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Author Topic: VCE Methods Question Thread!  (Read 4802434 times)  Share 

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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18645 on: July 26, 2020, 05:02:13 pm »
+5
Hey y'all,

Can someone please explain what I'm doing wrong, I don't really understand what the absolute function is and those two lines, and why they switch the signs in the answer

Thanks

I want to make this clear - I'm not a fan of answering an already answered question. I think it's rude to the people who have already put in an answer - particularly well explained and clear answers like the ones above. However, I feel that while the responses from Bri and james clearly explain what the modulus function is, they don't explain WHY it appears when integrating these kinds of functions.

To understand that, let's instead think about what integration is - it's an inverse operation of differentiation. So, we know that (-1/3)ln(4-3x) is an antiderivative (not necessarily the only one, but it's definitely one of them) of 1/(4-3x) because if you differentiate (-1/3)ln(4-3x), you get 1/(4-3x). In fact, let's do that working out:



Noting the application of the chain rule in the second step. Okay, so this leaves us with:



Which must mean that:



Well, for fun, what about when we differentiate (-1/3)ln(3x-4)? Well, as above:



Well hang on a second, this means that:



And that must mean that:



But, we already decided that the integral of 1/(4-3x) was equal to the ln(4-3x), not ln(3x-4)! This seems like a contradiction, but it's not - in truth, it's because the integral of 1/(4-3x) could be EITHER of the logarithmic functions, we just don't know which one, because it depends on whether 3x-4 is positive, or if 4-3x is positive. So, to say that it could be either one and we just don't know which, we put the modulus signs in there - because as other said, those modulus signs will MAKE the inside positive, so it will always work. Then, we can remove those modulus signs later if we know the initial conditions - but that's getting into specialist territory, and to memory you're not expected to know about removing the modulus signs in methods.

Hopefully this makes a bit more sense conceptually, instead of just being another "random" rule that you have to remember.

Sherlock.Holmes

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Re: VCE Methods Question Thread!
« Reply #18646 on: July 26, 2020, 08:03:37 pm »
0

Hopefully this makes a bit more sense conceptually, instead of just being another "random" rule that you have to remember.

Your answer is wrong because you don't have absolute values
You have However you should have had either Or
With those two options being equivalent remember,

Thank you all, makes much more sense now !!

rozmaaate

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Re: VCE Methods Question Thread!
« Reply #18647 on: July 30, 2020, 05:35:24 pm »
0
Hi guys can someone please help me out with this question as I have no idea how to begin approaching it


Calculate the exact values of v which satisfy 4( root2) cos(v) = (root)2 cos(v) + 3, −𝜋 ≤ v ≤ 5𝜋.

Thanks in advance
« Last Edit: July 30, 2020, 05:43:21 pm by rozmaaate »

The Cat In The Hat

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Re: VCE Methods Question Thread!
« Reply #18648 on: July 30, 2020, 06:15:39 pm »
+3
4√2 cos(v) = √2 cos(v) + 3, -π≤v≤5π
Is it reasonable simply to push it through the CAS?
Otherwise, I'd start by rearranging it (by subtracting √2 cos(v) from both sides) to
3√2 cos(v) = 3
which becomes
cos(v) = 1/√2
the answer to which you can refer to the exact values table, which is π/4.
Thus, v = π/4 as the first solution.
Then use symmetry to find subsequent solutions.
Please don't take my answer as necessarily right; I'm not very good at these things, which is part of the reason why I didn't try solving for subsequent ones :)
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rozmaaate

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Re: VCE Methods Question Thread!
« Reply #18649 on: July 30, 2020, 06:27:54 pm »
0
4√2 cos(v) = √2 cos(v) + 3, -π≤v≤5π
Is it reasonable simply to push it through the CAS?
Otherwise, I'd start by rearranging it (by subtracting √2 cos(v) from both sides) to
3√2 cos(v) = 3
which becomes
cos(v) = 1/√2
the answer to which you can refer to the exact values table, which is π/4.
Thus, v = π/4 as the first solution.
Then use symmetry to find subsequent solutions.
Please don't take my answer as necessarily right; I'm not very good at these things, which is part of the reason why I didn't try solving for subsequent ones :)

when subtracting cos(v) from each other does one of them cancel out ? besides that everything else makes sense , ps you are correct !!

« Last Edit: July 30, 2020, 06:31:25 pm by rozmaaate »

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Re: VCE Methods Question Thread!
« Reply #18650 on: July 30, 2020, 06:43:58 pm »
0
Hello,

Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.

Thanks.

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Re: VCE Methods Question Thread!
« Reply #18651 on: July 30, 2020, 07:04:15 pm »
0
when subtracting cos(v) from each other does one of them cancel out ? besides that everything else makes sense , ps you are correct !!
Yes, the one on the left.
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rozmaaate

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Re: VCE Methods Question Thread!
« Reply #18652 on: July 30, 2020, 07:06:20 pm »
0
Yes, the one on the left.

could you explain why if you don't mind sorry if it sounds dumb

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Re: VCE Methods Question Thread!
« Reply #18653 on: July 30, 2020, 07:32:01 pm »
+3
Hello,

Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.

Thanks.
Try a proof by contradiction; suppose (a/n) is irreducible and (a/n)^2 is reducible and see if you can arrive at the result that (a/n) must then be reducible.

