A function whose graph is symmetrical the origin is given by
A. f(x)=ex+e-x
B. f(x)=logex
C. f(x+y)=f(x)+f(y)
D. None of these

Hint: Here the question provides a number of functions and asks for the symmetry about the origin, and to prove this we have to use the even and odd function rule. If a function is even function then it will be symmetrical with respect to origin, even function states that:
 \[
   \Rightarrow for\,any\,function\,f(x) \\
   \Rightarrow for\,even\,function \to f(x) + f( - x) = 0 \;
 \]

Complete step by step solution:
The given question need to find the symmetry for the given function, for which we have to use the even function rule, for every equation we have to check for the even rule, and the functions which will satisfy the rule will be our required function which will show the symmetry along origin, on solving we get:
Even function rule:
 \[
   \Rightarrow for\,any\,function\,f(x) \\
   \Rightarrow for\,even\,function \to f(x) + f( - x) = 0 \;
 \]
Using this rule for our function, on solving we get:
 \[
   \Rightarrow for\,function\,f(x) = {e^x} + {e^{ - x}} \\
   \Rightarrow for\,even\,function \to f(x) + f( - x) = 0 \\
   \Rightarrow ({e^x} + {e^{ - x}}) + {e^{ - x}} + {e^{ - ( - x)}} = 0 \\
   \Rightarrow {e^x} + {e^{ - x}} + {e^{ - x}} + {e^x} = 0 \\
   \Rightarrow 2({e^x} + {e^{ - x}}) = 0 \;
 \]
Hence the first function is not an even function as well as not symmetric along origin.
 \[
   \Rightarrow for\,function\,f(x) = {\log _e}x \\
   \Rightarrow for\,even\,function \to f(x) + f( - x) = 0 \\
   \Rightarrow ({\log _e}x) + ({\log _e}( - x)) = 0 \\
   \Rightarrow {\log _e}x - {\log _e}x = 0 \\
   \Rightarrow 0 = 0 \;
 \]
Hence the second function is an even function and is symmetric along origin.
 \[
   \Rightarrow for\,function\,f(x + y) = f(x) + f(y) \\
   \Rightarrow for\,even\,function \to f(x) + f( - x) = 0 \\
   \Rightarrow (f(x) + f(y)) + (f( - x) + f( - y)) = 0 \\
   \Rightarrow f(x) + f(y) + f( - x) + f( - y) = 0 \;
 \]
Hence the third function is also not even a function and not symmetric along origin.
Formulae Used:
 \[
   \Rightarrow for\,any\,function\,f(x) \\
   \Rightarrow for\,even\,function \to f(x) + f( - x) = 0 \;
 \]
So, the correct answer is “Option B”.

Note: The above question can also be solved by plotting the curve for every function and then check for the symmetry, here to plot the curve you can assume some values of “x” and then find “f(x)” and then after plotting theses coordinates on graph you can get the curve and then check for the symmetry for each function.