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March 29, 2024, 05:10:55 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164552 times)  Share 

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Stick

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Re: Specialist 3/4 Question Thread!
« Reply #990 on: December 11, 2012, 10:12:22 am »
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I was looking at the implicit differentiation there and was all like O_O. :P But the second method and b^3's explanation has really covered it well. Well done, guys. :)
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barydos

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Re: Specialist 3/4 Question Thread!
« Reply #991 on: December 11, 2012, 11:34:07 am »
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Simple question here!

I'm just curious how you work out question b) part (i)
I can only get x = R\{2} but that's not right haha
« Last Edit: December 11, 2012, 11:57:16 am by Anonymiza »
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Homer

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Re: Specialist 3/4 Question Thread!
« Reply #992 on: December 11, 2012, 12:10:12 pm »
+1
I might have a slightly different interpretation of this, initally I went to use implicit differentiation too, then realised that they probably haven't come across it yet. But the way its worded, it may be asking for the values that m can be, which you can get from the second method, graph it and you can see that the gradient can be . To get this via another method, if you graph the original function, we can see it has asymptotes of and . And when we look at the graph, looking at the right arm, the gradient is going to always be greater than 2 when it approaches and always going to be less than (well more negative) than . For the left arm the gradient on the top part of the curve is under the asymptote and moving away from it, so that the gradient is always less than (more negative) than -2 and for the bottom half, always going to be greater than 2 (for the asymtptote). SO that is .

Anyways, thats just another way of looking at it, less mathematical though, but might help you understand it a little bit more.
 


the answer is E, which is any real number in the interval , would that be the same as .
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Re: Specialist 3/4 Question Thread!
« Reply #993 on: December 11, 2012, 12:21:48 pm »
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I'm just curious how you work out question b) part (i)
I can only get x = R\{2} but that's not right haha


Is the answer [-2,2)  by any chance ?
« Last Edit: December 11, 2012, 12:27:42 pm by argonaut »

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Re: Specialist 3/4 Question Thread!
« Reply #994 on: December 11, 2012, 01:10:45 pm »
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Is the answer [-2,2)  by any chance ?

-2 is not inclusive, but yes otherwise it is correct! TEACH ME YOUR WAYS
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pi

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Re: Specialist 3/4 Question Thread!
« Reply #995 on: December 11, 2012, 01:25:57 pm »
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the answer is E, which is any real number in the interval , would that be the same as .

Note that "R/[-2,2]" isn't a fraction :P It means all real numbers ("R") excluding ("\") the interval of -2 to 2 inclusive ("[-2,2]") :)

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Re: Specialist 3/4 Question Thread!
« Reply #996 on: December 11, 2012, 01:39:53 pm »
+1
I'm just curious how you work out question b) part (i)

you need the series to converge, which occurs if

so, first you find r (common ratio) in terms of x,


then you apply the conditions for a series to be convergent and therefore possible for an infinite sum to exist
« Last Edit: December 11, 2012, 01:45:43 pm by polar »

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Re: Specialist 3/4 Question Thread!
« Reply #997 on: December 11, 2012, 02:05:02 pm »
+1
you need the series to converge, which occurs if

so, first you find r (common ratio) in terms of x,


then you apply the conditions for a series to be convergent and therefore possible for an infinite sum to exist


Yes.

Also, using the formula for the sum of a geometric series with a=1, r=x/2

S = [1-(x/2)^n]/[1-(x/2)]

Therefore, if you dont want S to bugger off to infinity as n-> infinity,

|x/2| < 1

giving the result above.
I initially thought that x=-2 would be OK, but it can be seen that in this case S will oscillate between the valus of 0 and 1 as n-> infinity, which does not make the series convergent.

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Re: Specialist 3/4 Question Thread!
« Reply #998 on: December 11, 2012, 02:17:45 pm »
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AWESOME  thanks both of you :)
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Homer

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Re: Specialist 3/4 Question Thread!
« Reply #999 on: December 12, 2012, 10:39:06 am »
+1



The function has vertical asymptote where is equal to:
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Re: Specialist 3/4 Question Thread!
« Reply #1000 on: December 12, 2012, 10:43:21 am »
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If you think about it, the asymptotes of sec(x) occurs when cos(x) = 0.

Therefore,
implies


But the function f has been transformed. So we need to know the new value of x.

Let the inner function (the bracketed function) be equal to the value of x which gave the asymptote locations of the untransformed function (sec(x))
Let

Solving this will tell us which x value will give the location of the new asymptote.

« Last Edit: December 12, 2012, 10:46:48 am by Hancock »
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Re: Specialist 3/4 Question Thread!
« Reply #1001 on: December 12, 2012, 10:59:23 am »
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What would happen to the ?

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Re: Specialist 3/4 Question Thread!
« Reply #1002 on: December 12, 2012, 11:06:49 am »
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Yes, they would have a slight effect on it as well. If you need to, draw the regular cosine graph and find your x-intercepts. :)
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Re: Specialist 3/4 Question Thread!
« Reply #1003 on: December 12, 2012, 11:19:12 am »
+1
what would be the cosine equation? :/

cos(2x+pi/2)    or          root(3)cos(2x+pi/2)-2
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Re: Specialist 3/4 Question Thread!
« Reply #1004 on: December 12, 2012, 11:21:37 am »
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if you're trying to find the asymptotes, then use the first one



thus, asymptotes occur at the x-intercepts of
« Last Edit: December 12, 2012, 11:25:51 am by polar »