1st - easy
a,b,c All are real numbers such that
a+b+c >0 ab+bc+ca >0 abc >0 Prove a,b,c are all >0
2nd - medium
Find all functions f which follows the following equation -
f(x)+f(1/(1-x))=x
9for 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
9tapan i think uve made a small errror
im getting
2f(x) = x - 1/x - 1/(1-x) + 1 ( also verified )
method:
in gvn eq replace
x by 1/1-x and (x-1) /x
and ull find it solvable in f(x)
1you 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 .
341for 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
