Question:

<p><strong>Solution:</strong></p><p>The given function, <span class="math">$f(x) = \begin{cases} k_1(x - \pi)^2 - 1, & x \leq \pi \\ k_2 \cos x, & x > \pi \end{cases}$</span></p><p>is twice differentiable, so <span class="math">$f(x)$</span> must be continuous at <span class="math">$x = \pi$</span></p><p>So <span class="math">$\lim_{x \to \pi^+} f(x) = f(\pi)$</span></p><p>Substituting: <span class="math">$-k_2 = -1 \Rightarrow k_2 = 1$</span> ... (i)</p><p><span class="math">$f'(x) = \begin{cases} 2k_1(x - \pi), & x < \pi \\ -k_2 \sin x, & x > \pi \end{cases}$</span></p><p>At <span class="math">$x = \pi$</span>, for differentiability: <span class="math">$f'(\pi^-) = f'(\pi^+)$</span></p><p>This gives: <span class="math">$0 = -k_2 \sin \pi = 0$</span> (automatically satisfied)</p><p>For second derivative to be continuous: <span class="math">$f''(x) = \begin{cases} 2k_1, & x < \pi \\ -k_2 \cos x, & x > \pi \end{cases}$</span></p><p>At <span class="math">$x = \pi$</span>: <span class="math">$2k_1 = -k_2 \cos \pi = k_2$</span></p><p>With <span class="math">$k_2 = 1$</span>: <span class="math">$2k_1 = 1 \Rightarrow k_1 = \frac{1}{2}$</span></p><p>∴ Answer is <span class="math">$(\frac{1}{2}, 1)$</span></p>