If f(x) = [x]^{2} - 5[x] + 6 = 0,where [x] de | vedclass

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(C) Given the equation: $[x]^{2} - 5[x] + 6 = 0$. Let $t = [x]$. Then the equation becomes $t^{2} - 5t + 6 = 0$. Factoring the quadratic equation: $(t

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$[2, 4)$

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(C) Given the equation: $[x]^{2} - 5[x] + 6 = 0$. Let $t = [x]$. Then the equation becomes $t^{2} - 5t + 6 = 0$. Factoring the quadratic equation: $(t - 3)(t - 2) = 0$. This gives $t = 2$ or $t = 3$. Thus,$[x] = 2$ or $[x] = 3$. By the definition of the greatest integer function: If $[x] = 2$,then $x \in [2, 3)$. If $[x] = 3$,then $x \in [3, 4)$. Combining these intervals,we get $x \in [2, 3) \cup [3, 4) = [2, 4)$. Therefore,the correct option is $C$.

If $f(x) = [x]^{2} - 5[x] + 6 = 0$,where $[x]$ denotes the greatest integer function,then $x \in$

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