First slide

Functions (XII)

Question

Let f (x) be a real valued function such that $f (0) = \frac{1}{2}$ and $f (x + y) = f (x) f (a - y) + f (y) f (a - x)$ , $\forall x, y \in R$ then for some real a,

Moderate

A

f(x) is a periodic function
B

f(x) is a constant function
C

$f (x) = \frac{1}{2}$
D

$f (x) = \frac{\sin x}{2}$

Solution

$f (x + y) = f (x) f (a - y) + f (y) f (a - x) - - - - (i)$
Put x = y = 0, we get $f (a) = \frac{1}{2}$
Let y = 0
$\begin{array}{ll} \Rightarrow & f (x) = f (x) f (a) + f (0) \cdot f (a - x) \\ \Rightarrow & f (x) = \frac{1}{2} f (x) + \frac{1}{2} f (a - x) \\ \Rightarrow & f (x) = f (a - x) \end{array}$
Put y = a- x in Eq. (i),
$\begin{aligned} f (a) & = (f (x))^{2} + (f (a - x))^{2} \\ \Rightarrow (f (x))^{2} & = \frac{1}{4} \\ \Rightarrow f (x) & = \pm \frac{1}{2} \\ Hence, f (x) & = \frac{1}{2} \end{aligned}$

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