VCE Methods Question Thread!

[ArtyDreams]

Hi! I needddddd some help with implied/maximal domains, because I really feel like it comes up a lot with composite functions, and I'm horrible at limiting the domains on these!

I'll leave an example question:

So what is the implied domain of 1/√(x^2 - 3x)

So I know that you cant divide by 0 or have a negative under a positive sqaure root. So in this case, I understand that 0, and 3 have to be in my answer somewhere. So how do I know whethe it is R\(0,3) or isnt defined for another section or something like that? I hope it makes sense.

I just dont get how to do it, since I dont think we are meant to know what each of these graphs look like???

I hope my question makes sense - thanks so much!!! 

[studyingg]

Hi! I needddddd some help with implied/maximal domains, because I really feel like it comes up a lot with composite functions, and I'm horrible at limiting the domains on these!

I'll leave an example question:

So what is the implied domain of 1/√(x^2 - 3x)

So I know that you cant divide by 0 or have a negative under a positive sqaure root. So in this case, I understand that 0, and 3 have to be in my answer somewhere. So how do I know whethe it is R\(0,3) or isnt defined for another section or something like that? I hope it makes sense.

I just dont get how to do it, since I dont think we are meant to know what each of these graphs look like???

I hope my question makes sense - thanks so much!!! 

it helps to quickly sketch the quadratic x^2 -3x

it would have an x int at 0 and 3, and a minimum t.p between these roots. Because it is negative between 0 and 3, and =0 at x=3 and x=0, u can infer that the maximal domain is x<0 and x>3

[Matthew_Whelan]

So the denominator has to be >0 which you know
x^2 - 3x > 0
x > 3  but you can also recognise that when x < 0 the denominator is also positive.
The graph is 1/(expression) so it looks like a truncus but instead of an asymptote at one point, it covers the region [0,3].
When graphing I usually plug in a few easy x values like 0, 1, 2 and get an idea of the graph if unsure, although most graph shapes you should know.
I think this was what you were asking 

[ArtyDreams]

it helps to quickly sketch the quadratic x^2 -3x

it would have an x int at 0 and 3, and a minimum t.p between these roots. Because it is negative between 0 and 3, and =0 at x=3 and x=0, u can infer that the maximal domain is x<0 and x>3

So the denominator has to be >0 which you know
x^2 - 3x > 0
x > 3  but you can also recognise that when x < 0 the denominator is also positive.
The graph is 1/(expression) so it looks like a truncus but instead of an asymptote at one point, it covers the region [0,3].
When graphing I usually plug in a few easy x values like 0, 1, 2 and get an idea of the graph if unsure, although most graph shapes you should know.
I think this was what you were asking 

Ohhhhhhhh I think I fairly understand it now! Thank you both so much!!! I guess I'll do some more practice and see how I go.

Now, just a general question, does anyone have any ideas of some questions that may come up in the exam? Perhaps a recurring theme or questions that havent come up in a while - or even poorly answered questions that MAY come?

I'm fairly across all the content, just a little curious.

[Just another student]

Hi there,

Do I have to show absolutely every single algebra step for a 'show that' question? Can I just sub values into relevant formulas and then say therefore ..... = ...... and get full marks?

[KiNSKi01]

Hi there,

Do I have to show absolutely every single algebra step for a 'show that' question? Can I just sub values into relevant formulas and then say therefore ..... = ...... and get full marks?

Not exactly sure what you mean but-
If by these values you don't mean the answer they give you then yes. (e.g question is show that x=5, you can't have x=5 anywhere into the final line of working)

[Just another student]

Hi there, can someone pls help with VCAA 2014 E2 Q4/b?

The answer shows Pr(X less than 9) = 0.10565

However that would mean on CAS we enter normal Cdf from 0-8 right?

Because it is not less than or equal to.

However when I enter normal Cdf 0-8, I get 0.06681, it is only when I include 9 (0-9) I get the right answer.

Would appreciate if someone could help me out.

[persistent_insomniac]

How do you find the average value from a graph (without using integraration) - can someone please explain how they did it in the question below?

[randomnobody69420]

Draw a box with one side being x=-pi/8 , one side being x= pi/8 , one side being y=2. You're trying to find the area under the graph between -pi/8 and pi/8, but you can see that the graph is actually symmetrical within that box. So all you have to do is find the area of that box, then divide it by 2 since the graph is symmetrical. Now that you have the area, multiply it by 1/pi/4, and that's your average value

[randomnobody69420]

Can someone explain how to do the very last question of 2017 NHT Exam 1? I don't understand why there are 2 solutions for c (1,2]

[KiNSKi01]

Can someone explain how to do the very last question of 2017 NHT Exam 1? I don't understand why there are 2 solutions for c (1,2]

Yo this was actually discussed recently
here was my response:

Hey this is tricky to spot but you have to consider the domain of g(x) because this is what gives the solution of c>2.

When c=2, the graph of g(x) starts on the line y=x, (therefore there are two solutions for y=x as it also intercepts at (1,1) - this is what the examiners report is getting at

This means that for any value of c>2 there will only be one real solution because the domain of g(x) restricts it from intercepting its inverse twice

ngl would not have considered the c>2 under exam conditions

[randomnobody69420]

Thanks KiNSKi01 I kind of get it now.

[emceee]

Hi there, can someone pls help with VCAA 2014 E2 Q4/b?

The answer shows Pr(X less than 9) = 0.10565

However that would mean on CAS we enter normal Cdf from 0-8 right?

Because it is not less than or equal to.

However when I enter normal Cdf 0-8, I get 0.06681, it is only when I include 9 (0-9) I get the right answer.

Would appreciate if someone could help me out.

Hi! I think you're confusing binomial distributions with normal distributions. Remember that binomial is discrete, which normal is continuous. Therefore, for binomial distributions, you would be right in saying that Pr(X<9)=Pr(X=<. But because the question is about a continuous normal distribution, it's actually Pr(X<9)=Pr(X=<9), so you would enter normal Cdf 0-9 into your CAS.

[Just another student]

Hi! I think you're confusing binomial distributions with normal distributions. Remember that binomial is discrete, which normal is continuous. Therefore, for binomial distributions, you would be right in saying that Pr(X<9)=Pr(X=<. But because the question is about a continuous normal distribution, it's actually Pr(X<9)=Pr(X=<9), so you would enter normal Cdf 0-9 into your CAS.

that helps so much, THANKS!

[tigerclouds]

"At what time of the day is the temperature in the office decreasing most rapidly?"
How would you approach this?
Could you always find the derivative of the derivative of the graph and set that = 0? It worked in this case but I don't know if it always does. What also kind of worked is adding the stationary points and dividing them by 2.
Are these always true? If so, why?

Thank you in advance!