HSC 2016 Maths Marathon (archive) (1 Viewer)

[Drsoccerball]

Re: HSC 2016 2U Marathon

I'm not sure that's assumed knowledge for 2U...

My bad You would just draw the 2 graphs and add their ordinates separately.

 

[leehuan]

Re: HSC 2016 2U Marathon

Have to add ordinates. Only 2U method to do this - auxiliary is 3U.
__________________________

Instructive question:



Don't spoil it for no reason anyone who's not doing an HSC in some year later than 2015

 

[Nailgun]

Re: HSC 2016 2U Marathon

Have to add ordinates. Only 2U method to do this - auxiliary is 3U.
__________________________

Instructive question:



Don't spoil it for no reason anyone who's not doing an HSC in some year later than 2015

isn't this x=0?

 

[InteGrand]

Re: HSC 2016 2U Marathon

isn't this x=0?

 

[Nailgun]

Re: HSC 2016 2U Marathon

Ah I see

 

[leehuan]

Re: HSC 2016 2U Marathon

Yep. It's instructive because of the indeterminate form. Besides that it's just knowledge of index laws.

 

[Green Yoda]

Re: HSC 2016 2U Marathon

Next Question:
A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?

 

[Drsoccerball]

Re: HSC 2016 2U Marathon

Next Question:
A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?

Cases is the way to go.

 

[leehuan]

Re: HSC 2016 2U Marathon

Cases is the way to go.

Lol basically what I was thinking... cause the cases are so few.

@2016ers

Spoiler

If you think about it, it's already quite restricting that the die in his left hand has to be at least 3

 

[Speed6]

Re: HSC 2016 2U Marathon

Sketch y=2cos(x) - sin(x)

How would you sketch this using the auxiliary form method?

 

[Paradoxica]

Re: HSC 2016 2U Marathon

How would you sketch this using the auxiliary form method?

 

[v_k_9696]

Re: HSC 2016 2U Marathon


the inflexion point does not make sense
check the working out, its from a past paper. someone confirm it please or am i just coo coo


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[leehuan]

Re: HSC 2016 2U Marathon

View attachment 33037
the inflexion point does not make sense
check the working out, its from a past paper. someone confirm it please or am i just coo coo


Sent from my iPhone using Tapatalk

http://www.wolframalpha.com/input/?i=inflexion+points+x^3-6x^2+9x+4

You're cuckoo. Sorry

 

[v_k_9696]

Re: HSC 2016 2U Marathon

nevermind i did something wrong in my working out


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[Nailgun]

Re: HSC 2016 2U Marathon

lelel I tried this again and it was actually pretty easy

that being said this is only really a 2U question if you originally ask to find the derivative of x^(2x)

Next Question:
A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?

4/33?

 

[InteGrand]

Re: HSC 2016 2U Marathon

lelel I tried this again and it was actually pretty easy

that being said this is only really a 2U question if you originally ask to find the derivative of x^(2x)



4/33?

I don't think it's 4/33.

 

[Nailgun]

Re: HSC 2016 2U Marathon

I don't think it's 4/33.

whhops 5/33 added up wrong

 

[InteGrand]

Re: HSC 2016 2U Marathon

whhops 5/33 added up wrong

I don't think it's that either.

 

[Nailgun]

Re: HSC 2016 2U Marathon

I don't think it's that either.

rip

There are 11 possible sums for the right hand, and six possible numbers for left hand therefore 66 outcomes possible

0 successful outcomes if you get 1 in the left hand
0 successful outcome if you get 2 in the left hand
1 successful outcomes if you get 3 in the left hand
2 successful outcomes if you get 4 in the left hand
3 successful outcomes if you get 5 in the left hand
4 successful outcomes if you get 6 in the left hand

1+2+3+4=10

10/66 = 5/33

what am i doing wrong lol

edit: jokes i see it now
this is what happens when i dont use a probability tree lel

 

[InteGrand]

Re: HSC 2016 2U Marathon

rip

There are 11 possible sums for the right hand, and six possible numbers for left hand therefore 66 outcomes possible

0 successful outcomes if you get 1 in the left hand
0 successful outcome if you get 2 in the left hand
1 successful outcomes if you get 3 in the left hand
2 successful outcomes if you get 4 in the left hand
3 successful outcomes if you get 5 in the left hand
4 successful outcomes if you get 6 in the left hand

1+2+3+4=10

10/66 = 5/33

what am i doing wrong lol

edit: jokes i see it now
this is what happens when i dont use a probability tree lel

Yeah, with the 11 different sums, some of them can happen in more ways than others. So in actuality, there are 36 possibilities for the hand with two dice (Die 1 has 6 possibilities, and Die 2 has 6 possibilities).