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March 29, 2024, 02:02:29 am

Author Topic: VCE Methods Question Thread!  (Read 4802468 times)  Share 

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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14865 on: May 07, 2017, 08:48:20 pm »
+1
I'm not too confident with this, but I'll give it a try.


Note that there are going to be multiple ways to transform this graph to get the desired result, so this may not be the answer in the book. Also, someone should check my working, since I'm not too confident with transformations.

Looks like your working is good :)
Completed VCE 2016
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deStudent

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Re: VCE Methods Question Thread!
« Reply #14866 on: May 08, 2017, 05:02:44 pm »
0
http://imgur.com/a/Dcwsk
For a) is it wrong to say the function is discontinuous at x=3? The limits from both sides approaches 3 which suggests it is actually continuous but by observation, it seems logical to think that is discontinuous..
Thanks

Joseph41

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Re: VCE Methods Question Thread!
« Reply #14867 on: May 08, 2017, 05:09:33 pm »
0
Welcome to the Maths Methods Questions thread! ;D

To post a question or response, you'll first need to make a free ATAR Notes account. It should take about four seconds! Then, simply scroll down to the bottom of this thread and type in the "Quick Reply" box, as shown below!


Alternatively, feel free to browse 994 pages of previous questions and answers! Navigate the thread with these page number buttons, found at the top and bottom of each page.



All the best! :)

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zhen

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Re: VCE Methods Question Thread!
« Reply #14868 on: May 08, 2017, 05:47:34 pm »
+1
http://imgur.com/a/Dcwsk
For a) is it wrong to say the function is discontinuous at x=3? The limits from both sides approaches 3 which suggests it is actually continuous but by observation, it seems logical to think that is discontinuous..
Thanks
I would say that the function is discontinuous at x=3. But you should check the answer to see what the answers say.

deStudent

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Re: VCE Methods Question Thread!
« Reply #14869 on: May 08, 2017, 05:55:16 pm »
0
I would say that the function is discontinuous at x=3. But you should check the answer to see what the answers say.
Thanks. The answer doesn't say anything about x=3.

zhen

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Re: VCE Methods Question Thread!
« Reply #14870 on: May 08, 2017, 06:29:33 pm »
+2
Thanks. The answer doesn't say anything about x=3.
Actually I checked the answer and it says the function is discontinuous at x=3 and x=4

deStudent

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Re: VCE Methods Question Thread!
« Reply #14871 on: May 08, 2017, 07:06:57 pm »
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Actually I checked the answer and it says the function is discontinuous at x=3 and x=4
thanks man and soz about that, turns out the worked solutions suck ass.

ish708

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Re: VCE Methods Question Thread!
« Reply #14872 on: May 09, 2017, 08:59:36 am »
+1
Can anyone help me out with these questions?

Consider the function f:R→R, f(x)=m(x-a)s(x-b), where m, a, b ∈ R and s ∈ N.
What generalisations can be made?

and

Let f: R→R, f(x)=(x-a)s(x-b) where a, b ∈ R, a < b and s ∈ N. Find the values of p for which f(x)=p has zero, one, two or three solutions when s = 1, 2, 3, 4 and 5. What generalisations can be made?

simrat99

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Re: VCE Methods Question Thread!
« Reply #14873 on: May 09, 2017, 10:32:02 am »
+1
Can someone please help me with the following question?
Find the values of x for which the graph of y=f(x) has a stationary point and state the nature of each stationary point. Consider 0≤x≤2π only.
f(x)=2cosx+2sinxcosx
Thanks :)

lilhoo

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Re: VCE Methods Question Thread!
« Reply #14874 on: May 09, 2017, 06:49:27 pm »
+1
Hi guys I know this isn't a Methods 3/4 question but I am just asking for some tips.

Are there tips for sketching sine and cosine graphs with horizontal and vertical translations? I often find myself not knowing where to start on the axis because the horizontal scaling is different from the period. Do I just estimate?
 For example, if the equation is y=2sin3(x- (pi)/4)

While the period is 2(pi)/3 and the phase shift is pi/4 to the right, where would I even start on the axis because I can't seem to estimate.

