A) Continuous at all x, \[0\le x\le 2\]and differentiable at all x, except \[2/3\]in the interval [0,2]
B) Continuous and differentiable at all x in [0,2]
C) Not continuous at any point in [0,2]
D) Not differentiable at any point [0,2]
Correct Answer: A
Solution :
\[f(x)=\left\{ \begin{array}{*{35}{l}} x\text{ ,} & 0\le x\le 1 \\ 1\text{ ,} & 1<x\le 2 \\ \end{array} \right.\] \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(1-h)\]\[=\underset{h\to 0}{\mathop{\lim }}\,\,\,(1-h)=1\] \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(1+h)=1\] Hence function is continuous in (0, 2). Now \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,(0+h)=0=f(0)\] \[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,(2-h)=1=f(2)\] Hence function is continuous in [0, 2] Clearly, from graph it is not differentiable at \[x=1.\]You need to login to perform this action.
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