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March 30, 2024, 12:06:09 am

Author Topic: 4U Maths Question Thread  (Read 660021 times)  Share 

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myopic_owl22

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Re: 4U Maths Question Thread
« Reply #2130 on: December 08, 2018, 12:07:14 pm »
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Thanks Rui! Definitely clears things up. As a follow-up question, I was wondering how you'd go about approaching/ setting out an answer for a question like the one in 2017 - Q13(e)



"By using vectors" - does that mean demonstrate geometrically?


RuiAce

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Re: 4U Maths Question Thread
« Reply #2131 on: December 08, 2018, 02:55:33 pm »
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Thanks Rui! Definitely clears things up. As a follow-up question, I was wondering how you'd go about approaching/ setting out an answer for a question like the one in 2017 - Q13(e)

(Image removed from quote.)

"By using vectors" - does that mean demonstrate geometrically?


This question doesn't use vector addition directly but rather draws on a key consequence (in fact namely the \(z-w\) issue). Recognising that \(\overrightarrow{DC}\) represents \(c-d\) and etc. originate from the parallelogram, but it can be something that people just memorise.



Rearranging this equation gives what we wish to prove.
« Last Edit: December 08, 2018, 03:10:33 pm by RuiAce »

myopic_owl22

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Re: 4U Maths Question Thread
« Reply #2132 on: December 08, 2018, 06:26:03 pm »
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Thanks again! Sometimes I find it difficult to know how to explain myself in an exam - good to see that a diagram and drawn vectors weren't required. :)

not a mystery mark

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Re: 4U Maths Question Thread
« Reply #2133 on: December 08, 2018, 08:27:28 pm »
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I feel like we'll get to know each other a lot on this forum haha.
I'm unsure if I'm way too tired or if this is hard:

Find the locus of z if

a) is purely real
b) is purely imaginary
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2134 on: December 08, 2018, 08:44:26 pm »
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I feel like we'll get to know each other a lot on this forum haha.
I'm unsure if I'm way too tired or if this is hard:

Find the locus of z if

a) is purely real
b) is purely imaginary
It is quite tedious.



Similarly if it is purely real, setting the imaginary part equal to 0 gives the straight line \(x-6y-2 = 0\), but again we must exclude the point \((2,0)\) on the line.

david.wang28

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Re: 4U Maths Question Thread
« Reply #2135 on: December 09, 2018, 06:19:16 pm »
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Hello,
Can anyone please give me a detailed answer for Q8 ASAP (I have included some of my working out)? Thanks :)
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2136 on: December 09, 2018, 06:34:40 pm »
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Hello,
Can anyone please give me a detailed answer for Q8 ASAP (I have included some of my working out)? Thanks :)
Please be mindful when using comments like this. It creates unnecessary pressure to rush people who are taking their time to help out.

There's no question attached.

david.wang28

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Re: 4U Maths Question Thread
« Reply #2137 on: December 09, 2018, 07:36:14 pm »
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Please be mindful when using comments like this. It creates unnecessary pressure to rush people who are taking their time to help out.

There's no question attached.
Sorry about the ASAP part Rui. Here is question 8 (although I have already written it down on the paper).
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2138 on: December 09, 2018, 08:15:30 pm »
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Sorry about the ASAP part Rui. Here is question 8 (although I have already written it down on the paper).





david.wang28

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Re: 4U Maths Question Thread
« Reply #2139 on: December 09, 2018, 08:29:42 pm »
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Thank you for the help Rui! Much appreciated :)
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Sr-1425

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Re: 4U Maths Question Thread
« Reply #2140 on: December 12, 2018, 08:07:07 pm »
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Hi, I was wondering how one would go about proving that:
Sinx+sin2x+sin3x+......+sin(nx)=sin(x(n+1)/2)*sin(nx/2) / sin(x/2)

Knowing that De Moivre's theorem must be used.

RuiAce

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Re: 4U Maths Question Thread
« Reply #2141 on: December 12, 2018, 09:25:37 pm »
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Hi, I was wondering how one would go about proving that:
Sinx+sin2x+sin3x+......+sin(nx)=sin(x(n+1)/2)*sin(nx/2) / sin(x/2)

Knowing that De Moivre's theorem must be used.
This particular one was addressed in this thread

Sr-1425

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Re: 4U Maths Question Thread
« Reply #2142 on: December 12, 2018, 09:29:56 pm »
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Thank you.

Elias S

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Re: 4U Maths Question Thread
« Reply #2143 on: December 14, 2018, 06:47:34 pm »
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Hi :)
I have a problem that I can't seem to resolve in my head. Perhaps someone knows the answer. I want to compute the integral Of course, to an extension 2 student this is not too difficult, you just use the substitution However, when you eventually get to the line you encounter a problem, because However, everywhere I look I find the solution continue as Is it correct to continue like this, ignoring the absolute values? If yes, why? If not, how can we make it rigorous?

RuiAce

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Re: 4U Maths Question Thread
« Reply #2144 on: December 14, 2018, 06:58:05 pm »
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Hi :)
I have a problem that I can't seem to resolve in my head. Perhaps someone knows the answer. I want to compute the integral Of course, to an extension 2 student this is not too difficult, you just use the substitution However, when you eventually get to the line you encounter a problem, because However, everywhere I look I find the solution continue as Is it correct to continue like this, ignoring the absolute values? If yes, why? If not, how can we make it rigorous?

The simple fix to make it rigorous is to impose the condition that \( -\frac\pi2 \leq \theta \leq \frac\pi2 \). This covers all values in the range of \(y=\sin \theta\), that is \(-1\leq y\leq 1\).

And then, note that for any angle \(\theta\) in the first or negative first quadrants, \(\cos\theta\) is positive. Therefore \( |\cos\theta| = \cos\theta\) for this case.

In practice, it just tends to be assumed. It's nicer to use, because now we can assume \(\theta = \sin^{-1}x\) without justification, thereby avoiding any general formula tediosity

Edit: You made a typo I think, but it was clear enough to me that you substituted in \(\boxed{x=\sin\theta}\)
« Last Edit: December 14, 2018, 07:08:06 pm by RuiAce »