Thank you.  ;D

geminii

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Re: VCE Methods Question Thread!
« Reply #14875 on: May 09, 2017, 06:50:23 pm »
+1
Does the graph of (x-a)^4(x-b) ever have three solutions? How can I tell if it is three solutions? I think there is a point of inflection at (a,0)  (stationary or not???) but I am not sure...and if there was, would this mean there are three solutions at p=0 because there is also the other intercept of b at y=0??

Thanks in advance!
(I've attached a pic of the graph so what I've said makes more sense - I've used a = 1, b = 2 because the question says that a<b)

EDIT: Is there also a stationary point of inflection at (a,0) in the graph (x-a)^3(x-b)? And if so would there be three solutions at p=0 as well??
« Last Edit: May 09, 2017, 07:08:53 pm by geminii »
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geminii

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Re: VCE Methods Question Thread!
« Reply #14876 on: May 09, 2017, 07:29:20 pm »
+1

My immediate though is when ever a < b , in (x-a)^n (x-b) and n is 4 specifically, you won't have 3 solutions. Only when a>b.

Reason being when the graph is not reflected, and a>b the maximum turning point will be above the point of inflection when graphed. Allowing for 3 solutions.

When a < b the the contrary applies and the graphs maximum turning point won't be above the point of inflection. Hence 3 solutions are not possible.

*I hope someone corrects me on this* But perhaps in general it can said when a<b with respect to (x-a)^n(x-b), and where n is greater than or equal to three, the graph f (x)=p cannot have 3 solutions.

Awesome, thank you! That last sentence especially helped! so just to confirm it's only when y = (x-a)^2(x-b) that it has three solutions?
2016-17 (VCE): Biology, HHD, English, Methods, Specialist, Chemistry

2018-22: Bachelor of Biomedical Science @ Monash Uni

Max Kawasakii

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Re: VCE Methods Question Thread!
« Reply #14877 on: May 09, 2017, 07:49:44 pm »
+3
Awesome, thank you! That last sentence especially helped! so just to confirm it's only when y = (x-a)^2(x-b) that it has three solutions?

I deleted my reply because I was completly wrong, don't try do things in your head it never works.



Does have three solutions we can prove this by simply graphing it.






and


So it can be said that for f(x)=p ; p will have three solutions when
P is between -256/3215<P<0

EDIT: Just graph things my man, don't think there is a trick to it..
« Last Edit: May 09, 2017, 07:53:21 pm by Max Kawasakii »
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LPadlan

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Re: VCE Methods Question Thread!
« Reply #14878 on: May 10, 2017, 07:18:04 pm »
0
Find the y-coordinate and the gradient at the point on the curve corresponding to the given value of x:
y=x^2(2x+1)^1/2 at x=0
can someone please show me a step by step solution by using the quotient rule.
this is on page 399 question 2c on methods 3-4

Max Kawasakii

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Re: VCE Methods Question Thread!
« Reply #14879 on: May 11, 2017, 01:06:28 am »
+3
Find the y-coordinate and the gradient at the point on the curve corresponding to the given value of x:
y=x^2(2x+1)^1/2 at x=0
can someone please show me a step by step solution by using the quotient rule.
this is on page 399 question 2c on methods 3-4
This particular question can be differentiated using the product rule.

hence
furthermore
hence
The prodcut rule states that

As a result


To find the gradient at the point where x=0, you sub 0 into dy/dx everwhere you see x



To find the y coordinate you sub x=0 into f(x)



Hence the gradient of f(x) when x=0 is 0
The Y-Coordinate when x=0 is 0


Hope this Answers your question, for future reference the Quotient Rule is used when your function is in the form of

If g(x)≠0 then;


« Last Edit: May 11, 2017, 01:15:34 am by Max Kawasakii »
'Strive not to be a success, but rather to be of value.' - Albert Einstein