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Author Topic: 3U Maths Question Thread  (Read 1230312 times)  Share 

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fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4245 on: October 29, 2019, 06:27:01 pm »
+1
Hey there!

Your approach doesn't work because by spacing out the children with one adult, you're eliminating a few possibilities: what if we had two adults in between, or three?

Thus, to find the answer, we seat the adults first in 5! ways, then seat the children in one of the gaps ie. 6x5x4, since the gaps cannot be used for a child more than once. This equates to 14400.

Hope this helps :)
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spnmox

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Re: 3U Maths Question Thread
« Reply #4246 on: October 29, 2019, 07:10:23 pm »
0
Oh okay thank you so much!!! That makes sense.

An unbiased die is thrown 6 times. Calculate the probabilities that the six scores obtained will consist of exactly two 6' s and four odd numbers.

I did 6C2*(1/6)^2*6C4*(1/2)^4=25/64. The answer is 5/192. How do you get the answer and why does my approach not work?

myopic_owl22

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Re: 3U Maths Question Thread
« Reply #4247 on: October 30, 2019, 04:45:33 pm »
+4
Hi there, sorry I'm not fun_jirachi but hopefully I can help out :)

You're almost there with your working. There just isn't a need to multiply by 6C4 because you've already chosen 2 of the 6 dice to have the score of six, so the other 4 have already been 'chosen'.

I.e final solution is 6C2 • (1/6)^2 • (1/2)^4

myopic_owl22

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Re: 3U Maths Question Thread
« Reply #4248 on: October 30, 2019, 05:45:22 pm »
+1
I've got a question myself if anyone can help me out --

For a coordinate geometry rhombus proof, is it sufficient to prove 2 adjacent sides equal and perpendicular diagonals? Or would I have to prove that the diagonals bisect as well?

Thanks :)

RuiAce

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Re: 3U Maths Question Thread
« Reply #4249 on: October 30, 2019, 06:04:50 pm »
+2
I've got a question myself if anyone can help me out --

For a coordinate geometry rhombus proof, is it sufficient to prove 2 adjacent sides equal and perpendicular diagonals? Or would I have to prove that the diagonals bisect as well?

Thanks :)
You’d have to prove diagonals bisect or at least something else, for example opposite sides equal.

What you’ve given right now may unintentionally lead to a kite that’s not a rhombus. A kite doesn’t necessarily have all sides equal in length, but you may have found two adjacent sides that are equal in length by coincidence.
« Last Edit: October 30, 2019, 06:06:36 pm by RuiAce »

fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4250 on: October 30, 2019, 06:34:40 pm »
+2
Hi there, sorry I'm not fun_jirachi but hopefully I can help out :)

You're almost there with your working. There just isn't a need to multiply by 6C4 because you've already chosen 2 of the 6 dice to have the score of six, so the other 4 have already been 'chosen'.

I.e final solution is 6C2 • (1/6)^2 • (1/2)^4

Yep, this is essentially the gist of what's happening; you don't need to choose twice.

There's also another way of thinking about it, considering that P(event) = no. of favourable outcomes / total outcomes. It follows similar logic to the question above, with the favourable outcomes equalling 6C4 x 34 x 12 (getting odds, and sixes, then rearranging the identical objects) and the total number of ways being 66. You'll find that this also equals 5/192. :)

I've got a question myself if anyone can help me out --

For a coordinate geometry rhombus proof, is it sufficient to prove 2 adjacent sides equal and perpendicular diagonals? Or would I have to prove that the diagonals bisect as well?

Thanks :)

Just adding onto Rui's answer, usually two pairs of parallel lines indicates a parallelogram - having two of these sides being adjacent and equal ensures it's a rhombus. Diagonals bisecting at right angles should also be fine :)
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annabeljxde

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Re: 3U Maths Question Thread
« Reply #4251 on: November 03, 2019, 07:55:03 pm »
0
Hi!

Could someone please explain these two questions quickly thanks!
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lace02

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Re: 3U Maths Question Thread
« Reply #4252 on: November 03, 2019, 07:56:30 pm »
0
Hey all,

Just wondering how do you find the centre of motion for a simple harmonic question?

Thanks

RuiAce

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Re: 3U Maths Question Thread
« Reply #4253 on: November 03, 2019, 08:28:42 pm »
+1
Hi!

