Let f: ℝ → ℝ be a differentiable function that satisfies the...

Given,

Functional equation $f\left(x + y\right) = f\left(x\right) + f\left(y\right) - 1$

Now taking $x = 0 \text{ and } y = 0 in above equation we get,</p><p>$f\left(0 + 0\right) = f\left(0\right) + f\left(0\right) - 1 ⇒ f\left(0\right) = 1$</p><p>Now we know that,</p><p>${f}^{'}\left(x\right) = \lim_{h \to 0}\dfrac{f\left(x + h\right) - f\left(x\right)}{h}$</p><p>$⇒ {f}^{'}\left(x\right) = \lim_{h \to 0}\dfrac{f\left(x\right) + f\left(h\right) - 1 - f\left(x\right)}{h}$</p><p>$⇒ {f}^{'}\left(x\right) = \lim_{h \to 0}\dfrac{f\left(h\right) - 1}{h}$</p><p>Now let$\lim_{h \to 0}\dfrac{f\left(h\right) - 1}{h} = k$</p><p>So, the equation becomes${f}^{'}\left(x\right) = k$</p><p>Now putting$x = 0$in above equation we get,</p><p>${f}^{'}\left(0\right) = k ⇒ k = 2 \left\{\text{as given}{f}^{'}\left(0\right) = 2\right\}$</p><p>So,${f}^{'}\left(x\right) = 2$</p><p>Now integrate both side we get,</p><p>$f\left(x\right) = 2x + c$</p><p>Now again taking$x = 0$we get,</p><p>$f\left(0\right) = 2 \times 0 + c ⇒ c = 1\left\{\text{as}f\left(0\right) = 1\right\}$</p><p>So,$f\left(x\right) = 2x + 1$</p><p>And$f\left(- 2\right) = 2 \times \left(- 2\right) + 1 ⇒ f\left(- 2\right) = - 3$</p><p>Hence, \left|f\left(- 2\right)\right| = 3$