If the function

$f(x)$ is differentiable then, $f(x)$ is also continuous.

$\therefore \lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{-}} f(x)=f(\pi)$

$\Rightarrow-1=-K_{2} \Rightarrow K_{2}=1$

$\therefore f^{\prime}(x)=\left\{\begin{array}{rl}2 K_{1}(x-\pi): & x \leq \pi \\ -K_{2} \sin x & x>\pi\end{array}\right.$

Then, $\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{-}} f(x)=0$

$f^{\prime \prime}(x)=\left\{\begin{array}{ccc}2 K_{1} & ; & x \leq \pi \\ -K_{2} \cos x & ; & x>\pi\end{array}\right.$

Then, $\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{-}} f(x)$

$\Rightarrow 2 K_{1}=K_{2} \Rightarrow K_{1}=\frac{1}{2}$

$\operatorname{So},\left(K_{1}, K_{2}\right)=\left(\frac{1}{2}, 1\right)$