$ f(x) = |\ln x| - |x - 1| $
$ = \begin{cases} \ln x - (x - 1), & x \ge 1 \ -\ln x + (x - 1), & 0 < x < 1 \end{cases} $
$ = \begin{cases} \ln x - x + 1, & x \ge 1 \ -\ln x + x - 1, & 0 < x < 1 \end{cases} $
$ f'(x) = \begin{cases} \frac{1}{x} - 1, & x \ge 1 \ -\frac{1}{x} + 1, & 0 < x < 1 \end{cases} $
$ f'(1^+) = f'(1^-) = 0 \Rightarrow f(x) $ is differentiable $ \forall x > 0 $
$ f'(x) < 0 ; \forall x > 1 $
$ f'(x) < 0 ; \forall 0 < x < 1 $
$ \Rightarrow f(x) $ is decreasing $ \forall x \in (0, \infty) $
Online Test Series, Information About Examination,
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Online Test Series, Information About Examination,
Syllabus, Notification
and More.