Let f(x)=ln(1-x/1+x). Find x,y for which f(x)+f(y)=f(x+y/1+xy).

Asked by Anil | 11th May, 2017, 12:29: PM

Expert Answer:

begin mathsize 16px style straight f open parentheses straight x close parentheses equals ln open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses
Given space that space straight f left parenthesis straight x right parenthesis plus straight f left parenthesis straight y right parenthesis equals straight f open parentheses fraction numerator straight x plus straight y over denominator 1 plus xy end fraction close parentheses
ln open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses plus ln open parentheses fraction numerator 1 minus straight y over denominator 1 plus straight y end fraction close parentheses equals ln open parentheses fraction numerator 1 minus open parentheses straight x plus straight y close parentheses over denominator 1 plus straight x plus straight y end fraction close parentheses
ln open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction cross times fraction numerator 1 minus straight y over denominator 1 plus straight y end fraction close parentheses equals ln open parentheses fraction numerator 1 minus open parentheses straight x plus straight y close parentheses over denominator 1 plus straight x plus straight y end fraction close parentheses
rightwards double arrow fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction cross times fraction numerator 1 minus straight y over denominator 1 plus straight y end fraction equals fraction numerator 1 minus open parentheses straight x plus straight y close parentheses over denominator 1 plus straight x plus straight y end fraction
solving space we space get space
2 xy open parentheses straight x plus straight y close parentheses equals 0
xy equals 0 space or space straight x plus straight y equals 0
for space xy equals 0 space solution space is space but space obvious.
for space straight x plus straight y equals 0 rightwards double arrow straight x equals negative straight y
putting space in space given space equation comma
ln open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses plus ln open parentheses fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction close parentheses equals ln 1
ln open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction cross times fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction close parentheses equals 0
fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction cross times fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction equals 1
LHS space and space RHS space contradicts.
Hence comma space straight x equals 0 equals straight y.
Also comma
1 minus straight x greater than 0
rightwards double arrow straight x less than 1 space space but space straight x greater than 0
rightwards double arrow 0 less than straight x less than 1


end style

Answered by Sneha shidid | 11th May, 2017, 04:23: PM