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(B) For $f$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^-} f(x) = f(1) = 1$. First,consider the right-hand limit:

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(B) For $f$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^-} f(x) = f(1) = 1$. First,consider the right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (a - \frac{\sin [x-1]}{x-1})$. For $x > 1$ and $x$ very close to $1$,$0 Thus,$\lim_{x 	o 1^+} f(x) = a - \frac{\sin(0)}{x-1} = a - 0 = a$. Since $f(1) = 1$,we have $a = 1$. Next,consider the left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (b - [\frac{\sin [x-1] - [x-1]}{([x-1])^3}])$. For $x Thus,$\lim_{x 	o 1^-} f(x) = b - [\frac{\sin(-1) - (-1)}{(-1)^3}] = b - [\frac{-\sin(1) + 1}{-1}] = b - [\sin(1) - 1]$. Since $0 The greatest integer $[\sin(1) - 1] = -1$. So,$\lim_{x 	o 1^-} f(x) = b - (-1) = b + 1$. Setting this equal to $f(1) = 1$,we get $b + 1 = 1$,which implies $b = 0$. Therefore,$a + b = 1 + 0 = 1$.

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