VCE Specialist 3/4 Question Thread!

[S_R_K]

Just a quick question regarding question 3 in last years (2017) VCAA exam. I got the right answer by equating coefficients but was wondering how to do it using the sum and product of roots (seems to be more efficient). I have found the iTute solutions to be a bit brief and would appreciate a bit more in depth explanation. Thanks for your help!

I've attached both the question and the relevant iTute working.

Yes, it is a nice efficient method that is unfortunately not part of the course! It is also very simple to work out for yourself:

Consider a (monic) cubic polynomial with (complex) roots a, b, and c. Then it can be written as (z – a)(z – b)(z – c). Now expand brackets, and observe how the coefficients of each term can be written in terms of the roots a, b, and c. You'll find that the sum of the roots gives the coefficient of the z^2 term, but with opposite sign; the product of the roots gives the constant, but with opposite sign; (and the sum of each possible pairwise product of roots gives the coefficient of the z term).

The result for a polynomial of any degree is known as Viete's formulas.

[TheAspiringDoc]

Are graphs like hyperbolas, truncuses and ellipses still a part of the specialist 3/4 study design? I’m seeing them on older exams (e.g. 2013) but can’t find them in the study design?

Edit: what about stuff like cyclic quadrilaterals too? (E.g. alternate segment theorem)

[S_R_K]

Are graphs like hyperbolas, truncuses and ellipses still a part of the specialist 3/4 study design? I’m seeing them on older exams (e.g. 2013) but can’t find them in the study design?

Yes, they are mentioned as part of the area of study on vector-valued functions.

You will (probably) not get a question asking you to directly sketch a hyperbola from its cartesian equation; but you might be given a vector function whose cartesian equation is a hyperbola, and be asked to sketch that.

Edit: what about stuff like cyclic quadrilaterals too? (E.g. alternate segment theorem)

All high school geometry is presumed knowledge. This includes the theorems about angles in circles, etc.

[Unsplash]

Hi guys another question.

I'm confused between when to use the "linear combinations of random variables formulas" and when not to. In particular the 2016 VCAA Exam 2 MCQ 18, they do not use the formula. The only explanation I can find for it is:

"The oranges and lemons are all randomly selected so the masses of the fruits are independent. Hence Var(O1+O2+O3+L1+L2)=Var(O1)+Var(O2) …etc
The answer is A." (This is from the iTute download page to their solutions)

I assume this means that each fruit picked, must be treated as its own independent random variable?

But why is this not the case with the following exam question I have attached (not VCAA)? They consider the mass of males and the mass of females to be the random variables, but not each individual person.

[TheBigC]

Hi guys another question.

I'm confused between when to use the "linear combinations of random variables formulas" and when not to. In particular the 2016 VCAA Exam 2 MCQ 18, they do not use the formula. The only explanation I can find for it is:

"The oranges and lemons are all randomly selected so the masses of the fruits are independent. Hence Var(O1+O2+O3+L1+L2)=Var(O1)+Var(O2) …etc
The answer is A." (This is from the iTute download page to their solutions)

I assume this means that each fruit picked, must be treated as its own independent random variable?

But why is this not the case with the following exam question I have attached (not VCAA)? They consider the mass of males and the mass of females to be the random variables, but not each individual person.

REFER TO ATTACHMENT

[Unsplash]

That’s what i thought you would do, however they did the following.

That’s what I’m confused about, in what scenarios do you use each approach?

See the attachment.

[TheBigC]

That’s what i thought you would do, however they did the following.

That’s what I’m confused about, in what scenarios do you use each approach?

See the attachment.

The issue with their approach is that it is assumed that every male and every female espouse the same weight... however, each person will have a unique weight (with identical distributions, however)... so in a scenario where there are multiple items each with a different POSSIBLE weight, height or other continuous random variable you must use the approach I displayed. Whenever you encounter these questions, ask yourself: would every apple in the basket have the same mass? Would each person on the basketball team have the same height? Do I expect all of these items to have the same random variable value? If they are not the same, then use the approach I exemplified, if they are then use their approach.

