VCE Methods Question Thread!

[j.wang]

thank you so much

[QueenSmarty]

Hey everyone! Just a quick question on the bound reference. I've stuck in sections of practice exams, however some of the sections were too long so I folded it and stuck it in that way (the sections don't fold out of the book, they fold in the book, if that makes sense). Would that be okay or do I need to stick them in so that nothing folds at all?

[chrisjudd00]

Hi all. Just completed the Northern Hemisphere VCAA Methods Exam 1 for 2017. Was just wondering if anyone else has done the same and has found it to be significantly harder than previous VCAA methods exams; and is this to be expected. Thanks in advance.

[gnaf]

yeah me too chris esp the last bit of the last question

Some mcq questions I got very wrong

In pic 1, when I switch x and y and solve for y in the cas, I get 2 eqns. Why do we choose the 1st eqn (the 2nd eqn isn't even an option in the ans)

In pic 2, I found the total area to be a^2+ 0.5*a*2a=2a^2. So the avg value was the 1/(a-(-a)) * integral of 2a^2 from -a to a for dx= 2a^2... the ans is a/2 (D) though

In pic 3, I let f(x)= its inverse, so e^(kx)= (1/k)*loge(x) and solved for x, subbing in the values of k in the ans options aka guess and check. The ans is k=1/e (E) but when I sub k=1/e into e^(kx)= (1/k)*loge(x) and solve for x, I get 2 ans for x (x=2.71....846 and x=2.71....877) which means that there's 2 point of intersections, not one as the question requires?

In pic 4, I let f'(x)=0 and I got 3 ans (including x=2)
For the 2 ans, where x wasn't equal to 2, I let them equal x is less than or equal to 2
So, I got [-8,1] for one and (-infinity, 1] for the other
The ans is D, [-8, infinity) though...

[VanillaRice]

yeah me too chris esp the last bit of the last question

Some mcq questions I got very wrong

In pic 1, when I switch x and y and solve for y in the cas, I get 2 eqns. Why do we choose the 1st eqn (the 2nd eqn isn't even an option in the ans)

In pic 2, I found the total area to be a^2+ 0.5*a*2a=2a^2. So the avg value was the 1/(a-(-a)) * integral of 2a^2 from -a to a for dx= 2a^2... the ans is a/2 (D) though

In pic 3, I let f(x)= its inverse, so e^(kx)= (1/k)*loge(x) and solved for x, subbing in the values of k in the ans options aka guess and check. The ans is k=1/e (E) but when I sub k=1/e into e^(kx)= (1/k)*loge(x) and solve for x, I get 2 ans for x (x=2.71....846 and x=2.71....877) which means that there's 2 point of intersections, not one as the question requires?

In pic 4, I let f'(x)=0 and I got 3 ans (including x=2)
For the 2 ans, where x wasn't equal to 2, I let them equal x is less than or equal to 2
So, I got [-8,1] for one and (-infinity, 1] for the other
The ans is D, [-8, infinity) though...

1) Consider the domain and range of f - these must correspond to the range and domain of the inverse respectively - have a think (or do a quick sketch on your calc) as to which equation it should be. Remember the inverse is analogous to a reflection in the line y = x

2) You have attempted to find a function for the area under the graph. Remember that if the area is under the graph, it is negative.  However, the average value takes into account negative values/areas.

Essentially, the left hand side (the sloped line) of the function is cancelled out, because if you think about it, the average value in this region is 0!
Recall the average value is the value of y which forms a rectangle from x = -a to x = a, and the area of that rectangle is equivalent to the net signed area of f from x = -a to x = a.
So the average value can be given by y in:

Solving for y gives us the average value of a/2.
Another way of solving this question is to find a hybrid function which represents f, and solve using the formula for average value.

3) I can't seem to find the 3rd image, could you double check?

4) Your method looks correct - you should be looking for values of a such that the stationary point at x=2 is the one to the far right. You have found the required intervals, but remember both must true i.e. [-8,1] and (-inf,1] must both be true, and that is only for the interval [-8,1] (which is the intersection of the two). This is not in the list of solutions, but the closest one is D. How can we check D is actually true? Substitute in values greater than 1 and sketch the graph! I've also answered this same question in a different post, so please see my comments there for more information.

Hope this helps

[martinlil]

Yeah I found it tough in some areas too, do any of you guys think this will be similar to what we get this year? 

[Rieko Ioane]

Hi,

For the NHE E1 https://imgur.com/a/QRDC8

How do we also get c>2 as a solution too?

[snowisawesome]

How do you approach the difficult questions in methods? Also, is it true that in unit 3/4 methods exams, multiple choice questions are harder than extended response?

[VanillaRice]

Hi,

For the NHE E1 https://imgur.com/a/QRDC8

How do we also get c>2 as a solution too?

