A) f is discontinuous at\[x=1\]
B) f is differentiable at \[x=1\]
C) f is continuous but not differentiable at \[x=1\]
D) None of these
Correct Answer: C
Solution :
\[f(x)=\left\{ \begin{align} & \,\,\,x,\,\,\,\,\,\,\,\,0\le x\le 1 \\ & 2x-1,\,\,\,\,\,\,x>1 \\ \end{align} \right.\] \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(1-h)=\underset{h\to 0}{\mathop{\lim }}\,(1-h)=1\] \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(1+h)=\underset{h\to 0}{\mathop{\lim 2}}\,(1+h)-1=1\] \[\because \,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=1\] \ Function is continuous at \[x=1\]. \[Lf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1-h)-f(1)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{(1-h)-1}{-h}=1\] \[Rf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{2+2h-1-1}{h}=2\] \ \[Lf'(1)\ne Rf'(1)\] \ Function is not differentiation at \[x=1\]
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