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HSC Revision Thread 2006 (1 Viewer)

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Riviet

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bboyelement: I know you need the previous bit of the question as this continues on, it's not that hard actually, if you use one or two of the previous results in that question.
 

3li

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actualli u dont need previous parts, i just used a vector, pythag identity, and roots of unity (DMT)
 

Riviet

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vafa said:
Eventually one of my attach look nice!
Yeah, especially the first one! So much clearer, now I'm actually going to try reading. :p
 

Trebla

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Help with these questions please (preferably before tonight). Hope you can read my writing.....Thanks.
 

shimmerz_777

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bboyelement said:
help me with this question please
yea that question doesnt make sense to me either, i had a look at the answers posted, but even with the first bit its still confusing, i keep ending up with1 as the answer. like bar l1wl is how i see that, which just = 1? what am i missing or not getting. ill check over the answer again
 

shimmerz_777

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Trebla said:
Help with these questions please (preferably before tonight). Hope you can read my writing.....Thanks.
1

let ST = x, PS=b RS=a (althought typo says e in answer)

by alternate angles on parallel lines,
BST=STP=@ (@ = pheta)

tan@= b/x
x=bcot@

volume of water V= ((b^2)cot@)/2

when it is turned, V=ah
((b^2)cot@)/2 = ah
((b^2)cot@)/2a = h, as required.
 

shimmerz_777

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7i)
consider triangles MPV, WPM and VWP
PV=PW=s (radii of same circle)
hence VWP is iscololese
(similiar proof for OVW reveals that OVPW is a kite)
diagonals of a kite intersect at right angles, if that isnt a general proof u can use angle sum of triangles to prove that PMV=90

ii)OT^2 = OU^2 + TU^2 (pythagaras in OPV)
PT^2= TU^2 +PU^2 (pythagaras in PTV)
hence OT^2 - PT^2 = OU^2 - PU^2

OM^2= r^2 - MW^2 pythagaras.....
PM^2 = s^2- MW^2 pythagaras....
OM^2 - PM^2 = r^2 - S^2

iii) when the point T is moved onto the line WV, the point U lies on point M
hence M=U, substituting M for U
OU^2 – PU^2 = OM^2 – PM^2 = r^2 – s^2
Hence now
OT^2 – PT^2 = r^2 – s^2

havent got the last question though...
 

3li

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shimmerz_777 said:
yea that question doesnt make sense to me either, i had a look at the answers posted, but even with the first bit its still confusing, i keep ending up with1 as the answer. like bar l1wl is how i see that, which just = 1? what am i missing or not getting. ill check over the answer again
k my methods pretty fast
draw vectors of unity 5 with mod 1
use tip to tail then ull find that vector from P1 - P0 is w-1
so to find mod of w-1 let w = cis2pie/5 cos its roots of unity remember each root rotated by cis 2pie/5

then mod is square root of{ (cos2pie/5-1)^2 + sin^2 2pie/5}
use pythag, double angle and ull get the answer

similar for part ii u can draw more vectors, find more mod and times em together ....
its an alternative if u forgot cos rule like me
 

3li

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shimmerz_777 said:
7i)
consider triangles MPV, WPM and VWP
PV=PW=s (radii of same circle)
hence VWP is iscololese
(similiar proof for OVW reveals that OVPW is a kite)
diagonals of a kite intersect at right angles, if that isnt a general proof u can use angle sum of triangles to prove that PMV=90

ii)OT^2 = OU^2 + TU^2 (pythagaras in OPV)
PT^2= TU^2 +PU^2 (pythagaras in PTV)
hence OT^2 - PT^2 = OU^2 - PU^2

OM^2= r^2 - MW^2 pythagaras.....
PM^2 = s^2- MW^2 pythagaras....
OM^2 - PM^2 = r^2 - S^2

iii) when the point T is moved onto the line WV, the point U lies on point M
hence M=U, substituting M for U
OU^2 – PU^2 = OM^2 – PM^2 = r^2 – s^2
Hence now
OT^2 – PT^2 = r^2 – s^2

havent got the last question though...
i did the last 1 a while back
simultaneously solve for two of the 3 lines then realise that the third line satisfies the equation. so all three pass thru it
 

gamecw

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p( x)=3x^4-11x^3+14x^2-11x+3
solve p( x) over complex field + factorise P(x ) over real
do this one plz!
 

