Q16 (1 Viewer)

[nicholash747]

Worked solutions someone?

 

[Kimyia]

I completely BSed it until it happened to give me the answer but I had the equation from (i) and another equation and I simultaneously solved them, then made x^2 > 0 which happened to give me a half. No idea what the other equation was, though :\

 

[hsc2012la]

It would be awesome if someone could upload their worked solutions of Q16 C!!

 

[mytstephen]

i got pi/6 for the min question . was a bit hard and i left some parts out.

I left my answer in degrees... do you think this will affect anything?

 

[RishBonjour]

I completely BSed it until it happened to give me the answer but I had the equation from (i) and another equation and I simultaneously solved them, then made x^2 > 0 which happened to give me a half. No idea what the other equation was, though :\

i did that. but didnt remove r. fmllllllllllllll.


^above. yeah, degrees is wrong t think :/

 

[marinagee]

i hate you all

 

[Kimyia]

i got pi/6 for the min question . was a bit hard and i left some parts out.

Yay, I got pi/6 too

 

[vinhkn]

BOS trolled us. Gave most people the feeling they'll ace the exam and a "state rank" mentality. NEK MINUT Q16.. Raped in the as$$$ssh0l333zzz ymcmb yolo #fuckdapopoBOS

 

[ZOMGbirdy]

c)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!

 

[Kimyia]

BOS trolled us. Gave most people the feeling they'll ace the exam and a "state rank" mentality. NEK MINUT Q16..

Haha this.

 

[RishBonjour]

c)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!

nice work . did you do all other questions? (potential state rank)

 

[hsc2012la]

c)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!

If you move the 1/4 across you get a negative 1/4 and you can't root negatives :S

 

[Timske]

c)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!

did this too

 

[NBGHHS]

FROM THE OTHER THREAD: for those who couldn't figure out 16 C)
You had to use simultaneous for the two curves, and bring that equation reducible to quadratics
from there, since the circle was touching the sides of the curve at two points, you had to use the discriminant = 0 (using the reducible quadratic) to bring it down to the proven result.

 

[osak23]

16a) equiangular
ii) x=ab/a+b

b) iv) pie/6

c) discriminant = o
c>0.5 part no idea

 

[chillybags]

Hi! so i can do worked solutions if people want but basically here was how to do 16c i. took me a solid 30min to work out how to do it. Anyways.

equate the two graphs. so you get

y + (y-c)^2 = R^2.
y + y^2 - 2yc + c^2 - R^2 = 0
y^2 + y(1-2c) + (c^2 - R^2) = 0

now there are 2 equal and identical solutions to the above equation (because they intersect twice at the same location, just on oposite sides of the y axis)

hence the discriminate = 0

(1-2c)^2 - 4(c^2 - R^2) = 0 will then simplify down into the required answer, 4c = 1-4C^2

No idea how to do part ii though...

 

[alantuzza]

for 6b) i couldnt prove the eqn of the line, so i lost 2 marks there but im sure i got the next 2 parts right, so i should get 6/8
for 6c) i got 0/3 lol
and for 6a, i should get full.
lost one mark on 14 and 15.
sud get about 90-92% raw mark, do you guys have any idea what that would align to

 

[alantuzza]

do u think there would be 1 mark awarded for equating the two equations in 16 c) i, but not using the dsicriminant

 

[KierenTran]

Step by Step solutions for B and C please!

 

[Kimyia]

Hi! so i can do worked solutions if people want but basically here was how to do 16c i. took me a solid 30min to work out how to do it. Anyways.

equate the two graphs. so you get

y + (y-c)^2 = R^2.
y + y^2 - 2yc + c^2 - R^2 = 0
y^2 + y(1-2c) + (c^2 - R^2) = 0

now there are 2 equal and identical solutions to the above equation (because they intersect twice at the same location, just on oposite sides of the y axis)

hence the discriminate = 0

(1-2c)^2 - 4(c^2 - R^2) = 0 will then simplify down into the required answer, 4c = 1-4C^2

No idea how to do part ii though...

Beast Haha, yep think I'll definitely get zero for C.