If m and M are respectively minimum and maximum values of f(x) = |2- |x||, −3 ≤ x ≤ 3, then

The correct option is m = 0 and M = 2.

Let us analyze the function $f\left(x\right)=\left|2-\left|x\right|\right|$ defined on the closed interval $-3\le x\le 3$ to find its minimum value ($m$) and maximum value ($M$).

Step 1: Understand the absolute value components
The variable $x$ lies in the interval $\left[-3,3\right]$. Therefore, the absolute value $\left|x\right|$ ranges from a minimum of 0 to a maximum of 3:
$0\le \left|x\right|\le 3$

Step 2: Find the minimum value ($m$)
Because $f\left(x\right)$ is defined as an absolute value expression, $f\left(x\right)=\left|2-\left|x\right|\right|$, its value can never be negative. Thus, the absolute minimum possible value for $f\left(x\right)$ is 0.
This minimum value of 0 is achieved when the inner term is equal to 0:
$2-\left|x\right|=0\Rightarrow \left|x\right|=2$
Since $x=\pm 2$ lie within the given interval $\left[-3,3\right]$, we have:
$m=0$

Step 3: Find the maximum value ($M$)
Let us evaluate $f\left(x\right)$ at the critical points and the boundaries of the interval $\left[-3,3\right]$:
1. At $x=0$ (where $\left|x\right|$ is minimized):
$f\left(0\right)=\left|2-0\right|=2$
2. At the boundaries $x=\pm 3$ (where $\left|x\right|$ is maximized):
$f\left(\pm 3\right)=\left|2-3\right|=\left|-1\right|=1$
3. At the points where the function is zero, $x=\pm 2$:
$f\left(\pm 2\right)=0$

Comparing these values, the maximum value of $f\left(x\right)$ on the interval $\left[-3,3\right]$ is:
$M=2$

Conclusion:
The minimum value is $m=0$ and the maximum value is $M=2$.