A) Continuous at \[x=-1\]
B) Continuous at \[x=0\]
C) Discontinuous at \[x=\frac{1}{2}\]
D) All of these
Correct Answer: D
Solution :
\[f(x)=\left\{ \begin{align} & \frac{1-|x|}{1+x},\,x\ne -1 \\ & 1\,\,\,\,\,\,\,\,\,\,\,,\,\,x=-1 \\ \end{align} \right.\] and \[f(x)=\left\{ \begin{align} & 1\,\,\,\,\,\,\,\,\,,x<0 \\ & \frac{1-x}{1+x}\,,x\ge 0 \\ \end{align} \right.\] \[f(2x)=\left\{ \begin{align} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,x<0 \\ & \frac{1-[2x]}{1+[2x]},\,\,x>0 \\ \end{align} \right.\]Þ \[f(2x)=\left\{ \begin{align} & 1\,\,\,\,\,\,\,\,\,\,,\,\,x<0 \\ & 1\,\,\,\,\,\,\,\,\,\,,\,\,0\le x<\frac{1}{2} \\ & 0\,\,\,\,\,\,\,\,\,\,,\,\,\frac{1}{2}\le x\le 1 \\ & -\frac{1}{3}\,\,\,\,\,,\,\,\,1\le x<\frac{3}{2} \\ \end{align} \right.\] Þ \[f(x)\], for all values of x where \[x<\frac{1}{2}\]a continous function and for \[x=\frac{1}{2}\] and \[x=1\] \[f(x)\]be a discontinous function.
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