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June 26, 2026, 01:15:04 am

Author Topic: VCE Methods Question Thread!  (Read 6161039 times)  Share 

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pi

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Re: VCE Methods Question Thread!
« Reply #7470 on: December 24, 2014, 01:49:27 pm »
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This may be a bit off topic but how come guests can also post in these posts i thought they could only view as they havent created an account

I just tried as a guest and they can't post? I could only print and view the thread.

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #7471 on: December 24, 2014, 01:56:30 pm »
+2
Hey can someone help me with this question

A car travels half a distance of its journey at an average speed of 80km/h and half at an average speed of xkm/h
Define a function S which gives the average speed for the total journey as a function of x

Let's consider some function f, and for f, we can break it into a set of n different parts. If we want to find the average of f, what we can do is add up all of these separate parts multiplied by their own little segment. So, we'd see:



And that's what we're going to do here:


DSubShell

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Re: VCE Methods Question Thread!
« Reply #7472 on: December 24, 2014, 02:26:40 pm »
+1
This may be a bit off topic but how come guests can also post in these posts i thought they could only view as they havent created an account

I think the people you see are ones who got their account deleted (by choice) by an admin.
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SE_JM

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Re: VCE Methods Question Thread!
« Reply #7473 on: December 25, 2014, 01:33:33 pm »
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Hello,
I'm not getting these questions. Could someone please help me out? :)

find d/d(x2) [ (a+x2)(1/2)/(a+x2)2 ]

and also this one:
differentiate with respect to x

y= 3* 2x+1

and another one  :P:
differentiate:

|(1-3x)/2|


I know it's Christmas, and probs people won't answer but  ;D thanks  :)
« Last Edit: December 25, 2014, 02:47:23 pm by SE_JM »

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #7474 on: December 25, 2014, 04:28:08 pm »
+3
Hello,
I'm not getting these questions. Could someone please help me out? :)

find d/d(x2) [ (a+x2)(1/2)/(a+x2)2 ]

and also this one:
differentiate with respect to x

y= 3* 2x+1

and another one  :P:
differentiate:

|(1-3x)/2|


I know it's Christmas, and probs people won't answer but  ;D thanks  :)

Yeesh - no sleep for the wicked. :P I'm going to take this time to remind you all to have regular breaks - nothing wrong with a day or two off! Particularly if you spend months straight studying, it's not healthy, and you'll hurt your brain and be less effective for SACs/exams.

The first one, you could of course apply the quotient rule - but I'm gonna use exponential rules:



Now, we want to take the derivative with respect to x^2, not just x. This can be confusing, so let's make a substitution to make life easier. Let u=x^2:



For the second, we note that we can only differentiate exponentials of base e. That translation is going to be annoying, so I'll get rid of it. You can of course try this yourself without removing the translation:



Note that the last step should not be necesarry, however I feel that you should venture to leave your answers in the base given (unless asked to change the base, of course).


Now, the final one - |(1-3x)/2|. I'm going to break this up:



Now we can diff it, remembering what points are differentiable and which aren't (notice the domains for the hybrid!!):


SE_JM

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Re: VCE Methods Question Thread!
« Reply #7475 on: December 26, 2014, 09:38:06 am »
0
Yeesh - no sleep for the wicked. :P I'm going to take this time to remind you all to have regular breaks - nothing wrong with a day or two off! Particularly if you spend months straight studying, it's not healthy, and you'll hurt your brain and be less effective for SACs/exams.

The first one, you could of course apply the quotient rule - but I'm gonna use exponential rules:



Now, we want to take the derivative with respect to x^2, not just x. This can be confusing, so let's make a substitution to make life easier. Let u=x^2:



For the second, we note that we can only differentiate exponentials of base e. That translation is going to be annoying, so I'll get rid of it. You can of course try this yourself without removing the translation:



Note that the last step should not be necesarry, however I feel that you should venture to leave your answers in the base given (unless asked to change the base, of course).


Now, the final one - |(1-3x)/2|. I'm going to break this up:



Now we can diff it, remembering what points are differentiable and which aren't (notice the domains for the hybrid!!):



Hi! Thank you!

Happy boxing day  :) lol :P

Zues

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Re: VCE Methods Question Thread!
« Reply #7476 on: December 26, 2014, 06:03:48 pm »
0
why are logarithm graphs to the base 2 steeper?

the form of a log is klog_a_(x) = y

why do a and x need to be positive and is it y that needs to be positive as well? what are the restrictions so to speak

thanks

brightsky

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Re: VCE Methods Question Thread!
« Reply #7477 on: December 26, 2014, 06:18:00 pm »
+2
why are logarithm graphs to the base 2 steeper?

the form of a log is klog_a_(x) = y

why do a and x need to be positive and is it y that needs to be positive as well? what are the restrictions so to speak

thanks

y = log_a(x) is only considered a logarithmic function if two conditions are met:

(i) a is greater than 0
(ii) a is not equal to 1

To understand why a (the base) needs to be greater than 0, we must consider the inverse function y = a^x. Suppose a = -2. Then the function becomes y = (-1/2)^x. Let us run a few tests:

Let x = 0. Then y = (-1/2)^0 = 1.
Let x = 1. Then y = (-1/2)^1 = -1/2.
Let x = 2. Then y = (-1/2)^2 = 1/4.

