If the function f(x) = \ cases k _ 1 ( x - π ) ^ 2 - 1, & x ≤ π k _...

f ( x )$is differentiable then will also continuous then$f ( \pi ) = {k}_{1}{\left(\pi - \pi\right)}^{2} - 1 = - 1$</p><p>$f\left({\pi}^{+}\right) = \lim_{h \to 0}{k}_{2}\cos \left(\pi + h\right) = - {k}_{2}$</p><p>$f\left({\pi}^{+}\right) = f\left(\pi\right) ⇒ {k}_{2} = 1 \ldots \left(1\right)$,$f$is continuous.</p><p>Now,$f ' ( x ) = \left\{$\begin{cases}\dfrac{d}{dx}\left\{{k}_{1} ( x - \pi{)}^{2}\right\}, & x \le \pi \\ \dfrac{d}{dx}\left\{{k}_{2}\cos x\right\}, & x > \pi\end{cases}$\right)

f ' ( x ) = \left\{\begin{cases}2{k}_{1} ( x - \pi ) , & x \le \pi \\ - {k}_{2}\sin x, & x > \pi\end{cases}\right)</p><p>then,f ' \left({\pi}^{-}\right) = 2{k}_{1}\left(\pi - \pi\right) = 0$</p><p>$f ' \left({\pi}^{+}\right) = - {k}_{2}\sin \left(\pi\right) = 0$</p><p>$⇒ f ' \left({\pi}^{-}\right) = f ' \left({\pi}^{+}\right) = 0$</p><p>Similarly, we get</p><p>$f ' ' ( x ) = \left\{$\begin{cases}2{k}_{1}, & x \le \pi \\ - {k}_{2}\cos x, & x > \pi\end{cases}$\right)

As the function is twice differentiable, the second-order derivatives, $LHD = RHD$

⇒ f ' ' \left({\pi}^{-}\right) = f ' ' \left({\pi}^{+}\right) ⇒ 2{k}_{1} = - {k}_{2}cos \pi

$⇒ 2{k}_{1} = {k}_{2}$

$⇒ {k}_{1} = \dfrac{1}{2}$ (from \left(1\right))