[alsheriff]

Hi

I have a question regarding functions given to us in the exam. Say if we are given something like v=t for one leg of a journey, and then v=sqrt t for another, are we allowed to write "Let v1(t)=t" and "Let v2(t)=sqrt t" and use this definition throughout the question, making it real clear at the beginning of the question, or will we get penalised?

Thanks

[S_R_K]

Hi

I have a question regarding functions given to us in the exam. Say if we are given something like v=t for one leg of a journey, and then v=sqrt t for another, are we allowed to write "Let v1(t)=t" and "Let v2(t)=sqrt t" and use this definition throughout the question, making it real clear at the beginning of the question, or will we get penalised?

Thanks

Yes, you may use abbreviated notation for the rule of a function, as long as either (i) the exam has clearly defined it for you as part of the question, or (ii) you clearly define it yourself as part of your response.

[ezferns]

Hi everyone,

From the 2017 NHT Exam 2, question 6d, it asks for 'the largest value of the sample mean that could be observed before the bank’s claim was rejected at the 5% level of significance? Give your answer correct to the nearest 10 dollars'

When you do the calculations you get sample mean = 409869.12 and the answers say to round up to 409870
But wouldn't you have to round down as its the largest value *before* the bank's claim is rejected?
I'm confused

Thanks in advance!

[Jakeybaby]

Hi everyone,

From the 2017 NHT Exam 2, question 6d, it asks for 'the largest value of the sample mean that could be observed before the bank’s claim was rejected at the 5% level of significance? Give your answer correct to the nearest 10 dollars'

When you do the calculations you get sample mean = 409869.12 and the answers say to round up to 409870
But wouldn't you have to round down as its the largest value *before* the bank's claim is rejected?
I'm confused

Thanks in advance!

I agree with you, at $409870, the banks claim would have been rejected.
I'd be rounding down the $409860.

[S_R_K]

Hi everyone,

From the 2017 NHT Exam 2, question 6d, it asks for 'the largest value of the sample mean that could be observed before the bank’s claim was rejected at the 5% level of significance? Give your answer correct to the nearest 10 dollars'

When you do the calculations you get sample mean = 409869.12 and the answers say to round up to 409870
But wouldn't you have to round down as its the largest value *before* the bank's claim is rejected?
I'm confused

Thanks in advance!

No, don't round down. You would only round down if the context or question made it clear that the amounts of money were restricted to multiples of 10. An instruction to write an answer to a specified degree of precision does not imply that the amounts of money are restricted in this way. So the correct answer is 409870, because that is the closest integer multiple of 10 to the exact answer.

[Guideme]

can anyone help me with this question please
Thank you

[S_R_K]

can anyone help me with this question please
Thank you

From what you've written so far, it seems like you're on the right track. Let u = z^2, and the equation is quadratic in u: you get u^2 - 2u + 4 = 0.

This gives u = 1 + sqrt(3)i or 1 – sqrt(3)i.

Then, you have a couple of options for finding the square roots of u.

Option 1: Convert to polar form and use de Moivre's theorem to find the square roots, then convert back to cartesian.

For instance, 1 + sqrt(3)i = 2cis(pi/3), so its square roots will be of the form r*cis(theta), where r^2 = 2, and 2*theta = pi/3 + 2*pi*n (where n is an integer). Solving for r and theta gives the square roots as sqrt(2)*cis(pi/6) and sqrt(2)*cis(–5pi/6). Then convert back to cartesian form. Repeat this for u = 1 – sqrt(3).

Option 2: Express the square roots in cartesian form as (x + yi). Then expand (x + yi)^2, and equate the real and imaginary components with u = 1 + sqrt(3)i, and solve simultaneously for x and y. Repeat this for u = 1 – sqrt(3)i.

[hailstormb]

Is it worth doing any of the past VCAA exam papers from 2007-2012?