We analyze the piecewise function carefully on each interval and then check continuity and differentiability at all possible problematic points.

Given

where $[t]$ is the greatest integer function.

1. Simplify each branch

1.1 For $x\in(-2,-1)$

If $x\in(-2,-1)$, then $[x]=-2$. So $x-[x]=x+2 \in (0,1).$ Hence on $(-2,-1)$, $f(x)=\frac{\sin(x+2)}{x+2}.$ This is continuous and differentiable throughout $(-2,-1)$ because $x+2\neq 0$ there.

1.2 For $|x|<1$

Then $x\in(-1,1)$. Since $|x|<1$, we have $[|x|]=0$. So $f(x)=\max\{2x,0\}, \qquad -1<x<1.$ Thus

More precisely:

• for $x<0$, $2x<0$, so $\max\{2x,0\}=0$;
• for $x>0$, $2x>0$, so $\max\{2x,0\}=2x$;
• at $x=0$, both are $0$, so $f(0)=0$.

This is continuous on $(-1,1)$, but at $x=0$ we should check differentiability.

1.3 Otherwise

The “otherwise” region means all $x$ not in $(-2,-1)$ and not in $(-1,1)$. So on the remaining set, $f(x)=1.$ In particular, this includes $x=-2,-1,1$ and all $x\le -2$ and $x\ge 1$.

2. Candidate points for discontinuity / non-differentiability

Inside each open interval, the formula is smooth except possibly at $x=0$ due to the max-function. So only boundary/junction points need checking: $x=-2,-1,0,1.$

3. Continuity check

3.1 At $x=-2$

We have $f(-2)=1$ because $x=-2$ falls in the “otherwise” case.

Right-hand limit from $(-2,-1)$: $\lim_{x\to -2^+} \frac{\sin(x+2)}{x+2}=1.$ Left-hand limit from $x<-2$: since $f(x)=1$ there, $\lim_{x\to -2^-} f(x)=1.$ Thus both one-sided limits equal $f(-2)=1$.

So $f$ is continuous at $x=-2$.

3.2 At $x=-1$

We have $f(-1)=1$ ("otherwise" case).

Left-hand limit from $(-2,-1)$: $\lim_{x\to -1^-} \frac{\sin(x+2)}{x+2}=\frac{\sin 1}{1}=\sin 1.$ Right-hand limit from $(-1,1)$: for $-1<x<0$, $f(x)=0$, so $\lim_{x\to -1^+} f(x)=0.$ Since $\sin 1 \neq 0,$ and left and right limits are unequal, $f$ is not continuous at $x=-1$.

3.3 At $x=0$

From the simplified form:

• for $x<0$, $f(x)=0$;
• for $x>0$, $f(x)=2x$;
• and $f(0)=0$.

So $\lim_{x\to 0^-} f(x)=0, \qquad \lim_{x\to 0^+} f(x)=0,$ and both equal $f(0)=0$.

Hence $f$ is continuous at $x=0$.

3.4 At $x=1$

We have $f(1)=1$ ("otherwise").

Left-hand limit from $(-1,1)$: for $0<x<1$, $f(x)=2x$, so $\lim_{x\to 1^-} f(x)=2.$ Right-hand limit from $x>1$: $f(x)=1$, so $\lim_{x\to 1^+} f(x)=1.$ They are unequal, so $f$ is not continuous at $x=1$.

4. Number of discontinuity points

Discontinuity occurs at $x=-1,\ 1.$ So $m=2.$

5. Differentiability check

A function not continuous at a point cannot be differentiable there, so immediately:

• not differentiable at $x=-1$,
• not differentiable at $x=1$.

Now check the remaining candidate points.

5.1 At $x=-2$

On the left of $-2$, $f(x)=1$, so left derivative is $0$. On the right, $f(x)=\frac{\sin(x+2)}{x+2}.$ Let $t=x+2$. Then as $x\to -2^+$, $t\to 0^+$ and $f(x)=\frac{\sin t}{t}.$ Thus right derivative at $-2$ is