A Thread For Questions

[ngRISING]

hey guys, ima need a lot of mm help so i decided instead of making a thread everytime i need help, i'll have a thread that anyone  else can use as well. guess i'll start lol

what is the factor theorem? .

what is the expansion method that is really quick and has C's in it. something binomial expansions. i need a bit of help on that..

ty ty

[Ilovemathsmeth]

Factor theorem: If (x - a) is a factor of P(x) where p(x) = polynomial if and only if P(a) = 0.

It's very similar to the remainder theorem.

That is the binomial theorem. It's not on the current VCAA course. It's a little confusing but don't over think it, if you want to revise it just in case (maybe you should) then just write out the formula a couple of times.

[kdgamz]

i dont bother trying to memorise the binomial theorem....i always use Pascal's triangle

[m@tty]

You would need theorem if it was to the power of over 20.

[TrueTears]

The binomial theorem states:



Notice this only works for linear expressions.

This can be easily proved by deriving it from any simple linear expression to the power of 'low' n.

[TonyHem]

Few questions:
and for x e [0,infinite), then is defined over what domain?

I thought since they are both defined over [0,3], that would be the new domain but it is [0,9]
So I graphed the new function and it is from (-infinite,9] but why is that limited at 0? edit: is 0 the lowest because
How do I work out these questions?

2. and for x e R.
a) show g(f(x)) exists - done
b) write down the rule for g(f(x)) and state domain - done BUT - I got the domain by saying they were both defined over [0,pi] ? It worked for this Q but not for others.
c) Find the range - How do I get this? I tried to do what I did with the domain (-infinite,1] for one and [-3,3] for the other, giving me [-3,1]. range is [-8,1]
And say if you have the domain and sub it into the equation to find the y-value, is there any way to see whether this is the lowest/highest y-value?

3. Find all values of k such that equation has exactly one solution.

I tried discriminant and ended up with K = -0.25 but then also got k>0?
Basically, I let u = 2^x so I had

used discriminant to get 1+4k = 0, k = -1/4
and if I use the quadratic formula, I get
so doesn't   have to be greater than 0? If I solve it, then I get k > 0

[TrueTears]

1. You might want to have a look at ran g dom f. Then restrict ran g. That will solve your question.

2. . The domain is [I don't think it's meant to be as in your question, typo?]

Now, imagine the graph of , the max and min values of is 1 and 0 respectively.

Thus the range is

EDIT: Yes the domain is , what do you mean if worked for this question and not for others? The result that domain of = domain of holds for all functions f and g. Provided that ran g dom f

[TonyHem]

1. You might want to have a look at ran g dom f. Then restrict ran g. That will solve your question.

2. . The domain is [I don't think it's meant to be as in your question, typo?]

Now, imagine the graph of , the max and min values of is 1 and 0 respectively.

Thus the range is

3. Different question but same principle, have a read that will solve your dilemma: http://vcenotes.com/forum/index.php/topic,16831.msg173738.html#msg173738

1) Still not sure, range of g(x) is [0,infinite) and domain of f(x) is [-3,3]. So they are both defined over [0,3] and yeah I get that it's [0,9] when I graph it but why doesn't finding where both functions are defined work? What other way is there other than thinking about what the graph looks like?
Yeah made a typo on question 2. I get the range now but for the domain, isn't what I did for Q1, the same as Q2 but Q1 is wrong and Q2 wasnt?
3)I've read that thread before, still not sure + don't have the solutions to check the answers.
Is the answer k = 0 and k = -1/4?

[TrueTears]

1. Okay I'll guide ya, at the moment ran g is not a subset of dom f, so you need to restrict ran g so that it is.

2. You need to satisfy 2 conditions here

1.

2.

Thus this condition is that

[Gloamglozer]

1.   I know how supplementary angles work but can someone please explain to me how complementary angles work?

2.  How would I solve for the 1st quadrant angles in:

for

[TrueTears]

1. Sorry, it's probs just my understanding, but isn't complementary angles just 2 angles that add up to 90 degrees? I'm not quite sure what you mean by how they "work".

2. Let

So





Thus .

The period is

So by adding and subtracting to our 'basic' answer we will find all the solutions within the domain, namely:



Thus .

[Gloamglozer]

1.  If you use Essentials, it's on page 201.  I'm a bit confused as to how the book has explained it.

2.  Ah ok.  So you do it that way.  I just looked at and because it is positive, the answers can only lie in the first and third quadrants.  I thought that to find the first quadrant angle, you just add to the "basic" answer.

Oh and btw, with your answer, aren't you missing as an answer?

[TrueTears]

lol yeah is also a solution, thought it was , silly me !

[TrueTears]

1.  If you use Essentials, it's on page 201.  I'm a bit confused as to how the book has explained it.

2.  Ah ok.  So you do it that way.  I just looked at and because it is positive, the answers can only lie in the first and third quadrants.  I thought that to find the first quadrant angle, you just add to the "basic" answer.

Oh and btw, with your answer, aren't you missing as an answer?

And yeah your way is also fine, you can add to the basic answer for , but I do it after I get the basic answer for x. Whatever floats your boat




So this is just a matter of transformations, the angle 'x' is first reflected in the x axis and then it's rotated 90 degrees anti-clockwise. So looking at the RHS picture, after you've applied the transformations, the angle 'x' that the triangle makes with the positive y axis is the same as the angle it makes with the positive x axis. They are similar triangles because you have 2 sides the same and 1 right angle in both. Since cosine is defined as adjacent on hypotenuse, it is the same as .

The other 'formulas' can also be derived this way.

[Gloamglozer]

Ah I get the complementary stuff now.  Thanks TT.  Didn't realise that the angle was reflected in the x axis first.


Going back to the last question, I reckon your method is better.  But I don't understand how you got the period.  How does:



Equate to ?

I got zero.