VCE Methods Question Thread!

[knightrider]

Here are some hints, and a partial solution which leads to the answer. Please attempt this question yourself (without looking at solution), and don't try to randomly look at hints unless you need them. The first hint is free though.

Hint 1

We want to find the value of x from the point such that:

• , which represents the point is a stationary point.
• , indicating that the point is on the x-axis (ie. y = 0).

Thus we have the simultaneous equations:



Now we want to find the value of x such that these two equations are satisfied. We know that the point has y-value of 0. So our point, so far, is .

Hint 2

Hint 3

Now you may have two equations, the left hand side of both of those equations being polynomials of degree two.

You could solve both of them using the quadratic formula on both equations, and find the value x such that both equations are satisfied. But this could get ugly.

On the other hand, like in hint 2, we removed the damned term. What about this time, we remove the term?

Think about it.

End solution

The two equations, when using hints 1 and 2, are:



Following from hint 3, we could multiply the second equation by so  that we can remove altogether:



Hence:


Resulting in the desired answer:


So the desired turning point is therefore with .

Thanks so much e^1 
Really helped 

[Berimbolo King]

VCAA 2014 Exam 2 question 4 part a
having trouble with it can someone explain how they got that answer
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2014/2014mmcas2-w.pdf

[@#035;3]

Hey guys quick few q's; thanks in advance

1)   If cos(x) =8/17  and 270° < x < 360°, then  sin(x)  must equal:
A    -9/17
B    -15/17
C    5/17
D    -5/17
E    -8/17                                  I got B; is this correct?


2) The graph of y=-1/3*tan(a*pi*x/2)  has an asymptote at:
A   –a
B    pi/2
C    pi/a
D    1/a
E    a/pi                  I got D is this correct?

3) The solution (to 3 decimal places) to the equation 6*cos(x)-2=0  over the domain [-pi,pi] is:
A    1.231
B    -1.911
C    1.911
D    -1.231           How would I approach this one..

cheers!
 

[keltingmeith]

Hey guys quick few q's; thanks in advance

1)   If cos(x) =8/17  and 270° < x < 360°, then  sin(x)  must equal:
A    -9/17
B    -15/17
C    5/17
D    -5/17
E    -8/17                                  I got B; is this correct?


2) The graph of y=-1/3*tan(a*pi*x/2)  has an asymptote at:
A   –a
B    pi/2
C    pi/a
D    1/a
E    a/pi                  I got D is this correct?

Seem fine to me.

3) The solution (to 3 decimal places) to the equation 6*cos(x)-2=0  over the domain [-pi,pi] is:
A    1.231
B    -1.911
C    1.911
D    -1.231           How would I approach this one..

Use your calculator

[RazzMeTazz]

Hi, I'm finding it really hard to understand probability! For the question attached below, could someone please explain why the first step calculates the number of selections of taking 2 black shirt from a total of 4?

Thanks

[RazzMeTazz]

With conditional probability, for e.g. to find Pr ( A | B ) =  Pr (A∩B) / Pr (B) =[ Pr (A) * Pr(B) ] / Pr (B)

Does that mean events A and B would have to be independent, if you use this formula?

Since only with independent events you can multiply both the individual probabilities together to find the probability of when they will both occur?

Thanks

[keltingmeith]

VCAA 2014 Exam 2 question 4 part a
having trouble with it can someone explain how they got that answer
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2014/2014mmcas2-w.pdf

P(something is super)=0.1 (because 10% of the population is "super")
.: P(X>x)=0.1
1-P(X<x)=0.1
.: P(X<x)=0.9
Now, just inverse norm.

Hi, I'm finding it really hard to understand probability! For the question attached below, could someone please explain why the first step calculates the number of selections of taking 2 black shirt from a total of 4?

Thanks

They found that because they were using the law:
P(A)=(number of A)/(total number)

They wanted to know the probability of selecting 2 black shirts out of 7 - well, there are 4 black t-shirts, so that's two out of 4 selections are what we want. The totaly number of selections, though, is 2 out of 7. Just plug that in above, and ta-da.

A little unsure why you might be doing these questions, though - they're not really relevant to 3/4 methods.

With conditional probability, for e.g. to find Pr ( A | B ) =  Pr (A∩B) / Pr (B) =[ Pr (A) * Pr(B) ] / Pr (B)

Does that mean events A and B would have to be independent, if you use this formula?

Since only with independent events you can multiply both the individual probabilities together to find the probability of when they will both occur?

