If the function $f\left(x\right)=\left\{\begin{array}{l}{k}_1{\left(x-\pi \right)}^2,x\le \pi \\ k_2\cos x,x>\pi \end{array}\right.$ is twice the differentiable then the ordered pair $\left(k_{1,}{k}_2\right)$ is

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{l}{k}_1{\left(x-\pi \right)}^2,x\le \pi \\ k_2\cos x,x>\pi \end{array}\right. \\ \Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{l}{k}_{1}2\left(x-\pi \right),x<\pi \\ -k_2\sin x,x>\pi \end{array}\right. \\ \Rightarrow f\text{'}\text{'}\left(x\right)=\left\{\begin{array}{l}{k}_{1}2,x<\pi \\ -k_2\cos x,x>\pi \end{array}\right. \end{gathered}$$

The function is twice differentiable. $f\text{'}\text{'}\left({\pi }^{-}\right)=f\text{'}\text{'}\left({\pi }^{+}\right)$

$\begin{array}{c}2k_1=k_2\left(-\cos \pi \right)\\ =k_2\end{array}$

hence, $(k_1,k_2)=\left(\frac{1}{2},1\right)$