|x−y| ⇒|f′(y)|≤0 [using the definition of f′(y) ] ⇒f′(y)=0 [since, modulus value can never be less than 0 ] On integrating it, we get f(y)=c (constant) Given, f(0)=1 gives C=1 ∴f(y)=1∀y∈R From given options, f(x)>0∀x∈R is satisfied only. Hence, answer will be option (d).
