VCE Methods Question Thread!

[knightrider]

How would you find the equation of a trig graph if you were given point such as

x=50,y=27
x=40,y=23
x=23,y=12

[keltingmeith]

Hi, can someone explain how I would do this... And what if they asked for one solution, or an infinite number of solutions? It comes up in almost every prac exam I've done and I still get it wrong 

Consider the following simultaneous equations,

mx + y= 2
2x + (m-1)y =m

Find the value(s) of m for which there are no solution.

We can write simultaneous linear equations (such as this one) in the form of a matrix, like so:


For this matrix to have a solution, the determinant must be non-zero - from this, we can infer that there will be no solution or infinite solutions if the determinant is zero. So, let's find the determinant:


So, we have a zero determinant if m=0 or m=2. Let's sub this into the original lines:

For m=-1,
-x+y=2
2x-2y=-1 ----> -x + y = -1/2

For m=2,
2x+y=2
2x+y=2

So, for m=-1, our two lines will be parallel, and so there will be no solutions.
For m=2, the lines are exactly the same, and so there are infinite solutions.

Finally, we will have exactly one solution whenever m does not equal 2 or -1.

[#J.Procrastinator]

We can write simultaneous linear equations (such as this one) in the form of a matrix, like so:


For this matrix to have a solution, the determinant must be non-zero - from this, we can infer that there will be no solution or infinite solutions if the determinant is zero. So, let's find the determinant:


So, we have a zero determinant if m=0 or m=2. Let's sub this into the original lines:

For m=-1,
-x+y=2
2x-2y=-1 ----> -x + y = -1/2

For m=2,
2x+y=2
2x+y=2

So, for m=-1, our two lines will be parallel, and so there will be no solutions.
For m=2, the lines are exactly the same, and so there are infinite solutions.

Finally, we will have exactly one solution whenever m does not equal 2 or -1.

Thank you so much!

What if we were asked to find the value(s) of  two pronumerals where they have an infinite number of solutions, such as:

kx + 4y=2n
2x + (k+2)y= -1

I let the determinant of the matrix equal to 1 and solved for k. But what about the n? Can you give me a hint as to what to do?

[keltingmeith]

Thank you so much!

What if we were asked to find the value(s) of  two pronumerals where they have an infinite number of solutions, such as:

kx + 4y=2n
2x + (k+2)y= -1

I let the determinant of the matrix equal to 1 and solved for k. But what about the n? Can you give me a hint as to what to do?

So this situation, k will tell you one of two things. Either,

1. There is exactly one solution.
2. There are no solutions OR infinite solutions.

If we have the latter scenario, then n will determine one of two things:

1. Our lines are parallel.
2. The lines are the same.

Hint: one value of n will give option 2, every other value of n will give 1.

[IndefatigableLover]

Just have a question with b ii) and below (just some hints could be fine as well), I worked out a & b i though:

a):

b i):

[psyxwar]

I'm guessing they just want you to use linear approximation.

So delta y=dy/dx * delta x

[Reus]

Could some one please do "find the 4th term of (x + 5)7"? Thanks.

[Equilibriaas]

Could some one please do "find the 4th term of (x + 5)7"? Thanks.

First term = 7C0 x^7= x^7
Second term = 7C1 (x)^6 times (5)^1
Third Term= 7C2 (x)^5 times (5)^2
The fourth term is 7C3 (x)^4 times (5)^3= 35 x (x^4) x 125= 4375x^4

HOPE THIS HELPS

[BLACKCATT]

Could some one please do "find the 4th term of (x + 5)7"? Thanks.

Are these questions still on the current study design?

[Reus]

First term = 7C0 x^7= x^7
Second term = 7C1 (x)^6 times (5)^1
Third Term= 7C2 (x)^5 times (5)^2
The fourth term is 7C3 (x)^4 times (5)^3= 35 x (x^4) x 125= 4375x^4

HOPE THIS HELPS

Thank you!!

[keltingmeith]

Are these questions still on the current study design?

I don't think it is explicitly, but it's just binomial theorem which is covered in unit 1/2, so I'd call it fair-game.

[Yacoubb]

Hey,

Does anybody have a hold of the TSSM 2012 Exam 2 solutions for methods? If so, could you please PM me? Thank you

[LiquidPaperz]

is there any tricks you guys have with probability? do you rely on formulas or more so on common knowledge and logical thinking.

- Year 11 atm

[silverpixeli]

is there any tricks you guys have with probability? do you rely on formulas or more so on common knowledge and logical thinking.

- Year 11 atm

Combination of the following: (d/w if you arent familiar with the terms im using below for year 11)

knowing and getting the probability definitions and syntax like what Pr(X) means, what random events are, knowing and understanding intuitively mutual exclusiveness and independence and conditional probability, and formulas for things like transitions and binomial distributions

experience with lots of probability questions so that you find it easy to tell what type of probability question you're dealing with, though not in a 'okay this is binomial, step 1...' way, just in a broad way of classifying what concept a question is based around

being able to visualise the setup of a question, often with the aid of tree diagrams or venn diagrams, drawing a probability density function, or similar tools if necessary


with all of that, problem solving is very logical - a lot of the time the path to the answer is quite simple because the tricky part of probability is understanding how everything in the question fits together


What you can do at the year 11 curriculum level:

make sure you have a really good grasp of any formulas you're studying and why they work and make sense, and make sure you get a handle on the language used in probability like outcomes, events, random events, random variables, all that

[Yacoubb]

What is the difference between mutually exclusive and independent?