If ef(x)=10+x10−x,x∈(−10,10) and f(x)=kf200x100+x2then k =

ef(x)=10+x10−x⇒f(x)=log⁡10+x10−x∴f200x100+x2=log⁡10+200x100+x210−200x100+x2=log⁡1000+200x+10x21000−200x2+10x2=log⁡100+20x+x2100−20x+x2=log⁡10+x10−x2=2log⁡10+x10−x=2f(x) By the given condition f(x)=k.2f(x)⇒k=12