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April 04, 2026, 02:25:51 am

Author Topic: VCE Methods Question Thread!  (Read 5999317 times)  Share 

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alchemy

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Re: VCE Methods Question Thread!
« Reply #5445 on: July 22, 2014, 06:59:07 pm »
0
Can someone explain this problem and how they got the answer: http://rbmix.com/problem/pr/pr.php
I haven't been keeping in touch with methods recently...feeling behind :(

Zealous

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Re: VCE Methods Question Thread!
« Reply #5446 on: July 22, 2014, 07:08:33 pm »
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Can someone explain this problem and how they got the answer: http://rbmix.com/problem/pr/pr.php
I haven't been keeping in touch with methods recently...feeling behind :(
Application of the Chain Rule with a little bit of algebraic manipulation at the end to make it look nice.

Hint:



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Re: VCE Methods Question Thread!
« Reply #5447 on: July 22, 2014, 07:38:12 pm »
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how about this?

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #5448 on: July 22, 2014, 08:09:48 pm »
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How would I go about doing this question?

Consider the functions f: [0, infinity)  → R, f(x) = x-c and g: [-2, infinity)  → R, g(x)=x^2 + c, where c is a real number. How do we find all the possible values for c for which f composition g and g composition f exist?

Quick question, when we're given this for example,  f: [2, infinity)  → R , what does this mean exactly? And how would we read this?

Thanks! :D

I'll answer the second question before I get to the first - "f: X -->Y" can be read as "the function f where X is mapped onto Y". X is the domain, which you are hopefully familiar with. Y is the co-domain, which is what the range can exist in. I wouldn't worry too much about this, as methods hasn't ever seemed to give a question that requires you to have to compare the range to the co-domain. (feel free to correct me if anyone has seen a SAC that says otherwise)

Now, for the second question, let's first consider f(g(x)). For this function to exist, . The range of g is and the domain of f is . By comparing the two, we see that out initial statement is satisfied when - therefore, f(g(x)) exists if .

Now, let's consider g(f(x)). For this functino to exist, . Now, the range of f is going to be , and the domain of g is . Once again, by directly comparing the two, we see that the initial statement qualifies when . So, g(f(x)) exists if

#J.Procrastinator

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Re: VCE Methods Question Thread!
« Reply #5449 on: July 22, 2014, 09:35:41 pm »
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I'll answer the second question before I get to the first - "f: X -->Y" can be read as "the function f where X is mapped onto Y". X is the domain, which you are hopefully familiar with. Y is the co-domain, which is what the range can exist in. I wouldn't worry too much about this, as methods hasn't ever seemed to give a question that requires you to have to compare the range to the co-domain. (feel free to correct me if anyone has seen a SAC that says otherwise)

Now, for the second question, let's first consider f(g(x)). For this function to exist, . The range of g is and the domain of f is . By comparing the two, we see that out initial statement is satisfied when - therefore, f(g(x)) exists if .

Now, let's consider g(f(x)). For this functino to exist, . Now, the range of f is going to be , and the domain of g is . Once again, by directly comparing the two, we see that the initial statement qualifies when . So, g(f(x)) exists if

Thank you very much :)
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Mieow

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Re: VCE Methods Question Thread!
« Reply #5450 on: July 22, 2014, 09:47:11 pm »
+3
Who the fck is Tasmania Jones tho?
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #5451 on: July 22, 2014, 10:06:39 pm »
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Who the fck is Tasmania Jones tho?

Some random who uses pointless maths to help her in tackling challenges that are best done by actually doing them... :P

StupidProdigy

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Re: VCE Methods Question Thread!
« Reply #5452 on: July 23, 2014, 09:20:19 pm »
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Just a quick question, I got my results back for a sac on tuesday but was unhappy with one of the questions.
The question asked something like 'solve the following equations to find the times the wave was at .....', it related to a sin graph
I got the question right, but my teacher took 1 mark off out of 4 because I didn't specify two wave times, only one, even though there was no set domain in the question relating to two waves. So should I lose a mark for a question which was not written correctly by the teacher? Rant over hahah, and thanks for any replies
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #5453 on: July 24, 2014, 02:06:13 pm »
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hey can anyone see what i've done wrong with this question

" there are two blue socks and four red socks in a box, two socks are randomly pulled out simultaneously"

find probability that the two socks are of different colour.

i did:

[ncr(4,1) + ncr(2,1) ] / ncr (6,2)

anything wrong?

