3 Unit Maths HSC Exam Revision (2 Viewers)

[random-1006]

Totally didn't see that way.

HSC is in one month XD.$0.05 that everyone's super speedily studying.

ill be interested to see how this yrs maths papers look, i thought last yrs maths papers were pretty different from the past hsc, particularly the 2 unit one, i miss the days when they actually had a whole Q2 dedicated to co ordinate geometry.

 

[random-1006]

 

[hscishard]

I came up with a lot of solutions..wtf.
The ones that work are 2 and (-1 - root5) / 2

Others I came across was -1 and -1 + root5 / 2
Can you show working out?

x = 2 , y = root2
x = (-1 - root5) / 2, y = root(3-root5) / 2

 

[random-1006]

I came up with a lot of solutions..wtf.
The ones that work are 2 and (-1 - root5) / 2

Others I came across was -1 and -1 + root5 / 2
Can you show working out?

x = 2 , y = root2
x = (-1 - root5) / 2, y = root(3-root5) / 2

please dont tell me you were guessing and checking

there is only ONE way ( im pretty sure) to do this question, how did you do it

draw a picture, there shouldnt be anymore than 2

i made this one up myself, i think i didnt make a mistake when i wrote the question lol

 

[hscishard]

please dont tell me you were guessing and checking

there is only ONE way ( im pretty sure) to do this question, how did you do it

draw a picture, there shouldnt be anymore than 2

i made this one up myself, i think i didnt make a mistake when i wrote the question lol

I solved it algebraicly.
Prob because I squared both sides..

 

[random-1006]

I solved it algebraicly.
Prob because I squared both sides..

and then you got a quartic, put up the working lol

and intersections mean both x and y coordinates lol

 

[hscishard]

and then you got a quartic, put up the working lol

x+2 = (x^2 -2)^2
x+2 = x^4 -4x^2 + 4
x^4-4x^2 -x +2 = 0




Now focusing on the x^2(x+2) -1 factor..




Rewriting as a whole
= 0
Lol fucken hell. Now focusing on the x^2 + x -1
x=,-2 and 1

 

[hscishard]

and intersections mean both x and y coordinates lol

ya I realised 2mins after I posted.
Have a feeling I'm gonna forget in an exam

 

[random-1006]

x+2 = (x^2 -2)^2
x+2 = x^4 -4x^2 + 4
x^4-4x^2 -x +2 = 0




Now focusing on the x^2(x+2) -1 factor..




Rewriting as a whole
= 0
Lol fucken hell. Now focusing on the x^2 + x -1
x=http://latex.codecogs.com/gif.latex?\frac{-1\pm \sqrt{5}}{2}[/IMG[B]],-2 and 1[/B][/quote]

that was unexpected, mmm

hmmm, doesnt make sense geometrically

wrong signs for starters, now just need to think bout it little more

for you to set x^2 -2 = sqrt(x+2) , you are saying that x^2 -2 > 0

but if you use -1 + sqrt5 / 2 , you will see x^2 -2 < 0, so that solns out, now what to do about the other one

 

[random-1006]

lol, i was expecting someone to go

inverse functions intersection on the line y= x

hence solve x^2 -2 = x or sqrt (x+2) = x

i dont really like that 3rd soln that you found , doesnt really make sense when you sketch it, and it doesnt come out with the above method for some reason

 

[random-1006]

 

[random-1006]

how do you get the actual sqrt sign in latex, i cnt find it

 

[hscishard]

how do you get the actual sqrt sign in latex, i cnt find it

6+ln 16
Bottom left side iirc

 

[random-1006]

dont skip this question

lol i knew people would find it different

 

[hscishard]

dont skip this question

lol i knew people would find it different

I don't get it. The graph goes negative first then positive.

 

[random-1006]

I don't get it. The graph goes negative first then positive.

shit, the eqn is not a good one

change it to A= (t-3)^3 +2

 

[pooshwaltzer]

dA/dt = 4(t^3) - 12t

d^2A/dt^2 = 12(t^2) -12 = 0

t = |1|

Sorry, I've no bloody clue.

 

[random-1006]

dA/dt = 4(t^3) - 12t

d^2A/dt^2 = 12(t^2) -12 = 0

t = |1|

Sorry, I've no bloody clue.

lol, its correct, just someone else stated ( correctly) that A(1) < 0, ie negative concentration lol. I just made it up similar to a past trial i had seen

i thought it was a interesting question that a few people would get stuck on , i didnt get it the first time i saw it

 

[hscishard]

shit, the eqn is not a good one

change it to A= (t-3)^3 +2

Still doesn't work. As x--> infinity, gradient --> infinity.

But I think I know what you're looking for. Differentiating the derivative[treating it as a function] and finding it's maximum.

 

[random-1006]

Still doesn't work. As x--> infinity, gradient --> infinity.

But I think I know what you're looking for. Differentiating the derivative[treating it as a function] and finding it's maximum.

the concentrate in bloodstream is A, its derivative dA/ dt is the rate at which that changes , if we want it to " change most rapidly", we want to maximise dA/dt etc