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

We are given that $$f : \mathbb{R} \to \mathbb{R}$$ is differentiable, $$f(x+y) = f(x) + f(y) - 1$$ for all $$x, y \in \mathbb{R}$$, and $$f'(0) = 2$$.

Setting $$x = y = 0$$, we get $$f(0) = f(0) + f(0) - 1 = 2f(0) - 1$$, so $$f(0) = 1$$.

Now let $$g(x) = f(x) - 1$$. Then:

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

So $$g$$ satisfies Cauchy's functional equation: $$g(x + y) = g(x) + g(y)$$. Since $$f$$ is differentiable, $$g$$ is also differentiable, and the only differentiable solution of Cauchy's equation is $$g(x) = kx$$ for some constant $$k$$.

We have $$g'(0) = f'(0) = 2$$, so $$k = 2$$. Hence $$g(x) = 2x$$, giving $$f(x) = 2x + 1$$.

Now, $$f(-2) = 2(-2) + 1 = -3$$, so $$|f(-2)| = |-3| = 3$$.

So, the answer is $$3$$.