VCE Methods Question Thread!

[atar.notes.user]

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

[QueenSmarty]

Probably another stupid question but how do you do q3 from 2014 exam 2 (multiple choice)? I got B, but the answer is D

[VanillaRice]

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

Note these types of questions (involving related rates) are no longer in the study design.

Probably another stupid question but how do you do q3 from 2014 exam 2 (multiple choice)? I got B, but the answer is D

No such thing as stupid questions here

I suspect you have ignored the area bound by the graph between x = 0 and x = 4 (which is under the x -axis). Your answer of 20/3 is only the area of the graph between x = -2 and x = 0, you must also consider the area from x = 0 and x = 4. Have a go at this and post if you run into any trouble
WARNING: the area enclosed by the graph between x=0 and x=4 is negative.

[uhoh]

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 

Thanks so much VanillaRice, yeah thankfully it's not VCAA

Q7: I don't understand how the report did this question (1st image= question, 2nd image= ans from report)

Q18 (3rd image): The ans is B but I got this by hand. It doesn't work on the calc if I define g(x) since I don't get true when I type B in...

Q20 (4th image): the examiners report says the inverse transformation maps y = x^2 onto f(x). What does this mean?

For 2013 3di) Section 2, what does the examiners report say to find d/dx of f(x)? Shouldn't it be finding d/dx of length EF? I got the right answer, but just wondering about this line of working. Sorry I can't post an image- it's extended response

General question: If the average value of f(x) is 2, does that mean if we draw y=2, the areas above y=2 and below y=2 are the same?

Another general question: Do we need to add units to our ans? e.g. For to state the max temp, I wrote 27. Do I write 27 degrees Celsius? Same with period- if the x-axis represents hours, do I need to write the period in h? Also, if the equation is already defined in the question e.g. f(x)=x^2- when I show my working to find the x-int for examples, can I just f(x)=0 or do I have to write X^2=0?

[ZNormal]

Hi guys sorry for asking so many questions!
And thank you for vanillarice and atarnotesuser for answering my other question

Could someone please help me on the VCAA 2012 EXAM 2 SECTION 2 Question 2c?

I just have no idea how VCAA did it in the solutions and my numbers don’t factorise to what they have,

Thank you so much in advance!

Edit: sorry, to be precise, I don’t know how they got -2/(2p-4)^2 for the gradient

[derplerd]

Hey everyone!
My teacher told me that I could attach my textbook to my exercise book and use that as a bound reference (as long as it is taped together and then contacted over).
Is this true?? 

Thanks

[QueenSmarty]

Is q2a from 2015 exam 2 still part of the study design? If it is, is it likely to come up on the exam?

[Syndicate]

Is q2a from 2015 exam 2 still part of the study design? If it is, is it likely to come up on the exam?

Yes, it's still part of the course (It's basically using tan(theta) = m). As for it's likeliness to appear on the exam, I am not exactly sure about that (I don't think anyone would be except the chief assessor).

[plsbegentle]

can someone help explain the very last question for 2015 exam 1?

thanks in advance

[VanillaRice]

Thanks so much VanillaRice, yeah thankfully it's not VCAA

Q7: I don't understand how the report did this question (1st image= question, 2nd image= ans from report)

Q18 (3rd image): The ans is B but I got this by hand. It doesn't work on the calc if I define g(x) since I don't get true when I type B in...

Q20 (4th image): the examiners report says the inverse transformation maps y = x^2 onto f(x). What does this mean?

For 2013 3di) Section 2, what does the examiners report say to find d/dx of f(x)? Shouldn't it be finding d/dx of length EF? I got the right answer, but just wondering about this line of working. Sorry I can't post an image- it's extended response

General question: If the average value of f(x) is 2, does that mean if we draw y=2, the areas above y=2 and below y=2 are the same?

Another general question: Do we need to add units to our ans? e.g. For to state the max temp, I wrote 27. Do I write 27 degrees Celsius? Same with period- if the x-axis represents hours, do I need to write the period in h? Also, if the equation is already defined in the question e.g. f(x)=x^2- when I show my working to find the x-int for examples, can I just f(x)=0 or do I have to write X^2=0?

7) I would approach this question in a different way. I would sketch the function on my calculator, restricting my domain as required for each option. Check which are not one to one (i.e. will have no inverse), and eliminate those. You should be able to find one which gives a one to one with the maximum domain.
The way the report has approached this question is to find the maximum domain for which an inverse may exist, and then selected the option which fits within that domain.

18) You need to restrict your domain as per the question. Are you on the CAS ti-nspire?
You need to type in
2f(8x)=f(x²)+6|x>0
which will restrict that domain to x>0.
Alternatively, when you type in just 2f(8x)=f(x²)+6, select the output equation, and solve for x. You should get x>=0 as your answer, which verifies that the expression is only true when x is greater than or equal to zero, as per the question.

