Let [cdot] denote the greatest integer functi | vedclass

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(A) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.Thus,$-1 \le \frac{4x+2[x]}{3} \le 1$,which implies $-3 \le 4x+2[x] \le 3$.Let $[x] = n$,where $n$

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$6$

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(A) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.Thus,$-1 \le \frac{4x+2[x]}{3} \le 1$,which implies $-3 \le 4x+2[x] \le 3$.Let $[x] = n$,where $n$ is an integer. Then $x \in [n, n+1)$.The inequality becomes $-3 \le 4x + 2n \le 3$,which simplifies to $-3-2n \le 4x \le 3-2n$,or $x \in [\frac{-3-2n}{4}, \frac{3-2n}{4}]$.Since $x \in [n, n+1)$,we must have $[n, n+1) \cap [\frac{-3-2n}{4}, \frac{3-2n}{4}] 
eq \emptyset$.For $n = -1$: $x \in [-1, 0) \cap [\frac{-1}{4}, \frac{5}{4}] = [-\frac{1}{4}, 0)$.For $n = 0$: $x \in [0, 1) \cap [-\frac{3}{4}, \frac{3}{4}] = [0, \frac{3}{4}]$.Combining these,the domain is $[-\frac{1}{4}, \frac{3}{4}]$.Thus,$\alpha = -\frac{1}{4}$ and $\beta = \frac{3}{4}$.Then $\alpha + \beta = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2}$.Finally,$12(\alpha + \beta) = 12 	imes \frac{1}{2} = 6$.

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \cos^{-1} \left( \frac{4x+2[x]}{3} \right)$ is $[\alpha, \beta]$,then $12(\alpha + \beta)$ is equal to:

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