Let f(x) = [x]^2 - [x+3] - 3, x in mathbb{R}, | vedclass

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(B) Given $f(x) = [x]^2 - ([x] + 3) - 3 = [x]^2 - [x] - 6$. Factoring the expression, we get $f(x) = ([x] - 3)([x] + 2)$. $(1)$ For $f(x) > 0$, we nee

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$f(x) < 0$ only for $x \in [-1, 3)$

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(B) Given $f(x) = [x]^2 - ([x] + 3) - 3 = [x]^2 - [x] - 6$. Factoring the expression, we get $f(x) = ([x] - 3)([x] + 2)$. $(1)$ For $f(x) > 0$, we need $[x] > 3$ or $[x] This implies $x \in [4, \infty)$ or $x \in (-\infty, -2)$. Thus, option $A$ is incorrect. $(2)$ For $f(x) This implies $x \in [-1, 3)$. Thus, option $B$ is correct. $(3)$ Calculating the integral: $\int_0^2 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx$. For $x \in [0, 1)$, $[x] = 0$, so $f(x) = 0^2 - 0 - 6 = -6$. For $x \in [1, 2)$, $[x] = 1$, so $f(x) = 1^2 - 1 - 6 = -6$. Thus, $\int_0^2 f(x) dx = \int_0^1 (-6) dx + \int_1^2 (-6) dx = -6 - 6 = -12$. Option $C$ is incorrect. $(4)$ For $f(x) = 0$, we need $[x] = 3$ or $[x] = -2$. This implies $x \in [3, 4)$ or $x \in [-2, -1)$, which contains infinitely many values. Option $D$ is incorrect.

Let $f(x) = [x]^2 - [x+3] - 3, x \in \mathbb{R}$, where $[\bullet]$ is the greatest integer function. Then:

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