3u Mathematics Marathon V 1.1 (1 Viewer)

[haque]

Let k be an integer with which we start the consecutive integers thus
k(k+1)(k+2)...(k+r)
=r+kPr times k(note this is permutation notation)
divide this by r! u get r+kCr times k where C is the combination notation. this is an integer as we know thus it is divisible by r!.

 

[haque]

i couldn't edit my other post so here's the answer yip-
let k be an integer with which we start, so
k(k+1)....(k+r)=r+kPr times k
divide this by (r)! and u get r+kCr times k which is an integer. thus it is divisible by r! as it bla bla..

 

[Yip]

haque, arent u taking the product of (r+1) consecutive integers? k(k+1)(k+2)...(k+r) obviously has r+1 terms, and also, could u elaborate on ur method?
the solution i have is basically this:
let the r continuous integers be n,n-1,n-2.....(n-r+1)
n(n-1).......(n-r+1)=n!/(n-r)!
dividing by r! gives n!/(n-r)!r!=nCr which is an integer

 

[haque]

what is the probability that 2 particular players will meet in a knockout tournament in which the players are eliminated from each round, so if we started with eight then 4 would go through the quarterfinal and then 2 and so on-if there are 2^n players entering.

 

[haque]

ok say we take k as the starting number then the product is k(k+1)....(k+r-1). by definition this is (k+r-1)Pr where nPk is defined as n!/(n-k)!. if we divide the expression by r! then we get an expression in the form of n!/(n-k)!k! which by definition is nCk(note the k's are diffeent between my ecpression and the general form). we know the nCk is always integral, thus dividing by r! gives an integral result-thus it is divisible by r!

 

[haque]

yea that's the solution i gave in my second post-i just saw ur editetd one now-LOL i'm so slow at typing-but my answers basically the same as urs.

 

[Yip]

1982 4u Q8(ii)?....lol a little hard for 3u dont u think haque

p=(2^n)-1/(2^n)C2 [number of situations where u can be paired together/number of 2ppl combinations u can take from 2^n players]
=[(2^n)-1]/[2^n]!/[(2^n)-2]!2!
=2[(2^n)-1]/2^n[(2^n)-1]
=2^(1-n)

 

[haque]

Yip i got it from cambridge-but i believe u i think i saw it in the 4U as well-knew u'd get it so i put it in. LOL

 

[Riviet]

I was about to say that I remember (recently sometime) doing that question in the cambridge 3 unit book, BUT it was in the extension section.

Looks like a new question is needed.

 

[Yip]

New Q:

Let F₀(x)=x, and for each positive integer k,

F_k(x)=k[∫₀^x(F_k-1(t)dt+x∫₀^-1(F_k-1(t)dt)]

Prove that for each positive integer k,
(a) F_k(-1)=F_k(0)=0
(b) Use induction to prove that
(i) F_k(x) is a polynomial in x of degree k+1 whose constant term is 0 and the coefficients of x^k+1 and x^k are respectively 1/(k+1) and 1/2
(ii) F_k(x)-F_k(x-1)=x^k
(c) If n is any positive integer, show that

F_k(n)=1^k+2^k+3^k+.......+n^k

 

[haque]

YIp I just have to say I'll have to attempt it later-I just got a physics trial i gotta do-my parents just reconfiscated by differential equations and Real Analysis books and told me to do physics instead.

 

[haque]

Yip i got the answer out and it'll take ages to type it out-i'll give u the soluiton at school-i'm pretty sure its right cos the lecturer of maths at UNSW checked it(Tran's dad) and gave me some hints and so i didn't completely do it independently-he also said that there are restrictions on k and that these and other limit values should have been stated(hes pretty pedantic). so u wanna post a new question or should i or should someone else?

 

[Riviet]

Go for it haque.

 

[haque]

Riviet i would have scanned my solutions up for this one but my scanner's a bit so here's the next question(back to 3U standard): A particle is projected from a point O with initial velocity V that makes an angle @ with horizontal. show that there exists a point P on the projectile's path such that OP is perpendicular to the direction of the motion of the particle at P if tan@>2sqrt2

PS-for yip's one use induction for part b onwards-i'm still yet to do the last part but yip said it was once u got the earlier parts so yea.

 

[Yip]

Parametric equations of the trajectory:
dx/dt=vcosa,dy/dt=-gt+vsina
x=vtcosa, y=-0/5gt^2+vtsina

gradient OP=(-gt^2+2vtsina)/2vtcosa
gradient PM=dy/dt/dx/dt=(-gt+vsina)/vcosa where M is the point where the velocity vector at P intersects the horizontal axis
OPxPM=-1
after some bashing, u get:
2v^2(sin^2 a+cos^2 a)-3vgtsina+(gt)^2
We want real values of v, so we need discriminant>0
D=9(gt)^2sin^2 a-8(gt)^2(sin^2 a+cos^2 a)>0
sin^2 a>8cos^2 a
tan^2 a>8
tana>2sqrt2

btw haque, induction is required for b part i and ii, and part c uses (b) ii (no induction required for c)

Im not really sure about my method, its a little fudgy lol, theres probably a better way...

 

[Riviet]

Yip, do you think you could type up or scan your solutions for all of your last question?

 

[Yip]

the projectile? or the integral one?

 

[Riviet]

The integral one of course.

Thanks.

 

[Yip]

Solution attached (the working is a bit abridged but its enough to understand how to solve it)

 

[Riviet]

Next Question:

Solve for x: (x²+4x+5)² - 12(x²+4x) - 40 = 0