VCE Methods Question Thread!

[uhoh]

For this exam 1 question, how do I know what the graph looks like and whether the median is before or after the mode?
1st image: eqn, 2nd image part iv): question (not sure if you need to do ii and iii to find iv though) and 2nd image: Ans

And for sign tables, do we need to write the corresponding value of dy/dx for an x value (apart from when dy=dx=0)? Or can we just write dy/dx= positive/ negative? e.g. x=2, dy/dx= 3 (or can we just write dy/dx=+ve)

bump

[chantelle.salisbury]

can someone help with the 2012 vcaa paper Multi choice q. 8?

[VanillaRice]

For this exam 1 question, how do I know what the graph looks like and whether the median is before or after the mode?
1st image: eqn, 2nd image part iv): question (not sure if you need to do ii and iii to find iv though) and 2nd image: Ans

And for sign tables, do we need to write the corresponding value of dy/dx for an x value (apart from when dy=dx=0)? Or can we just write dy/dx= positive/ negative? e.g. x=2, dy/dx= 3 (or can we just write dy/dx=+ve)

Sorry, looks like this question was missed!

Is this a VCAA question? Looks quite difficult. In my opinion, this type of question is likely to come up in an exam.
I've been able to find solutions to part ii and iii, but am stuck on your specific question. I'll post what I've come up with and hopefully someone could see something I didn't

Spoiler

ii) Consider

Multiply both sides by 1/pi and integrate

You can now integrate f(x) from 0 to pi/4 to find out if the median lies before or after pi/4.

Solution

So now the median


Get into the same form as the question

So p = pi/2

iii) For part ii, the mode can usually be found by finding the maximum i.e. find the derivative and equate to zero. So:


Multiply both sides by pi/4 (to simplify the coefficients)

And we can eventually rearrange to get

My idea was to compare this with the expression from part ii)

Since

We can write

I'm not sure if this is along the right track though, so hopefully someone can provide a reason as to why M > m.

The only other way I can see of sketching the graph is to ignore the relative locations of m and M. We know m > pi/4, so the bulk of the graph is to the right of pi/4. If you draw y = x, and y = sin(2x), I guess you could approximate the graph of f(x) by using the product of functions method. But once again, I don't think this is the type of question they usually put on VCAA exams (especially exam 1 - you are usually asked to sketch functions you are familiar with).

RE: sign tables - are you referring to sign diagrams in the context of finding the nature of stationary points? If so, writing either positive, negative or zero should be sufficient.

can someone help with the 2012 vcaa paper Multi choice q. 8?

What does f(p) = f(q) = 0 mean? We have intercepts at x = p and x = q.
What does f ′(m) = f ′(n) = 0 mean? There are stationary point at x = m and x = n.
Since a is a negative number, we know f(x) is a negative cubic.
Have a go at combining all of this information, and sketch a potential graph, and post again if you get stuck

Hope this helps


EDIT: Oops, re-did my solution, accidentally looked at the specialist exam 

[Rieko Ioane]

Hi,

Could someone help with Q3e and Q4f from VCAA 2011 E2:

Q3e) what is the logic behind the working for this question? I'm a little confused why we need to multiply by 0.5 to each probability of T>3 respectively (I know we can pick either chocolate A or B) but for some reason it seems like we don't need to account for drawing A or B?

It's probably just me and probably my probability notation is causing me to neglect some probabilities I need..

Q4f) I don't understand the logic behind VCAA's working. It don't know how they got their value(s) for k by solving dT/dx =< 0, isn't this only possible if you have a value of k already?

Thanks

[Rusten]

Hi, I was wondering what the method is on the CAS for those multiple choice questions where you have to test the functions given to see if they're 'true' for the statement given. eg. f(2x) - 2f(x) = 0 for all x, and then they list a bunch of functions.
I hope that makes sense, thanks in advance

[MsEmilia]

Hey, so this has come up a few times, but 2010-1 exam has a good example. Question 11a says to use similar triangles, can someone please explain how I can apply this concept to get the answer? The explanation VCAA offers doesn't quite make sense to me either

[Max Kawasakii]

Hi, I was wondering what the method is on the CAS for those multiple choice questions where you have to test the functions given to see if they're 'true' for the statement given. eg. f(2x) - 2f(x) = 0 for all x, and then they list a bunch of functions.
I hope that makes sense, thanks in advance

My advice would be to use logic to eliminate some of the options. For example if the question states "For which function is the rule 2F(0)=2" and one of the subsequent options presents as "f(x)=ln(x)"; then you can eliminate that for the natural log of 0 is undefined.

It's a tedious process but how I approach them, do it enough and you can immediately recognize 1 or 2 that will not satisfy the condition.

[chantelle.salisbury]

Sorry, looks like this question was missed!

Is this a VCAA question? Looks quite difficult. In my opinion, this type of question is likely to come up in an exam.
I've been able to find solutions to part ii and iii, but am stuck on your specific question. I'll post what I've come up with and hopefully someone could see something I didn't

Spoiler

ii) Consider

Multiply both sides by 1/pi and integrate

You can now integrate f(x) from 0 to pi/4 to find out if the median lies before or after pi/4.

