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Author Topic: 4U Maths Question Thread  (Read 659891 times)  Share 

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classof2019

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Re: 4U Maths Question Thread
« Reply #2400 on: October 22, 2019, 03:04:33 pm »
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How do you go about this question?

What is the total number of different combinations that can be made using 3 of the letters of the word MADAGASCAR?

Thanks

fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2401 on: October 22, 2019, 03:37:07 pm »
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Hey there!

When we make a selection of letters then look at combinations, it's important to note that we need to choose the letters we arrange first, before arranging them to form combinations. The issue with this question is there are duplicate A's, so we need to consider cases in terms of how many A's we get. What does make it easier is that though there are four A's, there are only ever three letters chosen, which reduces the number of cases.

First note that there is ever only one way to pick any number of A's, since they're all the same. We just need to pick the 6 other filler letters depending on how many A's we have out of three ie. if there is one A, we need two filler letters.

Thus, the number of ways will just be the number of ways to pick the filler letters, then rearrange the whole set itself - 6C3 x 3! + 6C2 x 3! + 6C1 x 3 + 6C0 x 1.

Hope this helps :)
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rh45_21

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Re: 4U Maths Question Thread
« Reply #2402 on: October 24, 2019, 12:23:12 pm »
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Just a question that I came across that even stumped my teacher, would appreciate the help. Its pt. iv) of the first and ii) of the second


RuiAce

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Re: 4U Maths Question Thread
« Reply #2403 on: October 24, 2019, 01:24:42 pm »
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Just a question that I came across that even stumped my teacher, would appreciate the help. Its pt. iv) of the first and ii) of the second


For the first part, we just use a double angle formula on \(\cos \frac\pi5 \).
\begin{align*}
\cos\frac\pi5 \sin \frac\pi{10} &= \frac14\\
\left(1 - 2\sin^2 \frac\pi{10}\right)\sin \frac\pi{10} &= \frac14\\
\sin \frac\pi{10} - 2sin^3 \frac\pi{10} &= \frac14\\
4\sin \frac\pi{10} - 8\sin^3 \frac\pi{10} &= 1
\end{align*}
\[ \text{Hence }8\sin^3 \frac\pi{10} - 4\sin \frac\pi{10} + 1 = 0 \]
By definition, this literally means that \(x=\frac\pi{10}\) is a root of the equation \(8x^3-4x+1=0\).

fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2404 on: October 24, 2019, 03:03:41 pm »
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Hey there!

The diagram in the spoiler might come in handy :)
Spoiler

For i), note that at height h, we have the area being equal to \(\int_{-\sqrt{h}}^{\sqrt{h}} (h-x^2) \ dx \). This will equal the value in given in the question :)

For ii), we clearly want to find the volume as x changes, and as such we need to somehow relate h to x, since the variable slice is in terms of h. Refer to the diagram in the spoiler! From similar triangles, we can see that h=x/a + 1, and thus we can evaluate the volume of the solid as below:
Spoiler
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mani.s_

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Re: 4U Maths Question Thread
« Reply #2405 on: October 28, 2019, 07:19:42 pm »
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Hi can someone help me with Question 14???

Thanks :)

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Re: 4U Maths Question Thread
« Reply #2406 on: October 28, 2019, 07:38:56 pm »
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Hi can someone help me with Question 14???

Thanks :)

mani.s_

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Re: 4U Maths Question Thread
« Reply #2407 on: October 28, 2019, 08:22:55 pm »
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Hi Thanks so much for your reply!!! I just had one question, how did you deduce that a = 0 from a^2 + ai / a^2 + b^2 + 1 ???

louisaaa01

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Re: 4U Maths Question Thread
« Reply #2408 on: October 28, 2019, 08:32:21 pm »
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Hi Thanks so much for your reply!!! I just had one question, how did you deduce that a = 0 from a^2 + ai / a^2 + b^2 + 1 ???

Hi!

So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.

This essentially implies that the imaginary part of the expression is zero. After realising the denominator, we look for the coefficient of i and let this equal zero. In this case, the coefficient of i (the imaginary part) is; ai / (a2 +b2 +1). Since the denominator cannot equal zero, this means that a = 0.
« Last Edit: October 28, 2019, 08:36:58 pm by louisaaa01 »
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mani.s_

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Re: 4U Maths Question Thread
« Reply #2409 on: October 28, 2019, 08:37:47 pm »
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Hi!

So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.

This essentially implies that the imaginary part of the expression is zero, that is, after realising the denominator, the coefficient of i. Since the denominator cannot equal zero, this means the coefficient of i on the numerator, in this case 'a' = 0.
I'm so sorry, but I still don't understand how we can say a =0??? I'm sorry for troubling you guys :(

mani.s_

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Re: 4U Maths Question Thread
« Reply #2410 on: October 28, 2019, 08:46:22 pm »
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Hi!

So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.

This essentially implies that the imaginary part of the expression is zero. After realising the denominator, we look for the coefficient of i and let this equal zero. In this case, the coefficient of i (the imaginary part) is; ai / (a2 +b2 +1). Since the denominator cannot equal zero, this means that a = 0.
Wait I think i understand it. So basically we're solving a / a^2 + b^2 +1  =  0, since the imaginary part is 0. from this we get a = 0

louisaaa01

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Re: 4U Maths Question Thread
« Reply #2411 on: October 28, 2019, 08:47:33 pm »
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Wait I think i understand it. So basically we're solving a / a^2 + b^2 +1  =  0, since the imaginary part is 0. from this we get a = 0

Yes, that's exactly right! We let the imaginary part = 0, since, if a number has no imaginary part, it must be real.
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mani.s_

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Re: 4U Maths Question Thread
« Reply #2412 on: October 28, 2019, 09:03:00 pm »
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I'm sorry for bothering you guys again, but can someone help me with this question?

THANK YOU SO MUCH FOR YOUR HELP!!!

RuiAce

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Re: 4U Maths Question Thread
« Reply #2413 on: October 28, 2019, 09:07:43 pm »
+1
I'm sorry for bothering you guys again, but can someone help me with this question?

THANK YOU SO MUCH FOR YOUR HELP!!!
\begin{align*}
z^2 &= \overline{z}^2\\
z^2 - \overline{z}^2 &= 0\\
(z+\overline{z})(z-\overline{z} )&= 0\tag{diff. of two squares}
\end{align*}
\[ \text{Let }z=x+iy.\\ \text{If }z+\overline{z}=0\text{, then }(x+iy)+(x-iy)=0\\ \text{and hence }2x = 0 \implies x = 0 \implies z\text{ is purely imaginary.} \]
The other case is handled similarly.

Alternatively, just sub \(z=x+iy\) at the start and deduce that \(2xy = 0\), so either \(x=0\) or \(y=0\).

In the future, please post any attempts you have also made in the question. Although there's nothing wrong with us helping, it's quite redundant if we're always answering the whole thing - it doesn't show us where the gap in learning is, and doesn't necessarily help it sink in either.
« Last Edit: October 28, 2019, 09:10:14 pm by RuiAce »

mani.s_

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Re: 4U Maths Question Thread
« Reply #2414 on: November 06, 2019, 09:41:54 pm »
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How would I solve this question? I know that if you multiply a complex number by i, you get a rotation of pi/2. I also know that arg(zw)=arg(z)+arg(w), but how would I use these to show the working out??

Any help is appreciated :)