A) ? 2
B) 0
C) 2
D) Undefined
Correct Answer: B
Solution :
\[f(x)=|x-1|+|x-3|\] \[f(x)=\left\{ \begin{matrix} -(x-1)-(x-3), & x<1 \\ (x-1)-(x-3), & x>1 \\ (x-1)-(x-3), & x<3 \\ (x-1)+(x-3), & x>3 \\ \end{matrix} \right.\]\[=\left\{ \begin{matrix} 4-2x, & x<1 \\ 2\,\,\,\,\,\,\,\,\,, & 1<x<3 \\ 2x-4, & x>3 \\ \end{matrix} \right.\] At\[x=2\], \[f(x)=\]2. Hence \[\,{f}'(x)=0\].You need to login to perform this action.
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