VCE Methods Question Thread!

[keltingmeith]

Not exactly related to methods but can someone who does both spesh and methods end up with below 40 raw for methods? Or do they have a much higher chance of getting 40+ raw in methods than someone who only does methods? (feel free to move this to appropriate subforum if needed mods)

Adding to Shadowxo:

You're asking two different questions, believe it or not, because it sounds like you're only asking one.

Is it possible for someone who is studying, or has studied, specialist to get below 40 in methods? Of course - only the top 8% get above 40. In 2016, there were about 2,800 students in specialist, and about 8,800 students in methods. That means that, at most, only 700 of those students could get above 40 in methods.

Is it guaranteed that those 700 spots would be taken by spec students?  Of course not. They're two completely different subjects, made up of different content and different styles of questions.

Is it likely that a lot of them had done spec? Yes, because typically the best students will do spec, plus there is some overlap between the two subjects (mainly in calculus). It's not a direct result of doing specialist, though, and more a coincidence.

[snowisawesome]

I know i've asked this question quite a few times already and I apologise for that, however i'm still confused with it
if you're asked to find the general solution for sin(x) = -1/2,
and you let the base angle = pi + pi/6 = 7pi/6 would it be correct?
Because my book said that the base angle for this question would be -pi/6
Thanks

[Bri MT]

I know i've asked this question quite a few times already and I apologise for that, however i'm still confused with it
if you're asked to find the general solution for sin(x) = -1/2,
and you let the base angle = pi + pi/6 = 7pi/6 would it be correct?
Because my book said that the base angle for this question would be -pi/6
Thanks

The base angle is there purely to help you get your answer. The textbook is trying to teach you a specific method and way of thinking about it which is why they have given that q & a.  VCAA isn't going to ask you "what is the base angle?". As long as you can use a mathematically valid method to solve trig equations they really don't care.

[snowisawesome]

The base angle is there purely to help you get your answer. The textbook is trying to teach you a specific method and way of thinking about it which is why they have given that q & a.  VCAA isn't going to ask you "what is the base angle?". As long as you can use a mathematically valid method to solve trig equations they really don't care.

But woudn't 7pi/6 and -pi/6 produce different answers each?

[VanillaRice]

I know i've asked this question quite a few times already and I apologise for that, however i'm still confused with it
if you're asked to find the general solution for sin(x) = -1/2,
and you let the base angle = pi + pi/6 = 7pi/6 would it be correct?
Because my book said that the base angle for this question would be -pi/6
Thanks

miniturtle pretty much summed it up
I however will add something to clarify why your textbook chooses the 4th quadrant instead of the 3rd. What I'm about to say is beyond the scope of VCE Methods, but we generally want the base angle for sine to be in the domain [-pi/2,pi/2], which is the principle range of the inverse sine function.

However, both are technically correct. Depending on how you write your general solution, you may also need to consider the fact that there are two solutions in one revolution of the unit circle. I'll also quote what I mentioned in a previous comment.

EDIT: To be more precise, you should consider the base angle to be pi/6. This is because sin has two solutions in the unit circle: x = pi +pi/6 = 7pi/6 (like you said), and 0 - pi/6 = -pi/6 (or 2pi - pi/6 = 11pi/6).

So if we're really going to be precise about the "base angle", it should be pi/6, as it forms the basis for generating the two solutions in one period, which we can then apply as a general solution.

Hope this helps

EDIT:

But woudn't 7pi/6 and -pi/6 produce different answers each?

sin(7pi/6) = sin(-pi/6) = -1/2. Remember that sine has two solutions per period.

[keltingmeith]

But woudn't 7pi/6 and -pi/6 produce different answers each?

If you're having trouble believing what people in this thread are saying, try putting their answers into practice. Use this base angle for a general solution, along with your base angle, and sub in different values of n to see what values x is equal to. If they have the same values of x (even if they're for different values of n - it doesn't matter what n gives you what x, as long as an n gives you the x), then both are fine.

[snowisawesome]

Thanks guys for the help

Stuck on a question from my textbook though 
maths quest 11 mm
exercise 2.4
11.A hybrid function is defined by
f(x) = x+a, for (-infinity, -8]
         x^(1/3)+2 for (-8, 8]
         b/x for (8, infinity)
a. determine the values of a and b so that the function is continuous for x Ɛ R, and for these values, sketch the graph of y = f(x)
Use the values of a and b from part a for the remainder of this question
i've found a = 8, and b = 32, and sketched the graph

b. Deteremine the values of k for which the equation f(x) =k has:
i. no solutions
ii. one solution
iii. two solutions
c. find {x: f(x) = 1}

I'm not sure how to do part b and c, would appreciate if someone can help out

EDIT: it's from maths quest 12 mm textbook
as if i'm still in year 11 anymore , year's 12's the big year

[Shadowxo]

Thanks guys for the help

Stuck on a question from my textbook though 
maths quest 11 mm
exercise 2.4
11.A hybrid function is defined by
f(x) = x+a, for (-infinity, -8]
         x^(1/3)+2 for (-8, 8]
         b/x for (8, infinity)
a. determine the values of a and b so that the function is continuous for x Ɛ R, and for these values, sketch the graph of y = f(x)
Use the values of a and b from part a for the remainder of this question
i've found a = 8, and b = 32, and sketched the graph

b. Deteremine the values of k for which the equation f(x) =k has:
i. no solutions
ii. one solution
iii. two solutions
c. find {x: f(x) = 1}

