4U Maths Question Thread

[beau77bro]

and where can i find harder binomial theorem Qs, harder sequences and series? - not really major parts but i have major holes.

[aryak]

Hey,
How do you find the minor and major axis of a locus of an ellipse using complex numbers?
Ex: Iz-3I + Iz+3I = 12

Thanks

[sudodds]

HELPPP, i can't work out this question. and i am struggling with my reasoning. i have all the principles for Probability, but i can't apply them and im struggling, like in the last question how i didn't see how you didnt arrange them, well it was because it was an arrangement and was set. but now im sturggling with a lot of probability and was wondering how i could brush up or revise - like just arranging things properly or knowing when not to, or when it's just repeats or basic stuff i am over complicating.


i cant do the first part, i would like to attempt the second after you guys give me hints, thankyou.

hey! thought i'd give Rui a bit of a break and have a go aha

So this isn't too bad if you break it down a bit. So let's just put one of that pair in the left-most slot in the first rung (call them Player A). Let's look at the possibilities they verse Player B, depending on where Player B is.

If Player B is next to them in the same bracket (a 1 in 3 chance), the condition is already satisfied. So, 1/3 there. If Player B is in one of the other slots (a 2 in 3 chance), two things need to happen. Player A needs to win their game, and Player B does too. So multiply 2/3 by 1/2, and then 1/2 again, and you get 1/6.

Add these together!

1/3+1/6=1/2

(sorry I don't know how to use the fancy math text like Rui but hopefully this makes sense aha)

[RuiAce]

hey! thought i'd give Rui a bit of a break and have a go aha

So this isn't too bad if you break it down a bit. So let's just put one of that pair in the left-most slot in the first rung (call them Player A). Let's look at the possibilities they verse Player B, depending on where Player B is.

If Player B is next to them in the same bracket (a 1 in 3 chance), the condition is already satisfied. So, 1/3 there. If Player B is in one of the other slots (a 2 in 3 chance), two things need to happen. Player A needs to win their game, and Player B does too. So multiply 2/3 by 1/2, and then 1/2 again, and you get 1/6.

Add these together!

1/3+1/6=1/2

(sorry I don't know how to use the fancy math text like Rui but hopefully this makes sense aha)

OI TRUUUUUUUU.

Now, hints for the second part:

(whereas for two we only had \(\binom42\binom22 \frac{1}{2!} = 3 \) )
Try splitting the cases this way before going into subcases:
Case 1 - The pairings are split something like (A,B) (C,D) || (E,F) (G,H), so A and C can meet by the second round. This will be the case that best reflects the computations in part (i) (although they won't be an exact replica)
Case 2 - The pairings are split something like (A,B) (E,F) || (C,D) (G,H), so A and C are forced to meet in the third round.

That allows you to generalise it upwards

[RuiAce]

and where can i find harder binomial theorem Qs, harder sequences and series? - not really major parts but i have major holes.

Your best bet is to try papers from before 2000 (or even 1990) if you want some of these ones. (Or some of the extension level questions in the Cambridge textbook; like they're not really examinable but they definitely are hard.)

[RuiAce]

Hey,
How do you find the minor and major axis of a locus of an ellipse using complex numbers?
Ex: Iz-3I + Iz+3I = 12

Thanks

I presume you knew that the foci were at 3 and -3.



(Major axis = 12, minor axis = 6√3)

[Checkmate123]

in finding the nth root of -1, all textbooks and solutions say it is cis(((2k+1)pi)/n). However isn't the general solution cis((2kpi +/- pi)/n)? Do we ignore the plus/minus for some reason?

[RuiAce]

in finding the nth root of -1, all textbooks and solutions say it is cis(((2k+1)pi)/n). However isn't the general solution cis((2kpi +/- pi)/n)? Do we ignore the plus/minus for some reason?

You can check this by listing out a few cases.
For the ± case, you would have cis(π/n), cis(3π/n), cis(5π/n) and so on, and also the other way.
For the + case, you would also have cis(π/n), cis(3π/n), cis(5π/n) and the rest.

This happens more or less because
1. They both go both ways: you can just sub k=-1 instead of k=1.
2. The trig functions are periodic: \( \sin (2\pi + x ) = \sin x\) and \( \cos (2\pi+x) = \cos x\).

[samanthachoy]

Hey Rui!
Just wondering if you had any general tips for graphing functions when they're part of tan inverse ((tan^-1) f(x) )?

[Checkmate123]

You can check this by listing out a few cases.
For the ± case, you would have cis(π/n), cis(3π/n), cis(5π/n) and so on, and also the other way.
For the + case, you would also have cis(π/n), cis(3π/n), cis(5π/n) and the rest.

This happens more or less because
1. They both go both ways: you can just sub k=-1 instead of k=1.
2. The trig functions are periodic: \( \sin (2\pi + x ) = \sin x\) and \( \cos (2\pi+x) = \cos x\).

Thanks. So for any question, I can safely use only + and sub in k=1?

[RuiAce]

Hey Rui!
Just wondering if you had any general tips for graphing functions when they're part of tan inverse ((tan^-1) f(x) )?

If it's the limiting behaviour, i.e. as \(x\to \infty\) we had f(x) exploding, then we would end up with a horizontal asymptote
If it's just an asymptote, i.e. as \(x\to a\) we had f(x) exploding, anticipate a hole in the graph because that's gonna lead to a discontinuity.

[RuiAce]

Thanks. So for any question, I can safely use only + and sub in k=1?

I've never found any problem with only using +

Although at the same time, I've never solved it by inspection in high school (unless it was a multiple choice). I preferred doing the long method just in case I'd get called out for skipping steps and thus lose marks.

[samanthachoy]

Thanks Rui, that really cleared it up for me! Also, would you be able to share any general tips on attacking 4u probability questions? (like how to answer certain types of questions - whether to use binomial probability, combinations, the box method or something else)

[frog1944]

Now, hints for the second part:

Hi RuiAce, for the first round, what is the probability the players will meet? I understand that the number of ways to pick the 8 players is 105, but what is the number of ways that they will meet? I have the book, it says 1/7 but I'm not sure how to arrive at that using your method (I originally was attempting to do it this way)

Thanks

[RuiAce]

Thanks Rui, that really cleared it up for me! Also, would you be able to share any general tips on attacking 4u probability questions? (like how to answer certain types of questions - whether to use binomial probability, combinations, the box method or something else)

Not too sure what the box method is.

If you suspect a binomial probability then you're suspecting the same thing you saw in 3U. You're looking out for if you need to consider the number of ways k success can happen out of n trials, given a success probability p.

There's some extra counting techniques from the trial survival lectures in the notes section of this website