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#### TrueTears

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« on: November 26, 2011, 10:04:52 pm »
+28

If you have general questions about the VCE Maths Methods course or how to improve in certain areas, this is the place to ask!

Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you, including TuteSmart tutors! So you may even get multiple answers from different people offering their insights - very cool.

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OTHER METHODS RESOURCES

Original post.
So for those of you keen to indulge yourself in the wonders of methods 3/4 throughout the holiday coz u have nothing better to do, post away your questions in this thread. Everyone can discuss and benefit from this, i'll personally try to answer all the questions that's posted too

the more discussion the better, once a question is posted feel free to post all kinds of solutions, talk about other things related to the question and feel free to go on mathematically related tangents, that's how one improves at maths!

Please keep ALL discussion on this thread relevant & specific to Maths Methods. Any general discussion regarding study habits etc should be directed to the appropriate board. Ask yourself whether your post relates to Methods specifically. Posts that do not fit the scope of this thread will be removed without notice.
« Last Edit: February 26, 2020, 03:22:24 pm by PhoenixxFire »
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#### Hutchoo

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##### Re: VCE Methods Question Thread!
« Reply #1 on: November 26, 2011, 10:21:13 pm »
+5
I'm ready to TT it up.

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##### Re: VCE Methods Question Thread!
« Reply #2 on: November 27, 2011, 08:05:07 pm »
+1
hey TT when you did the Methods book over the holidays, how many exercises/chapters did you cover a day

#### monkeywantsabanana

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##### Re: VCE Methods Question Thread!
« Reply #3 on: November 27, 2011, 08:39:18 pm »
0
$f: R^{+}\rightarrow R , f(x)=x^{\frac{-1}{2}}$
$g: R\rightarrow R , g(x)=3-x$

By restricting the domain of $g$, obtain a function $g*$ such that $f o g*$ exists

I've done this so far:

For $f o g*$ to exist:
$ran g* \subseteq dom f$
$ran g* \subseteq (0,infinity)$

I then drew $g(x)$ in full domain. Can anyone suggest where to go from here?

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#### b^3

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##### Re: VCE Methods Question Thread!
« Reply #4 on: November 27, 2011, 08:43:21 pm »
+5
So the range of g has to be (0, infinity)
Now find the x values for the extreme y values.
So 3-x goes to infinity for x approching -infinity.
3-x=0
then x=3
So this domain restirction will give the required range.
i.e. dom g*=(-infinity,3)
Then use that to obtain the fog* function.
fog*(x)=(3-x)^(-1/2) for x is a member of (-infinity,3)

EDIT:small edit
EDIT2: that was a bad explanantion sorry. If you want me to clear it up, just ask.
« Last Edit: November 27, 2011, 08:45:52 pm by b^3 »
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#### monkeywantsabanana

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##### Re: VCE Methods Question Thread!
« Reply #5 on: November 27, 2011, 08:46:18 pm »
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AH YES YES I SEE WHERE I WENT WRONG, 'cus i subbed (-infinity,3) into ran g* to check if it's a subset, now i realise i was subbing in the domain!

Thanks b^3

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#### TrueTears

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##### Re: VCE Methods Question Thread!
« Reply #6 on: November 27, 2011, 09:25:10 pm »
+7
hey TT when you did the Methods book over the holidays, how many exercises/chapters did you cover a day
hey, i cant rmb exactly now, but i rmb i just did as much as i cud everyday, time seemed to fly pass lol

$f: R^{+}\rightarrow R , f(x)=x^{\frac{-1}{2}}$
$g: R\rightarrow R , g(x)=3-x$

By restricting the domain of $g$, obtain a function $g*$ such that $f o g*$ exists

I've done this so far:

For $f o g*$ to exist:
$ran g* \subseteq dom f$
$ran g* \subseteq (0,infinity)$

I then drew $g(x)$ in full domain. Can anyone suggest where to go from here?
yup pretty much what b^3 said, let's further this, try out this following question: it should test your understanding

Let a be a positive number. Let $f : [2, \infty) \to \mathbb{R}, f(x)=a-x$  and let  $g:(-\infty,1] \to \mathbb{R}, g(x)= x^2 +a$. Find all values of 'a' for which f(g(x)) and g(f(x)) both exist

[this question is mainly directed towards monkeywantsabanana, think about it first, then others can post some solutions etc]
« Last Edit: November 27, 2011, 10:08:27 pm by TrueTears »
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#### monkeywantsabanana

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##### Re: VCE Methods Question Thread!
« Reply #7 on: November 27, 2011, 10:37:07 pm »
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I DON'T HAVE THE SOLUTION

I've got this far:

$f :[2, \infty) \to \mathbb{R}, f(x)=a-x$
$dom = [2,\infty)$
$ran = (-\infty,a-2]$

$g:(-\infty,1] \to \mathbb{R}, g(x)= x^2 +a$
$dom = (-\infty,1]$
$ran = [a,\infty)$

For $f(g(x))$ to exist:$ran g \subseteq dom f$

For $g(f(x))$ to exist: $ran f \subseteq dom g$

I've stated the obvious... Not sure where to go from here.

