Let f: R rightarrow R be defined as f(x) = be | vedclass

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(B) To check for continuity,we examine the points $x = 0, 1, 2$.At $x = 0$:$f(0^-) = \lim_{x 	o 0^-} [e^x] = 0$ (since $e^x < 1$ for $x < 0$).$f(0^+)

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

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(B) To check for continuity,we examine the points $x = 0, 1, 2$.At $x = 0$:$f(0^-) = \lim_{x 	o 0^-} [e^x] = 0$ (since $e^x $f(0^+) = a e^0 + [0 - 1] = a - 1$.For continuity at $x = 0$,$a - 1 = 0 \implies a = 1$.At $x = 1$:$f(1^-) = a e^1 + [1 - 1] = a e + 0 = a e$.$f(1^+) = b + [\sin(\pi)] = b + 0 = b$.Since $a = 1$,$f(1^-) = e \approx 2.718$ and $f(1^+) = b$. Since $e$ is not an integer,$f$ is discontinuous at $x = 1$ for any $b$.At $x = 2$:$f(2^-) = b + [\sin(2\pi)] = b + 0 = b$.$f(2^+) = [e^{-2}] - c = 0 - c = -c$.For continuity at $x = 2$,$b = -c \implies b + c = 0$.Thus,$f$ is discontinuous at $x = 1$ regardless of $a, b, c$. If we set $a = 1$ and $b + c = 0$,$f$ is discontinuous only at $x = 1$. In this case,$a + b + c = 1 + (b + c) = 1 + 0 = 1$.

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