So firstly, suppose (a/n)^2 is reducible; what does this say about a and n? Remember that the fraction a^2 / n^2 is reducible if a^2 and n^2 have a gcd > 1 but that does mean that a and n must have a gcd > 1. I'm not too sure how thorough your teacher wants you to be. So I'm just going to jump to the implication that a and n must have a gcd > 1. So if a and n have a gcd > 1 what does this say about the fraction a/n? (btw, gcd > 1 means that they have a common factor greater than 1) This just means that the fraction is "reducible". So we've assumed at the beginning that a/n is irreducible but we've just arrived at the conclusion that a/n must be reducible (the more formal way to do this would be to use contraposition, but i think this should be good enough) so since we've arrived at contradicting results, our initial assumption that (a/n)^2 is reducible must be false.

This is just to help you an idea. The more formal way to do this would be to use contraposition, but I think this should be good enough. So since we've arrived at contradicting results, our initial assumption that (a/n)^2 is reducible must be false.

However, this proof is technically a proof by contraposition; if we prove that (a/n)^2 is reducible implies (a/n) is reducible, this also proves that if (a/n) is irreducible, (a/n)^2 is irreducible.

But we can also frame it in a way where it's contradiction, but its a little bit iffy especially since we never actually used the assumption that (a/n) is irreducible. Framing it in a contradiction would be instead of saying that a/n must be reducible at the end, after we get to the point where gcd(a, n) > 1 we would get stuck. Because we've assumed that a/n is irreducible at the beginning, gcd(a, n) must be 1. But since we've just arrived at the conclusion that gcd(a, n) > 1, this contradicts our assumptions, meaning that (a/n)^2 must not be reducible.

Hope this helps! If you have any further questions don't hesitate to ask! :)

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18654 on: July 31, 2020, 01:15:48 am »
+6
Hello,

Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.

Thanks.

This feels so much like a specialist 1/2 question lol

could you explain why if you don't mind sorry if it sounds dumb

Don't forget that cos(v) is also a number. It's just like any pronumeral. Remember how when solving equations like:



You would move the 7x to the other side?



By this point, you might be on auto-pilot - but let's stop and think for a second. You COULD just work like you normally do, where you know that you can ignore the x and just do -4-7, and you would get:



But, how does this step work? And why can we do -4x-7x but not -4x-7y? Well, you can think of this is as factorising out the x, like so:



And this is why we can't do the same step with with -4x-7y - because there's nothing to factor out!

So, what does this have to do with cos(v)? Well, cos(v) is just a number - in fact, you can think of cos(v) AS x, and you could solve an equation like so:



But, that's a lot of work! So, why don't we do this WITHOUT the substitution?



In the third step, notice that all I'm doing is factoring out that cos(v) - just like I did earlier with x. Feel free to skip this step - but if doing it helps your understanding of WHY you can "cancel out" (which is a bad description btw - things aren't cancelling out) in the way that Cat did, then power to you

rozmaaate

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Re: VCE Methods Question Thread!
« Reply #18655 on: July 31, 2020, 09:07:46 am »
+1
This feels so much like a specialist 1/2 question lol

Don't forget that cos(v) is also a number. It's just like any pronumeral. Remember how when solving equations like:



You would move the 7x to the other side?



By this point, you might be on auto-pilot - but let's stop and think for a second. You COULD just work like you normally do, where you know that you can ignore the x and just do -4-7, and you would get:



But, how does this step work? And why can we do -4x-7x but not -4x-7y? Well, you can think of this is as factorising out the x, like so:



And this is why we can't do the same step with with -4x-7y - because there's nothing to factor out!

So, what does this have to do with cos(v)? Well, cos(v) is just a number - in fact, you can think of cos(v) AS x, and you could solve an equation like so:



But, that's a lot of work! So, why don't we do this WITHOUT the substitution?



In the third step, notice that all I'm doing is factoring out that cos(v) - just like I did earlier with x. Feel free to skip this step - but if doing it helps your understanding of WHY you can "cancel out" (which is a bad description btw - things aren't cancelling out) in the way that Cat did, then power to you

Thanks so much!!

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18656 on: July 31, 2020, 11:00:28 pm »
+4
wait how did you get the left hand side to equal 2 cos(v) ?? in regards to my question of  solving x for 4√2 cos(v) = √2 cos(v) + 3, -π≤v≤5π

2cos(v)=cos(v) is an equation I made up, it has nothing to do with your equation. But you should be able to apply the same principles to see how it works with your equation

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Re: VCE Methods Question Thread!
« Reply #18657 on: August 03, 2020, 06:53:44 pm »
+1
Hello.

find the general solution to the equation.

I did it but I found two solutions, one with (5pi/3) and one with (2pi/3). The textbook says only (2pi/3). Help

james.358

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Re: VCE Methods Question Thread!
« Reply #18658 on: August 03, 2020, 07:18:40 pm »
+6
Hey a weaponized ikea chair!

Because the period of tan(x) is π, 2π/3 + nπ is actually the same as 5π/3 + nπ, where n ∈ Z

Hope this helps!
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Re: VCE Methods Question Thread!
« Reply #18659 on: August 03, 2020, 07:20:53 pm »
0
Hey a weaponized ikea chair!

Because the period of tan(x) is π, 2π/3 + nπ is actually the same as 5π/3 + nπ, where n ∈ Z

Hope this helps!
James

Hello,

Thanks for your reply.

Is giving both solutions technically correct or would it be marked wrong in an exam? Thanks