Let f : ℝ → ℝ be a differentiable function that satisfies the relation f ( x + y ) = f ( x ) + f ( y ) - 1 , ∀ x , y ∈ ℝ . If f ' ( 0 ) = 2 , then | f ( - 2 ) | is equal to

Step-by-step solution

Given, Functional equation f x + y = f x + f y - 1 Now taking x = 0   &   y = 0 in above equation we get, f 0 + 0 = f 0 + f 0 - 1 ⇒ f 0 = 1 Now we know that, f ' x = lim h → 0 f x + h - f x h ⇒ f ' x = lim h → 0 f x + f h - 1 - f x h ⇒ f ' x = lim h → 0 f h - 1 h Now let lim h → 0 f h - 1 h = k So, the equation becomes f ' x = k Now putting x = 0 in above equation we get, f ' 0 = k ⇒ k = 2 as given  f ' 0 = 2 So, f '