If the function $\mathrm{f}(\mathrm{x})= \begin{cases}\mathrm{k}_{1}(\mathrm{x}-\pi)^{2}-1, & \mathrm{x} \leq \pi \\ \mathrm{k}_{2} \cos \mathrm{x}, & \mathrm{x}>\pi\end{cases}$
is twice differentiable, then the ordered pair $\left(\mathrm{k}_{1}, \mathrm{k}_{2}\right)$ is equal to :
Correct Option: 1
$f(x)$ is continuous and differentiable
$f(\pi)=f(\pi)=f\left(\pi^{+}\right)$
$-1=-k_{2}$
$\mathrm{k}_{2}=1$
$f^{\prime}(x)= \begin{cases}2 k_{1}(x-\pi) ; & x \leq \pi \\ -k_{2} \sin x & ; x>\pi\end{cases}$
$f^{\prime}(\pi)=f^{\prime}\left(\pi^{+}\right)$
$0=0$
so, differentiable at $x=0$
$f^{\prime \prime}(x)=\left\{\begin{array}{cc}2 k_{1} \quad ; x \leq \pi \\ -k_{2} \cos x ; x>\pi\end{array}\right.$
$f^{\prime \prime}\left(\pi^{-}\right)=f^{\prime \prime}\left(\pi^{+}\right)$
$2 \mathrm{k}_{1}=\mathrm{k}_{2}$
$\mathrm{k}_{1}=\frac{1}{2}$
$\left(\mathrm{k}_{1}, \mathrm{k}_{2}\right)=\left(\frac{1}{2}, 1\right)$
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