got [1,∞]. My logic (probs not logical lol) is that (x-1)/(x-2)≥0 and x+2≠0. So x≥1 and x≠-2, which means that the implied domain is [1,∞]. 

5. Just a general question: when drawing piece-wised/ hybrid functions, do the endpoints and start points of each "part"/ different equation of the function need to be labelled with open/ closed circles?

[One Step at a Time]

Sorry for posting again, couldn't fit these attachments in my previous one 

For 14d), e) and f) (question in 1st pic and answer in 2nd pic), I don't understand how the domain was calculated.

e.g. For d), we sub a+3 for x, but to find the new domain, we minus 3 from the original domain (x>3 for the 1st equation and x≤3 for the 2nd equation) to get a>0 for the 1st equation and a≤0 for f(a+3)?

Hope my question made sense haha thank you! 

[RuiAce]

Sorry for posting again, couldn't fit these attachments in my previous one 

For 14d), e) and f) (question in 1st pic and answer in 2nd pic), I don't understand how the domain was calculated.

e.g. For d), we sub a+3 for x, but to find the new domain, we minus 3 from the original domain (x>3 for the 1st equation and x≤3 for the 2nd equation) to get a>0 for the 1st equation and a≤0 for f(a+3)?

Hope my question made sense haha thank you!