!~CsSa TrAiL SoLuTiOnS~! (1 Viewer)

[Dash]

Does anyone have the CSSA trial solutions???

I need them asap!

 

[ND]

Ok i posted the solutions to Q8 in another post:

http://www.boredofstudies.org/community/showthread.php?s=&threadid=10875

I'll do Q7 now...

a)i) Factorise using difference of squares then perfect square, the diff... etc. Comes out to -(a+b-c)(b+a+c)(-b+a+c)(-b+a-c)
ii) -(a+b-c)(b+a+c)(-b+a+c)(-b+a-c) is > 0 because a,b,c are sides of a triangle. So 4*b^2*c^2 - (b^2+c^2-a^2) > 0
(b^2+c^2-a^2) < 4*b^2*c^2

b)i) f(x) = arccosx = y
Let f(-x) = arccos(-x) = pi-y
E(x) = f(x) + f(-x) ...[1]
= y+pi-y
=pi which is obviously even.

O(x) = f(x) - f(-x) ...[2]
= y-pi+y
= 2*arccosx - pi which is obviously odd.

ii) From [1] amd [2]:
f(x) = (E(x)+O(x))/2
then sketch.

c)i) Just use relationship U_n = U_n-1 + U_n-2 for U2.
ii) I'll just do the n=k+1 part:
For n=k+1 we need to prove:
u_1*u_2 + u_2*u_3 +...+ u_2n+1*u_2n+2 + u_2n+2*u_2n+3 = (u_2n+3)^2 - (u_1)^2
LHS = u_1*u_2 + u_2*u_3 +...+ u_2n+1*u_2n+2 + u_2n+2*u_2n+3
= (u_2n+1)^2 - (u_1)^2 + u_2n+1*u_2n+2 + u_2n+2*u_2n+3
Ok now use u_n = u_n-1 + u_n-2 with u_2n+2 for u_n-1, and it will cancel down to:
(u_2n+3)^2 - (u_1)^2 = RHS

Attached is the question.

 

[McLake]

As soon as they are in the DB I will upload them (but someone needs to donate them ...)

 

[Ragerunner]

I've uploaded it to bos but if you can't wait then get it there

http://users.on.net/unix/4U_Maths_CSSA_2003sol.pdf

Its not really in order though

 

[Dash]

OH THANX A MILLION!!!

 

[Ragerunner]

no problems

I don't do 4U maths so i wouldn't be able to answer any question ou have on it

 

[Dash]

LOL thats cool..... By the way, how does immolation plus infero equal raising the dead???

 

[Ragerunner]

haha. I dunno. I just made that up

It actually equals Death Coil, not raise the dead