A real valued function \[f\left( x \right)\] satisfies the functional equation \[f\left( {x - y} \right) = f\left( x \right)f\left( y \right) - f\left( {a - x} \right)f\left( {a + y} \right)\] where \[a\] is a given constant and \[f\left( 0 \right) = 1,f\left( {2a - x} \right)\] is equal to
A. \[f\left( { - x} \right)\]
B. \[f\left( a \right) + f\left( {a - x} \right)\]
C. \[f\left( x \right)\]
D. \[ - f\left( x \right)\]

Hint: First of all, substitute \[x = y = 0\] so that we can use the value of \[f\left( 0 \right) = 1\] to simplify the given function and to get the value of \[f\left( a \right)\]. Then use the value of \[f\left( {x - y} \right)\] to simplify and get the solution of \[f\left( {2a - x} \right)\]. So, use this concept to reach the solution of the given problem.

Complete step by step solution:
Given \[f\left( {x - y} \right) = f\left( x \right)f\left( y \right) - f\left( {a - x} \right)f\left( {a + y} \right)......................................................\left( 1 \right)\]
Also, given that \[f\left( 0 \right) = 1\]
Substituting \[x = y = 0\] in equation \[\left( 1 \right)\], we have
\[
   \Rightarrow f\left( {0 - 0} \right) = f\left( 0 \right)f\left( 0 \right) - f\left( {a - 0} \right)f\left( {a + 0} \right) \\
   \Rightarrow f\left( 0 \right) = f\left( 0 \right)f\left( 0 \right) - f\left( a \right)f\left( a \right){\text{ }}\left[ {f\left( 0 \right) = 1} \right] \\
   \Rightarrow 1 = 1 \times 1 - {\left( {f\left( a \right)} \right)^2} \\
   \Rightarrow 1 = 1 - {\left( {f\left( a \right)} \right)^2} \\
   \Rightarrow {\left( {f\left( a \right)} \right)^2} = 1 - 1 = 0 \\
  \therefore f\left( a \right) = 0 \\
\]
Now, consider
\[
   \Rightarrow f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right) \\
   \Rightarrow f\left( {2a - x} \right) = f\left( a \right)f\left( {x - a} \right) - f\left( {a - a} \right)f\left( {a + x - a} \right){\text{ }}\left[ {{\text{from equation }}\left( 1 \right)} \right] \\
   \Rightarrow f\left( {2a - x} \right) = (0)f\left( {x - a} \right) - f\left( 0 \right)f\left( {a + x - a} \right){\text{ }}\left[ {f\left( a \right) = 0} \right] \\
   \Rightarrow f\left( {2a - x} \right) = 0 - 1 \times f\left( x \right){\text{ }}\left[ {f\left( 0 \right) = 1} \right] \\
  \therefore f\left( {2a - x} \right) = - f\left( x \right) \\
\]
Thus, the correct option is D. \[ - f\left( x \right)\]


Note: In mathematics, a real-valued function is a function whose values are real numbers. In other words, it is a function that assigns a real number to each member of its domain.