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(B) Given the function $f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}$.For $x 	o 0^-$,we have $[x] = -1$ and ${x} = 1+x$ (since $x = -1 + (

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$3$

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(B) Given the function $f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}$.For $x 	o 0^-$,we have $[x] = -1$ and ${x} = 1+x$ (since $x = -1 + (1+x)$ where $0 Also,$[1+x] = 0$ for $x \in (-1, 0)$.Substituting these into the limit:$\lim_{x 	o 0^-} f(x) = 0 + \frac{\alpha^{2(-1) + (1+x)} + (-1) - 1}{2(-1) + (1+x)} = \frac{\alpha^{-1+x} - 2}{-1+x}$.As $x 	o 0^-$,the limit becomes $\frac{\alpha^{-1} - 2}{-1} = 2 - \frac{1}{\alpha}$.Given that the limit is equal to $\alpha - \frac{4}{3}$,we set up the equation:$2 - \frac{1}{\alpha} = \alpha - \frac{4}{3}$.Rearranging the terms:$\alpha + \frac{1}{\alpha} = 2 + \frac{4}{3} = \frac{10}{3}$.Solving $\alpha + \frac{1}{\alpha} = \frac{10}{3}$,we get $\alpha = 3$ or $\alpha = \frac{1}{3}$.Since $\alpha$ must be an integer,the value is $\alpha = 3$.

Let $[t]$ denote the greatest integer $\leq t$ and ${t}$ denote the fractional part of $t$. Then the integral value of $\alpha$ for which the left-hand limit of the function $f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}$ at $x=0$ is equal to $\alpha-\frac{4}{3}$ is

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