Answer with explanation:
For a function f(x), it is given that following Statement hold.
\(f(x_{1})+f(x_{2})=f(x_{1}+x_{2})\)
Coming to the options
(A)f(x)=x+1
\(f(x_{1})=x_{1}+1\\\\f(x_{2})=x_{2}+1\\\\f(x_{1}+x_{2})=x_{1}+x_{2}+1
eq f(x_{1})+f(x_{2})\)
This function does not define f.
(B)f(x)=2x
\(f(x_{1})=2x_{1}\\\\f(x_{2})=2x_{2}\\\\f(x_{1}+x_{2})=2(x_{1}+x_{2})\\\\=2x_{1}+2x_{2}=f(x_{1})+f(x_{2})\)
This function defines f.
\((C)\rightarrow f(x)=\frac{1}{x}\\\\f(x_{1})=\frac{1}{x_{1}}\\\\f(x_{2})=\frac{1}{x_{2}}\\\\f(x_{1})+f(x_{2})=\frac{1}{x_{2}}+\frac{1}{x_{1}}\\\\f(x_{1}+x_{2})=\frac{1}{x_{1}+x_{2}}\\\\f(x_{1})+f(x_{2})
eq f(x_{1}+x_{2})\)
This function does not define f.
\((D)\rightarrow f(x)=e^x\\\\f(x_{1})=e^{x_{1}}\\\\f(x_{2})=e^{x_{2}}\\\\f(x_{1})+f(x_{2})=e^{x_{2}}+e^{x_{1}}\\\\f(x_{1}+x_{2})=e^{x_{1}+x_{2}}
eq f(x_{1})+f(x_{2})\)
This function does not define f.
\((E)\rightarrow f(x)=x^2\\\\f(x_{1})=(x_{1})^2\\\\f(x_{2})=(x_{2})^2\\\\f(x_{1})+f(x_{2})=(x_{1})^2+(x_{2})^2\\\\f(x_{1}+x_{2})=[x_{1}+x_{2}]^2
eq f(x_{1})+f(x_{2})\)
This function does not define f.
Option B: f(x)=2x , defines f.
