Question

Let f(x) = sgn(x).sin(x), where sgn(x) is the signum of ‘x’. Which of the following is incorrect?
f(x) is continuous everywhere.
f(x) is an even function
f(x) is non-periodic.
f(x) is differentiable for all x except x = 0.
f(x) is non-monotonic.

Answer 1

Hint: In the given question, we have been asked to find the statement which is correct and it is given that\[f\left( x \right)=sgn x.\sin x\]. In order to solve the question, first we know the values of the signum function of x at different intervals. Later we put these values in the given function and find the value of f(x). Then we can infer from the answer that the function f(x) = 0 at x = 0.

Complete step by step answer:
We have given that,
\[f\left( x \right)=sgn x.\sin x\]
As we know that sgn(x) is the signum function of ‘x’.
\[sgn (x)=\left\{ \begin{matrix}
   1\ ,x>0 \\
   0\ ,x=0 \\
   -1\ ,x<0 \\
\end{matrix} \right.\]
Therefore,
Putting the value of sgn(x) in the above given equation;
When x > 0, so sgn(x) = 1
\[f\left( x \right)=sgn x.\sin x=1.\sin x=\sin x\]
When x < 0, so sgn(x) = -1
\[f\left( x \right)=sgn x.\sin x=-1.\sin x=-\sin x\]
When x = 0, so sgn(x) = 0
\[f\left( x \right)=sgn x.\sin x=0.\sin x=0\]
\[f\left( x \right)=\left\{ \begin{matrix}
   -\sin x\ ,x<0 \\
   0\ ,x=0 \\
   \sin x\ ,x>0 \\
\end{matrix} \right.\ \]
Thus,
The function f(x) is continuous for all the values of ‘x’ and we can see below that the graph of the given function is symmetric to y-axis.

Graph is not continuous for the value of x = 0.
Thus the given function is not differentiable at x equal to 0.
Therefore,The option (d) i.e. f(x) is differentiable for all x except x = 0 is the correct answer.

Note: In order to answer the question, students need to know about the concept of the signum function of ‘x’ as otherwise if they didn’t know they would not be able to solve the question. The signum function represents the derivation of the absolute value of function. The range of the signum function is always -1, 1 and 0.