f(x)=1−h(x)....(i)g(x)=1−k(x)......(ii)h(x)=f(x)+1......(iii)j(x)=g(x)+1......(iv)k(x)=j(x)+1.......(v) From Eqs. (i) and (iii), we get f(x)+h(x)=1−h(x)+f(x)+1⇒h(x)=1.....(vi)⇒f(x)=0.....(vii) From Eqs. (ii) and (iv), we get g(x)=1−k(x)=1−(j(x)+1)g(x)=−j(x)......(viii) From Eqs. (iv) and (viii), we get j(x)−g(x)=1 and g(x)=−j(x)⇒j(x)=21​.....(ix)∴g(x)=−21​......(x)∴k(x)=j(x)+1=23​......(xi) Hence , f(x) = 0 g(x)=−21​h(x)=1j(x)=21​k(x)=23​ Thus all the functions are constant