2 blue socks and 4 red socks out of 6 socks?

Pr(socks are different colour) + Pr(socks are the same colour) = 1
Try using that
Pr(socks are the same colour) = Pr(blue, blue) + Pr(red, red)
Which should be straightforward
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knightrider

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Re: VCE Methods Question Thread!
« Reply #5454 on: July 24, 2014, 05:23:01 pm »
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how would you do these questions

If sin (0.7) = 0.644, cos (0.7 ) = 0.765 and tan (0.7 ) = 0.842, find the value of each of the following.(Hint: pie = 3.142, approximately.)

sin (2.442)=
cos (3.842)=
tan (5.584)=
sin (−0.7)=



paper-back

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Re: VCE Methods Question Thread!
« Reply #5455 on: July 24, 2014, 06:19:41 pm »
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how would you do these questions

If sin (0.7) = 0.644, cos (0.7 ) = 0.765 and tan (0.7 ) = 0.842, find the value of each of the following.(Hint: pie = 3.142, approximately.)

sin (2.442)=
cos (3.842)=
tan (5.584)=
sin (−0.7)=

sin(2.442)=sin(3.14-0.7)
=sin(0.7)
=0.644

cos(3.842)=cos(3.142+0.7)
=-cos(0.7)
=-0.765

tan(5.584)=tan(2pi - 0.7)
=-tan(0.7)
=-0.842

sin(-0.7)=sin(2pi - 0.7)
=-sin(0.7)
=-0.644

knightrider

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Re: VCE Methods Question Thread!
« Reply #5456 on: July 24, 2014, 07:09:43 pm »
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sin(2.442)=sin(3.14-0.7)
=sin(0.7)
=0.644

cos(3.842)=cos(3.142+0.7)
=-cos(0.7)
=-0.765

tan(5.584)=tan(2pi - 0.7)
=-tan(0.7)
=-0.842

sin(-0.7)=sin(2pi - 0.7)
=-sin(0.7)
=-0.644

Thanks paper back  :) but how do you determine if you have to minus or plus 0.7

StupidProdigy

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Re: VCE Methods Question Thread!
« Reply #5457 on: July 24, 2014, 07:17:25 pm »
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Its determined by symmetry. So the number (theta) in quadrant 1 stays as is, in quadrant 2 it is pi-theta, quadrant 3 is pi+theta and quadrant 4 is 2pi-theta :)
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knightrider

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Re: VCE Methods Question Thread!
« Reply #5458 on: July 24, 2014, 07:22:45 pm »
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Its determined by symmetry. So the number (theta) in quadrant 1 stays as is, in quadrant 2 it is pi-theta, quadrant 3 is pi+theta and quadrant 4 is 2pi-theta :)

Thanks for that :)
For these numbers how do you know which quadrant they are in
sin (2.442)=
cos (3.842)=
tan (5.584)=
sin (−0.7)=


paper-back

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Re: VCE Methods Question Thread!
« Reply #5459 on: July 24, 2014, 09:00:39 pm »
+3
Thanks for that :)
For these numbers how do you know which quadrant they are in
sin (2.442)=
cos (3.842)=
tan (5.584)=
sin (−0.7)=


This may be of some help


For the first one we know that 2.442 is less than yet more than
Hence it has to be in the second quadrant
Since sine is positive in the first quadrant and it's asking sin(2.442) the answer will be positive

For the second it's greater than but less than
Therefore it is in the 3rd quadrant
Since only  tan is positive in the 3rd quadrant we end up with a negative answer

I'll let you do the third one

For the last one, it's gone backwards
It would be less than but greater than so it's in the 4th quadrant
Only cos is positive in this quadrant and therefore the answer is negative

I hope this made sense

EDIT: It's meant to be in the picture
« Last Edit: July 24, 2014, 09:02:51 pm by paper-back »