Question

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. $- f\left( x \right)$

Answer 1

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.