First slide

Functions (XII)

Question

$f is a real-valued function not identically zero, satisfying f (x + y) + f (x - y) = 2 f (x) \cdot f (y) \forall x, y \in R \cdot f (x)$ $is definitely$

Moderate

A

odd
B

even
C

neither even nor odd
D

none of these

Solution

$Putting x = 0, y = 0 in the functional equation we get 2 f (0) = 2 f^{2} (0) \Rightarrow f (0) = 0, 1 . But if f (0) is equal to zero then$
$f (x) = 0 \forall x \in R . Thus, f (0) = 1 . Now putting x = 0 in the functional equation we get, f (y) + f (- y) = 2 f (0) \cdot f (y)$
$\Rightarrow f (y) = f (- y) . Thus, f (x) is even.$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App