1.If the function f(x)={k1(π−x)2−1,x≤πk2cosx,x>πf(x) = \begin{cases} k_1(\pi - x)^2 - 1, & x \leq \pi \\ k_2 \cos x, & x > \pi \end{cases}f(x)={k1(π−x)2−1,k2cosx,x≤πx>π is twice differentiable, then the ordered pair (k1,k2)(k_1, k_2)(k1,k2) is equal to:Login to continueOnly logged in users canattempt or see the solution.
