A) \[g(x)\] is continuous and derivable in (0,2)
B) \[g(x)\] is discontinuous at finite number of points in (0,2)
C) \[g(x)\] is non-derivable at 2 points
D) \[g(x)\] is continuous but non-derivable at one point
Correct Answer: D
Solution :
\[g(g)=\left\{ \begin{align} & {{x}^{3}}-{{x}^{2}}+x+1\,;\,0\le x\le 1 \\ & 3-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;\,1<x\le 2 \\ \end{align} \right.\] Now, \[g(x)\] is continuous at \[x=1\] but non-derivative at\[x=1\].You need to login to perform this action.
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