If the function f x = k 1 ( x - π ) 2 - 1 , x ≤ π k 2 cos x , x > π is twice differentiable, then the ordered pair k 1 , k 2 is equal to:

Step-by-step solution

f ( x ) is differentiable then will also continuous then f ( π ) = k 1 π - π 2 - 1 = - 1 f π + = lim h → 0 k 2 cos π + h = - k 2 f π + = f π ⇒ k 2 = 1 … 1 , f is continuous. Now, f ' ( x ) = d d x k 1 ( x - π ) 2 , x ≤ π d d x k 2 cos x , x > π f ' ( x ) = 2 k 1 ( x - π ) , x ≤ π - k 2 sin x , x > π then, f ' π - = 2 k 1 π - π = 0 f ' π + = - k 2 sin π = 0 ⇒ f ' &