A) 0
B) 1
C) 2
D) Does not exist
Correct Answer: D
Solution :
\[L\,{f}'\,(1)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(1-h)-f(1)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{m\,{{(1-h)}^{2}}-m}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{m\,[1+{{h}^{2}}-2h-1]}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,m\,(2-h)=2m\] and \[R\,{f}'\,(1)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f\,(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{2\,(1+h)-m}{h}\]. For differentiability, \[L\,{f}'(1)=R\,{f}'\,(1)\] . But for any value of \[m,\,\,R\,\,{f}'\,(1)=L\,{f}'(1)\] not possible.
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