This question requires some careful thought about what happens to both g and its inverse as c changes.
Consider your answer of c = 1. In part a, you had c < 1. What happens when c < 1? The graphs do not intersect each other. What happens as c approaches 1 from the negative direction (i.e. as we move from c = -2 to -1 to 0....). The graphs will get closer and closer to each other (g(x) moves to the left, while its inverse moves downwards). The first time that they touch is c = 1 (have a quick sketch of the two graphs to see what this looks like). However, what happens when we keep going? If we go slightly above c = 1 (say, c = 1.1), the graphs will intersect at two points. Let's move to the case where c = 2. Here, the two graphs intersect twice, one of which is at their turning points (at the point (-1, -1), on the line y=x).  Now, we must consider the domain restriction. Since c = 2 represents where the two graphs intersect at their 'turning points', and we only have half of a 'parabola' anything beyond c = 2 will only have 1 solution (intersection) between the two graphs, since the second half of the 'parabola' doesn't actually exist (due to domain restrictions)!

I hope this makes sense, please let me know if there's any problems

[Rieko Ioane]

This question requires some careful thought about what happens to both g and its inverse as c changes.
Consider your answer of c = 1. In part a, you had c < 1. What happens when c < 1? The graphs do not intersect each other. What happens as c approaches 1 from the negative direction (i.e. as we move from c = -2 to -1 to 0....). The graphs will get closer and closer to each other (g(x) moves to the left, while its inverse moves downwards). The first time that they touch is c = 1 (have a quick sketch of the two graphs to see what this looks like). However, what happens when we keep going? If we go slightly above c = 1 (say, c = 1.1), the graphs will intersect at two points. Let's move to the case where c = 2. Here, the two graphs intersect twice, one of which is at their turning points (at the point (-1, -1), on the line y=x).  Now, we must consider the domain restriction. Since c = 2 represents where the two graphs intersect at their 'turning points', and we only have half of a 'parabola' anything beyond c = 2 will only have 1 solution (intersection) between the two graphs, since the second half of the 'parabola' doesn't actually exist (due to domain restrictions)!

I hope this makes sense, please let me know if there's any problems

Ah, I see thanks. So there's no algebraic way of getting this part I assume?

[VanillaRice]

Ah, I see thanks. So there's no algebraic way of getting this part I assume?

You might be able to do it by equating g(x) and its inverse, and you'll get a quartic equation of sorts. However, it seems like a bit too much effort for an extra mark, so I think this would be the best method (unless of course someone else has a better solution).

EDIT:

How do you approach the difficult questions in methods? Also, is it true that in unit 3/4 methods exams, multiple choice questions are harder than extended response?

Probably the most important thing to remember is that they can't examine you on what isn't on the study design. Every question is a technique you would (or should) have learnt and practised. The difficult questions are only difficult because they are asking you to apply familiar techniques in unfamiliar circumstances (for example, by throwing a heap of different variables at you). Break down each part of the question - what is it really asking you to do?

Multiple choice questions are sometimes viewed as 'difficult' with respect to the effort required to obtain a single mark. I would argue that extended response questions have more avenue to be harder, but that's just my opinion

Hi all. Just completed the Northern Hemisphere VCAA Methods Exam 1 for 2017. Was just wondering if anyone else has done the same and has found it to be significantly harder than previous VCAA methods exams; and is this to be expected. Thanks in advance.

Yeah I found it tough in some areas too, do any of you guys think this will be similar to what we get this year? 

It's quite difficult to predict the difficulty of exams each year. There have definitely been some exams which have been notably more difficult than others. Regardless of this year's exam difficulty, I don't that the distribution of marks and grades shouldn't change too much, since everyone is doing the same exam after all. It's best not to worry about something you can't control, and start thinking about ways you might tackle difficult questions (especially by practising doing some), as well as making sure you are able to do the 'easy' questions

[sertani]

This question was in the 2015 methods exam 2 MC question 10. The assessors report didn't explain how to do it.

The binomial random variable X, has E(X)=2 and Var(X)=4/3.
PR(X=1) is equal to

a) (1/3)^6
b) (2/3)^6
c)1/3*(2/3)^5
d)6*(1/3)*(2/3)^5
e)6*(2/3)*(1/3)^5

Thanks in advance!

[plsbegentle]

This question was in the 2015 methods exam 2 MC question 10. The assessors report didn't explain how to do it.

The binomial random variable X, has E(X)=2 and Var(X)=4/3.
PR(X=1) is equal to

a) (1/3)^6
b) (2/3)^6
c)1/3*(2/3)^5
d)6*(1/3)*(2/3)^5
e)6*(2/3)*(1/3)^5

Thanks in advance!

So remembering that to find the variance of a binom distribution, the formula is np(1-p) and np is the expected value (E(X)), so just sub into the values and solve for p. 2(1-p)=4/3 which is 1/3. Now we know that the p is 1/3, we can also find the n: 1/3n=2 therefore n=6

The last step is to use the binom formula which is i think is somewhere in the formula sheet but 6Cr1=6 1/3^1 is 1/3 and (1-1/3)^5=2/3^5, so the answer is D

[sertani]

So remembering that to find the variance of a binom distribution, the formula is np(1-p) and np is the expected value (E(X)), so just sub into the values and solve for p. 2(1-p)=4/3 which is 1/3. Now we know that the p is 1/3, we can also find the n: 1/3n=2 therefore n=6

The last step is to use the binom formula which is i think is somewhere in the formula sheet but 6Cr1=6 1/3^1 is 1/3 and (1-1/3)^5=2/3^5, so the answer is D

THANKS!!

[syubi]

To be safe, is it better to use un-rounded figures for things such as probability or standard deviation in the following questions, even when in a previous question you have to work out a rounded figure? Thanks!!