Let f:​ S→S where S=0,∞ be a twice differentiable function such that fx+1=xfx. If f:S→R be defined as gx=logefx, then the value of g"5−g"1 is equal to :

g(x+1)=ln(f(x+1))=ln(xf(x))=lnx+ln(f(x))=lnx+g(x)⇒g(x+1)-g(x)=lnx  ⇒g'(x+1)-g'(x)=1x⇒g''(x+1)-g''(x)=-1x2…… (i)  Putting x=1,2,3,4 in (i), we get ⇒g''(2)-g''(1)=-1 , g''(3)-g''(2)=-14g''(4)-g''(3)=-19,   g''(5)-g''(4)=-116 Adding.   g''(5)-g''(1)=-1-14-19-116=205144∴g''(5)-g''(1)=205144