May 31, 2020, 02:45:42 pm

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RuiAce

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« Reply #2460 on: February 17, 2020, 09:32:48 pm »
+4
I think my coaching taught us some out of syllabus content for vectors- they said that for completeness of this topic, learning the cross product is necessary. I'd still love to know how it would be done though, but I'm not too fussed about it since it won't be tested

Also, could someone explain this proofs question: Q10d from the cambridge textbook? I've forgotten how roots work- am I meant to use the sum and product of roots?
(Image removed from quote.)

thanks so much
Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:

1. Take one point on each line and find a direction vector through these points. So for example, (1,3,1) to (0,2,1). This vector is then a vector going from one of the lines, to the other. (Doesn't matter which line it goes from and to, since ultimately we'll be considering a distance, and hence ignore the direction.) Call this vector $\vec{a}$.

2. Find the direction vectors of each of the lines.

3. Take the cross product of the vectors in step 2 to find a vector $\vec{n}$ perpendicular to both of the lines.

4. The projection of the vector joining the two lines $\vec{a}$, onto the normal vector $\vec{n}$, that is the vector
$\operatorname{proj}_{\vec{n}}\vec{a},$
will have the shortest distance between the two lines.
________________________________________________________

Within the scope of the new syllabus, sums and products of roots would've been the first thing I thought of. You should find that the equivalence is indeed true, but I only prove the reverse implication. The forward implication is left as your exercise.
$\text{Let }\frac1\alpha\text{ and }\frac1\beta\text{ be the roots of}\\ cx^2 + bx + a.$
\text{Then it follows from the sums and products of roots that}\\ \begin{align*} \frac{\alpha+\beta}{\alpha\beta} = \frac1\alpha+\frac1\beta &= -\frac{b}{c},\\ \frac{1}{\alpha\beta} &= \frac{a}{c} \end{align*}
$\text{The second equation immediately rearranges to }\alpha\beta = \frac{c}{a}.\\ \text{Subbing into the first equation,}\\ \frac{\alpha+\beta}{\frac{c}{a}} = -\frac{b}{c} \implies \alpha+\beta = -\frac{b}{a}.$
\text{Thus a quadratic equation with roots }\alpha\text{ and }\beta\text{ is}\\ \begin{align*}(x-\alpha)(x-\beta) &= 0\\ \implies x^2-(\alpha+\beta)x + \alpha\beta &= 0\\ \implies x^2 + \frac{b}{a}x + \frac{c}{a} &= 0\\ \implies ax^2+bx+c &= 0 \end{align*}\\ \text{as required.}

milie10

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« Reply #2461 on: February 17, 2020, 10:48:47 pm »
0
Hi, back again

Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:
...
Thanks heaps again Rui!

This proofs topic is so confusing for me (it's such a new concept!). Hopefully it will get a lot easier with practice

Last one for the day- I asked a few of my friends this and we are all quite mind boggled by this question (cambridge enrichment question). Does anyone know how to do this?

Thanks
« Last Edit: February 17, 2020, 11:07:20 pm by milie10 »

RuiAce

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« Reply #2462 on: February 18, 2020, 04:29:39 pm »
+3
Hi, back again
Thanks heaps again Rui!

This proofs topic is so confusing for me (it's such a new concept!). Hopefully it will get a lot easier with practice

(Image removed from quote.)
Last one for the day- I asked a few of my friends this and we are all quite mind boggled by this question (cambridge enrichment question). Does anyone know how to do this?

Thanks
Part i:
From (4) there exists a passenger that earns exactly 3 times as much as the guard. Therefore immediately from (3) Mr Ward cannot be this passenger, because the guard would have to earn \$33333.3333333... a year which is impossible (because money is only correct to the nearest cent.)

Therefore the passenger, who's neighbours with the guard from (4), must be Dr Pender or Mr Sadler. However in order to be neighbours with the guard, they must live halfway between Melbourne and Sydney as well, from (2). This eliminates Mr Sadler because of (1).

Conclusion: Dr Pender lives halfway between Melbourne and Sydney.

