Question

If $f\left(x\right)$ is continuous at $f\left(x\right)=\begin{array}{c}\frac{1-\cos kx}{x^2},\\ =\frac{1}{2},\end{array}\begin{array}{c}forx\ne 0\\ forx=0\end{array}$
Find $k$.

Answer 1

$f\left(x\right)=\frac{1-\cos kx}{x^2}=\frac{2{\sin }^2\frac{kx}{2}}{{\left(\frac{kx}{2}\right)}^2}\times {\left(\frac{k}{2}\right)}^2$

$f\left(x\right)$ is continuous at $x=0$

$\Rightarrow {lim}_{x\to 0}f\left(x\right)=f\left(0\right)$

$\therefore {lim}_{x\to 0}\frac{1-\cos kx}{x^2}=\frac{1}{2}$

$\Rightarrow {lim}_{x\to 0}\frac{2{\sin }^2\frac{kx}{2}}{{\left(\frac{kx}{2}\right)}^2}\times {\left(\frac{k}{2}\right)}^2=\frac{1}{2}$

$\Rightarrow 2\times 1\times \frac{4}{k^2}=\frac{1}{2}$

$\Rightarrow k^2=1$

$\therefore k=\pm 1$