Could someone please explain these two questions quickly thanks!
\[ \text{Let }\theta = 2x \]
\begin{align*}
\sin 2x &= -\frac12\\
\sin \theta &= -\frac12\\
\theta &= n\pi + (-1)^n \sin^{-1}\left(- \frac12\right)\\
\theta &= n\pi + (-1)^n \left(-\frac\pi6\right)\\
\theta &= n\pi + (-1)^{n+1} \left(\frac\pi6\right)\\
2x &= n\pi + (-1)^{n+1} \frac\pi6\\
x &= \frac{n\pi}2 + (-1)^{n+1} \frac\pi{12}
\end{align*}
Now with the other one, after experimenting with a little, I found that the better option in my opinion is to complete the square. It turns out that the particle initially being at \(x=k\) is a red herring, since they all start at \(x=k\)!

Completing the square allows us to convert into the \(v^2 = n^2\left[ a^2 - (x-x_0)^2\right]\) form, from which we can immediately deduce the period, amplitude and centre of motion.
\begin{align*}
v^2 &= n^2 \left(2kx - x^2\right)\\
&= n^2 \left[  k^2 - \left(x^2 - 2kx + k^2 \right)\right]\\
&= n^2 \left[ k^2 - (x-k)^2 \right]
\end{align*}
This shows that the amplitude is \(k\) and the centre of motion is also \(k\). The only correct answer here is B.
Hey all,

Just wondering how do you find the centre of motion for a simple harmonic question?

Thanks
Through multiple ways depending on the context.
- If you know the extremities of motion, simply just take the midpoint of them.
- If you know the lower extreme point and the amplitude, simply compute lower extreme + amplitude
- If you know the upper extreme point and the amplitude, simply compute upper extreme - amplitude
- Use the method of the previous question if you have \(v^2\) in terms of \(x\).
- Solve some kind of equation involving \(x=a\cos(nt+\alpha)\), where \(a\) is the amplitude
« Last Edit: November 03, 2019, 09:30:17 pm by RuiAce »

r1ckworthy

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Re: 3U Maths Question Thread
« Reply #4254 on: November 03, 2019, 09:11:55 pm »
0
Hey RuiAce, you left the 1/2 as positive in the first part of your answer, which should be negative. I believe the answer is D?
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4255 on: November 03, 2019, 09:29:15 pm »
+1
Hey RuiAce, you left the 1/2 as positive in the first part of your answer, which should be negative. I believe the answer is D?

Whoops, good pickup!

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Re: 3U Maths Question Thread
« Reply #4256 on: January 01, 2020, 04:48:26 pm »
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Hi, I'm stuck on this vectors question (from Cambridge):
Suppose that OABC is a parallelogram. Let M be the midpoint of OA and let P be the point of intersection of MC and OB. Prove, using vectors, that OP = 1/3 * OB.

RuiAce

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Re: 3U Maths Question Thread
« Reply #4257 on: January 01, 2020, 07:37:25 pm »
+3
Hi, I'm stuck on this vectors question (from Cambridge):
Suppose that OABC is a parallelogram. Let M be the midpoint of OA and let P be the point of intersection of MC and OB. Prove, using vectors, that OP = 1/3 * OB.
This question drove me insane for the past hour because I genuinely cannot think of any MX1 way of doing it. Even the MX2 approach I took feels outside of the scope of the HSC. I have no idea what the intended method the authors of the Cambridge textbook had on my mind.

Regardless, this was the closest to MX1 scope I managed to find. It's adapted from other sources. Let \(\overrightarrow{OA} = \mathbf a\) and \(\overrightarrow{OC} = \mathbf c\). Then \(\overrightarrow{OB} = \mathbf a + \mathbf c\).

Note that \(P\) obviously lies on \(OB\), so \(\overrightarrow{OP}\) is automatically parallel to \(\overrightarrow{OB}\).