Part ii:
There is a passenger who lives in Melbourne from (6). Well we know it instantly ain't Mr Sadler because that would contradict (1). But we just singled out where Dr Pender lives.

Conclusion: Mr Ward is the passenger who lives in Melbourne. Consequently, we now know that Ward is the guard, again using (6).

Part iii:
We're told from (5) that apparently Pender beats the fireman at pool. But the only roles left are now the fireman and the driver. If Pender were the fireman, he'd have to beat himself in pool?!

Conclusion: Pender is the driver, because he'd have to be a different person to the fireman.
__________________________________________________________________________________________

Alright. This took me a while because I'm not the best at logic puzzles. Honestly didn't expect one to show up in the Cambridge textbook even as enrichment! Perfectly understandable because the enrichment section is known to push students to think beyond syllabus expectations.

I could argue that it does lie in the syllabus expectations because you are expected to know how logic works, and use a sequence of implications. But there were lots of implications to think through here; all of them requiring you to think beyond the basic facts given in the question.

To be totally honest, I'm not 100% convinced by my own solution yet. I checked and it does match the answer given, but I would advise you to repeatedly scan each step and be able to say "I can agree with your logic" at each point. Only if you can confidently say you agree with all of the logic, can the solution be regarded as very likely correct. (If not, well better ask haha.)

Also, note that this may not be the most optimal route, even if it ends up being correct!
« Last Edit: February 18, 2020, 05:48:30 pm by RuiAce »

milie10

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« Reply #2463 on: March 05, 2020, 12:24:24 pm »
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Hi!!

I have reached the dreaded topic of proving ~inequalities~ :')

I have a general question about working out layout:
e.g. RTP $(a+b)( \frac{1}{a} + \frac{1}{b} ) \ge 4$
I like to work backwards by assuming that the proof is correct, shuffling it around until it gives me a property that I know is correct (e.g. in this case I ended up with $(a-b)^2 \ge 0$) and then writing the proof out backwards in my final answer- if I do this, where should I put my original assumption? I know we're not meant to write it as an inequalities since that means that you're assuming that it's true
I was thinking about doing LHS= and RHS= but then that doesn't let me move numerals from either side

orrr is there a better method to solving these that doesn't involve me working backwards?

Thanks so much
« Last Edit: March 05, 2020, 12:48:43 pm by milie10 »

fun_jirachi

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« Reply #2464 on: March 05, 2020, 01:15:35 pm »
+3
Hey there!

Working backwards is often the best way to do this - after you're done working backwards to something you know is true (ie. an identity!), literally write your working backwards. It sounds confusing, but essentially working backwards twice gets your working to go forwards from the identity to the expression you wanted to prove in the first place. The identity usually goes at the top. You can't put 'your original assumption' anywhere - assuming something is true then proving it is a huge no-no. This should only be a tool to point you in the correct direction!

In this case, we'd have that:

$(x-y)^2 \geq 0 \ (\forall x, y \ \in \mathbb{R}) \\ \text{Hence, we have that } \ x^2 + y^2 \geq 2xy \\ \frac{x^2+y^2}{2} \geq xy$

which is our identity.

Now, we then use this identity such that $x = \sqrt{\frac{a}{b}} \text{ and } y = \sqrt{\frac{b}{a}}$ to have that
$\frac{a}{b} + \frac{b}{a} \geq 2$
Adding two to both sides we have
$\frac{a}{a}+\frac{a}{b}+\frac{b}{a}+\frac{b}{b} \geq 4$
Then factorising the left hand side, we have our expression.

You can do LHS = xxxx and RHS = yyyy and make them meet in the middle (making sure you only work on one side at a time!) but this is often tacky and is best used as a last resort.

Hope this makes sense!
Failing everything, but I'm still Flareon up.

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esteban

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« Reply #2465 on: March 06, 2020, 06:55:20 am »
+2
Hi!!

