A) continuous at\[x=-2\]
B) not continuous at\[x=-2\]
C) differentiable at\[x=-2\]
D) continuous but not derivable at\[x=-2\].
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{matrix} \frac{|x+2|}{{{\tan }^{-1}}(x+2)}, & x\ne -2 \\ 2, & x=-2 \\ \end{matrix} \right.\] \[\therefore \] \[\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(-2-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|-2-h+2|}{{{\tan }^{-1}}(-2-h+2)}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-h}{{{\tan }^{-1}}h}=-1\] and \[\underset{x\to -{{2}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(-2+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\left[ \frac{|-2+h+2|}{{{\tan }^{-1}}(-2+h+2)} \right]\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{{{\tan }^{-1}}h}=1\] \[\because \] \[\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to -{{2}^{+}}}{\mathop{\lim }}\,f(x)\] So,\[f(x)\]is not continuous as well as not differentiable at\[x=-2\].
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