First slide

Functions (XII)

Question

If $f (x + y) = f (x) . f (y)$ for all real x, y and $f (0) \neq 0,$ then the function $g (x) = \frac{f (x)}{1 + {\{f (x)\}}^{2}}$ is

Moderate

A

even function
B

odd function
C

odd if $f (x) > 0$
D

neither even nor odd

Solution

Given $f (x + y) = f (x) f (y) .$
Put $x = y = 0, t h e n f (0) = 1.$
Put $y = - x,$ then
$f (0) = f (x) f (- x) \Rightarrow f (- x) = \frac{1}{f (x)}$
Now, $g (x) = \frac{f (x)}{1 + {\{f (x)\}}^{2}}$
$\Rightarrow g (- x) = \frac{f (- x)}{1 + {\{f (- x)\}}^{2}}$
= $\frac{\frac{1}{f (x)}}{1 + {(\frac{1}{f (x)})}^{2}}$
$= \frac{f (x)}{1 + {\{f (x)\}}^{2}} = g (x)$

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