8-5-09 Two problems

Rohan!!

answers yaar!!

for 1)

assume a -ve exists

WLOG take a,b -ve and c +ve

c>|a| + |b|

ab>c(|a| + |b|) > (|a| + |b|)²

ie 0 >a²+b²+ab which is impossible as ab also >0

for 2) is f : R→R ???????

you wont believe but the second one is from IMO !!! :OOOO

though it is easily solvable in 3 steps ;D

f(x)+f(1/(1-x))=x

f(1/(1-x))+f(1/1-(1/1-x))=1/(1-x) = > f(1/(1-x))+f((x-1)/x)) = 1/(1-x)

f((x-1)/x)+f(1/(1-(x-1)/x))=(x-1)/x

but 1/(1-(x-1)/x)= x

=> Last equation = f((x-1)/x)+f(x)=1-(1/x)

so three equations and three variables !

we get the answer as

f(x)= [ x - 1/x - (1/(1-x)) + 1]/2

so celestine your answer is correct .

for the 1st one, i guess the easiest way to conclude that a,b,c are +ve is to see that a,b, and c are roots of the cubic x³-px²+qx-r where p,q,r>0

This cubic cannot have any negative roots as for x<0, the expression is strictly negative. This implies in turn that a,b, and c are all +ve