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March 29, 2024, 01:49:29 am

Author Topic: 4U Maths Question Thread  (Read 659830 times)  Share 

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RuiAce

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Re: 4U Maths Question Thread
« Reply #2055 on: October 21, 2018, 11:51:13 pm »
+2
« Last Edit: October 21, 2018, 11:54:14 pm by RuiAce »

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Re: 4U Maths Question Thread
« Reply #2056 on: October 21, 2018, 11:52:42 pm »
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In theory, I would give you the marks for it.

Although in practice, usually when you aren't told anything about what \(z_1\) and \(z_2\) are, I like to keep my answer 'as arbitrary as possible' in a sense. Placing \(z_1\) and \(z_2\) in places that almost look like you're just chucking them at random places in thin air feels less 'restricted' in a sense, as opposed to doing what that other one did. Copying what they did can unintentionally give the impression that \(z_1\) and \(z_2\) have to lie on the \(x\)-axis no matter what, when really they don't. So whilst it's technically not incorrect to do it that way, it probably isn't the best habit to go towards.

Especially since, potentially in the exam they could give you something like \( \arg \left( \frac{z-3+4i}{z+2-i} \right) = \frac\pi3\) instead.

hmm true that makes sense. thank you!!!

justwannawish

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Re: 4U Maths Question Thread
« Reply #2057 on: October 22, 2018, 02:16:10 pm »
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Let w be a non real cube of unity. Show the other non-real cube root is w^2.

What would be the best method for this?

envisagator

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Re: 4U Maths Question Thread
« Reply #2058 on: October 22, 2018, 03:24:38 pm »
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Hi Rui, with the following question attached, can you please explain why you do the part highlighted and its significance. Thank You!!!

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Opengangs

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Re: 4U Maths Question Thread
« Reply #2059 on: October 22, 2018, 03:29:51 pm »
+1
Let w be a non real cube of unity. Show the other non-real cube root is w^2.

What would be the best method for this?
For cube root of unity, we have \(z^3 - 1 = 0\), giving us \(z^3 = 1\). So clearly \(z = 1\) is a root. Note that \(z^3 - 1\) can be factorised in the form: \((z - 1)(z^2 + z + 1)\). Finding the roots of the irreducible quadratic is fairly straight forward. We have:
\[ \begin{align*}z &= \frac{-1 \pm \sqrt{3}i}{2}\end{align*}\]

Taking \(\omega\) to be \(\frac{-1 + \sqrt{3}i}{2}\) and squaring it gives us:
\[ \begin{align*}\omega^2 &= \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 \\ &= \frac{1 - 2\sqrt{3}i - 3}{4} \\ &= \frac{-1 - \sqrt{3}i}{2}\end{align*}\]

which is the other cube root of unity.

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Re: 4U Maths Question Thread
« Reply #2060 on: October 22, 2018, 03:57:27 pm »
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Hi Rui, with the following question attached, can you please explain why you do the part highlighted and its significance. Thank You!!!
With multiple zeroes, we know that at least \(P(x) = P'(x) = 0\). To show that there are no multiple zeroes, we essentially want to show that for any \(x\), \(P(x) = 0 \neq P'(x)\). Notice that
\[P(x) = P'(x) + \frac{x^n}{n!}\],

which is also where the first highlighted part comes from. The goal now is to show that: \(\frac{x^n}{n!}\) is non-zero for \(n > 1\).

Suppose that, for some \(\alpha\), we have \(P(\alpha) = P'(\alpha)\). Then clearly that must occur if:
\[\frac{\alpha^n}{n!} = 0 \Rightarrow \alpha = 0\].

Then we have \(P(0) = P'(0) = 0\), which is equivalent to the second highlighted part.

justwannawish

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Re: 4U Maths Question Thread
« Reply #2061 on: October 22, 2018, 07:34:25 pm »
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For cube root of unity, we have \(z^3 - 1 = 0\), giving us \(z^3 = 1\). So clearly \(z = 1\) is a root. Note that \(z^3 - 1\) can be factorised in the form: \((z - 1)(z^2 + z + 1)\). Finding the roots of the irreducible quadratic is fairly straight forward. We have:
\[ \begin{align*}z &= \frac{-1 \pm \sqrt{3}i}{2}\end{align*}\]
Taking \(\omega\) to be \(\frac{-1 + \sqrt{3}i}{2}\) and squaring it gives us:
\[ \begin{align*}\omega^2 &= \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 \\ &= \frac{1 - 2\sqrt{3}i - 3}{4} \\ &= \frac{-1 - \sqrt{3}i}{2}\end{align*}\]

which is the other cube root of unity.

Hey, I forgot to mention that the next part of the question was that using the above, prove w^2 + w + 1=0

So would I still be able to use this method and then requote it for part ii?