Mill

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divide through by x^2 since x != 0

then use some incredibly sexy complex results (let x = cis @)
 

toadstooltown

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These ones are a bit weird in the way you have to approach them. You know it's one of them when the coëfficients 'rise and fall' but aren't a binomial expansion.
3x^4-11x³+14x²-11x+3
x²(3x²-11x+14-11/x+3/x²)
x²[3(x²+1/x²) -11(x+1/x) +14]
x²[3({x+1/x}²-2) -11(x+1/x) +14] - completing the square to make a quadratic
x²[3(x+1/x)²-11(x+1/x)+8]
x+1/x = [11±5]/6 - from quadratic formula
x+1/x = 8/3, 1
So,
x²(x+1/x-8/3)(x+1/x-1)
x²(3x+3/x-8)(x+1/x-1) - as we note the leading coëfficient
(3x²+3-8x)(x²+1-x) - multiplying each bracket by x
x = (4±sqrt7)/3, (1±isqrt3)/2 , x over C- from solving each quadratic
P(x)=3(x²-x+1)(x-4/3-sqrt7/3)(x-4/3+sqrt7/3) factorised over R. (NB sqrt7/3 = (sqrt7)/3)

If you sub the reals in at least you get 0, can't be bothered checking the Im answers. Let me know if I've made a mistake. :)
 

gamecw

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toadstooltown said:
These ones are a bit weird in the way you have to approach them. You know it's one of them when the coëfficients 'rise and fall' but aren't a binomial expansion.
3x^4-11x³+14x²-11x+3
x²(3x²-11x+14-11/x+3/x²)
x²[3(x²+1/x²) -11(x+1/x) +14]
x²[3({x+1/x}²-2) -11(x+1/x) +14] - completing the square to make a quadratic
x²[3(x+1/x)²-11(x+1/x)+8]
x+1/x = [11±5]/6 - from quadratic formula
x+1/x = 8/3, 1
So,
x²(x+1/x-8/3)(x+1/x-1)
x²(3x+3/x-8)(x+1/x-1) - as we note the leading coëfficient
(3x²+3-8x)(x²+1-x) - multiplying each bracket by x
x = (4±sqrt7)/3, (1±isqrt3)/2 , x over C- from solving each quadratic
P(x)=3(x²-x+1)(x-4/3-sqrt7/3)(x-4/3+sqrt7/3) factorised over R. (NB sqrt7/3 = (sqrt7)/3)

If you sub the reals in at least you get 0, can't be bothered checking the Im answers. Let me know if I've made a mistake. :)
thanks
 

Trebla

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3li said:
i did the last 1 a while back
simultaneously solve for two of the 3 lines then realise that the third line satisfies the equation. so all three pass thru it
I know that's what you have to do. I just don't know how to do it.
 

toadstooltown

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Nah, first time I came across one of those I spent half an hour blindly guessing roots untill I finally gave up and asked my teacher. It was frustrating so wouldn't wanna leave someone else equally as frustrated right before the exam!
 

Trebla

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Omfg!!!!!! If You Read The Attachment I Posted Earlier, You Would Realise That Question 2 On That Attachment Appeared In This Year's Paper In A Similar Form!!!!!!
 
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Riviet

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Trebla said:
Omfg!!!!!! If You Read The Attachment I Posted Earlier, You Would Realise That Question 2 On That Attachment Appeared In This Year's Paper In A Similar Form!!!!!!
Well I guess that gives you a slight unfair advantage. :p

Closing the thread since the exam's over.
 
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