- Is discontinuous at finitely many points.
- Is continuous everywher.
- Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0
- None of these.
Solution:
Given function is
$\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$
Consider,
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^-}\text{f(x)}=\frac{1}{\text{n}^2}$
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}}1=1$
Hence, fnuction is discontinuous at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0
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$f(x) = sin^{-1} \left( {\frac{{\,\,1 - \,\,\left| x \right|}}{3}} \right) + cos^{-1}\left( {\frac{{\left| x \right|\,\, - \,\,3}}{5}} \right)$ .
Then domain of $f(x)$ is given by :
(where $C$ is a constant of integration.)