Hence, we can write \(\overrightarrow{OP} = \lambda (\mathbf{a}+\mathbf{c})\) for some scalar \(\lambda>0\).
________________________________________________________________________________
\[ \text{Inevitably, }\overrightarrow{MP}\text{ and }\overrightarrow{CP}\text{ have to be involved.}\\ \text{But we can just note that }\overrightarrow{MP}\text{ and }\overrightarrow{PC}\text{ point in the same direction, so we can write}\\ \overrightarrow{MP} = \mu\overrightarrow{PC}\text{ for some scalar }\mu > 0. \]
________________________________________________________________________________

We now derive expressions for \(\overrightarrow{MP}\) and \(\overrightarrow{PC}\) in terms of \(\mathbf a\) and \(\mathbf c\).
\begin{align*}
\overrightarrow{MP} &= \overrightarrow{MO} + \overrightarrow{OP}\\ &= -\frac12 \mathbf{a} + \lambda(\mathbf{a}+\mathbf{c})\\
&= \left( \lambda - \frac12 \right) \mathbf{a} + \lambda \mathbf{c}
\end{align*}
\begin{align*}
\overrightarrow{PC} &= \overrightarrow{PO}+\overrightarrow{OC}\\
&= -\lambda(\mathbf{a}+\mathbf{c}) + \mathbf{c}\\
&= -\lambda \mathbf{a} + (1-\lambda)\mathbf{c}.
\end{align*}
________________________________________________________________________________

So by subbing these expressions in, we now obtain
\[ \left( \lambda - \frac12 \right) \mathbf{a} + \lambda \mathbf{c} = -\lambda\mu \mathbf{a} + (1-\lambda)\mu\mathbf{c}. \]
This is where I believe the problem goes outside of MX1. How we can continue here is supposedly by equating the scalars on each of the vectors:
\begin{align*}
\lambda - \frac12 &= -\lambda\mu\\
\lambda &= (1-\lambda) \mu
\end{align*}
In theory, to justify this rigorously we'd need to use linear independence. As a theorem, two vectors \(\mathbf{v}\) and \(\mathbf{w}\) are 'linear independent' when they are non-parallel, as it turns out. This is obviously true in the case of the parallelogram.

But the actual definition is that two vectors are linear independent, if whenever \(\lambda \mathbf{v} +\mu \mathbf{w}=\mathbf{0}\), the scalars \(\lambda\) and \(\mu\) must be equal to \(0\) themselves. The consequence of this result is that when two vectors are linearly independent, as they are here, extracting and equating coefficients suddenly becomes acceptable.
________________________________________________________________________________

Assuming that ramble above, we can then solve. The second equation becomes \( \mu = \frac{\lambda}{1-\lambda} \), which when subbed in gives
\begin{align*}
\lambda-\frac12 &= -\frac{\lambda^2}{1-\lambda}\\
(2\lambda-1)(1-\lambda) &= -2\lambda^2\\
2\lambda-1-2\lambda^2 +\lambda &= -2\lambda^2\\
\lambda &= \frac13.
\end{align*}
So \(\overrightarrow{OP} = \frac13 (\mathbf a+\mathbf b) = \frac13 \overrightarrow{OB}\) as required.

Obviously, I'm highly unconvinced that this was the intended approach. I am watchful for if there's any more MX1-like responses that show up, but can't make any promises.

Edit: So I've asked some of my friends on this problem as well. They're not too sure how this can be done either. (At least, not without linear independence.) I'm somewhat convinced now that this question shouldn't be in the exercise altogether, since linear independence is not a part of the HSC syllabus. I'm not sure what the author's intended methods were here.

(It may be worth that with similar triangles, the problem gets made a lot easier. But obviously, Euclidean geometry has been wiped from the syllabus.)
« Last Edit: January 02, 2020, 10:39:30 pm by RuiAce »

JkbC

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Re: 3U Maths Question Thread
« Reply #4258 on: January 02, 2020, 08:46:36 pm »
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Hi, I'm finding some vector questions difficult from Cambridge (2019/20 Y12 3U, Ex 8F).

I'm really struggling with the whole exercise but if you could show me how to do question 6 and 14 (attached) that would help me grasp the concept.

Thanks! ;D




RuiAce

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Re: 3U Maths Question Thread
« Reply #4259 on: January 02, 2020, 09:43:16 pm »
0
Hi, I'm finding some vector questions difficult from Cambridge (2019/20 Y12 3U, Ex 8F).

I'm really struggling with the whole exercise but if you could show me how to do question 6 and 14 (attached) that would help me grasp the concept.

Thanks! ;D




Are we supposed to assume that \(g = 9.8\) or something here?