HSC Extension 1 Mathematics Predictions / Thoughts (2 Viewers)

[Xanthi]

I'm gonna guess the questions at b6 performance will be judged as follows:

MC 9/10
Q11: 14/15
Q12: 13/14
Q13 : 14/16
Q14: Binomial 2/8 Trig 6/7

Overall 58/70 cutoff

 

[Trebla]

In lieu of someone posting the paper, would anyone be able to outline (from memory) what each of those harder questions asked you to do or prove?

 

[shashysha]

I'm gonna guess the questions at b6 performance will be judged as follows:

MC 9/10
Q11: 14/15
Q12: 13/14
Q13 : 14/16
Q14: Binomial 2/8 Trig 6/7

Overall 58/70 cutoff

58 seems a bit high considering last years was 52 and the tests were around same difficulty... although new syllabus so no idea lol

 

[Xanthi]

58 seems a bit high considering last years was 52 and the tests were around same difficulty... although new syllabus so no idea lol

It was just from adding the individual q estimated cutoffs. You could probably deduct another mark or two for careless and misc errors making it 56-57/70, which sounds reasonable (I think last year's was somewhat harder, but not extremely so).

 

[thomson2354]

What would 48/70 scale to?? Given that mid 50s would be in the 90s.. would this be in the 80s?

 

[Xanthi]

What would 48/70 scale to?? Given that mid 50s would be in the 90s.. would this be in the 80s?

For sure

 

[Xanthi]

And...there we have it. Ext 1 may very well do better than ext 2. 56-57 is 80-81%. Wow.

 

[black.mamba]

In lieu of someone posting the paper, would anyone be able to outline (from memory) what each of those harder questions asked you to do or prove?

14a

prove 2ncn =(nc0)^2 + (nc1)^2 + (nc2)^2 + ... +(ncn)^2

using (1+X)^2n= (1+x)^n(1+x)^n

14a ii

in a group of 2n people with n men and n women prove that the number of groups with an equal number of men and women is equal to rhs in 14a

14a iii

now choose one leader for the men and one for the women; show that the number of arrangements is 1^2 * (nc1)^2 + 2^2 *(nc2)^2 + ... +n^2 * (ncn)^2

14a iv

now choose the leaders first from the n men and n women, and then find the number of groups that can be formed (ie reverse of method in ii and iii)
Find a simple expression equal to the value in 14c using the result in i

(not in question but value is n^2 * (2n-2)c(n-1) )

 

[black.mamba]

soz if badly formatted I’m on my phone

 

[Fabrizio]

In lieu of someone posting the paper, would anyone be able to outline (from memory) what each of those harder questions asked you to do or prove?

Thats what i can remember

 

[Fabrizio]

14a

prove 2ncn =(nc0)^2 + (nc1)^2 + (nc2)^2 + ... +(ncn)^2

using (1+X)^2n= (1+x)^n(1+x)^n

14a ii

in a group of 2n people with n men and n women prove that the number of groups with an equal number of men and women is equal to rhs in 14a

14a iii

now choose one leader for the men and one for the women; show that the number of arrangements is 1^2 * (nc1)^2 + 2^2 *(nc2)^2 + ... +n^2 * (ncn)^2

14a iv

now choose the leaders first from the n men and n women, and then find the number of groups that can be formed (ie reverse of method in ii and iii)
Find a simple expression equal to the value in 14c using the result in i

(not in question but value is n^2 * (2n-2)c(n-1) )

Lol i started typing realised how.lomg it would take and just wrote on piece of paper

 

[thomson2354]

How did we do that volume question? I just broke it up for the 2 curves and subtracted their volumes between x=0 to Root3/2 i think it was..

 

[black.mamba]

How did we do that volume question? I just broke it up for the 2 curves and subtracted their volumes between x=0 to Root3/2 i think it was..

yes but between x= 0 and x=Pi/6

 

[Fabrizio]

How did we do that volume question? I just broke it up for the 2 curves and subtracted their volumes between x=0 to Root3/2 i think it was..

Was that the sin@ and cos2@ one?

 

[Ooklei]

yes but between x= 0 and x=Pi/6

Lol. That question i did everything right until i subbed the values

 

[thomson2354]

yes but between x= 0 and x=Pi/6

yep thats what it was lol.. good memory xD

 

[MrGresh]

Here's a sketchy copy of the exam, missing some multis and last part of Q11

 

[shashysha]

and by this i mean like two separate integrals

yeah I think you'll lose a mark or two for that

 

[thomson2354]

Was that the sin@ and cos2@ one?

yeah

 

[Fabrizio]

yes but between x= 0 and x=Pi/6

^

 

[Fabrizio]

What did people do for 5 and 9