Let

f\left(x\right)=\{\begin{array}{cc}\frac{\left|\sin x\right|}{x}& \text{, x ≠ 0}\\ 1& \text{, x = 0}\end{array}

Total number of critical points in $x\in (-2\pi ,2\pi )\; is$

The correct option is 5.

A critical point of a function $f\left(x\right)$ is a point in its domain where either the derivative is zero ($f\text{'}\left(x\right)=0$) or the derivative does not exist.

Let the given function be:
$f\left(x\right)=\{\begin{array}{cc}\frac{\left|\sin x\right|}{x}& ,x\ne 0\\ 1& ,x=0\end{array}$
We need to find the number of critical points in the interval $x\in (-2\pi ,2\pi )$.

Step 1: Check differentiability at $x=0$
Let's find the left-hand limit and right-hand limit of $f\left(x\right)$ as $x\to 0$:
- For $x\to 0^{+}$, we have $\sin x>0$, so:
$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x}$
- For $x\to 0^{-}$, we have $\sin x<0$, so:
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{-\sin x}{x}$
Since the left-hand limit is not equal to the right-hand limit, $f\left(x\right)$ is discontinuous at $x=0$. Therefore, the derivative $f\text{'}\left(0\right)$ does not exist, making $x=0$ the first critical point.

Step 2: Check differentiability where $\left|\sin x\right|$ changes sign
In the interval $x\in (-2\pi ,2\pi )$ excluding $x=0$, $\sin x=0$ at:
$x=-\pi$ and $x=\pi$
At these points, the function is continuous, but the left-hand derivative and right-hand derivative are unequal due to the sharp change in sign of the absolute value function. Thus, the derivative does not exist at these points.
This gives us two more critical points: $x=-\pi$ and $x=\pi$.

Step 3: Find points where $f\text{'}\left(x\right)=0$
Let us analyze the sub-intervals where $f\left(x\right)$ is differentiable:
1. For $x\in (0,\pi )$:
$f\left(x\right)=\frac{\sin x}{x}\Rightarrow f\text{'}\left(x\right)=\frac{x\cos x-\sin x}{x^2}$
Setting $f\text{'}\left(x\right)=0$ yields:
$\tan x=x$
Since $\tan x>x$ for $x\in (0,\frac{\pi }{2})$ and $\tan x<0<x$ for $x\in [\frac{\pi }{2},\pi )$, there is no solution in this interval.

2. For $x\in (\pi ,2\pi )$:
$f\left(x\right)=-\frac{\sin x}{x}\Rightarrow f\text{'}\left(x\right)=-\frac{x\cos x-\sin x}{x^2}$
Setting $f\text{'}\left(x\right)=0$ gives:
$\tan x=x$
In the interval $x\in (\pi ,2\pi )$, the graph of $y=\tan x$ intersects $y=x$ at exactly one point in the third quadrant (specifically in the range $(\pi ,\frac{3\pi }{2})$). This gives one critical point.

3. For $x\in (-\pi ,0)$:
$f\left(x\right)=-\frac{\sin x}{x}\Rightarrow f\text{'}\left(x\right)=0\Rightarrow \tan x=x$
There is no solution in this interval.

4. For $x\in (-2\pi ,-\pi )$:
$f\left(x\right)=\frac{\sin x}{x}\Rightarrow f\text{'}\left(x\right)=0\Rightarrow \tan x=x$
Similarly, the curves $y=\tan x$ and $y=x$ intersect at exactly one point in the range $(-\frac{3\pi }{2},-\pi )$. This gives one more critical point.

Conclusion:
Summing up all the identified critical points:
- Points where derivative does not exist: $x=0,-\pi ,\pi$ (3 points)
- Points where $f\text{'}\left(x\right)=0$: 1 point in $(\pi ,2\pi )$ and 1 point in $(-2\pi ,-\pi )$ (2 points)
Total number of critical points = $3+2=5$.