HSC Tips - Harder 3U (1 Viewer)

[sen00]

try this inequality:

Prove that for all integers n >= 1

(n+1)^n >= (2^n) * (n!)

 

[HappyFeet]

By AM-GM:

(1+2+3+.......+n)/n >= \[n]sqrt {n!}

so LHS= (n+1)/2 >= \[n]sqrt {n!}

Hence (n+1)^n >= (2^n) * (n!)

 

[sen00]

nice soln, thats the way.

its like the first inequality they chuck at you every maths camp.

i asked my school maths teacher and he solved it really quick. james ruse gets higher UAIs than every other school because james ruse is overall better than every other school at maths by a fair margin. mainly because teachers are brilliant at james ruse.

 

[HappyFeet]

Go through a 4 unit text book and you can find heaps of inequalities where you can elegantly solve (AM-GM, Cauchy, Rearrangement) or else bash (Expand everything out, Majorisation).

 

[sen00]

lol i dont think you are allowed rearrangment/c-s/majorisation. i dont even think am-gm-hm is allowed unless you prove them.

 

[BackCountrySnow]

Circle geometry:
Go through arnold.

arnold?
can someone please explain..

 

[tacogym27101990]

its a textbook

 

[BackCountrySnow]

ohok, i thought it was a way to remember something.
like sohcahtoa

 

[jkwii]

pfft. dont even need that one anymore.

 

[tommykins]

回复: Re: HSC Tips - Harder 3U

Inequalities - first part is often simple, second part often uses the first part, try to connect them together.

Induction - try to manipulate your answer to the expected answer, constantly refer to it and you should get there due time.

Circle geo - always look out for cyclic quads or the alternate segment theorem, these are most common in the HSC exams I've seen.

 

[jkwii]

Re: 回复: Re: HSC Tips - Harder 3U

ye all the harder Qs on inequalties are based on a known result and are usually seen as a follow up Q.

 

[adosh]

hello everyone,,, im having a problem wtih permuts combins,,,,now geha answers a questions as such "in how many ways can 8 people be arranged into two sets of four for tennis??" : (8C4x4C4)/2!= 35,,,now i understand this but heres a similar fitzpatrick question

find the number of wats in which 6 women and 6 men can be arranged in three sets of four for tennis wihcout restrictions??? and the asnwer is 155925.....now why cant we apply gehas method to this questuion, that is

(12C4x8C4x4C4)/3!=5775 but this is way off from 155925,,,,couls someone please help..thanks in advance

 

[mick135]

Wow,
is probability seriously the hardest thing you guys find in harder 3 unit???

probability is basic and repetitive to me. Look at a text book, and remember different tricks and USE YOUR BRAIN
ie: if you take 1 marble in a bag of 50, there's now only 49 left.
And with those "at least questions" = always add or subtract from 1.

the binomial theorm is a real killer to me. like 3/4 page responses for a 1 mark question...

 

[shanks27]

3. If a question asks you for example to find the number of three digit codes with the digits in ascending order from the numbers of 1-9, and each number can only be chosen once, remember that for each selection of numbers, only one is in ascending order, hence the answer would be 9C3.

can you explain this one, i battle with probability

 

[hurikai]

can you explain this one, i battle with probability

Your numbers are 1,2,3,4,5,6,7,8,9. Now think of randomly picking any three numbers. Let's say 3, 9 and 1. There's only one way you can put that into an ascending order, and that is 1,3,9.

So with that in mind, you can rephrase the question. Instead of every possible SEQUENCE of numbers pressed in ascending order, you are now just looking for every possible GROUP of 3 numbers (order not important), because they'll rearrange themselves. Thus 9C3.

 

[umaycallmealex]

does anyone know a good proof for AM/GM inequality with 3 numbers?
i.e. (abc)^1/3 </= (a+b+c)/3?

 

[Undermyskin]

hello everyone,,, im having a problem wtih permuts combins,,,,now geha answers a questions as such "in how many ways can 8 people be arranged into two sets of four for tennis??" : (8C4x4C4)/2!= 35,,,now i understand this but heres a similar fitzpatrick question

find the number of wats in which 6 women and 6 men can be arranged in three sets of four for tennis wihcout restrictions??? and the asnwer is 155925.....now why cant we apply gehas method to this questuion, that is

(12C4x8C4x4C4)/3!=5775 but this is way off from 155925,,,,couls someone please help..thanks in advance

Shouldn't it be 34650?

12C4*8C4

Their orders don't matter so with 12, you have 12C4 combinations. With the rest 8, you have 8C4. Multiply, you get 34650. (???)