Question

If $\operatorname{sgn}(x)$ denotes signum function then, no. of points of discontinuities of $\operatorname{sgn}\left(x^{3}-x\right)$ is discontinuous at $x=$
A.3
B.1
C.0
D.2

Answer 1

Hint: A continuity equation is the mathematical way to express this kind of statement.
- The function is expressed at $x=a$.
- The limit of the function as the approaching of $x$ takes place, a exists.
- The limit of the function as the approaching of $\mathrm{x}$ takes place, a is equal to the function value $f(a)$

Complete step-by-step answer:
In Maths, a function $f(x)$ is said to be discontinuous at a point 'a' of its domain $D$ if it is not continuous there. The point 'a' is then called a joint of discontinuity of the function. In limits and continuity, we must have learned a continuous function can be traced without lifting the end on the graph. The discontinuity may arise due to any of the following situation:
1. The right-hand limit or the left-hand limit or both of a function
may not exist.
2. The right-hand limit and the left-hand limit of function may exist but are unequal.
3. The right-hand limit, as well as the left-hand limit of a function, may exist, but either of the two or both may not be equal to $\mathrm{f}(\mathrm{a})$.
$\operatorname{sgn}\left(x^{3}-x\right)=\dfrac{\left|x^{3}-x\right|}{x^{3}-x}$
Now it can be written as:
$\operatorname{sgn}\left(\mathrm{x}^{3}-\mathrm{x}\right)=-1 \quad$ if $\quad \mathrm{x}^{3}-\mathrm{x}<0$
$=0 \quad$ if $\quad x^{3}-x=0$
$=1 \quad$ if $\quad x^{3}-x>0$
So, the function is discontinuous at those points where $x^{3}-x=$
$\mathrm{x}^{3}-\mathrm{x}=0$
Or, $\mathrm{x}\left(\mathrm{x}^{2}-1\right)=0$
Or, $x(x+1)(x-1)=0$
Therefore, $\mathrm{x}=0,1,-1$
Hence the correct option is A.

Note: The common applications of continuity equation are used in pipes, tubes and ducts with flowing fluids or gases, rivers, overall procedure as diaries, power plants, roads, logistics in general, computer networks and semiconductor technologies and some other fields. A function is continuous if it is defined for all values, and equal to the limit at that point for all values (in other words, there are no undefined points, holes, or jumps in the graph.)
1. $f(c)$ must be defined. The function must exist at an $x$ value (c), which means we can't have a hole in the function (such as a 0 in the denominator).
2. The limit of the function as x approaches the value c must exist.
3. The function's value at $c$ and the limit as x approaches c must be the same.