If the function f(x) = begin{cases} k_{1}(x-p | vedclass

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(A) For $f(x)$ to be twice differentiable,it must be continuous and differentiable at $x = \pi$.$1$. Continuity at $x = \pi$:$f(\pi^{-}) = f(\pi) = f(

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$(\frac{1}{2}, 1)$

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(A) For $f(x)$ to be twice differentiable,it must be continuous and differentiable at $x = \pi$.$1$. Continuity at $x = \pi$:$f(\pi^{-}) = f(\pi) = f(\pi^{+})$$k_{1}(\pi-\pi)^{2} - 1 = k_{2} \cos(\pi)$$-1 = -k_{2} \implies k_{2} = 1$.$2$. First derivative $f'(x)$:$f'(x) = \begin{cases} 2k_{1}(x-\pi), & x \leq \pi \ -k_{2} \sin x, & x > \pi \end{cases}$At $x = \pi$,$f'(\pi^{-}) = 2k_{1}(\pi-\pi) = 0$ and $f'(\pi^{+}) = -k_{2} \sin(\pi) = 0$.Since $0 = 0$,the function is differentiable at $x = \pi$ for any $k_{1}, k_{2}$.$3$. Second derivative $f''(x)$:$f''(x) = \begin{cases} 2k_{1}, & x \leq \pi \ -k_{2} \cos x, & x > \pi \end{cases}$For $f''(x)$ to be continuous at $x = \pi$:$f''(\pi^{-}) = f''(\pi^{+})$$2k_{1} = -k_{2} \cos(\pi)$$2k_{1} = -k_{2}(-1) = k_{2}$Since $k_{2} = 1$,we have $2k_{1} = 1 \implies k_{1} = \frac{1}{2}$.Thus,$(k_{1}, k_{2}) = (\frac{1}{2}, 1)$.

If the function $f(x) = \begin{cases} k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi \end{cases}$ is twice differentiable,then the ordered pair $(k_{1}, k_{2})$ is equal to

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