\(\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x^2}, & \text { if } x \leq 0 \ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text { if } x>0\end{array}\right.\) is continuous at \(x=0\), then the value of \(\mathrm{k}\) is

\(\mathrm{f}(x)\) is continuous at \(x=0\)
\(\begin{aligned}
\therefore \quad \text { L.H.L. } & =\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{\mathrm{k} x}{2}}{x^2} \\
& =2 \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{\mathrm{k} x}{2}}{\frac{\mathrm{k}^2 x^2}{4} \times 4} \cdot \mathrm{k}^2 \\
& =\frac{1}{2} \mathrm{k}^2 \lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{k} x}{2}}{\frac{\mathrm{k} x}{2}}\right)^2 \\
\therefore \quad \text { L.H.L. } & =\frac{1}{2} \mathrm{k}^2
\end{aligned}\)
\(\begin{aligned}
\therefore \quad \text { R.H.L. } & =\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{x})(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} \\
& =\sqrt{16+\sqrt{0}+4}
\end{aligned}\)
\(\therefore \quad\) R.H.L. \(=8\)
\(\mathrm{f}(x)\) is continuous at \(x=0\)... [Given]
\(\begin{array}{ll}
\therefore \quad & \text { L.H.L. }=\text { R.H.L } \\
\therefore \quad & \frac{1}{2} \mathrm{k}^2=8 \\
\therefore \quad & \mathrm{k}^2=16 \\
\therefore \quad \mathrm{k}= \pm 4
\end{array}\)