Solution

So now the median


Get into the same form as the question

So p = pi/2

iii) For part ii, the mode can usually be found by finding the maximum i.e. find the derivative and equate to zero. So:


Multiply both sides by pi/4 (to simplify the coefficients)

And we can eventually rearrange to get

My idea was to compare this with the expression from part ii)

Since

We can write

I'm not sure if this is along the right track though, so hopefully someone can provide a reason as to why M > m.

The only other way I can see of sketching the graph is to ignore the relative locations of m and M. We know m > pi/4, so the bulk of the graph is to the right of pi/4. If you draw y = x, and y = sin(2x), I guess you could approximate the graph of f(x) by using the product of functions method. But once again, I don't think this is the type of question they usually put on VCAA exams (especially exam 1 - you are usually asked to sketch functions you are familiar with).

RE: sign tables - are you referring to sign diagrams in the context of finding the nature of stationary points? If so, writing either positive, negative or zero should be sufficient.
What does f(p) = f(q) = 0 mean? We have intercepts at x = p and x = q.
What does f ′(m) = f ′(n) = 0 mean? There are stationary point at x = m and x = n.
Since a is a negative number, we know f(x) is a negative cubic.
Have a go at combining all of this information, and sketch a potential graph, and post again if you get stuck

Hope this helps


EDIT: Oops, re-did my solution, accidentally looked at the specialist exam 

awesome.. thanks that makes so much sense now i shlda seen that before..
thanks again for your help

[zhen]

Hey, so this has come up a few times, but 2010-1 exam has a good example. Question 11a says to use similar triangles, can someone please explain how I can apply this concept to get the answer? The explanation VCAA offers doesn't quite make sense to me either

Here’s my explanation

[chantelle.salisbury]

umm... just really not sure on how to do antidifferentiation when say there is x*f(x). i seem to always miss a step but im not sure what.
for example how do i antidifferentiate x*sin(2x)?
thankyou

[Rusten]

Hey, so this has come up a few times, but 2010-1 exam has a good example. Question 11a says to use similar triangles, can someone please explain how I can apply this concept to get the answer? The explanation VCAA offers doesn't quite make sense to me either

If you draw a diagram it might be easier; they're trying to say that because the cylinder (and the 'triangle' you can create within it) is within the cone, the triangles will have the same height:radius ratio, and thus you can equate the two to find h in terms of r.

[GloriousHeights]

Hey! Does anyone know if we can put past VCAA exams in our bound references? Like bind them with anything else we take in?
Thanks! 

[VanillaRice]

Hi,

Could someone help with Q3e and Q4f from VCAA 2011 E2:

Q3e) what is the logic behind the working for this question? I'm a little confused why we need to multiply by 0.5 to each probability of T>3 respectively (I know we can pick either chocolate A or B) but for some reason it seems like we don't need to account for drawing A or B?

It's probably just me and probably my probability notation is causing me to neglect some probabilities I need..

Q4f) I don't understand the logic behind VCAA's working. It don't know how they got their value(s) for k by solving dT/dx =< 0, isn't this only possible if you have a value of k already?

Thanks

I can't seem to find Q3e, I'm assuming you mean 2e?
2e) We multiply each probability by 0.5 because the question states that there is an equal number of chocolates from A and B i.e. the probability of selecting A = probability of selecting B = 1/2.

4f) We get k = 5sqrt(37)/74 by substituting x = sqrt(7)/2 into T'(x) = 0 and solving for k.
Now, recall that we have restricted the domain of T(x) to x<=sqrt(7)/2.
Have a think about what it actually means for T'(x) to be less than or equal to 0 over the domain. It means that T(x) will be a decreasing function over that whole interval. Which means that the right end-point will always be the minimum as long as T'(x) < 0 over the interval. We find that this is true for k >= 5sqrt(37)/74 (by solving T'(x) <= 0, where x = sqrt(7)/2 for k). Hope that makes sense

Hi, I was wondering what the method is on the CAS for those multiple choice questions where you have to test the functions given to see if they're 'true' for the statement given. eg. f(2x) - 2f(x) = 0 for all x, and then they list a bunch of functions.
I hope that makes sense, thanks in advance

The way that I do it is (note this is on the ti-nspire) define f(x) to be whatever the function is for that option. Then, simply type each option out what you want to prove to be true e.g. type out f(2x) - 2f(x). You are looking for the option which gives you 0.

Hey! Does anyone know if we can put past VCAA exams in our bound references? Like bind them with anything else we take in?
Thanks! 

Yes - you can put anything, and do anything to your bound references, provided your bound reference still follows the rules set out by VCAA.

Hope this helps

[GloriousHeights]

Thank you so much VanillaRice! I'm definitely gonna need them 

[CarrymetoUni]

Can someone please show me how to do question 6 from the 2009 VCAA Exam 1