General Thoughts: Mathematics Extension 2 (1 Viewer)

iStudent

Well-Known Member
Joined
Mar 9, 2013
Messages
1,158
Gender
Male
HSC
2014
Did anyone use 15cii for 15civ? :/
Can't see the connection in 15cii with the rest of the q. Unless you're not meant to use calculus for 15ciii o.o
 

SquareZ

New Member
Joined
Oct 7, 2014
Messages
14
Location
Sydney
Gender
Male
HSC
2014
Was there even an end of question for Question 11
I didn't see the words End of question 11 and asked the person but she said there wasn't one... ?
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
This was a good paper, harder than last years.

Idk why there so many graph questions, it pissed me off because of how much time wasted. Hopefully they're not anal in marking it, also what did everyone get for the volume with the trapezium slices?

Also, I think a raw mark of 65ish would align to 90
 

iStudent

Well-Known Member
Joined
Mar 9, 2013
Messages
1,158
Gender
Male
HSC
2014
How did u guys prove Q16(a)(ii)? Collinear one
use xpq = ypr and the 2 right angles in the semicircle. Do a similar process to i for c2 and you end up with angle apx=angle rpc
 

GOsie

Active Member
Joined
May 9, 2013
Messages
101
Gender
Male
HSC
2014
This was a good paper, harder than last years.

Idk why there so many graph questions, it pissed me off because of how much time wasted. Hopefully they're not anal in marking it, also what did everyone get for the volume with the trapezium slices?

Also, I think a raw mark of 65ish would align to 90
Does 64/5 or something like that, maybe with a pi sound reasonable?
 

SquareZ

New Member
Joined
Oct 7, 2014
Messages
14
Location
Sydney
Gender
Male
HSC
2014
How did u guys prove Q16(a)(ii)? Collinear one
You prove that one side had all the same angles with the other side and then you can say it is 180 degrees thus making <APC a straight line therefore collinear
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
You could have been a smart-ass for Q14 (a) (ii) and said that since x=1 is a root of multiplicity 3, three of the roots are 1, 1 and 1.

So two complex roots of P(x) are 1 and 1, since all real numbers are complex numbers =)
 

njweno

New Member
Joined
May 29, 2013
Messages
4
Gender
Male
HSC
2014
- Didn't even bother attempting 16b, got home and realised how easy i and ii were haha
- The plane mechanics question, parts 1 and ii, WTF?? Anyone get them out?
- Got cooked by induction
- Bad paper
 

mreditor16

Well-Known Member
Joined
Apr 4, 2014
Messages
3,169
Gender
Male
HSC
2014
This was a good paper, harder than last years.

Idk why there so many graph questions, it pissed me off because of how much time wasted. Hopefully they're not anal in marking it, also what did everyone get for the volume with the trapezium slices?

Also, I think a raw mark of 65ish would align to 90
I got 24 root 3 all over 5
 

mreditor16

Well-Known Member
Joined
Apr 4, 2014
Messages
3,169
Gender
Male
HSC
2014
You could have been a smart-ass for Q14 (a) (ii) and said that since x=1 is a root of multiplicity 3, three of the roots are 1, 1 and 1.

So two complex roots of P(x) are 1 and 1, since all real numbers are complex numbers =)
just no.
 

iStudent

Well-Known Member
Joined
Mar 9, 2013
Messages
1,158
Gender
Male
HSC
2014
You could have been a smart-ass for Q14 (a) (ii) and said that since x=1 is a root of multiplicity 3, three of the roots are 1, 1 and 1.

So two complex roots of P(x) are 1 and 1, since all real numbers are complex numbers =)
Lol. Would you get the marks for that?
 

mreditor16

Well-Known Member
Joined
Apr 4, 2014
Messages
3,169
Gender
Male
HSC
2014
- Didn't even bother attempting 16b, got home and realised how easy i and ii were haha
- The plane mechanics question, parts 1 and ii, WTF?? Anyone get them out?
- Got cooked by induction
- Bad paper
wtf? since was there an induction Q?
 

GOsie

Active Member
Joined
May 9, 2013
Messages
101
Gender
Male
HSC
2014
You could have been a smart-ass for Q14 (a) (ii) and said that since x=1 is a root of multiplicity 3, three of the roots are 1, 1 and 1.

So two complex roots of P(x) are 1 and 1, since all real numbers are complex numbers =)
I thought about that. But it specified the two complex roots
 

sallythai96

New Member
Joined
Apr 25, 2014
Messages
1
Gender
Undisclosed
HSC
N/A
thought the paper was harder than last year and couldn't do a lot especially the mechanics and the second conics question. Wasn't that easy and screwed up badly -_-
 

mreditor16

Well-Known Member
Joined
Apr 4, 2014
Messages
3,169
Gender
Male
HSC
2014
okay here are some of my answers. one disclaimer - I did not get time to check answers.

q11 a) I) root 2 cis ( - pi /4) ii) -4 / 5 - 2/5 i

b) pi - 6 all over 2 (pi)^2

e) 216 pi units cubed

Q12. b) ii) pi / 18 , 11pi / 18, 13 pi /18
d) iii) 3 pi - 8 all over 12

Q13 a) 2 (root 3) - 3 all over 6
b) 24 root 3 all over 5

Q14. ii) -3/ 2 plus minus (root 15)/2 i

b) ii) pi /4

c ii) 150 M ln (5) all over F hours

Q16 c) x all over (1 + ln x) + C

again, reiterating these are merely my answers :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top