Let $$f: [-1, 3] \rightarrow R$$ be defined as
$$f(x) = \begin{cases} x + x, & -1 \le x < 1 \\ x + x, & 1 \le x < 2 \\ x + x, & 2 \le x \le 3 \end{cases}$$
where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at:

We have a piece-wise definition of $$f$$ on the closed interval $$[-1,3]$$:

$$ f(x)= \begin{cases} |x|+[x], & -1\le x<1,\\[4pt] x+|x|, & 1\le x<2,\\[4pt] x+[x], & 2\le x\le 3. \end{cases} $$

Possible points of discontinuity arise

(i) at the internal junctions $$x=1$$ and $$x=2$$ (because the formula itself changes), and

(ii) at those interior points where the greatest-integer function $$[x]$$ jumps, namely at every integer. Inside our domain these integers are $$-1,0,1,2,3$$. The point $$-1$$ is the left boundary and $$3$$ is the right boundary; the integers $$0,1,2$$ lie inside.

We now examine each of these candidates one by one, computing the left-hand limit (LHL), the right-hand limit (RHL) and the actual value of the function.

1. Point $$x=0$$ (lies in the first formula)

For $$x\to0^-$$ we have $$x<0$$, so $$|x|=-x$$ and $$[x]=-1$$. Hence $$f(x)=-x-1$$ and the LHL is $$ \lim_{x\to0^-}(-x-1)= -0-1=-1. $$

For $$x\to0^+$$ we have $$x>0$$, so $$|x|=x$$ and $$[x]=0$$. Hence $$f(x)=x$$ and the RHL is $$ \lim_{x\to0^+}x=0. $$

The function value at $$x=0$$ is obtained from the same first formula: $$ f(0)=|0|+[0]=0+0=0. $$

Since $$\text{LHL}=-1\neq0=\text{RHL}=\!f(0)$$, we get a jump. Therefore $$f$$ is discontinuous at $$x=0$$.

2. Point $$x=1$$ (junction of the first and second formula)

LHL: For $$x\to1^-$$ we are still in the first formula. Here $$x>0$$, so $$|x|=x$$ and $$[x]=0$$. Thus $$f(x)=x+0=x$$, giving $$ \lim_{x\to1^-}f(x)=1. $$

RHL: For $$x\to1^+$$ we use the second formula. Again $$|x|=x$$, hence $$f(x)=x+|x|=x+x=2x$$, so $$ \lim_{x\to1^+}f(x)=2\cdot1=2. $$

Function value at $$x=1$$ is governed by the second piece (since $$1\le x<2$$): $$ f(1)=1+|1|=1+1=2. $$

Because $$\text{LHL}=1\neq2=\text{RHL}=f(1)$$, a jump occurs. Hence $$f$$ is discontinuous at $$x=1$$.

3. Point $$x=2$$ (junction of the second and third formula)

LHL: For $$x\to2^-$$ (still in the second piece) we have $$|x|=x$$, so $$ f(x)=x+|x|=x+x=2x,\qquad \lim_{x\to2^-}f(x)=2\cdot2=4. $$

RHL: For $$x\to2^+$$ we use the third piece. In the neighbourhood $$2<x<3$$ we have $$[x]=2$$, giving $$ f(x)=x+2,\qquad \lim_{x\to2^+}(x+2)=2+2=4. $$

Value at $$x=2$$ (third formula because $$2\le x$$): $$ f(2)=2+[2]=2+2=4. $$

Since $$\text{LHL}=4=\text{RHL}=f(2)$$, the function is continuous at $$x=2$$.

4. Point $$x=3$$ (right end-point of the domain)

We check the left limit because no points lie to the right.

For $$x\to3^-$$ we are in the third piece with $$[x]=2$$, so $$ f(x)=x+2,\qquad \lim_{x\to3^-}(x+2)=3+2=5. $$

The actual value at the boundary is $$ f(3)=3+[3]=3+3=6. $$

The left limit $$5$$ does not equal the function value $$6$$, so $$f$$ is discontinuous at $$x=3$$.

5. Point $$x=-1$$ (left end-point)

Only the right limit matters. For $$x\to-1^+$$ we use the first piece; here $$|x|=-x$$ and $$[x]=-1$$, so $$ f(x)=-x-1,\qquad \lim_{x\to-1^+}(-x-1)=1-1=0. $$

The function value at the boundary is $$ f(-1)=|{-1}|+[{-1}]=1+(-1)=0. $$

Right limit equals the function value, therefore $$f$$ is continuous at $$x=-1$$.

Collecting the results, discontinuities occur exactly at

$$x=0,\;x=1,\;x=3.$$

Thus there are only three points of discontinuity in the interval $$[-1,3]$$.

Hence, the correct answer is Option D.