A) differentiable everywhere in R
B) except\[x=1\]and\[x=2\]differentiable everywhere in R
C) not continuous at\[x=1\]and\[x=2\]
D) increasing in\[R\]
Correct Answer: B
Solution :
Given, \[f(x)=|x-1|+|x-2|\] \[=\left\{ \begin{matrix} x-1+x-2,x\ge 2 \\ x-1+2-x,1\le x<2 \\ 1-x+2-x,x<1 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} 2x-3 & ,x\ge 2 \\ 1 & ,1\le x<2 \\ 3-2x & ,x\le 1 \\ \end{matrix} \right.\] \[\Rightarrow \] \[f'(x)=\left\{ \begin{matrix} 2 & ,x\le 2 \\ 0 & ,1\le x<2 \\ -1 & ,x\le 1 \\ \end{matrix} \right.\] Hence, except\[x=1\]and\[x=2,\text{ }f(x)\]is differentiable everywhere in R. Alternate method: We know\[|x|\]is not differentiable at\[x=0\]. So,\[|x-1|\]and\[|x-2|\]is not differentiable at\[1\]and 2. Hence,\[f(x)\]is differentiable everywhere except at\[x=1\]and\[x=2\].
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