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The function $f\left( x \right)=2\ln \left| x \right|-x\left| x \right|$ is increasing on the interval:
(a) (0, 1)
(b) $\left( 0,\infty \right)$
(c) (-1, 1)
(d) (-1, 0)

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Last updated date: 21st Jun 2024
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Answer
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Hint: First, before proceeding for this, we must know that for calculating the increasing range of any function f(x), we have the condition as ${f}'\left( x \right)>0$. Then, by using the chain rule and product rule as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, we get the derivative of the given function. Then, by using the condition for the increasing function and by using options, we get the desired range.

Complete step-by-step solution
In this question, we are supposed to find the interval for which the function $f\left( x \right)=2\ln \left| x \right|-x\left| x \right|$ is increasing.
So, before proceeding for this, we must know that for calculating the increasing range of any function f(x), we have the condition as:
${f}'\left( x \right) > 0$
Now, by using the chain rule and product rule as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, we get the derivative of the given function as:
$\begin{align}
  & {f}'\left( x \right)=2\dfrac{d}{dx}\ln \left| x \right|-\dfrac{d}{dx}\left( x\left| x \right| \right) \\
 & \Rightarrow {f}'\left( x \right)=2\times \dfrac{1}{\left| x \right|}\dfrac{d}{dx}\left| x \right|-\left( x\dfrac{d}{dx}\left| x \right|+\left| x \right|\dfrac{d}{dx}x \right) \\
 & \Rightarrow {f}'\left( x \right)=2\times \dfrac{1}{\left| x \right|}\times \dfrac{x}{\left| x \right|}-\left( x\times \dfrac{x}{\left| x \right|}+\left| x \right|\times 1 \right) \\
 & \Rightarrow {f}'\left( x \right)=\dfrac{2x}{{{\left| x \right|}^{2}}}-\left( \dfrac{{{x}^{2}}}{\left| x \right|}+\left| x \right| \right) \\
 & \Rightarrow {f}'\left( x \right)=\dfrac{2x}{{{\left| x \right|}^{2}}}-\dfrac{{{x}^{2}}}{\left| x \right|}-\left| x \right| \\
\end{align}$
Now, by using the condition for the increasing function, we get:
$\dfrac{2x}{{{\left| x \right|}^{2}}}-\dfrac{{{x}^{2}}}{\left| x \right|}-\left| x \right|>0$
Then, by solving the above expression, we get the range of x where function is increasing as:
$\begin{align}
  & \dfrac{2x-{{x}^{2}}\left| x \right|-{{\left| x \right|}^{3}}}{{{\left| x \right|}^{2}}}>0 \\
 & \Rightarrow 2x-{{x}^{2}}\left| x \right|-{{\left| x \right|}^{3}}>0 \\
\end{align}$
Now, we have to check the condition for each option by taking the value from that range as starting for the first option (0, 1) and test for the value of x as 0.5, we get:
$\begin{align}
  & 2\left( 0.5 \right)-{{\left( 0.5 \right)}^{2}}\left| 0.5 \right|-{{\left| 0.5 \right|}^{3}} \\
 & \Rightarrow 1-0.125-0.125 \\
 & \Rightarrow 0.75 \\
\end{align}$
So, it gives the values as positive which is greater than 0, so this option is correct.
Now, we have to check the condition for second option $\left( 0,\infty \right)$ and test for the value of x as 1, we get:
$\begin{align}
  & 2\left( 1 \right)-{{\left( 1 \right)}^{2}}\left| 1 \right|-{{\left| 1 \right|}^{3}} \\
 & \Rightarrow 2-1-1 \\
 & \Rightarrow 0 \\
\end{align}$
So, it gives the value as negative which is less than 0, so this option is incorrect.
Now, we have to check the condition for second option (-1, 1) and test for the value of x as 0, we get:
$\begin{align}
  & 2\left( 0 \right)-{{\left( 0 \right)}^{2}}\left| 0 \right|-{{\left| 0 \right|}^{3}} \\
 & \Rightarrow 0-0-0 \\
 & \Rightarrow 0 \\
\end{align}$
So, it gives the value as negative which is less than 0, so this option is incorrect.
Now, we have to check the condition for second option (-1, 0) and test for the value of x as -0.5, we get:
$\begin{align}
  & 2\left( -0.5 \right)-{{\left( -0.5 \right)}^{2}}\left| -0.5 \right|-{{\left| -0.5 \right|}^{3}} \\
 & \Rightarrow -1-0.125-0.125 \\
 & \Rightarrow -1.25 \\
\end{align}$
So, it gives the value as negative which is less than 0, so this option is incorrect.
Hence, option (a) is correct.

Note: Now, to solve these types of the questions we need to know some of the basics of differentiation so that we can get the answer easily. So, the basic formula required for this question is as:
$\dfrac{d}{dx}\left| x \right|=\dfrac{x}{\left| x \right|}$