Let f : be a differentiable function that satisfies the relation f ( x + y ) = f ( x ) + f ( y ) - 1 ,x , y . If f ' ( 0 ) = 2 , then | f ( - 2 ) | is equal to _______ . [2023]

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Solution

Q.
Let $f : ℝ \to ℝ$ be a differentiable function that satisfies the relation $f (x + y) = f (x) + f (y) - 1, \forall x, y \in ℝ .$ If $f' (0) = 2$ , then $| f (- 2) |$ is equal to _______ . [2023]

Ans.
(3)

$Given f (x + y) = f (x) + f (y) - 1$

$Putting x = y = 0 \Rightarrow f (0) = 1$

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f' (0) = \lim_{h \to 0} \frac{f (h) - f (0)}{h} \Rightarrow f' (0) = 2$

$f' (x) = 2 \Rightarrow f (x) = 2 x + C \Rightarrow y = 2 x + C$

$Now, f (0) = 1$

$∴$ $1 = 2 (0) + C \Rightarrow C = 1$

So, $f (x) = 2 x + 1$

$| f (- 2) |$ $=$ $| 2 (- 2) + 1 |$ $=$ $| - 3 |$ $= 3$

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