The function f(x) = 2ln ,|x| -x|x| is increasing on the interval - Vidyadip

MCQ

The function $f(x) = 2ln\,|x| -x|x|$ is increasing on the interval

✓
$(0,1)$
B
$(0,\infty)$
C
$(-1,1)$
D
$(-1,0)$

✓

Answer

Correct option: A.

$(0,1)$

a
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{2\ln x - {x^2}}&{{\rm{if}}}&{x > 0}\\
{2\ln \left( { - x} \right) + {x^2}}&{{\rm{if}}}&{x < 0}
\end{array}} \right.$

$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{ll}{\frac{2}{x}-2 x} & {\text { if } x>0} \\ {\frac{2}{x}+2 x} & {\text { if } x<0}\end{array}\right.$

$\therefore f^{\prime}(x)>0$ only for $x \in(0,1)$

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