Given that, #e^(f(x))=((10+x)/(10-x)), x in (-10,10).#
#:. lne^(f(x))=ln((10+x)/(10-x))#.
#:. f(x)*lne=ln((10+x)/(10-x)),#
# i.e., f(x)=ln((10+x)/(10-x))..........................(ast_1)#.#,
# or, f(x)=ln(10+x)-ln(10-x)#.
Plugging in #(200x)/(100+x^2)# in place of #x#, we get,
# f((200x)/(100+x^2))#,
#=ln{10+(200x)/(100+x^2)}-ln{10-(200x)/(100+x^2)}#,
#=ln{(1000+10x^2+200x)/(100+x^2)}-ln{(1000+10x^2-200x)/(100+x^2)}#,
#=ln[{10(100+x^2+20x)}/(100+x^2)]-ln[{10(100+x^2-20x)}/(100+x^2)]#,
#=ln[{10(100+x^2+20x)}/(100+x^2)-:{10(100+x^2-20x)}/(100+x^2)]#,
#=ln{(100+x^2+20x)/(100+x^2-20x)}#,
#=ln{((10+x)/(10-x))^2}#.
Thus, #f((200x)/(100+x^2))=ln{((10+x)/(10-x))^2}...........(ast_2)#.
Now, utilising #(ast_1) and (ast_2)# in
#f(x)=k*f((200x)/(100+x^2))......................."[Given]"#, we get,
#ln((10+x)/(10-x))=k*ln{((10+x)/(10-x))^2}#,
# i.e., ln((10+x)/(10-x))=ln((10+x)/(10-x))^(2k)#.
#:. 1=2k, or, k=1/2=0.5," which is the option "(1).#
