Ifef(x)=10+x10−x,x∈(−10,10) and  f(x)=k.f200x100+x2 then k = ______

ef(x)=10+x10−x,⇒f(x)= log10+x10-x f(x)=k.f200x100+x2         =klog10+200x100+x210-200x100+x2          =klog10100+x2+200x10100+x2-200x           =klog100+x2+20x100+x2-20x           =k log10+x10-x2f(x) = 2k f(x)