Answer
414.9k+ 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:
$\dfrac{d}{dx}\left| x \right|=\dfrac{x}{\left| x \right|}$
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 many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)