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Question
Function f(x) = | x | − | x − 1 | is monotonically increasing when
Options
x < 0
x > 1
x < 1
0 < x < 1
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Solution
0 < x < 1
\[f\left( x \right) = \left| x \right| - \left| x - 1 \right|\]
\[\text { Case 1: Let }x < 0 \]
\[\text { If x < 0 , then }\left| x \right| = - x\]
\[ \Rightarrow \left| x - 1 \right| = - \left( x - 1 \right)\]
\[\text { Now,}\]
\[f\left( x \right) = \left| x \right| - \left| x - 1 \right|\]
\[ = - x - \left( - x + 1 \right)\]
\[ = - x + x - 1\]
\[ = - 1\]
\[f'\left( x \right) = 0\]
\[\text { So,f }\left( x \right) \text { is not monotonically increasing when x< 0.}\]
\[\text { Case 2: Let }0 < x < 1\]
\[\text { Here,} \]
\[\left| x \right| = x\]
\[ \Rightarrow \left| x - 1 \right| = - \left( x - 1 \right)\]
\[\text { Now,}\]
\[f\left( x \right) = \left| x \right| - \left| x - 1 \right|\]
\[ = x + x - 1\]
\[ = 2x - 1\]
\[f'\left( x \right) = 2 > 0\]
\[\text { So },f\left( x \right) \text { is monotonically increasing when }0 < x < 1 . \]
\[\text { Case 3: Let x > 1} \]
\[\text { Ifx > 0, then }\left| x \right| = x\]
\[ \Rightarrow \left| x - 1 \right| = \left( x - 1 \right)\]
\[\text { Now,}\]
\[f\left( x \right) = \left| x \right| - \left| x - 1 \right|\]
\[ = x - x + 1\]
\[ = 1\]
\[f'\left( x \right) = 0\]
\[\text { So },f\left( x \right)\text { is not monotonically increasing when x >1 }.\]
\[\text { Thus },f\left( x \right) \text { is monotonically increasing when 0 < x < 1} . \]
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