Thanks!

RuiAce

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Re: 4U Maths Question Thread
« Reply #2062 on: October 22, 2018, 08:02:04 pm »
+2
Hey, I forgot to mention that the next part of the question was that using the above, prove w^2 + w + 1=0

So would I still be able to use this method and then requote it for part ii?

Thanks!



You can, of course, plug the value of \(\omega\) in there manually if you wish to. The computations should still fall out.

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Re: 4U Maths Question Thread
« Reply #2063 on: October 23, 2018, 10:44:15 am »
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Hey so uuuhh I need a lil help with question 7b)ii) from 2011 paper.
I got part i) easily, it was just a subsitution but then when I was looking at the solutions at the end of part i) they “relabelled” u as x so they could use it in their solution for part ii). Why is it that they can just relabel it as x straight after going through that whole process of substiting u=4-x?? Btw I tried to post screenshots of the question I was talking about but it said it was too large or something?

RuiAce

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Re: 4U Maths Question Thread
« Reply #2064 on: October 23, 2018, 10:49:49 am »
+1
Hey so uuuhh I need a lil help with question 7b)ii) from 2011 paper.
I got part i) easily, it was just a subsitution but then when I was looking at the solutions at the end of part i) they “relabelled” u as x so they could use it in their solution for part ii). Why is it that they can just relabel it as x straight after going through that whole process of substiting u=4-x?? Btw I tried to post screenshots of the question I was talking about but it said it was too large or something?
With the screenshots being too large, when that happens you'll need to upload it to an image hosting website like imgur and extract the link instead.
\[ u\text{ and }x\text{ are ultimately, nothing more than 'dummy variables'}\\ \text{used for the sake of definite integration.} \]
\[ \text{In general, it is always true that }\int_a^b f(x)\,dx = \int_a^b f(u)\,du.\\ \text{As an example, you can try computing }\int_0^1 x^2\,dx\\ \text{and check that it does equal }\int_0^1 u^2\,du.\]
Note that the fact they're equal is facilitated by the fact that \(x\) and \(u\) are 'doomed' to disappear, because once you sub the boundary values in, you don't care at all about what the variables you had at the start anymore were. This is the reason for why they're called 'dummy' variables in integration. Note that you don't run into the same luck when dealing with indefinite integrals, because the variables never disappear then.
\[ \text{So for that one, by the same logic,}\\ \int_1^3 \frac{\sin^2 \frac{\pi u}{8}}{u(4-u)}\,du = \int_1^3 \frac{\sin^2 \frac{\pi x}{8}}{x(4-x)}\,dx\]

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #2065 on: October 23, 2018, 03:57:10 pm »
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hey rui plz need some help, a friend gave me and some other mates this 2u que but I find it really hard.
When i first graphed x^2+xy+y^2=0, on desmos, it was just a dot on the origin.
So i just tried a blodge method by letting y=1 and used quadratic formula to get x=(1+sqrt[3] i)/2 and plugged in both values of x&y in the other eq.
After that i got it simplified to 2cos(2015pi/3)=1

But im not sure if this is a full proof method. Plz give it a try.

envisagator

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Re: 4U Maths Question Thread
« Reply #2066 on: October 23, 2018, 04:36:25 pm »
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Hi Rui, this was a question posted on here a few days ago. I'm not sure how to do part c of it, i know its got to do with product and sum of roots but just cant grasp wats happening. Thank You!!
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2067 on: October 23, 2018, 05:40:56 pm »
+2
hey rui plz need some help, a friend gave me and some other mates this 2u que but I find it really hard.
When i first graphed x^2+xy+y^2=0, on desmos, it was just a dot on the origin.
So i just tried a blodge method by letting y=1 and used quadratic formula to get x=(1+sqrt[3] i)/2 and plugged in both values of x&y in the other eq.
After that i got it simplified to 2cos(2015pi/3)=1

But im not sure if this is a full proof method. Plz give it a try.
Yeah I know the source of this question. I've tried to avoid their own solution here though:






Your method would only work if this were a multiple choice question and you just wanted to figure out the correct answer. Here, it's not ok, because until we got to the very end we could never confirm whether the value was constant, or if it depended on \(x\) and \(y\).
« Last Edit: October 23, 2018, 06:00:46 pm by RuiAce »

RuiAce

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Re: 4U Maths Question Thread
« Reply #2068 on: October 23, 2018, 05:47:50 pm »
+1
Hi Rui, this was a question posted on here a few days ago. I'm not sure how to do part c of it, i know its got to do with product and sum of roots but just cant grasp wats happening. Thank You!!


kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #2069 on: October 24, 2018, 04:53:09 pm »
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hey guys some help here plz