$\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x^2}, & \text { if } x \leq 0 \\…

$\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x^2}, & \text { if } x \leq 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text { if } x>0\end{array}\right.$ is continuous at $x=0$, then the value of $\mathrm{k}$ is
  1. 4
  2. 2
  3. -1
  4. -3

Solution

$\mathrm{f}(x)$ 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] $\begin{array}{ll} \therefore \quad & \text { L.H.L. }=\text { R.H.L } \\ \therefore \quad & \frac{1}{2} \mathrm{k}^2=8 \\ \therefore \quad & \mathrm{k}^2=16 \\ \therefore \quad \mathrm{k}= \pm 4 \end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

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