3U Maths Question Thread

[twelftholmes]

heyo... again! sorry for the continuous differential equation questions but on holidays right now I can't ask my teacher. this page has been amazing and really helpful so far though, so thank you.

Here is the link (once again haha): https://imgur.com/Qq8AybY

Okay so with this one I have absolutely no idea what they're doing?? Where did the a and b come from? I do understand that they are working with what they have since there is no condition (like a coordinate), but after that I am confusion.

[fun_jirachi]

The a and b are arbitrary constants that they use to simplify the expression. Not sure if partial fractions gets taught in 3U, but basically you can't integrate something of the form \(\frac{A}{(ax+b)(cx+d)}\), where A, a, b, c, d are real constants. We can simplify this sort of expression to two terms of the form \(\frac{A}{ax+b}\) which is easily integrable as we know \(\int \frac{f'(x)}{f(x)} \ dx = \ln |f(x)| + C\). They run through a whole process of finding out what the constants are so the integrals are equivalent, then integrate each term separately as a log integral.

As for the actual process of partial fractions, you usually take something of this form \(\frac{A}{(ax+b)(cx+d)}\), then split it up. For example, we could have the following:


Notice how each term on the right hand side has a denominator that is a factor of the denominator on the left hand side. You then multiply everything out such that there is a common denominator ie. in this case we'd end up with \(42 = A(x-1) + B(x+1)\), then either substitute values of x to find the constants A and B, or you can equate the coefficients of x to the right hand side (A+B = 0, etc) then solve a simultaneous equation. It gets more complicated if powers of linear terms exist in the original integral.

However, feel free to ignore the second part (that was for your interest) - if you weren't explicitly taught this (if you weren't it's highly unlikely that you'd need something like this, except perhaps in the most simplistic of cases ie. in the question (which should be somewhat familiar now you've done this )). I can check up later on this (still got a few things to do right now) if you're not sure.

Hope this helps

[twelftholmes]

your efforts are honestly a gift to the forum, thanks so much for taking the time to help math noobs like me

do you think this type of question where we have to assign random constants because we are asked to solve it even if we aren't given a point would come in hsc? it seems like a higher level to me but then again I got 33% in my 3unit trial so perhaps everything seems higher level to me XD)

[fun_jirachi]

Hey, that's really nice of you to say - but I hope more than anything you stay positive (you can do this!! you are a winner and will ace hsc)

The constants are fixed (not random, even though it seems that way). Partial fractions used to be in the old 4U course - they're not in the current 3U course (on a Ctrl-F search of the 2U and 3U syllabus) - but check with peers and maybe someone with a more informed opinion might be able to say. I can't think of anything right now that would involve assigning temporary constants - it becomes quite common with manipulating integrals into a better form amongst other things, but differential equations of this form are something you should definitely practice (and differential equations in general) if you're struggling. Theoretically simple differential equations (more the first two questions) should be questions you should be able to see and say to yourself 'I know how to do this' and then execute your knowledge and methods correctly. If you improve with this some of the more abstract questions as well as this will probably come a bit easier

[twelftholmes]

your encouragement is really hyping me up !!!
I hope my questions aren't too much haha but I'm riding this wave of extension 1 maths motivation before it goes 

question: https://imgur.com/RhlWsaX
answer: https://imgur.com/ubkUrZ3
I understand basic questions involving the pigeonhole principle but for this one I took a look at the worked solutions and my understanding flew out the window...

Mainly I'm having difficulty with the thought process associated with it, where do you start thinking about a question like this? The way they did it in the worked solutions doesn't make sense to me, honestly my initial thought was 20C11 and then from there no idea, but I see that's so far off.

If you could please explain this to me in super basic terms that would be amazing.
Once again, thank you!!

[fun_jirachi]

Hey

The way they did it in the answers is actually the most logical way of doing the question - I think it's pretty standard to come up with that once you realise 'okay, I've got to use PHP'. The harder part is actually realising that.

