A) \[f\]and\[f\] are continuous for \[x+1>0\]
B) \[f\]is continuous but \[f\] is not continuous for \[x+1>0\]
C) \[f\] and\[f\] are not continuous at \[x=0\]
D) \[f\]is continuous at \[x=0\]but\[f\] is not so
Correct Answer: A
Solution :
Given function can be denned as, \[f(x)=\left\{ \begin{matrix} -1/2({{x}^{2}}-1),-1\le x<0 \\ 1/2({{x}^{2}}+1),x\ge 0\,\,\,\,\,\,\, \\ \end{matrix} \right.\] \[\Rightarrow \] \[f(x)\left\{ \,\,\,\,\,\begin{matrix} -x,-1<x<0\,\,\,\,\,\,\, \\ 0,x=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ x,x>0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{matrix} \right.\] At\[x=0f({{0}^{-}})=\frac{1}{2},f({{0}^{+}})=\frac{1}{2},f(0)=\frac{1}{2}\] \[\therefore \] \[f\] is continuous at \[x=0.\] For \[f,f({{0}^{-}}),f(0),f({{0}^{+}})=0\] \[\therefore \]\[f\] is also continuous at \[x=0.\] Thus, both \[f\] and\[f\] are continuous at \[x=0.\] Hence, both are continuous for \[x>-1.\]
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