A) \[1\]
B) \[2\]
C) \[3\]
D) \[0\]
Correct Answer: B
Solution :
Given function can be rewritten as \[f(x)=\left\{ \begin{matrix} 6-3x, & x<1 \\ 4-x, & 1\le x<2 \\ x, & 2\le x<3 \\ 3x-6, & x\ge 3 \\ \end{matrix} \right.\] \[\Rightarrow \] \[f'(x)=\left\{ \begin{matrix} -3, & x<1 \\ -1, & 1<x<2 \\ 1, & 2<x<3 \\ 3, & x>3 \\ \end{matrix} \right.\] Thus, \[f(x)\] is increasing for \[x<2\] and increasing for\[x>2\]. Hence, \[f(x)\] is minimum for\[x=2\] and \[{{[f(x)]}_{\min }}=f(2)=2\]
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