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March 19, 2024, 04:12:34 pm

Author Topic: VCE Methods Question Thread!  (Read 4795813 times)  Share 

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michaeljacksonftw

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Re: VCE Methods Question Thread!
« Reply #16485 on: May 19, 2018, 05:43:12 pm »
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$$ \mbox{Let y = }x^{2}\sin(\frac{2\pi}{x}) $$
Product rule:
$$ \frac{dy}{dx} = \frac{d}{dx}[x^{2}]\sin(\frac{2\pi}{x}) + x^{2}\frac{d}{dx}[\sin(\frac{2\pi}{x})] $$
Chain rule:
$$  = 2x\sin(\frac{2\pi}{x}) + x^{2}(\cos(\frac{2\pi}{x})\frac{d}{dx}[\frac{2\pi}{x}])  $$
$$  = 2x\sin(\frac{2\pi}{x}) + x^{2}(\cos(\frac{2\pi}{x})\frac{-2\pi}{x^{2}})  $$
$$  = 2x\sin(\frac{2\pi}{x}) + (-2\pi\cos(\frac{2\pi}{x}))  $$

EDIT: Sorry for the delayed post, I presumed page 1101 was the most recent page and that this question hadn't been answered yet.
(2pix), NOT(2pi/x)

RuiAce

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Re: VCE Methods Question Thread!
« Reply #16486 on: May 19, 2018, 05:46:10 pm »
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michaeljacksonftw

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Re: VCE Methods Question Thread!
« Reply #16487 on: May 19, 2018, 08:05:09 pm »
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for the cambridge textbook for unit 3/4 methods, are we allowed to use a calculator for the multiple choice and extended response questions in the chapter review?

TheBigC

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Re: VCE Methods Question Thread!
« Reply #16488 on: May 20, 2018, 11:57:07 am »
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(2pix), NOT(2pi/x)

oops. Sorry - I misread there

michaeljacksonftw

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Re: VCE Methods Question Thread!
« Reply #16489 on: May 20, 2018, 02:15:10 pm »
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oops. Sorry - I misread there
All good

Seno72

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Re: VCE Methods Question Thread!
« Reply #16490 on: May 22, 2018, 07:35:40 am »
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Hey guys. Have any of you guys used the exampro maths trial sacs and examinations? Is it worth it?
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Lear

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Re: VCE Methods Question Thread!
« Reply #16491 on: May 23, 2018, 06:00:54 pm »
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If f(x)=x^2 and x is between 0 and 5. When writing the interval over which it is strictly increasing, do we include the end point. I.e is this function strictly increasing for values [0,5) or [0,5].
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heelloo112233

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Re: VCE Methods Question Thread!
« Reply #16492 on: May 24, 2018, 04:23:03 am »
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When given three points:
(8,2) (4,5) (2,10)
and the rule: y=aX10to the power of -(t+b divided by C)
how do you find the variables

heelloo112233

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Re: VCE Methods Question Thread!
« Reply #16493 on: May 24, 2018, 04:23:47 am »
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When given three points:
(8,2) (4,5) (2,10)
and the rule: y=aX10to the power of -(x+b divided by C)
how do you find the variables

S200

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Re: VCE Methods Question Thread!
« Reply #16494 on: May 24, 2018, 09:09:53 am »
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When given three points:
(8,2) (4,5) (2,10)
and the rule: y=aX10to the power of -(x+b divided by C)
how do you find the variables
Just to clarify, is the equation you are trying to solve

???

Cause it was kind of obscure from your comment...
« Last Edit: May 24, 2018, 09:12:21 am by S200 »
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heelloo112233

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Re: VCE Methods Question Thread!
« Reply #16495 on: May 24, 2018, 06:05:49 pm »
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yes but the negative is in the front of the whole power. Sorry about that

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Re: VCE Methods Question Thread!
« Reply #16496 on: May 24, 2018, 06:45:42 pm »
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If f(x)=x^2 and x is between 0 and 5. When writing the interval over which it is strictly increasing, do we include the end point. I.e is this function strictly increasing for values [0,5) or [0,5].
wouldn't it be (0,5] because f'(0)=0 and f'(5)=10
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RuiAce

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Re: VCE Methods Question Thread!
« Reply #16497 on: May 24, 2018, 07:08:12 pm »
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wouldn't it be (0,5] because f'(0)=0 and f'(5)=10
It depends on what you interpret to be "strictly increasing".

The usual convention is that if f'(0) = 0 (i.e. the curve is stationary) at a single point, but not along an entire interval or something, then it's ok. A function that would "not" be ok in this context could be something like
\[ f(x) = \begin{cases}x^2 +1& x \in [0,\infty)\\ 1 & x \in (-\infty,0) \end{cases}\]
because the derivative is 0 along the entire interval \( (-\infty, 0)\). The reason why this is ok is because of the actual definition of monotonicity: \( x_1 < x_2 \implies f(x_1) < f(x_2) \).

Of course, in the VCE they might not do this. But personally I don't see why you wouldn't
« Last Edit: May 24, 2018, 07:11:24 pm by RuiAce »

Lear

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Re: VCE Methods Question Thread!
« Reply #16498 on: May 24, 2018, 08:36:27 pm »
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How about if the end point (5) was not included? I.e what if the values of x were between [0,5)? Would it still strictly increase over the interval [0,5]?
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RuiAce

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Re: VCE Methods Question Thread!
« Reply #16499 on: May 24, 2018, 08:39:54 pm »
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How about if the end point (5) was not included? I.e what if the values of x were between [0,5)? Would it still strictly increase over the interval [0,5]?
Provided 5 is in the domain of the function (which, I guess we've implied it already), being monotonic increasing on [0,5) and [0,5] would be considered equivalent in my books