20) It means that when the transformations are applied to one of the options, you will get y = x² as your transformed equation. The 'inverse' transformation means we are moving in the opposite direction.

3di) If you read carefully, they are actually finding the derivative of the equation of the length of EF (which is y-f(x)), as you have stated.

RE: average value. The average value is the y value which forms a rectangle in the domain, which has an area equal to the area under the function in that same domain.

RE: units in answers. Not sure if you're deducted marks for no units, but it's always good practise to do so.

Hi guys sorry for asking so many questions!
And thank you for vanillarice and atarnotesuser for answering my other question

Could someone please help me on the VCAA 2012 EXAM 2 SECTION 2 Question 2c?

I just have no idea how VCAA did it in the solutions and my numbers don’t factorise to what they have,

Thank you so much in advance!

Edit: sorry, to be precise, I don’t know how they got -2/(2p-4)^2 for the gradient

How did you go about finding an equation for the tangent? I'm assuming you're not familiar with the method used in the solution.
Did you finding the derivative of the curve at x=p, and formed an equation in the form y = mx+c? You should have gotten:

Spoiler

Substitute (p,f(p)) and solve for c gives

So now our tangent equation is


Multiplying both sides by (2p-4)² gives


as required.

Hey everyone!
My teacher told me that I could attach my textbook to my exercise book and use that as a bound reference (as long as it is taped together and then contacted over).
Is this true?? 

Thanks

Yes, provided that the bound reference has one, single binding, and follows all other rules set out by VCAA, it should be fine.
That being said, I would not recommend bringing in your textbook. You have very limited time in Exam 2, and you should not be wasting that time flicking through your textbook. If you have topics that you're unsure about, I would recommend that you summarise them in your notebook. The choice is up to you, however.


Hope this helps

[VanillaRice]

can someone help explain the very last question for 2015 exam 1?

thanks in advance

You need to be able to recognise the trapezium in question, formed with vertices BCED.
The length BC = b and DE = d (as per the question), and the 'height' CE is 2.
So the area of the trapezium is

Now recall expressions for b and d from parts ci and ii



The final part that you need to do is find the derivative of A (A'), solve for when A'(theta)=0, and then substitute the corresponding value of theta (which matches the domain given in the stem) into A to find the minimum value of the trapezium.

Hope this helps Post if you run into any trouble.

[Bri MT]

can someone help explain the very last question for 2015 exam 1?

thanks in advance

a)

Lets drop an imaginary line straight down from T. It meets the x-axis at a 90 degree angle, and it's height is y. The horizontal distance from where this meets 2 is x-2.
We now have a triangle with height y and width x-2.  We know that the length of the hypotenuse is 2, because the stem of the question says that the circle has a radius of 2. We now use our knowledge that cos(theta) = a/h   -> 2*cos(theta)= x-2     -> 2cos(theta) + 2 = x
We also know that sin(theta) = o/h  -> 2sin(theta) = y 
We then just write this into co-ordinate form   (2cos(theta)+2, 2sin(theta))

For c) , as XY is the tangent, we just sub in the provided points & rearrange.
i) cos(θ)2 + sin(θ)b = 2 + 2cos(θ)  ->  sin(theta)b=2 -> b= 2/sin(theta)
ii) cos(θ)4 + sin(θ)d = 2 + 2cos(θ) ->  sin(theta)d = 2 - 2cos(theta) -> (2-2cos(theta))/sin(theta)

For d) We use the formula for area of a trapezium, sub in our values. Then because we want to know the max, we differentiate and set it equal to 0. This gives us the angle when the trapezium is at a max.
Now that we know the angle, we sub it back into the area formula to get the max area.

First step:
The trapezium area formula is on the summary sheet, and we do this:  0.5* (b+d) * (4-2)   = (b+d). 
Sub in b & d, and simplify before differentiating. 
The following steps are a bit messy to type onto the computer, but let me know if you have difficulties




Beaten by VanillaRice while typing. their steps are 100% correct but I'll post this incase you have difficulty with the preceding parts a) or c)

[GloriousHeights]

Does anyone know what type of study score I might get if I'm averaging 70-80% in my SACs and get below average in my exams?

[zhen]

Does anyone know what type of study score I might get if I'm averaging 70-80% in my SACs and get below average in my exams?

It’s impossible for us to know how you’ll do cause we don’t know the strength of your cohort and SAC difficulty. Also, I’m not sure what your definition of below average is. I think if you really wanted a good prediction of your study score, you should tell the range of scores you’re getting for your practice exams. Also, I honestly feel like worrying about this right now is a waste of time that you could dedicate to improving your study score. Just try your hardest to improve your marks and don’t worry about this type of thing. 

[GloriousHeights]

This might seem obvious, but how do I find the derivative in question 4a in the 2015 VCAA exam 1?
Also, how would I know the shape of the graph for 4b?
Thank you!