Question
Consider the following three statements for the function $f:(0,\infty) \rightarrow \mathbb{R}$ defined by
$$f(x) = \left| \log_{e}x \right| - |x - 1|: $$(I) f is differentiable at all $x > 0$.
(II) f is increasing in ( 0,1 ).
(III) f is decreasing in ( $1,\infty$ ).
Then.
Options
Solution
$f(x) = |\mathcal{l}nx| - |x - 1|$
$${= \left\{ \begin{matrix} \mathcal{l}nx - (x - 1) & x \geq 1 \\ - \mathcal{l}nx + (x - 1) & 0 < x < 1 \end{matrix} \right.\ }{= \left\{ \begin{matrix} lnx - x + 1 & x \geq 1 \\ - lnx + x - 1 & 0 < x < 1 \end{matrix} \right.\ }{f'(x) = \left( \begin{matrix} \frac{1}{x} - 1 & x \geq 1 \\ - \frac{1}{x} + 1 & 0 < x < 1 \end{matrix} \right.\ }$$$f'\left( 1^{+} \right) = f'\left( 1^{-} \right) = 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 $\forall x \in (0,\infty)$
Option (4)
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