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(C) Given the functional equation $f(x + y) = f(x) + f(y) - 1$. To find $f(0)$,put $x = 0$ and $y = 0$: $f(0 + 0) = f(0) + f(0) - 1 \Rightarrow f(0) =

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$3$

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(C) Given the functional equation $f(x + y) = f(x) + f(y) - 1$. To find $f(0)$,put $x = 0$ and $y = 0$: $f(0 + 0) = f(0) + f(0) - 1 \Rightarrow f(0) = 2f(0) - 1 \Rightarrow f(0) = 1$. Now,use the definition of the derivative: $f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$. Using the given relation $f(x + h) = f(x) + f(h) - 1$: $f'(x) = \lim_{h \rightarrow 0} \frac{f(x) + f(h) - 1 - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(h) - 1}{h}$. Since $f(0) = 1$,we can write $1 = f(0)$: $f'(x) = \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = f'(0)$. Given $f'(0) = 2$,so $f'(x) = 2$. Integrating both sides with respect to $x$: $f(x) = 2x + C$. Using $f(0) = 1$: $1 = 2(0) + C \Rightarrow C = 1$. Thus,$f(x) = 2x + 1$. Now,calculate $f(-2)$: $f(-2) = 2(-2) + 1 = -4 + 1 = -3$. Therefore,$|f(-2)| = |-3| = 3$.

Let $f: R \rightarrow R$ be a differentiable function that satisfies the relation $f(x + y) = f(x) + f(y) - 1$ for all $x, y \in R$. If $f'(0) = 2$,then $|f(-2)|$ is equal to:

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