Answer

Verified

404.1k+ views

**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:

$\dfrac{d}{dx}\left| x \right|=\dfrac{x}{\left| x \right|}$

Recently Updated Pages

How do you find slope point slope slope intercept standard class 12 maths CBSE

How do you find B1 We know that B2B+2I3 class 12 maths CBSE

How do you integrate int dfracxsqrt x2 + 9 dx class 12 maths CBSE

How do you integrate int left dfracx2 1x + 1 right class 12 maths CBSE

How do you find the critical points of yx2sin x on class 12 maths CBSE

How do you find the general solution to dfracdydx class 12 maths CBSE

Trending doubts

The provincial president of the constituent assembly class 11 social science CBSE

Gersoppa waterfall is located in AGuyana BUganda C class 9 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The hundru falls is in A Chota Nagpur Plateau B Calcutta class 8 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE