VCE Methods Question Thread!

[TheAspiringDoc]

Thanks VR! (Edit: is an asymptote considered a discontinuity?)

Last question for the day - I promise!!

In Cookiedream's "how I got a raw 44 in methods" it says "For general solutions, never forget to write k ∈Z in each step of your working out"

Whereas in TrueTears' explanation of how to find general solutions, the "n∈Z" is only added in the last step.

Spoiler

General solutions to circular functions.
TrueTears

Many people just simply apply a formula (similar to the one provided in the Essentials 3/4 text), some knows where it comes from but most just mindlessly apply the formulas without knowing where it comes from and if the question gets tricky then they will get stuck. Hopefully this tutorial help you understand how we go about finding general solutions to circular functions in a more systematic fashion.

We will consider the general solutions to each of our 3 main circular functions, , and . First we will look at the function.

Example 1.

Find the general solution to

First notice why does this question say find the GENERAL solution? This is because no domain is specified, if no domain is specified then there are infinitely many solutions to the above equation. [What this means is that if you sketch the graph of and draw the horizontal line , then there are infinitely many x values which gives a y value of ]

So here is how I would go about solving this question.

Let

Now we have the equation so let us find the 2 basic solutions to then we will use to use the 2 basic solutions to to find the 2 basic solutions for .

Solving is quite trivial. This is an "exact value" question.



Now if you do not know how I solved the above equation, you need to review your circular function fundamentals ASAP.

Substituting the 2 basic value into [1] yields:





Now the next step is finding the GENERAL solutions for x.

Look at the following graph of

(Image removed from quote.)

The red line is the line , as you can see, since a domain is not specified, it crosses the sin graph infinitely many times.

Now why did we find TWO basic solutions and not just one? As you can see the purple lines represent the solutions obtained from and to get the other purple line solutions we simply have to add and subtract periods away from our basic solution of .

But as you can see from the graph, no matter how many periods we add or subtract we will never end up on the green lines and this is what the other basic is for!

If we add and subtract periods away from then we reach all the other green solutions.

So what is the period of the graph? Well it's , again go back and review your fundamentals if you don't know how to calculate periods.

So our general solution is:

where

Now notice some of you might go, "wait what? Didn't you say we must SUBTRACT periods as well as adding them?"

This is another common mistake students often make, look at my definition of in my answer. I said is an INTEGER which means ITSELF can take on negative values, eg, n = ...-3,-2,-1,0,1,2,3...

So for example say n = 1

Then we have So here we are adding periods.

But if n = -1 then we have:



Which is equivalent to:

So here we are subtracting periods.

So that is why we don't write our solution as:

where

Because the subtracting periods is already taken into account due to the restriction on

However some of you like to have the in the middle and another way of writing the answer is this: (Note the difference!)

where

Why do we need in the middle here? This is because n is now an element of NATURAL numbers or 0, which means n = 0, 1, 2, 3 ...

So the 'subtracting' periods is NOT taken into account from our restriction on n, that is why we need to put in the middle since we need to 'manually' take into consideration ADDING and SUBTRACTING periods.

Both way of presenting the answer is fine, pick one and stick to it

Example 2.

Find the general solution to

NO DIFFERENCE, APPROACH THIS QUESTION THE EXACT SAME WAY AS EXAMPLE 1, TRY IT YOURSELF!

Example 3.

Find the general solution to

Now the tan function is a tiny bit different in that we only need to find ONE basic solution and not TWO. The rest of the principles of adding and subtracting periods is all the same.

Here is how I would solve this question:

Let us first sketch the graph of below:

(Image removed from quote.)

The red line is the line and the purple lines are the solutions to the equation.

As you can see from the graph, by finding any value of the x value that corresponds to the purple line and then adding and subtracting periods from that x value we get, we will be able to find all the solutions! So we don't need to solve for TWO basic solutions, ONE will be enough! (You might ask why do we need to solve for just one basic solution, graphically I have explained it, but algebraically this is because is a one to one function while sin and cos are not. You don't really need to know this though)

So let us solve it!





Thus the general solution is:

  where (Note the period of is and not )

OR another way of writing it is:

where


Well that's it folks, enjoy and post any questions if you don't understand!

So, they must be using different methods of working out. Which I way is quicker and safer and more commonly used by VCAA?

[snowisawesome]

I've heard that they're better than the chapter questions but still slightly off vcaa questions. Ultimately practice exams mimic the Vcaa Exam the most.

Thanks, do you by any chance know if sketching sin, cos, and tan graphs and also finding the general solution can appear in exam 1? Or will they only appear in exam 2?
Thanks

[VanillaRice]

Thanks VR!

Last question for the day - I promise!!

In Cookiedream's "how I got a raw 44 in methods" it says "For general solutions, never forget to write k ∈Z in each step of your working out"

Whereas in TrueTears' explanation of how to find general solutions, the "n∈Z" is only added in the last step.

Spoiler

So, they must be using different methods of working out. Which I way is quicker and safer and more commonly used by VCAA?

Safer way? Write it in every step. It doesn't take long to add it to the end of every line where it's necessary. You can also risk forgetting if you leave it to the last step. VCAA doesn't always have a rule book about how everything should be done, but it's convention in mathematics to define any variables you are using (e.g. k or n as an integer).

Thanks, do you by any chance know if sketching sin, cos, and tan graphs and also finding the general solution can appear in exam 1? Or will they only appear in exam 2?
Thanks

Those types of questions can and have appeared in Exam 1.

[snowisawesome]

Safer way? Write it in every step. It doesn't take long to add it to the end of every line where it's necessary. You can also risk forgetting if you leave it to the last step. VCAA doesn't always have a rule book about how everything should be done, but it's convention in mathematics to define any variables you are using (e.g. k or n as an integer).
Those types of questions can and have appeared in Exam 1.

Do you know which year they appeared in?
Thanks

[VanillaRice]

Do you know which year they appeared in?
Thanks

From a quick look, graphing of trig functions has appeared in the 2017 NHT, and 2006. I can't seem to find any past questions regarding general solutions (so I may have been wrong here), however I would not exclude learning the general solution to trig equations, only because it hasn't appeared previously. For example, it could come up in your SACs. I've found it mentioned in a supplement for the previous study design. IIRC the dot point for general solutions hasn't really changed between study designs, so I would say general solutions do have the potential to come up.

[snowisawesome]

From a quick look, graphing of trig functions has appeared in the 2017 NHT, and 2006. I can't seem to find any past questions regarding general solutions (so I may have been wrong here), however I would not exclude learning the general solution to trig equations, only because it hasn't appeared previously. For example, it could come up in your SACs. I've found it mentioned in a supplement for the previous study design. IIRC the dot point for general solutions hasn't really changed between study designs, so I would say general solutions do have the potential to come up.

But wouldn't they still be pretty unlikely to show up again if the last vcaa exam they showed up was in 2006? And it's rather confusing learning how to sketch trig graphs by hand, especially with transformations?

[Lear]

I would recommend against relying on probabilities and learn it anyway. They are likely to be on your SAC and possibly exam. Why take the risk?

[Sine]

Write down nEz the moment you introduce n and it doesnt hurt to write it down repeatedly.

Way too early in the year to be thinking abiut not studying portions of the subject l. If you actively choose not to revise a section you only have yourself to blame if it comes up on the exam.

[snowisawesome]

Thanks guys for the advice.
Is it possible for someone to show me how to sketch these graphs? I've read my textbook but it still doesn't really make sense
1. y = 6+4*cos(π/6*x-π/3) for 0<x<24
2. y = -2*sin(x+π/3)+1 (for one cycle)

And can someone please show me how to find the general solution to this equation? Again i've read my textbook but still don't understand it
cos2(x+π/3)=1/2

A = 21-3*cos(π*x/12),
when I let x = 4,
i get A = 21-3*cos(4*π/12)
Does anyone know how to simplify the above without a calculator?
Thanks

-(22-5cos((pi*x/12))
= - 22+5cos-((pi*x/12))
is this correct?

[DBA-144]

Thanks guys for the advice.
Is it possible for someone to show me how to sketch these graphs? I've read my textbook but it still doesn't really make sense
1. y = 6+4*cos(π/6*x-π/3) for 0<x<24
2. y = -2*sin(x+π/3)+1 (for one cycle)

And can someone please show me how to find the general solution to this equation? Again i've read my textbook but still don't understand it
cos2(x+π/3)=1/2
A = 21-3*cos(π*x/12),
when I let x = 4,
i get A = 21-3*cos(4*π/12)
Does anyone know how to simplify the above without a calculator?
Thanks

For the second one, you need to identify transformations of the graph. There is reflection in the x axis horizontal tranlation pi over 3 units left and then vertical translation up one unit. So now we know the transformations, we also know the mean position is y=1. As the graph of sine has been dilated by factor of 2 from x axis the amplitude changes to 2. So range is [1-2,1+2]. Period is the same. Also keep in mind the horizontal trnaslation.

So...

Sin x starts at origin. Under the transformation, it becomes (0-pi over 3, 0+ 1). Or (-pi/3, 1)
Try using these for the rest of the 5 main points. Remember as the graph is inverted now, it will go to minimum point now, not maximum.
Edit: sorry its rest of the 4 points. This is bcos the graph starts at mean goes to min then back to mean, then max and finally the mean again. This is the one cycle you need.

[snowisawesome]

Is there a list of skills/topics/questions that will DEFINITELY show up on exam 1?

[Lear]

Anything in the study can show up on exam 1. It’s unwise to play the prediction game, however, there are some trends that can be noticed by going through a few exams. I’d recommend looking for yourself so you get a better understanding

[snowisawesome]

Anything in the study can show up on exam 1. It’s unwise to play the prediction game, however, there are some trends that can be noticed by going through a few exams. I’d recommend looking for yourself so you get a better understanding

Thanks!
Do you by any chance know how 125a^3 - 27b^3 would be factorised?
And is it a good idea to rote learn how each type of polynomial would be factorised?
Thanks

[Opengangs]

Thanks!
Do you by any chance know how 125a^3 - 27b^3 would be factorised?
And is it a good idea to rote learn how each type of polynomial would be factorised?
Thanks

[snowisawesome]

Thanks!
There is a question in my textbook, which i referred to the worked solutions to but still seems confusing. Could you please explain it if possible?

If kx^2 + 4x - k+2=0, show that the equation has a solution for all values of k
this is the worked solutions from the textbook answers
discriminant = 16-4*k*(-k+2)
=16+4k^2-8k
=4(k^2-2k+4)
=4(k^2-2k+1^2 - 1^2+4)
=4[(k+1)^2+3]
4(k+1)^2 + 12
why does it change to 4[(k+1)^2+3] when the previous lines has 4(k^2-2k+1^2 - 1^2+4)
like how does the minus become a plus?
the rest of the answer had
as(k+1)^2 >0
4(k+1)^2>0
and 4(k+1)^2 +12 >0
discriminant is always greater than zero, therefore the equation will always have a solution for all values of k
could you please also explain this part of the solutions?

Thanks