I did a series of sketches and I came up with this:

$2-x^{2}\leq a\leq x+1$ which does not look correct at all.

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#### brightsky

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##### Re: VCE Methods Question Thread!
« Reply #8 on: November 27, 2011, 10:46:06 pm »
0
i got 2 =< a =< 3
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#### luffy

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##### Re: VCE Methods Question Thread!
« Reply #9 on: November 27, 2011, 11:00:22 pm »
+5
I DON'T HAVE THE SOLUTION

I've got this far:

$f :[2, \infty) \to \mathbb{R}, f(x)=a-x$
$dom = [2,\infty)$
$ran = (-\infty,a-2]$

$g:(-\infty,1] \to \mathbb{R}, g(x)= x^2 +a$
$dom = (-\infty,1]$
$ran = [a,\infty)$

For $f(g(x))$ to exist:$ran g \subseteq dom f$

For $g(f(x))$ to exist: $ran f \subseteq dom g$

I've stated the obvious... Not sure where to go from here.

I did a series of sketches and I came up with this:

$2-x^{2}\leq a\leq x+1$ which does not look correct at all.

Well, you actually did the right thing in your first couple of steps. You just need confidence in yourself.

f(x) = a - x. Dom = [2, infinity). Range = (-infinity, a - 2]

g(x) = x^2 + a. Dom = (-infinity, 1], Range = [a, infinity)

For f(g(x)) to exist, range of g(x) is a subset or equal to the domain of f(x). Hence, [a,infinity) must be a smaller set than [2,infinity). How is that possible? a must be greater than or equal to 2! In mathematical notation, $a \geq 2$.

Similarly, for g(f(x)) to exist, the range of f(x) must be a "smaller set" than the domain of g(x). Hence, (-infinity, a - 2] is 'smaller' as a group than (-infinity, 1]. How is that possible? a - 2 must be less than or equal to 1! In maths notation, $a - 2 \leq 1$. Or, in other words $a\leq 3$

So now you just find when both of them hold true, i.e. their 'intersection', which will be when $2 \leq a \leq 3$.

Hope I helped.

#### monkeywantsabanana

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##### Re: VCE Methods Question Thread!
« Reply #10 on: November 28, 2011, 04:32:53 pm »
0
Yeah that helped a lot, thanks luffy!

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#### jaydee

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##### Re: VCE Methods Question Thread!
« Reply #11 on: November 28, 2011, 11:19:50 pm »
0
hey i have an m/c question from the essential book. ch 10 q6
The graph of the function with rule y = f (x) has a local maximum at the point with
coordinates (a, f (a)). The graph also has a local minimum at the origin but no other
stationary points. The graph of the function with rule y = −2 f(x/2) + k where k is a
positive real number has:
A   a local maximum at the point with coordinates (2a,−2 f (a) + k)
B   a local minimum at the point with coordinates (a/2, 2 f (a) + k
C a local maximum at the point with coordinates (a/2 ,−2 f (a) + k
D a local maximum at the point with coordinates (2a,−2 f (a) − k)
E a local minimum at the point with coordinates (2a,−2 f (a) + k)

this ones a tricky one.. cant get my head round it
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#### b^3

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##### Re: VCE Methods Question Thread!
« Reply #12 on: November 28, 2011, 11:32:05 pm »
+3
hey i have an m/c question from the essential book. ch 10 q6
The graph of the function with rule y = f (x) has a local maximum at the point with
coordinates (a, f (a)). The graph also has a local minimum at the origin but no other
stationary points. The graph of the function with rule y = −2 f(x/2) + k where k is a
positive real number has:
A   a local maximum at the point with coordinates (2a,−2 f (a) + k)
B   a local minimum at the point with coordinates (a/2, 2 f (a) + k
C a local maximum at the point with coordinates (a/2 ,−2 f (a) + k
D a local maximum at the point with coordinates (2a,−2 f (a) − k)
E a local minimum at the point with coordinates (2a,−2 f (a) + k)

this ones a tricky one.. cant get my head round it
Firstly you need to work out the transformations to get y = −2 f(x/2) + k from y=f(x).
So firstly rearrange it to look like y=f(x)
(k-y)/2=f(x/2)
so $y=\frac{k-y'}{2}$ and $x=\frac{x'}{2}$
Solve for the images
2y=k-y'     x'=2x
y'=k-2y
so (x,y)-->(2x,k-2y)
So that is a dilation of factor 2 from the y-axis and a dilation of factor 2 from the x-axis, reflection in the x-axis and a translation of k units in the positive direction of the y-axis.
So the point (a, f(a)) becomes
(a,f(a))-->(2a,k-2f(a)). So that is where the original local maximum was, but since we reflected in the x-axis it is now a local minimum, so the answer will be E.
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#### jaydee

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##### Re: VCE Methods Question Thread!
« Reply #13 on: November 28, 2011, 11:42:51 pm »
0
oh thanks! i get it. now it seems so obvious > <
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#### Dominatorrr

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##### Re: VCE Methods Question Thread!
« Reply #14 on: December 02, 2011, 08:05:43 pm »
+1
Is there ever a question that pops up where you use the binomial theorem to expand an expression?