A) \[-1\]
B) \[1\]
C) \[0\]
D) None of these
Correct Answer: D
Solution :
Given, \[f(x)=\left\{ \begin{matrix} x, & if & 0<x<1 \\ 2-x, & if & 1\le x<2 \\ x-\frac{1}{2}{{x}^{2}}, & if & x=2 \\ \end{matrix} \right.\] At \[x=1,\] \[RHD=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2-(1+h)-(2-1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2-1-h-1}{h}=-1\] \[LHD=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(1-h)-f(1)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{(1-h)-(2-1)}{-h}=1\] \[\therefore \] \[LHD\ne RHD\]
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