VCE Methods Question Thread!

[lzxnl]

This is a pretty cool question... And definitely not VCE level. Requires multivariate calculus or knowledge of Fourier series.

Let's see if it's VCE level.
x^2 + ax + b^2 = 0
Real solutions iff a^2 - 4b^2 >= 0, a and b both real
So the question boils down to
What is the chance of picking two real numbers a and b in [0,1] such that the above inequality is satisfied?
As a, b >= 0, we have a >= 2b

This question does actually suck.
For a given b, define the probability that a is chosen such that a>2b to be P(b).
Then, for any given b, a is restricted to the domain [2b, 1]. Clearly a<=1, b<=1/2.
The probability is then given by the length of this interval divided by the original interval length, 1, so it's just 1 - 2b
P(b) = 1 - 2b

Let's interpret the meaning of this probability. It is the chance of picking a to meet the constraint given that this b value is chosen. Now this is a continuous probability question, so you have to state the question as 'chance b is chosen within some tiny interval'.
The chance that a is picked and b is picked to meet the constraint is then equal to the chance that a meets the constraint * chance that value of b is picked. If b is in a differential range [b, b+ db], then the chance b is picked in this range is just db/1 = db. So the total differential probability dp = (1 - 2b) db
Integrating this from 0 to 1/2 gives p = b - b^2 -> 1/4

Is that something like what the answers say?

[schooliskool]

What's the question? I couldn't see it because LaTex died.

Two real numbers, a and b, are randomly chosen from the interval [0,1]. Find the probability that the equation x^2+ax+b^2  has no real solutions

[odeaa]

Let's see if it's VCE level.
x^2 + ax + b^2 = 0
Real solutions iff a^2 - 4b^2 >= 0, a and b both real
So the question boils down to
What is the chance of picking two real numbers a and b in [0,1] such that the above inequality is satisfied?
As a, b >= 0, we have a >= 2b

This question does actually suck.
For a given b, define the probability that a is chosen such that a>2b to be P(b).
Then, for any given b, a is restricted to the domain [2b, 1]. Clearly a<=b<=1/2.
The probability is then given by the length of this interval divided by the original interval length, 1, so it's just 1 - 2b
P(b) = 1 - 2b

Let's interpret the meaning of this probability. It is the chance of picking a to meet the constraint given that this b value is chosen. Now this is a continuous probability question, so you have to state the question as 'chance b is chosen within some tiny interval'.
The chance that a is picked and b is picked to meet the constraint is then equal to the chance that a meets the constraint * chance that value of b is picked. If b is in a differential range [b, b+ db], then the chance b is picked in this range is just db/1 = db. So the total differential probability dp = (1 - 2b) db
Integrating this from 0 to 1/2 gives p = b - b^2 -> 1/4

Is that something like what the answers say?

yeah thats correct

i'm in a whole new world of confusion, give me a minute to get my head around that ahah

[keltingmeith]

Clearly a<=b<=1/2.

You sure about that one?

[toodles]

yeah thats correct

i'm in a whole new world of confusion, give me a minute to get my head around that ahah

(^dont know if that works. havent posted before)

Anyways, in response to the "Two real numbers, a and b are chosen from the interval [0,1]...", what lzxnl said is basically right, however the answer to the question is 3/4, not 1/4.

But thats not super important. Whats important is to have an easy way to solve questions like this, because, while they are very difficult, they are still within math method's level (even if I doubt there will be one on the upcoming exam).
In this case, the first step is finding the determinant and setting it to <0      a^2 - 4*b^2 < 0
Therefore,       a^2 < 4*b^2    ->    a<2*b      (discard -2*b as a>=0)

So now we have   a<2*b  which is hard to make sense of in terms of probability, but there is a quick and (fairly) easy way to do it. (I can't really draw a diagram here, so ill try my best to explain it)

Create a Cartesian plane with 'a' as the x-axis and 'b' as the y-axis (or the other way round, personal preference) and create a square from (0,1) to (1,1) to (1,0) back to (0,0), The area of this square (A = 1) denotes all possible combinations of real 'a' and 'b' over [0,1], where the total area = 100% of possibilities. (Think of this square as the sample space)
Now we know that    x^2 + a*x + b^2 = 0    has no real solutions for    a<2*b   so graph that relation onto your square.
Assuming 'a' is on the horizontal, you will get a straight line going from (0,0) to (1,1/2), and the Required Region will be above the line and contained within the square.
Now, to work out the required region, you take (Total area) - (area below line) =  (1) -  1/2 * 1/2 * 1 = 3/4    , giving you the final answer of 3/4.

To solve other questions of this type, its fairly easy to construct a sample space as a visual aid and work from there.

[babushka818]

(^dont know if that works. havent posted before)

Anyways, in response to the "Two real numbers, a and b are chosen from the interval [0,1]...", what lzxnl said is basically right, however the answer to the question is 3/4, not 1/4.

