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(B) For $f$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$\lim_{x 	o 0^-} f(x) = \sin(0) - e^0 = 0

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$3$

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(B) For $f$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$\lim_{x 	o 0^-} f(x) = \sin(0) - e^0 = 0 - 1 = -1$.$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (a + [-x]) = a + (-1) = a - 1$.Equating these,$a - 1 = -1 \implies a = 0$.For $f$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (a + [-x]) = a + (-1) = a - 1$.$\lim_{x 	o 1^+} f(x) = 2(1) - b = 2 - b$.Equating these,$a - 1 = 2 - b \implies 0 - 1 = 2 - b \implies b = 3$.Therefore,$a + b = 0 + 3 = 3$.

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