Given f(x)=[[ex],aex+[x−1],b+[sin(πx)],[e−x]−c,x<00≤x<11≤x<2x≥2
i.e. f(x)={\begin{matrix}0, & x<0 \\ a{e}^{x}-1, & 0\leq x<1 \\ b, & x=1 \\ b-1, & 1<x<2 \\ -c, & x\geq 2\end{matrix}
Here, we know f(x) is discontinuous at x=1 as b=b−1
For f(x) to be continuous at x=0;a=1
For f(x) to be continuous at x=2;b+c=1
Hence, a+b+c=2