Continuity and Differentiability — Class 12 Maths Solution

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{1 - \cos kx}}{{x\sin x}},}&{{\rm{ if }}x \ne 0}\\{\frac{1}{2},}&{{\rm{ if }}x = 0}\end{array}} \right.$at $x = 0$.
At $x = 0,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{1 - \cos kx}}{{x\sin x}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos k(0 - h)}}{{(0 - h)\sin (0 - h)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos ( - kh)}}{{ - h\sin ( - h)}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{1 - 1 + 2{{\sin }^2}\frac{{kh}}{2}}}{{h\sin h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{2{{\sin }^2}\frac{{kh}}{2}}}{{h\sin h}}$

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

$= \frac{{2{k^2}}}{4} = \frac{{{k^2}}}{2}$

Also, $f(0) = \frac{1}{2} \Rightarrow \frac{{{k^2}}}{2} = \frac{1}{2} \Rightarrow k = \pm 1$

Therefore the value of k =$\pm 1$