Find the value of $k$ so that the function $f(x) = \begin{cases} \frac{1 - \cos kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases}$ is continuous at $x=0$.

$ \pm 1$

The correct answer is Option (3) → $ \pm 1$ ##

We have,

$f(x) = \begin{cases} \frac{1 - \cos kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases} \text{ at } x = 0$$

At $x = 0$,

$\text{LHL} = \lim\limits_{x \to 0^-} \frac{1 - \cos kx}{x \sin x}$

Put $x = 0 - h$,

$= \lim\limits_{h \to 0} \frac{1 - \cos k(0 - h)}{(0 - h) \sin(0 - h)} = \lim\limits_{h \to 0} \frac{1 - \cos(-kh)}{-h \sin(-h)}$

$= \lim\limits_{h \to 0} \frac{1 - \cos kh}{h \sin h} \quad [∵\cos(-\theta) = \cos \theta, \sin(-\theta) = -\sin \theta]$

$= \lim\limits_{h \to 0} \frac{1 - (1 - 2 \sin^2 \frac{kh}{2})}{h \sin h} \quad \left[ ∵\cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} \right]$

$= \lim\limits_{h \to 0} \frac{2 \sin^2 \frac{kh}{2}}{h \sin h} = \lim_{h \to 0} \frac{2 \sin \frac{kh}{2}}{\frac{kh}{2}} \cdot \frac{\sin \frac{kh}{2}}{\frac{kh}{2}} \cdot \frac{1}{\frac{\sin h}{h}} \cdot \frac{k^2 h/4}{h}$

$= \frac{2k^2}{4} = \frac{k^2}{2} \quad \left[ ∵\lim_{h \to 0} \frac{\sin h}{h} = 1 \right]$

Also, $f(0) = \frac{1}{2} \Rightarrow \frac{k^2}{2} = \frac{1}{2} \Rightarrow k = \pm 1 \quad [∵\text{LHL} = \text{RHL} = f(0)]$