A) \[f(x)=\left\{ \begin{matrix} 1+x & , & x\le -1 \\ 2 & , & -1<x<1 \\ 1-x & , & x\ge 1 \\ \end{matrix} \right.\]
B) \[f(x)=\left\{ \begin{matrix} 1-x & , & x\le -1 \\ 2 & , & -1<x<1 \\ 1+x & , & x\ge 1 \\ \end{matrix} \right.\]
C) \[f(x)=\left\{ \begin{matrix} 1-x & , & x\le -1 \\ 1 & , & -1<x<1 \\ 1+x & , & x\ge 1 \\ \end{matrix} \right.\]
D) None of the above
Correct Answer: B
Solution :
Given that, \[f(x)=\max \{(1-x),\,\,2,\,\,(1+x)\}\] For\[x\le -1\], we find that\[1-x\ge 2\] and\[1-x\ge 1+x\] \[\therefore \]\[\max \{(1-x),\,\,2(1+x)\}=1-x\] For\[-1<x<1\], we find that \[0<1-x<2\]and\[0<1+x<2\] \[\therefore \]\[\max \{(1-x),\,\,2,\,\,(1+x)\}=1+x\] For\[x\ge 1\], we find that \[1+x\ge 2,\,\,1+x>1-x\] \[\therefore \]\[\max \{(1-x),\,\,2,\,\,(1+x)\}=1+x\] Hence,\[f(x)=\left\{ \begin{matrix} 1-x, & x\le -1 \\ 2, & -1<x<1 \\ 1+x, & x\ge 1 \\ \end{matrix} \right.\]
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