I have reached the dreaded topic of proving ~inequalities~ :')

I have a general question about working out layout:
e.g. RTP $(a+b)( \frac{1}{a} + \frac{1}{b} ) \ge 4$
I like to work backwards by assuming that the proof is correct, shuffling it around until it gives me a property that I know is correct (e.g. in this case I ended up with $(a-b)^2 \ge 0$) and then writing the proof out backwards in my final answer- if I do this, where should I put my original assumption? I know we're not meant to write it as an inequalities since that means that you're assuming that it's true
I was thinking about doing LHS= and RHS= but then that doesn't let me move numerals from either side

orrr is there a better method to solving these that doesn't involve me working backwards?

Thanks so much

You can also just start with the LHS and manipulate it directly. This doesn't involve writing down any inequalities that you have not yet proven.

Eg/

LHS = (a+b)(1/a+1/b) = (a+b)^2/ab = ((a-b)^2+4ab)/ab >= 4ab/ab = 4 = RHS, where the inequality follows from non-negativity of squares.

milie10

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« Reply #2466 on: March 07, 2020, 01:24:10 am »
0
hii back again

really struggling with these two - I made a start on the first one but not sure where to go from there

thank you!

fun_jirachi

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« Reply #2467 on: March 08, 2020, 07:23:37 pm »
+2
Hey there!

Just notice for the first question when transitioning from the second line to the third, you've made a small computational error: the third line should read $\frac{1}{n} < \ln (n) - \ln (n-1) < \frac{1}{n-1}$ instead of what you've got. From here, we can manipulate the expressions with log laws amongst other things to obtain the result in the question. Perhaps work from these lines here:

$\frac{1}{n} < \ln \left(\frac{n}{n-1}\right) < \frac{1}{n-1}
\\ -1 > n\ln \left(\frac{n-1}{n}\right) > -\frac{n}{n-1}$

For the second question, consider a similar idea where instead of upper rectangles and lower rectangles, we have a lower rectangle, an integral and a trapezium.

Hope this helps
Failing everything, but I'm still Flareon up.

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milie10

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« Reply #2468 on: March 19, 2020, 09:48:03 pm »
0
hi!

I did this using X~bin(24, 1/3) - does this method work?

These are the answers and I'm not sure how they got the 16/81

Thank you!

fun_jirachi

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« Reply #2469 on: March 19, 2020, 10:51:44 pm »
+1
Hey there!

'I did this using X~bin(24, 1/3) - does this method work?' - yes, for X=0, since we want there to be no tagged sheep.

However, doing parts (a) and (b) are supposed to lead you on to the answer given! Multi-step questions are supposed to lead you into discovering a result - make sure you look to incorporate earlier results in the later parts of the question

Having said that, from (a) you would have found that the probability of having no tagged sheep on a day is $\left(\frac{2}{3}\right)^4 = \frac{16}{81}$, which is where that comes from. They simply did it on a per day basis, which makes more intuitive sense given they asked for no sheep on six consective days, not 24 consecutive selections.

Hope this makes sense
Failing everything, but I'm still Flareon up.

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shekhar.patel

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« Reply #2470 on: March 21, 2020, 10:13:07 am »
0
Hi. How do I do Q7
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fun_jirachi

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« Reply #2471 on: March 21, 2020, 05:37:24 pm »
+1
Hey there!

If I'm not mistaken, an assumption should be made that $a, b \in \mathbb{Z}$. Notice that if a and b are integers, and they have an even difference, they must both be odd or both be even. Consider now what happens to the sum of a and b, and thus the difference of the squares.

Hope this helps
Failing everything, but I'm still Flareon up.

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mrsc

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« Reply #2472 on: May 24, 2020, 01:13:13 am »
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Hey guys, just needed help with this complex number question part (d). Thanks

fun_jirachi

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« Reply #2473 on: May 24, 2020, 06:50:26 am »
+1
Hey there!

Consider what happens if you take $(\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)$, where $\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}$.

We have that $(\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})$.

We also have that by expanding the brackets and simplifying using part (b), $(\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1$.

Hence, we have that $(2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1$.

What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to $(2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1$. From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!

Hope this helps
Failing everything, but I'm still Flareon up.

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