A) is continuous at x=1
B) is discontinuous at x=1 since \[f({{1}^{+}})\]does not exist though \[f({{1}^{-}})\]exists
C) is discontinuous at x=1 since \[f({{1}^{-}})\]does not exist thought \[f({{1}^{+}})\]exists
D) is discontinuous at x=1 since neither \[f({{1}^{+}})\]nor \[f({{1}^{-}})\]exists.
Correct Answer: D
Solution :
[d] We have \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(1-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\log (4+{{h}^{2}})}{\log (1-4h)}=-\infty \] And \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,f(1+h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\log (4+{{h}^{2}})}{\log (1+4h)}=\infty \]So, \[f({{1}^{-}})\]and \[f({{1}^{+}})\]do not exist.
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