So far so good. Now:

Let x = 1/2. Then y = (-1/2)^(1/2) = sqrt(-1/2). Woops! We get a complex number!

It should be clear now why we must apply such a restriction to a. If a were not greater than 0, then function would be undefined a certain points, and the graph of the function would look nothing like that of the logarithmic function with which we are all familiar. Note that the range of the basic logarithmic function y = log_a(x), where a is greater than 0 and not equal to 1, is R. Hence, the "y" to which you refer does not necessarily need to be positive; it can take any real number.
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Re: VCE Methods Question Thread!
« Reply #7478 on: December 26, 2014, 06:56:04 pm »
0
thanks brightsky!

just a questiont hat came to my mind, even though im not doing this topic its annoying me

y = -1/(1-2x)^2 ---- graph this

i then rearrange it

y= -1 / (-2x+1)^2
y= -1 / -(2x-1)^2

minuses cancel out, thus doesnt this make it a positive truncus? why is it negative, i remember doing a question similar earlier, albeit a hyperbola , and it worked. i think is it because..

y= -1 / (-2x+1)^2
y= -1 / [-(2x-1)]^2
y = -1/ (2x-1)^2 ?
thanks

brightsky

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Re: VCE Methods Question Thread!
« Reply #7479 on: December 26, 2014, 07:15:20 pm »
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thanks brightsky!

just a questiont hat came to my mind, even though im not doing this topic its annoying me

y = -1/(1-2x)^2 ---- graph this

i then rearrange it

y= -1 / (-2x+1)^2
y= -1 / -(2x-1)^2

minuses cancel out, thus doesnt this make it a positive truncus? why is it negative, i remember doing a question similar earlier, albeit a hyperbola , and it worked. i think is it because..

y= -1 / (-2x+1)^2
y= -1 / [-(2x-1)]^2
y = -1/ (2x-1)^2 ?
thanks

...and you would be correct. :)
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #7480 on: December 27, 2014, 01:25:03 am »
0
thanks brightsky!

just a questiont hat came to my mind, even though im not doing this topic its annoying me

y = -1/(1-2x)^2 ---- graph this

i then rearrange it

y= -1 / (-2x+1)^2
y= -1 / -(2x-1)^2

minuses cancel out, thus doesnt this make it a positive truncus? why is it negative, i remember doing a question similar earlier, albeit a hyperbola , and it worked. i think is it because..

y= -1 / (-2x+1)^2
y= -1 / [-(2x-1)]^2
y = -1/ (2x-1)^2 ?
thanks

They're all the same. Why? Well, a 'negative truncus' is something of the form y = -1/x^2 or similar
y = -1/x^2 = -1/(-x)^2
Geometrically, reflecting the graph of y = -1/x^2 in the x axis does nothing. That's why you can have a negative or positive coefficient of the x in the fraction and still get the same thing
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Re: VCE Methods Question Thread!
« Reply #7481 on: December 27, 2014, 01:02:18 pm »
0
with exponential graphs to any base vs logarithm graphs to any base vs exponentional graphs to base e, why are the 1st and 2nd similar looking graphs whereas the middle one in shaped the way it is, i know its a reflection in the line y=x ? but why?

also with the first and third, a constant c (vertical translation) acts as a horizontal asymptote, but in the second why is it not? and only the value inside () eg. log10(x-3)+ 5 , x=3 is a asymptopte because cant have negative, but why is a constant c not an asymptote?

thanks

Zues

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Re: VCE Methods Question Thread!
« Reply #7482 on: December 27, 2014, 02:23:05 pm »
0
and exactly how is maximal domain different to domain. i just treat those questions as a domain question, is there an example where they may be different?

Zues

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Re: VCE Methods Question Thread!
« Reply #7483 on: December 27, 2014, 04:32:50 pm »
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for c why do you need to times whole thing by 4? and not just the b value? also, how did they change the signs? if they divided by -1 wouldnt it be -4a + ...

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Re: VCE Methods Question Thread!
« Reply #7484 on: December 27, 2014, 05:31:28 pm »
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How would you do this question?

The three matrices A, B and C have orders (a × b), (a × c) and (b × a) respectively.

For which pairs of matrices does the sum exist? State the order of the answer(s).