f , g : R → R be two real valued function defined as f x = - x + 3 , x < 0 e x , x ≥ 0 and g x = x 2 + k 1 x , x < 0 4 x + k 2 , x ≥ 0 , where k 1 and k 2 are real constants. If g o f is differentiable at x = 0 , then g o f - 4 + g o f 4 is equal to

Step-by-step solution

Here f x = x + 3 ; x < - 3 - x + 3 ; - 3 ≤ x < 0 e x ; x ≥ 0 and g x = x 2 + k 1 x ; x < 0 4 x + k 2 ; x ≥ 0 Now g f x = f x 2 + k 1 f x ; f x < 0 4 f x + k 2 ; f x ≥ 0 i.e. g f x = x + 3 2 + k 1 x + 3 ; x < - 3 x + 3 2 - k 1 x + 3 ; - 3 ≤ x < 0 4 e x + k 2 ; x ≥ 0 For continuity at x = 0 , gof 0 = g f 0 - = g f 0 + i.e. 4 + k 2 = 9 - 3 k 1 = 4 + k 2 ⇒ 3 k 1 + k 2 = 5           ⋯ i Now differentiating, we get g f x &#39