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Q4
If f(Xf(Y))f(Y)=f(X+Y) for X,Y>=0
f(2)=0
f(X) non zero for 0=<X<2
I)f(1)=?
II)f(X)=?
 one year ago
 one year ago
Q4 If f(Xf(Y))f(Y)=f(X+Y) for X,Y>=0 f(2)=0 f(X) non zero for 0=<X<2 I)f(1)=? II)f(X)=?
 one year ago
 one year ago

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mahmit2012Best ResponseYou've already chosen the best response.1
dw:1337470247066:dw
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
let x=0, y=1 to get: f(0*f(1))*f(1)=(f(0+1) f(0)*f(1)=f(1) therefore f(0)=1 let y=2 to get: f(x*f(2))*f(2) = f(x+2) f(0)*0=f(x+2) therefore f(x+2)=0 let x = y = 1 to get: f(1*f(1))*f(1) = f(1+1) f(f(1))*f(1) = f(2) = 0 therefore f(f(1)) = 0 since f(1) cannot be zero from the question statement. summary: f(0) = 1 f(x+2) = 0 f(f(1)) = 0 therefore: f(x) = 1 for x=0 f(x) = ? for x=1 f(x) = 0 for x >= 2 f(f(1)) = 0 => f(1) >= 2 ?
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
ok one step to approach to answer.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
its VERY late where I am  so I am off to catch some sleep  I'll come back sometime tomorrow...
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
Ok but i would like you find it out before others.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1337533173527:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1337533652527:dw
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
dw:1337535999280:dw why is this zero?
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1337536022862:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1337536076711:dw
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
ah! of course  nice  I didn't spot that.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes  I followed the rest of the steps  it was just that bit that I circled that I could not understand. :)
 one year ago
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