But thats not super important. Whats important is to have an easy way to solve questions like this, because, while they are very difficult, they are still within math method's level (even if I doubt there will be one on the upcoming exam).
In this case, the first step is finding the determinant and setting it to <0      a^2 - 4*b^2 < 0
Therefore,       a^2 < 4*b^2    ->    a<2*b      (discard -2*b as a>=0)

So now we have   a<2*b  which is hard to make sense of in terms of probability, but there is a quick and (fairly) easy way to do it. (I can't really draw a diagram here, so ill try my best to explain it)

Create a Cartesian plane with 'a' as the x-axis and 'b' as the y-axis (or the other way round, personal preference) and create a square from (0,1) to (1,1) to (1,0) back to (0,0), The area of this square (A = 1) denotes all possible combinations of real 'a' and 'b' over [0,1], where the total area = 100% of possibilities. (Think of this square as the sample space)
Now we know that    x^2 + a*x + b^2 = 0    has no real solutions for    a<2*b   so graph that relation onto your square.
Assuming 'a' is on the horizontal, you will get a straight line going from (0,0) to (1,1/2), and the Required Region will be above the line and contained within the square.
Now, to work out the required region, you take (Total area) - (area below line) =  (1) -  1/2 * 1/2 * 1 = 3/4    , giving you the final answer of 3/4.

To solve other questions of this type, its fairly easy to construct a sample space as a visual aid and work from there.

I get that these individual concepts might be in the course but is this really the level of knowledge/application ability required in methods?! I'm so dead.. Thanks for explaining that though!

[lzxnl]

You sure about that one?

I typed that up at like 3 AM. Cut me some slack please
All fixed now

(^dont know if that works. havent posted before)

Anyways, in response to the "Two real numbers, a and b are chosen from the interval [0,1]...", what lzxnl said is basically right, however the answer to the question is 3/4, not 1/4.

But thats not super important. Whats important is to have an easy way to solve questions like this, because, while they are very difficult, they are still within math method's level (even if I doubt there will be one on the upcoming exam).
In this case, the first step is finding the determinant and setting it to <0      a^2 - 4*b^2 < 0
Therefore,       a^2 < 4*b^2    ->    a<2*b      (discard -2*b as a>=0)

So now we have   a<2*b  which is hard to make sense of in terms of probability, but there is a quick and (fairly) easy way to do it. (I can't really draw a diagram here, so ill try my best to explain it)

Create a Cartesian plane with 'a' as the x-axis and 'b' as the y-axis (or the other way round, personal preference) and create a square from (0,1) to (1,1) to (1,0) back to (0,0), The area of this square (A = 1) denotes all possible combinations of real 'a' and 'b' over [0,1], where the total area = 100% of possibilities. (Think of this square as the sample space)
Now we know that    x^2 + a*x + b^2 = 0    has no real solutions for    a<2*b   so graph that relation onto your square.
Assuming 'a' is on the horizontal, you will get a straight line going from (0,0) to (1,1/2), and the Required Region will be above the line and contained within the square.
Now, to work out the required region, you take (Total area) - (area below line) =  (1) -  1/2 * 1/2 * 1 = 3/4    , giving you the final answer of 3/4.

To solve other questions of this type, its fairly easy to construct a sample space as a visual aid and work from there.

Actually, your working just proves my answer correct because your equation for the discriminant is wrong. You've solved for a negative discriminant, so you've solved for the complement of the case that I proved. You're meant to take the area above that line which would yield 1/4.
But your explanation is easier to follow than mine.

[grannysmith]

Just did 2011 Exam 2 GG'ed

[knightrider]

How would you do this question attached ?

For this question attached ?

The answer says as follows

A dilation of factor 2 from the y-axis and a reflection in the y-axis  followed by translations of 4 units to the left and 2 upwards.

Shouldn't the answer be

A dilation of factor 2 from the y-axis and a reflection in the x-axis  followed by translations of 4 units to the left and 2 upwards.

Who is right?

Can anyone help with these ?
 

[Cosec]

Just did 2011 Exam 2 GG'ed

How did you go?

[grannysmith]

How did you go?

Dropped 6 marks ((
4f was just lol

[Cosec]

Dropped 6 marks ((
4f was just lol

Dropped 15. GG 40 raw

[knightrider]

Just a question about the methods exam.

as GA1 refers to sacs

GA2 refers to exam 1

GA3 refers to exam 2

Say for example you get a C+ on GA2

and then you get a A+ on GA3

Do these scores ever change like get moderated.
Or what you get on the exam stays as your score?

Many thanks 

[alex1234]

Can anyone clarify this please.
 

yeah i think you're right

[Cosec]

Can anyone confirm this binding for my bound refrence is acceptable?
The little black bits can be pulled out and pages removed or added but other than that. You can remove or add them and its still secured together.
image: (too big to upload here) http://imgur.com/ukgn4c0
 
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