Question

Q1.Find the value(s) of k so that the following function is continuous at x = 0.

Answer 1

Step-by-step explanation:

We have to find the value of k if the given function is continuous at x = 0.

Now,

$\rm \large\displaystyle\lim_{ \rm x \to0}{\rm \: f(x)} =\rm f(0)$

$\begin{gathered}\rm\longrightarrow\displaystyle\lim_{ \rm x \to0}{\rm \: f(x)} = \displaystyle\lim_{ \rm x \to0}{\rm \: \dfrac{1 - cos \: kx}{x \: sin \: x} } \\ \\ \\ \rm\longrightarrow \displaystyle\lim_{ \rm x \to0}{\rm \: \dfrac{ \dfrac{(1 - cos \: kx) \times {(kx)}^{2} }{ {(kx)}^{2} } }{ {x}^{2} \: \bigg(\dfrac{sin \: x}{x} \bigg) } } \\ \\ \\ \rm\longrightarrow \displaystyle\lim_{ \rm x \to0}{\rm \: \dfrac{ \dfrac{1 }{ {2} } \times {(kx)}^{2} }{ {x}^{2} } }\\ \\ \\ \rm\longrightarrow \displaystyle {\dfrac{ \rm {k}^{2} }{ 2 } }...(1)\end{gathered}$

Now, from (1) and (2) :-

$\begin{gathered} \rm \rightarrow \dfrac{ {k}^{2} }{2} = \dfrac{1}{2} \\ \\ \\\rm \rightarrow { {k}^{2} } = 1\\ \\ \\ \rm \rightarrow { {k} } = \pm 1\end{gathered}$