VCE Methods Question Thread!

[Redoxify]

Why did you disregard the negative solution of k? Is this meant to be included? Why/why not? Cheers

you do get -+root(3)/2, but it's an equalities question

root(3)/2 > -root(3)/2
so yes in reality you can put it as

k > root(3)/2 > -root(3)/2

[keltingmeith]

-4k^2<-3
k^2>3/4
k>root(3)/2

Sub k=-2 into the original equation, and tell me what's wrong with your solution.

[cosine]

you do get -+root(3)/2, but it's an equalities question

root(3)/2 > -root(3)/2
so yes in reality you can put it as

k > root(3)/2 > -root(3)/2

But the answer is

which doesn't seem to match with what you have said, any ideas?

[cosine]

But the answer is

which doesn't seem to match with what you have said, any ideas?

Saying that, does anyone know how they actually get to the above statement? Like i understand how we solve for k, in which we get
but how do we know to make it in the format above?

[keltingmeith]

Saying that, does anyone know how they actually get to the above statement? Like i understand how we solve for k, in which we get
but how do we know to make it in the format above?

Hint 1: You can't take a square root across an inequality like that.
Hint 2: Have you tried drawing a graph?

[cosine]

Hint 1: You can't take a square root across an inequality like that.
Hint 2: Have you tried drawing a graph?

I know, that's why I am asking why that doesn't work, and how could you draw a graph for it if it's just a straight line? (unless it's supposed to be a straight line). But seriously though, why can't we just take the square root across an inequality like that? And if we can't do it this way, how are we meant to do it? Which takes me back to my original question

Thanks redoxify and eulerfan101, your help is  greatly appreciated.

[keltingmeith]

I know, that's why I am asking why that doesn't work, and how could you draw a graph for it if it's just a straight line? (unless it's supposed to be a straight line). But seriously though, why can't we just take the square root across an inequality like that? And if we can't do it this way, how are we meant to do it? Which takes me back to my original question

Thanks redoxify and eulerfan101, your help is  greatly appreciated.

I don't think that what you've given us is a straight line - looks like a quadratic to me.
Also, why would a straight line make it hard to draw a graph?

As for why we can't take the square root across an inequality - it's for the same reason we can't take it across an equality, the square root function breaks down for negative numbers. (notice how the positive side of the parabola still gives us the correct section for the inequality, and it's just the negative portion that's wrong?)

[cosine]

I don't think that what you've given us is a straight line - looks like a quadratic to me.
Also, why would a straight line make it hard to draw a graph?

As for why we can't take the square root across an inequality - it's for the same reason we can't take it across an equality, the square root function breaks down for negative numbers. (notice how the positive side of the parabola still gives us the correct section for the inequality, and it's just the negative portion that's wrong?)

Haha, you're right..
So, in saying that, how do we conclude the answer? I get now how the inequality sign changes, but still a bit unsure as to how to get it into the final format a<k<b

[keltingmeith]

Haha, you're right..
So, in saying that, how do we conclude the answer? I get now how the inequality sign changes, but still a bit unsure as to how to get it into the final format a<k<b

These may be of assistance.

[cosine]

These may be of assistance.

That is a wonderful set of notes, but how do we conclude the answer? I get now how the inequality sign changes, but still a bit unsure as to how to get it into the final format a<k<b

[keltingmeith]

That is a wonderful set of notes, but how do we conclude the answer? I get now how the inequality sign changes, but still a bit unsure as to how to get it into the final format a<k<b

I'm curious if you actually read through the notes? Because at the end, they go through solving inequalities just like this.

Also, your answer should not be of the form a<k<b.

[knightrider]

For questions like these attached.

is the 3 on the end included with the sin or not ?

What would we do if we come across situations like these in exams?

[knightrider]

For this question attached.

Give the domain of .

I got R\{0} and the answers have

Who is right?

[nerdgasm]

For your first question, if someone just walked up to me on the street and asked me if the 3 was included with the sine or not, I would say 'no'. Normally, I would place brackets around whatever is meant to go in the sine function - I guess sometimes people take shortcuts and write "sin 2x" instead of "sin(2x)". If this occurs in an exam, I would also assume that the 3 is not part of the sine. I'm sure that VCAA exam writers will go over their exams and check for possible ambiguities, and they would write sin(3x+3) if that is what they want.

For your second question, I agree with the book's answer. The key is to note the original domain you were given (the bit in brackets before the arrow). What that means is 'x, such that x is greater than 1/3'. In other words, you need to only consider the parts of 1/(3x-1) where x is greater than 1/3. This is only the right branch of the hyperbola. Hence, in this case, f(x) is only the right branch. Now, it turns out that when we restricted the domain of the original function, we also restricted the range of the original function too. The range of the right branch of the hyperbola is (0, inf), and therefore this is the domain of f^-1(x).

[knightrider]

For your first question, if someone just walked up to me on the street and asked me if the 3 was included with the sine or not, I would say 'no'. Normally, I would place brackets around whatever is meant to go in the sine function - I guess sometimes people take shortcuts and write "sin 2x" instead of "sin(2x)". If this occurs in an exam, I would also assume that the 3 is not part of the sine. I'm sure that VCAA exam writers will go over their exams and check for possible ambiguities, and they would write sin(3x+3) if that is what they want.

For your second question, I agree with the book's answer. The key is to note the original domain you were given (the bit in brackets before the arrow). What that means is 'x, such that x is greater than 1/3'. In other words, you need to only consider the parts of 1/(3x-1) where x is greater than 1/3. This is only the right branch of the hyperbola. Hence, in this case, f(x) is only the right branch. Now, it turns out that when we restricted the domain of the original function, we also restricted the range of the original function too. The range of the right branch of the hyperbola is (0, inf), and therefore this is the domain of f^-1(x).

Thanks so much nerdgasm