If the function $ f(x)=\left\{\begin{array}{cc}k_{1}(x-\pi)^{2}-1, & x \leq \pi \ k_{2} \cos x, & x\pi\end{array}\right. $ is twice differentiable, then the ordered pair $ \left(\mathrm{k}_{1}, \mathrm{k}_{2}\right) $ is equal to : [JEE Main 2020, 5 Sep (Shift 1)]

Question

If the function $ f(x)=\left\{\begin{array}{cc}k_{1}(x-\pi)^{2}-1, & x \leq \pi \ k_{2} \cos x, & x>\pi\end{array}\right. $ is twice differentiable, then the ordered pair $ \left(\mathrm{k}_{1}, \mathrm{k}_{2}\right) $ is equal to : [JEE Main 2020, 5 Sep (Shift 1)]

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If the function \( f(x)=\left\{\begin{array}{cc}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi\end…

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If the function \( f(x)=\left\{\begin{array}{cc}k_{1}(x-\pi)^{2}-1, &amp; x \leq \pi \\ k_{2} \cos x, &amp; x&gt;\pi\end…

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If the function \( f(x)=\left\{\begin{array}{cc}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi\end…