Thanks

No - to use this formula, P(A∩B) has to be known /without/ using the multiplicative law that you've used. Notice that in your last step, you can simplify the fraction even more? This happens because the events are independent. Truthfully, using the conditional probability rule for independence events (as you have) is quite useless.

[garytheasian]

Here are some hints, and a partial solution which leads to the answer. Please attempt this question yourself (without looking at solution), and don't try to randomly look at hints unless you need them. The first hint is free though.

Hint 1

We want to find the value of x from the point such that:

• , which represents the point is a stationary point.
• , indicating that the point is on the x-axis (ie. y = 0).

Thus we have the simultaneous equations:



Now we want to find the value of x such that these two equations are satisfied. We know that the point has y-value of 0. So our point, so far, is .

Hint 2

Hint 3

Now you may have two equations, the left hand side of both of those equations being polynomials of degree two.

You could solve both of them using the quadratic formula on both equations, and find the value x such that both equations are satisfied. But this could get ugly.

On the other hand, like in hint 2, we removed the damned term. What about this time, we remove the term?

Think about it.

End solution

The two equations, when using hints 1 and 2, are:



Following from hint 3, we could multiply the second equation by so  that we can remove altogether:



Hence:


Resulting in the desired answer:


So the desired turning point is therefore with .

Hey do you mind explaining line 2 please and how that helps with the solution? I'm extremely confused and not sure how to process what you did and the following steps.
Thanks

[e^1]

Hey do you mind explaining line 2 please and how that helps with the solution? I'm extremely confused and not sure how to process what you did and the following steps.
Thanks

Hid response in spoiler to avoid giving away hints.

Spoiler

Well, we have two equations from hint 1. The second equation is a polynomial of degree 3, and I don't know the formula (or method) to solving that. So I simply rewrote the equation so that it also includes the first equation; I can then make that equal to zero so that the new equation is now just polynomial of degree 2.

In that case I played a little bit on the second equation to see if it could be simplified, with all the information I currently knew.

Hint 3 suggests to do a similar approach by removing the x^2 term, rather than using the quadratic equations on both of them. That way, we only get a linear polynomial (look in the 'end solution' spoiler) and that is easy to solve, for x.

[cosine]

When we are given the domain of a really tedious graph, y = 1-9sin^2(x) and are asked to find the range, would it be feasible to just sub in the domain endpoints and assume that is the range of the graph? (Mind you this is a non-calc question)

The domain of y= 1-9sin^2(x) is [0, pi], how would you personally find the range of this graph?

[keltingmeith]

When we are given the domain of a really tedious graph, y = 1-9sin^2(x) and are asked to find the range, would it be feasible to just sub in the domain endpoints and assume that is the range of the graph? (Mind you this is a non-calc question)

Give it a shot, then tell us if you can/can't.

The domain of y= 1-9sin^2(x) is [0, pi], how would you personally find the range of this graph?

Personally? I'd start with finding the range of sin(x) over that domain, and then slowly transforming that to 1-9sin^2(x).

[cosine]

Give it a shot, then tell us if you can/can't.

I am not sure, hence why I asked here...

[StupidProdigy]

I am not sure, hence why I asked here...

The basic function f(x)=sin(x) has a minimum output of -1 and a maximum output of 1. What happens when you square either of these two outputs? You get positive one, the maximum output. The minimum of sin(x)^2 is a bit trickier, and requires a logical approach. The minimum input of sin(x) was -1 but this became 1 when squared, so we know we have to find another value which will give the minimum of sin(x)^2. Think about it logically, what is the specific value of x, where -1<=x<=1 which will give the lowest output for sin(x)^2? Well any negative input becomes positive, so the lowest positive value will give the minimum of sin(x)^2. In our case the lowest positive number will be zero (can I even say zero is positive  ). So putting it all together the maximum output of sin(x)^2 is 1, minimum output is 0, therefore the range is 1-9*1-->-8 and 1-9*0-->1, so [-8,1]
Luckily the minimum also happens to occur at the end of the domain restriction in the question, otherwise you'd normally need to check the minimum occurs somewhere in the restricted domain

[keltingmeith]

I am not sure, hence why I asked here...

Reason I asked you to check the endpoints yourself:

f(x)=1-9sin^2(x)
f(0)=1-9sin^2(0)=1
f(pi)=1-9sin^2(pi)=1

So, this clearly doesn't say much about the range at all. Don't be scared to try things out! That's a part of the journey.

[bulbasaur]

Can someone help me with part (f) please??
Thanks in advance