Question

Function f(x) = | x | − | x − 1 | is monotonically increasing when
(a) x < 0
(b) x > 1
(c) x < 1
(d) 0 < x < 1

Answer 1

(d) 0 < x < 1

$$\begin{gathered} f\left(x\right)=\left|x\right|-\left|x-1\right| \\ \text{Case 1: Let}x<0 \\ \mathrm{If}x<0,\mathrm{t}\text{hen}\left|x\right|=-x \\ \Rightarrow \left|x-1\right|=-\left(x-1\right) \\ \text{Now,} \\ f\left(x\right)=\left|x\right|-\left|x-1\right| \\ =-x-\left(-x+1\right) \\ =-x+x-1 \\ =-1 \\ f\text{'}\left(x\right)=0 \\ \text{So,}f\left(x\right)\text{is not monotonically increasing when}\mathit{\text{x}}\text{< 0}\mathit{\text{.}} \\ \text{Case 2: Let}0\mathit{<}x\mathit{<}1 \\ \mathrm{Here}, \\ \left|x\right|\mathit{=}x \\ \Rightarrow \mathit{}\left|x-1\right|\mathit{=}\mathit{-}\left(x-1\right) \\ \text{Now,} \\ f\left(x\right)=\left|x\right|\mathit{-}\left|x\mathit{-}1\right| \\ =x\mathit{+}x-1 \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}2x-1 \\ f\mathit{\text{'}}\left(x\right)\mathit{=}2>0 \\ \text{So,}f\left(x\right)\mathit{}\text{is monotonically increasing when}\mathit{0}\mathit{<}x\mathit{<}\mathit{1}\mathit{.} \\ \text{Case 3: Let}x\mathit{>}\mathit{1}\mathit{} \\ \mathrm{If}x\mathit{>}0,\mathrm{t}\text{hen}\left|x\right|\mathit{=}x \\ \Rightarrow \mathit{}\left|x\mathit{-}1\right|\mathit{=}\left(x\mathit{-}1\right) \\ \text{Now}\mathit{\text{,}} \\ f\left(x\right)\mathit{=}\left|x\right|\mathit{-}\left|x\mathit{-}\mathit{1}\right| \\ \mathit{=}x\mathit{-}x\mathit{+}\mathit{1} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}1 \\ f\mathit{\text{'}}\left(x\right)=0 \\ \text{So}\mathit{\text{,}}f\left(x\right)\mathit{}\text{is not monotonicallyincreasing when}\mathit{\text{x >}}\text{1}\mathit{\text{.}} \\ \text{Thus,}f\left(x\right)\mathit{}\text{is monotonically increasing when}0\mathit{<}x\mathit{<}1. \end{gathered}$$