A) Equal to 0
B) Equal to 1
C) Equal to ?1
D) Indeterminate
Correct Answer: A
Solution :
If f is continuous at \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,{f}'(0+h)=\underset{h\to 0}{\mathop{\lim }}\,2(0+h)=0\], then \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)\]\[\Rightarrow \,f(0)=\,\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\] \[k=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \frac{\pi }{2}\,[0-h]}{[0-h]}\] \[k=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \frac{\pi }{2}\,[-h]}{[-h]}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \frac{\pi }{2}\,[-h-1]}{[-h-1]}\] \[k=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \,\left( -\frac{\pi }{2} \right)}{-1}\]; \[k=0\].You need to login to perform this action.
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