VCE Methods Question Thread!

[Gogo14]

Can end points also be turning points?

[Quantum44]

Can end points also be turning points?

I don't see why not. For instance you could restrict sin(x) to [0,pi/2]and pi/2 is a turning point and endpoint.

[Gogo14]

I don't see why not. For instance you could restrict sin(x) to [0,pi/2]and pi/2 is a turning point and endpoint.

According to our teacher, end points cannot be turning points. I lost marks on a test for it last term, and wanted to check if my teacher is right

[Quantum44]

According to our teacher, end points cannot be turning points. I lost marks on a test for it last term, and wanted to check if my teacher is right

Should it be an issue though, when you graph a function you label the coordinates of endpoints, you don't have to state whether it is a turning point or endpoint.

[Gogo14]

Should it be an issue though, when you graph a function you label the coordinates of endpoints, you don't have to state whether it is a turning point or endpoint.

the question asked for the turning points in a restricted domain

[zhen]

Can end points also be turning points?

I don't know whether your teacher is right or not, but I think that if turning points aren't endpoints this is how I would explain it. Suppose you have a function y=x^2, where the domain is [-4,0]. You don't know what will happen when x>0. Suppose this was actually a hybrid function, and y=x, when x>0. Then, at that point x=0 it is no longer a turning point. Basically since we don't know if the function will continue normally we can't call the endpoint point a turning point. So y=x^2, where the domain is [-4,0], then since we don't know what will happen after x=0, we can't call it a turning point. Attached is the graph of what I'm talking about

[LPadlan]

For which values of m does the equation mx^2-2mx+3=0 have:
a) two solutions for x
b) one solution for x?

[Quantum44]

For which values of m does the equation mx^2-2mx+3=0 have:
a) two solutions for x
b) one solution for x?

Discriminant = 4m^2 - 12m

For two solutions, solve discriminant > 0 for the values of m
For one solution, the discriminant = 0 for the value of m

[Shadowxo]

Can end points also be turning points?

Also want to add, a turning point is usually defined as a change in the sign of the gradient, which cannot happen it it's an end point, because it has no points after the end point that could have a different gradient sign. Therefore, end points cannot be turning points (but they can be stationary points as the gradient is zero).

[Ahmad_A_1999]

Could I please get some help with these two Multiple Choice please!

http://imgur.com/a/72K3s

(27) I don't know how to approach this :/

(28) I would like to know how to correctly approach this, I usually fluke tan graph questions 

Cheers

[zhen]

Could I please get some help with these two Multiple Choice please!

http://imgur.com/a/72K3s

(27) I don't know how to approach this :/

(28) I would like to know how to correctly approach this, I usually fluke tan graph questions 

Cheers

Pretty sure that 27 is B and 28 is C. Someone should correct me if I'm wrong, since I'm not too sure about these.
For question 27 the number in the log has to be greater than 0. So (x-2)^2 >0. This holds true for any value except x=2. For the range, the range of a logarithm is generally all real values unless it's restricted. This makes my answer B.
For question 28 the period is pi, so it has to be dilated by a factor of 1/2 from the y-axis, since the period of tan(x) is 2pi, meaning there needs to be a 2 in front of the x. Also, since it's translated pi/4 units it needs to be x-pi/4, making C my answer.

[Ahmad_A_1999]

Pretty sure that 27 is B and 28 is C. Someone should correct me if I'm wrong, since I'm not too sure about these.
For question 27 the number in the log has to be greater than 0. So (x-2)^2 >0. This holds true for any value except x=2. For the range, the range of a logarithm is generally all real values unless it's restricted. This makes my answer B.
For question 28 the period is pi, so it has to be dilated by a factor of 1/2 from the y-axis, since the period of tan(x) is 2pi, meaning there needs to be a 2 in front of the x. Also, since it's translated pi/4 units it needs to be x-pi/4, making C my answer.

Thanks!  I've got it for 28.

Although with 27 why am I not allowed to move the power '2' to the front of the logarithm? If I do this step and then graph on my calculator I get a different function than if the 2 was not moved to the front of the log 

[Shadowxo]

Could I please get some help with these two Multiple Choice please!

http://imgur.com/a/72K3s

(27) I don't know how to approach this :/

(28) I would like to know how to correctly approach this, I usually fluke tan graph questions 

Cheers

Edit: started writing this but paused for a bit so Zhen beat me to it
27.
so for any log, eg log_ab, b≠0, b must be positive aka b>0. This is because a (positive) number to a power can never equal zero or go negative, so what's inside the log therefore cannot be less than zero either. So we know (x-2)²>0 and therefore x≠2 but can be anything else. So, maximal or implied domain is x belongs to R \{2}. And (x-2)² can be any number between 0 and infinity not inclusive (can't be zero) so the log range is R (you can see this using a standard log graph). So answer is B

Also, 2logx is different to logx² only because you can take the negative values of x when it's squared as it'll end up as a positive number (squaring things can be annoying in maths)

28.
Not sure if you were taught this (don't think you have to know this, just a bit of additional info), but if you line a ruler up to the right side of a unit circle, the value of the tan of an angle is just the the ruler measurement where the line extends out to the ruler. That's always how I remembered it (image attached). So tan(0)=0. From this you can also tell that the period of tan x is π without having to memorise it, just imagine the angles changing (worked for me and just a little trivia - you don't need to know this but it may help)
From the graph you're given you know that the period is π/2, half of what a regular tan graph would be so you know as period = π/n for tan, n=2. First thing you do is dilate it accordingly, to tan(2x). You also know it's translated π/4 (or negative π/4 or 3π/4 etc, you can just read off the graph for various possible translations) because you know tan(0)=0 and the x intercept/s have moved along by π/4. So, you replace x with (x-π/4), which makes the equation tan(2(x-π/4)), C

[zhen]

Edit: started writing this but paused for a bit so Zhen beat me to it
27.
so for any log, eg log_ab, b≠0, b must be positive aka b>0. This is because a (positive) number to a power can never equal zero or go negative, so what's inside the log therefore cannot be less than zero either. So we know (x-2)²>0 and therefore x≠2 but can be anything else. So, maximal or implied domain is x belongs to R \{2}. And (x-2)² can be any number between 0 and infinity not inclusive (can't be zero) so the log range is R (you can see this using a standard log graph). So answer is C

Wasn't the answer to the first one B not C. That's what I'm getting from your working out.

[Shadowxo]

Wasn't the answer to the first one B not C. That's what I'm getting from your working out.

Oops yes, spent all the time working out and didn't thoroughly look through the question's answers