Let f be a real valued function satisfying f x - y - f x - f y for all x - y - If f 1-1-2- then the value of f 16 isA- 16B- 8C- 4D- 2

Let f be a re...

Question

Let $f$ be a real valued function satisfying $f (x + y) = f (x) + f (y)$ for all $x, y .$ If $f (1) = \frac{1}{2}$ , then the value of $f (16)$ is

16

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8

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4

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2

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Solution

The correct option is B $8$
Assuming $m$ be an integers.
$f (x + (m - 1) x) = f (x) + f ((m - 1) x)$
$= f (x) + f (x) + f ((m - 2) x)$
$= f (x) + f (x) + f (x) + f ((m - 3) x)$
$.$
$.$
$= m f (x) + f (0) . . . . . (i)$
Using $f (x + y) = f (x) + f (y)$
Putting $x = 0, y = 0 \Rightarrow f (0) = 2 f (0) \Rightarrow f (0) = 0$
$∴ f (m x) = m f (x) where m is an integers \Rightarrow f (16 \times 1) = 16 f (1) \Rightarrow f (16) = 16 \times \frac{1}{2} = 8$

Alternate Solution:
$f (2) = f (1) + f (1) = 1 \Rightarrow f (4) = f (2) + f (2) = 2 \Rightarrow f (8) = f (4) + (4) = 4 \Rightarrow f (16) = f (8) + f (8) = 8$

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