A) Continuous as well as differentiable at x = 0
B) Continuous but not differentiable at x = 0
C) Differentiable but not continuous at x = 0
D) Neither continuous nor differentiable at x = 0
Correct Answer: A
Solution :
| [a] We have, |
| \[Lf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-h\log \cosh }{-h\log (1+{{h}^{2}})}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\log \cosh }{\log (1+{{h}^{2}})}\left( \frac{0}{0}form \right)\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-\tan \,\,h}{2h/(1+{{h}^{2}})}=-1/2\] |
| \[Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\log \,\,\cosh }{h\,\,\log (1+{{h}^{2}})}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\log \,\,\cosh }{\log (1+{{h}^{2}})}\left( \frac{0}{0}form \right)\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-tan\,h}{2h/(1+{{h}^{2}})}=\frac{-1}{2}\] |
| Since \[Lf'(0)=Rf'(0),\] therefore \[f(x)\] is differentiable at \[x=0\] |
| Since differentiability \[\Rightarrow \]continuity, therefore f(x) is continuous at x = 0. |
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