Let $f(x) = \lbrack x\rbrack^{2} - \lbrack x + 3\rbrack - 3,x \in \mathbb{R}$ where $\lbrack \bullet

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

Accepted Answer

$f(x) = \lbrack x\rbrack^{2} - \lbrack x\rbrack - 6 = (\lbrack x\rbrack + 2)(\lbrack x\rbrack - 3)$

(1) $f(x) > 0 \Rightarrow \lbrack x\rbrack \in ( - \infty, - 2) \cup (3,\infty)$

$$\Rightarrow x \in ( - \infty, - 2) \cup \lbrack 4,\infty) $$(2) $f(x) < 0 \Rightarrow \lbrack x\rbrack \in ( - 2,3)$

$$\Rightarrow x \in \lbrack - 1,3) $$option (2) is correct

(3) $\int_{0}^{2}\mspace{2mu} f(x)dx = \int_{0}^{1}\mspace{2mu}(0 - 0 - 6)dx + \int_{1}^{2}\mspace{2mu}(1 - 1 - 6)dx$

$${= - 6 - 6 }{= - 12 }$$(4) $f(x) = 0 \Rightarrow \lbrack x\rbrack = 3$ or $\lbrack x\rbrack = - 2$

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