I'm not sure how to do part b and c, would appreciate if someone can help out

EDIT: it's from maths quest 12 mm textbook
as if i'm still in year 11 anymore , year's 12's the big year

for b) I think the best way to do it is simply find it from the graph. 
i) For the part where f(x) = x+8, as x goes to negative infinity so does y, so the min y value is -infinity. You can see from the graph that the maximum y value is at the intersection between x^1/3+2  and \( \frac{32}{x}\), where x=8 hence y=4 is the max y value.
So There are no solutions for k (ie the y value) belongs to (4, infinity)
ii) The graph should help you out here a lot
When x=8, y=4 and as x approaches infinity, y approaches 0.
As x approaches -8 (from the direction of x=8), y approaches 0.
So we can see there are two solutions for x belongs to (-8, infinity) ie k belongs to (0, 4]
And one solution for everything else; x belongs to (-infinity, -8] ie k belongs to (-infinity, 0]

c) I'd simply solve for each of the parts of the function and eliminate any impossible answers
eg for the first part, solve x+8=1
x=-7. Discard this answer as it is outside the range of (-infinity, -8]
And continue for the other parts

[peachxmh]

Hey guys, would really appreciate it if you helped me with this question because these distance, speed and time questions are doing my head in!

A man on a bicycle rides one-third of the way from town A to town B at a speed a km/h and the remainder of the way at 2b km/h. If the distance between the two towns is 9 km, find the time taken to ride from A to B.

I have worked this out to be t = 3/a + 3/b and am fine with this part of the question - the next part of the question is what I'm having trouble with:

If the man had travelled at a uniform rate of 3c km/h, he could have ridden from A to B and back again in the same time.
Show that 2/c = 1/a + 1/b.

If he rode from A to B and back again then the total distance would be 18km, so wouldn't the equation be 18/3c = 3/a +3/b (since t = d/s and he rides from A to B and back again at 3c km/h in the same time as when he rode from A to B at a km/h and 2b km/h)

In that case it would simplify to 6/c = 3/a +3/b and I fail to see how this is related to 2/c = 1/a + 1/b

Please help me understand what I'm doing wrong! Huge thanks in advance and sorry if my explanations are a little convoluted

[A TART]

Hey guys, would really appreciate it if you helped me with this question because these distance, speed and time questions are doing my head in!

A man on a bicycle rides one-third of the way from town A to town B at a speed a km/h and the remainder of the way at 2b km/h. If the distance between the two towns is 9 km, find the time taken to ride from A to B.

I have worked this out to be t = 3/a + 3/b and am fine with this part of the question - the next part of the question is what I'm having trouble with:

If the man had travelled at a uniform rate of 3c km/h, he could have ridden from A to B and back again in the same time.
Show that 2/c = 1/a + 1/b.

If he rode from A to B and back again then the total distance would be 18km, so wouldn't the equation be 18/3c = 3/a +3/b (since t = d/s and he rides from A to B and back again at 3c km/h in the same time as when he rode from A to B at a km/h and 2b km/h)

In that case it would simplify to 6/c = 3/a +3/b and I fail to see how this is related to 2/c = 1/a + 1/b

Please help me understand what I'm doing wrong! Huge thanks in advance and sorry if my explanations are a little convoluted

Take out a common factor
https://i.imgur.com/n8Mkpmo.jpg
If we chuck in numbers, both will still hold true.
eg: Let a=1 b=3

6/c = 3/a +3/b
6/c=3/1+3/3
6/c=3+1
c=6/4
=3/2
 
2/c = 1/a + 1/b
2/c=1/1+1/3
2/c=4/3
c=3/2

[chooby]

Hey guys, would really appreciate it if you helped me with this question because these distance, speed and time questions are doing my head in!

A man on a bicycle rides one-third of the way from town A to town B at a speed a km/h and the remainder of the way at 2b km/h. If the distance between the two towns is 9 km, find the time taken to ride from A to B.

I have worked this out to be t = 3/a + 3/b and am fine with this part of the question - the next part of the question is what I'm having trouble with:

If the man had travelled at a uniform rate of 3c km/h, he could have ridden from A to B and back again in the same time.
Show that 2/c = 1/a + 1/b.

If he rode from A to B and back again then the total distance would be 18km, so wouldn't the equation be 18/3c = 3/a +3/b (since t = d/s and he rides from A to B and back again at 3c km/h in the same time as when he rode from A to B at a km/h and 2b km/h)

In that case it would simplify to 6/c = 3/a +3/b and I fail to see how this is related to 2/c = 1/a + 1/b

Please help me understand what I'm doing wrong! Huge thanks in advance and sorry if my explanations are a little convoluted

Dividing both sides by 3 (or multiplying both sides by 1/3) would get you to the right answer.

[snowisawesome]

Should I be concerned if i cannot do most if not all of the questions on past vcaa methods exams at this stage of the year?

[chooby]

Should I be concerned if i cannot do most if not all of the questions on past vcaa methods exams at this stage of the year?

What. It's only like a few weeks into term 1, it's okay if you don't know everything by now. There's definitely a lot of concepts (e.g. binomial distribution and a bunch of key skills in he study design that we haven't even touched on yet, so there's still plenty of time to learn.

[snowisawesome]

What. It's only like a few weeks into term 1, it's okay if you don't know everything by now. There's definitely a lot of concepts (e.g. binomial distribution and a bunch of key skills in he study design that we haven't even touched on yet, so there's still plenty of time to learn.

Thanks
I'm attempted a few exam 2's and got like max ~10/80 I think

[Sine]

Should I be concerned if i cannot do most if not all of the questions on past vcaa methods exams at this stage of the year?

Thanks
I'm attempted a few exam 2's and got like max ~10/80 I think

have you learnt all the content? - if yes there is a problem

but for a large proportion of the state many are just starting to learn the course and there is no point testing yourself on something you haven't learnt yet.