The main tip-off you should've gotten that you didn't need nPr or nCr is that nowhere in the question does it ask you to find the number of ways numbers can be selected. It only tells you that 11 numbers are selected.

From here, there are a few ways to check it is a PHP question. Try identifying any possible pigeons and holes (in this case, pigeons = {set of numbers between 1 and 20 inclusive} and holes = {x, 21-x} where x is an integer from 1 to 20 inclusive). Other key tip-offs include the fact that you're told 11 numbers are selected (which just so happens to be one more than pigeons/holes).

Once you get to that point, the question becomes a lot more formulaic. You can start to ask yourself 'why do two numbers have to add up to 21?' Typically you look at things like the converse (is it possible to pick 11 numbers from the 20 such that none add up to 21?). Try starting with facts and build up with small logical steps until you can soundly reason/prove/disprove the statement in the question. This goes for any other proof question as well.

[mrsc]

Hey, just needed some help on this question-part (b). Not really sure what other equation I can make for the chain rule. Thanks in advance.

[fun_jirachi]

Hey there

Helpful diagram

Always helpful to draw a diagram, see above. Note that the question asks for the rate of change of the radius ie. \(\frac{dr}{dt}\) and from part a, we can write \(\frac{dV}{dt} = -K\pi r^2 = k\pi r^2\) (compressing \(K\pi\) into a single constant). We need a expression that connects the two - simplest one you can see is \(\frac{dV}{dr}\).

We also note that via similar triangles, \(h = 3r\). Substituting in the volume formula given, we have that \(V = \pi r^3 \implies \frac{dV}{dr} = 3\pi r^2\).

We now have \(\frac{dV}{dr}\) and \(\frac{dV}{dt}\). We want to find \(\frac{dr}{dt}\). Try taking it from here!

Hope this helps

[Ookei]

Confused on how to start the question because I'm too bad at probability. Here it is: https://imgur.com/3cKs4Wv

[fun_jirachi]

Hey there!

Note that a lot of the introduction to the question is pretty irrelevant. All you need to know from the intro is that there are 5 classrooms, each with two aircons, and the information given in the table. Please use the hints to solve the questions yourself before opening the spoilers for the answers

a) If every air con is turned off, that means all five classrooms have two air cons off. First, what is the chance that one classroom has both air cons off, then what is the chance that all five have them all off?

Answer

\(0.25^5\)

b) If two classrooms have no air cons off, and there is at most one classroom with one off, we have two cases:
- Two rooms with two air cons off, one room with one air con off, two rooms with no air cons off
- Three rooms with two air cons off, two rooms with no air cons off

Consider why this is an exhaustive list, and check that they both satisfy the criteria in the question. From here, you can calculate the probabilities of each case individually in a similar manner to a), then sum them up.

Spoiler

\(0.45^2\times 0.3\times 0.25^2 + 0.45^2\times 0.25^3\)

c) The 'at least' part of the question should be tipping you off to use the complement in some way, where possible and here, we can do exactly that. The probability that at least one classroom has no air cons off is the complement of no classroom has no air cons off. This is a rather more difficult question, but have a go nevertheless.

Hint 1

What is one arrangement of every classroom having at least one air con on?

Hint 2

Have you considered every way in which you can have every classroom having at least one air con on?

Answer

\(1 - (5\times 0.25\times 0.3^4)\)

[twelftholmes]

hello! here is the q and worked solution: https://imgur.com/a/upqabO1

my question here is based on confusion with the worked solution. i know there is another way of working this out by using 6P and 6Q, which I would rather use but don't remember how. i don't mind learning the way they did it in the worked solution but would rather know the 6P and 6Q way as that makes more sense to me (if you guys even know what I am talking about haha)

thanks in advance!

[fun_jirachi]

All you really have to change is the middle part where they make the assumption. Also, and this is very important, change the part where they say 'Assume the statement is true for n=k+1' - this is completely wrong and you cannot under any circumstances assume what you're trying to prove.



And continue the proof as provided. It's effectively the exact same proof, only they don't explicitly mention an integer constant p/q.