We analyze continuity of the piecewise function on each interval and at the junction points.

Given

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

1. Continuity inside each interval

(i) For $x<0$

Here $f(x)=[e^x]$. Since $x<0$, we have $0<e^x<1$, so $[e^x]=0 \quad (x<0).$ Thus on $(-\infty,0)$, $f(x)=0$, which is continuous.

(ii) For $0\le x<1$

Here $f(x)=ae^x+[x-1].$ For $0\le x<1$, we have $-1\le x-1<0$, so $[x-1]=-1 \quad (0\le x<1).$ Hence $f(x)=ae^x-1,$ which is continuous on $[0,1)$.

(iii) For $1\le x<2$

Here $f(x)=b+[\sin(\pi x)].$ Now for $1<x<2$, we have $\pi x\in(\pi,2\pi)$, so $\sin(\pi x)<0$ except at the endpoints. Hence:

• for $x=1$, $\sin(\pi)=0$, so $[\sin(\pi)]=0$;
• for $1<x<2$, $-1<\sin(\pi x)<0$, so $[\sin(\pi x)]=-1$.

Therefore, $f(1)=b, \qquad f(x)=b-1 \text{ for }1<x<2.$ So there is always a jump at $x=1$ from the right side of this piece itself.

(iv) For $x\ge 2$

Here $f(x)=[e^{-x}]-c.$ Since $x\ge 2$, we have $0<e^{-x}\le e^{-2}<1$, so $[e^{-x}]=0.$ Thus $f(x)=-c \quad (x\ge 2),$ which is continuous on $[2,\infty)$.

2. Possible discontinuity points

Since each piece is continuous in its own interval, discontinuities can occur only at the joining points: $x=0,\;1,\;2.$

3. Check continuity at $x=0$

Left limit at $0$

For $x<0$, $f(x)=0$, so $\lim_{x\to 0^-}f(x)=0.$

Right limit and value at $0$

For $0\le x<1$, $f(x)=ae^x-1$, so $f(0)=a-1, \qquad \lim_{x\to 0^+}f(x)=a-1.$

For continuity at $x=0$, $0=a-1 \implies a=1.$

So:

• continuous at $0$ iff $a=1$,
• otherwise discontinuous at $0$.

4. Check continuity at $x=1$

For $0\le x<1$, $f(x)=ae^x-1$, so $\lim_{x\to 1^-}f(x)=ae-1.$

At $x=1$, $f(1)=b+[\sin \pi]=b+0=b.$

For $1<x<2$, $f(x)=b-1$, so $\lim_{x\to 1^+}f(x)=b-1.$

For continuity at $x=1$, we need $ae-1=b=b-1,$ which is impossible because $b\ne b-1.$

Hence $f$ is always discontinuous at $x=1$, for every choice of $a,b,c$.

5. Check continuity at $x=2$

For $1<x<2$, $f(x)=b-1$, so $\lim_{x\to 2^-}f(x)=b-1.$

At $x=2$, $f(2)=[e^{-2}]-c=0-c=-c.$ Also, $\lim_{x\to 2^+}f(x)=-c.$

For continuity at $x=2$, $b-1=-c \implies b+c=1.$

So:

• continuous at $2$ iff $b+c=1$,
• otherwise discontinuous at $2$.

6. Count discontinuities

We have established:

• $x=1$ is always a discontinuity.
• $x=0$ is continuous iff $a=1$.
• $x=2$ is continuous iff $b+c=1$.

Therefore, to have exactly one discontinuity, we must make $x=0$ and $x=2$ continuous: $a=1, \qquad b+c=1.$ Then $a+b+c=1+(b+c)=1+1=2.$ So if $f$ is discontinuous at exactly one point, then $a+b+c=2 \ne 1.$

Thus statement C is true.

7. Check all options

A: There exists $a,b,c$ such that $f$ is continuous on $\mathbb R$

False, because $x=1$ is always a discontinuity.

B: If $f$ is discontinuous at exactly one point, then $a+b+c=1$

False, because exactly one discontinuity gives $a+b+c=2$.

C: If $f$ is discontinuous at exactly one point, then $a+b+c\ne 1$

True, since actually $a+b+c=2$.

D: $f$ is discontinuous at at least two points, for any values of $a,b,c$

False, because choosing $a=1$ and $b+c=1$ makes discontinuity only at $x=1$.

Final Answer

The correct option is $\boxed{\text{C}}$