Q16 (2 Viewers)

[remoteman213]

How do you get 1/2 for c) ii)?

When I try it I keep getting c>1/4

 

[RishBonjour]

Big deal mate. What are you trying to prove by coming on here and dissing everyone?

hes gonna come 1st in state. all hail.

 

[Stevm]

Solution to Q16) (B) (ii)

 

[christopher8827]

Oh shit. For question 16)a - did anyone assign the parallel lines for the corresponding angle wrong? I think I might have switched them around.... Anyone know the correct answer?

 

[Seal7]

How do you get 1/2 for c) ii)?

When I try it I keep getting c>1/4

Same here, btw I just discovered this forum, here is the proof I quoted to my friend for i)

"I did some work on Question 16 c

i) x^2 + (y-c)^2 = r^2

And y=x^2

Substitute 2 into 1

y + (y-c)^2= r^2


Rearrange (expand) to make a quadratic, y as the variable

y+y^2 -2yc + c^2 -r^2 = 0

Grouping terms

y^2 + y(1-2c) + (c^2-r^2) = 0

We know that there are two equal (in terms of y) solutions to this, because the circle and the parabola touch symmetrically around the y axis at the same value of y (given in the question). Therefore the discriminant of this quadratic = 0

discriminant = b^2-4ac

substituting in

(1-2c)^2 - 4(c^2-r^2) = 0
Expand and cancel (easy enough) to get 4c = 1 + 4r^2

In part ii) I got c>1/4, which is incorrect (we were meant to get c>1/2). Until I can figure this out I won't go any further. "

 

[maknaedori]

Does anyone know how many marks we lose in a 3 mark question if we just right the answer? E.g. if we did trial and error.

 

[mikaa]

OMG same! i ended up with c>1/4 hahaha got me so pissed!i

but then it was only a one marker so i gave up in the end.. tehe

 

[Jashua_Long]

Guess again RealiseNothing.

Big deal mate. What are you trying to prove by coming on here and dissing everyone?

You are so cool, dude.

 

[Rawf]

Re: The real solution to Q16. (B) (ii)

Erm strangely you can't see that pic... heres the link:
http://s11.postimage.org/oqv5788ch/20121022_154457.jpg

Tbh, I don't get it but thats simply cause i dont really care anymore and i cbs to understand it -__-

 

[talisman]

This is my answer to part 16c i)

 

[EinstenICEBERG]

Re: The real solution to Q16. (B) (ii)

Since there's many BS answers flying everywhere, here's (one of) the real answers:

alot of people does it a different method, but that seems cool. However, you didnt use the use 4c = 1 + 4r^2 as the starting line, as used the original equation, so im not sure if that's fully correct.

 

[talisman]

Attached is my solution to 16c ii). Tell me what you think of it.

 

[George121]

All this controversy over question 16 is hinting that general maths tomorrow is going to have controversy with a few questions.

#inb4fail

 

[cookeemonstah]

Re: The real solution to Q16. (B) (ii)

its part C

 

[mikaa]

^agreed.

he makes me feel so unwelcomed for uni :'(

 

[eileend]

Q16 a)i) i used the parrallel lines and how the two sides had the same x sides... and ended up with the proof of AAS

ii) i have no idea what i did but i think my final answer was x=(a+b)/4

b) i)I had no idea how to do it so i just wrote crap..
and (ii) to (iv) i was good with it

c)... that was horrible =.=

 

[duckysd]

The whole papaer was pretty easy. EXCEPT question 16(after c). Thought I do know how to answer now.
For C. i.
we have x^2 + (y-c)^2 = r^2 and y = x^2

Solving simultaneously (FOR y - or for x but y is easier). We get y^2 + y(2c-1) - r^2 + c^2 = 0
From the graph and according to the question, there is only one y solution where the circle and the parabola intersect. Hence they give you the clue that the points are symmetrical ((-x)^2 = (+x)^2). This means that the discriminant i.e. b^2 - 4ac must equal 0 since only one y root.

Now, (2c - 1)^2 - 4(c^2 - r^2) = 0
4c^2 - 4C + 1 - 4c^2 + 4r^2 = 0
4r^2 - 4c +1 = 0
FINALLY 4c = 1 +4r^2

For the second part, I was told by my fellow asians (who do 3/4 unit and sat the 2U test) that if we solve for x, we would get an inequality since the discriminant (for x this time) would be more than 0 (2 intersections on either side of the circle.). Next year I will sit 2U again and we'll see things from there.

 

[siobhanobrien]

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

for the win... totally just tried that and failed. cant take a negative square root. i say the question is wrong.

 

[lance687876]

a) was easy. l just hope they can read my handwriting!

b) was difficult to understand, l only attempted the last part. l'm glad the last part was pi/6 though!!!

c) all l did was make the 2 y's equal to each other so that both sides had x's and c's. would that get me a mark?

hopefully l'll get 7/15 for this section. so l still have a chance to get band 5 :/

 

[duckysd]

For 16 c)i) I spent more than an hour and wrote more than two pages. And still did not get the answer. Though that was a step.