Questions & Answers
Question
Answers
Related Questions
If $f(x)={{x}^{3}}sgn \ (x)$ then.
A. f is differentiable at $x=0$
B. f is continuous but not differentiable at $x=0$
C. $f'\left( {{o}^{-}} \right)=1$
D. none of these
Answer
![Verified]()
Verified
Verified
Hint: The sgn function written in question is called sign function or signum function. It is defined as
$ sgn \left( x \right)=\left\{ \begin{matrix}
-1 & \text{at} & x<0 \\
0 & \text{at} & x=0 \\
1 & \text{at} & x>0 \\
\end{matrix} \right.$
Complete step-by-step answer:
we have a function
$f\left( x \right)={{x}^{3}}sgn \ (x)$
$f\left( x \right)$ can be defined similarly as signum function
\[\Rightarrow f\left( x \right)={{x}^{3}}sgn \left( x \right)=\left\{ \begin{matrix}
-{{x}^{3}} & \text{at} & x<0 \\
0 & \text{at} & x=0 \\
{{x}^{3}} & \text{at} & x>0 \\
\end{matrix} \right.\]
Now the above function may be discontinuous at $x=0$
So let us check its continuity at $x=0$
$\underset{x\to 0}{\mathop{\lim }}\,\ \ f\left( x \right)=$ $\underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( 0-\text{h} \right)=$ $\underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( -\text{h} \right)$
$\Rightarrow \underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( -\text{n} \right)=-{{\text{n}}^{3}}=0$
Similarly we can prove that
$\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$
$\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\ \ f\left( x \right)=\ \ \ \ \ \ \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\ \ f\left( x \right)=\ \ \ \ \ \ f\left( 0 \right)=0$
$\therefore f\left( 0 \right)=0$
Therefore $f\left( x \right)$ is continuous at $x=0$
Now if we differentiate the above function we will get
$f'\left( x \right)=\left\{ \begin{matrix}
-3{{x}^{2}} & \text{at} & x<0 \\
0 & \text{at} & x=0 \\
3{{x}^{2}} & \text{at} & x>0 \\
\end{matrix} \right.$
Now according to the option we need to check its differentiable at $x=0$. For the function to be differentiable at $x=0$ it should satisfy
From above relation we can say that
$f'\left( {{o}^{-}} \right)=f'\left( {{o}^{+}} \right)=o$
$\therefore f\left( x \right)$ is differentiable at $x=0$
Since the function is continuous at $x=0$ and it is also differentiable at $x=0$
$\therefore $ Option B is eliminated similarly $f'\left( {{o}^{-}} \right)=o$, therefore, option C is also eliminated.
$\therefore $ correct option is A
Note: Sign um function re $ sgn \left( x \right)$ is alone not continuous at $x=0$ but when multiplied by ${{x}^{3}}$ the function $f\left( x \right)={{x}^{3}}sgn \left( x \right)$ is continuous at $x=0$ therefore, be careful while checking the continuity of a function when sign um function is involved.
$ sgn \left( x \right)=\left\{ \begin{matrix}
-1 & \text{at} & x<0 \\
0 & \text{at} & x=0 \\
1 & \text{at} & x>0 \\
\end{matrix} \right.$
Complete step-by-step answer:
we have a function
$f\left( x \right)={{x}^{3}}sgn \ (x)$
$f\left( x \right)$ can be defined similarly as signum function
\[\Rightarrow f\left( x \right)={{x}^{3}}sgn \left( x \right)=\left\{ \begin{matrix}
-{{x}^{3}} & \text{at} & x<0 \\
0 & \text{at} & x=0 \\
{{x}^{3}} & \text{at} & x>0 \\
\end{matrix} \right.\]
Now the above function may be discontinuous at $x=0$
So let us check its continuity at $x=0$
$\underset{x\to 0}{\mathop{\lim }}\,\ \ f\left( x \right)=$ $\underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( 0-\text{h} \right)=$ $\underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( -\text{h} \right)$
$\Rightarrow \underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( -\text{n} \right)=-{{\text{n}}^{3}}=0$
Similarly we can prove that
$\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$
$\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\ \ f\left( x \right)=\ \ \ \ \ \ \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\ \ f\left( x \right)=\ \ \ \ \ \ f\left( 0 \right)=0$
$\therefore f\left( 0 \right)=0$
Therefore $f\left( x \right)$ is continuous at $x=0$
Now if we differentiate the above function we will get
$f'\left( x \right)=\left\{ \begin{matrix}
-3{{x}^{2}} & \text{at} & x<0 \\
0 & \text{at} & x=0 \\
3{{x}^{2}} & \text{at} & x>0 \\
\end{matrix} \right.$
Now according to the option we need to check its differentiable at $x=0$. For the function to be differentiable at $x=0$ it should satisfy
From above relation we can say that
$f'\left( {{o}^{-}} \right)=f'\left( {{o}^{+}} \right)=o$
$\therefore f\left( x \right)$ is differentiable at $x=0$
Since the function is continuous at $x=0$ and it is also differentiable at $x=0$
$\therefore $ Option B is eliminated similarly $f'\left( {{o}^{-}} \right)=o$, therefore, option C is also eliminated.
$\therefore $ correct option is A
Note: Sign um function re $ sgn \left( x \right)$ is alone not continuous at $x=0$ but when multiplied by ${{x}^{3}}$ the function $f\left( x \right)={{x}^{3}}sgn \left( x \right)$ is continuous at $x=0$ therefore, be careful while checking the continuity of a function when sign um function is involved.
×
Sorry!, This page is not available for now to bookmark.
Like this
Liked
-
LIVECourses
-
Crash Courses 2021
-
Vedantu Pro
-
Vedantu Assist
-
Class 12
-
-
JEE
-
JEE Main & Advanced 2021
-
JEE Main & Advanced 2022
-
-
NEET
-
NEET 2021
-
NEET 2022
-
-
NDA
-
CBSE
-
Commerce
-
Class 11
-
Class 12
-
-
ICSE
-
Maharashtra Board
-
Class 9
-
Class 10
-
-
Micro Courses
-
-
FREEMaster Classes
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
