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Let [x] denote the greatest integer ≤ x, where x R ∈ . If the domain of the real valued function $f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}} is (- \infty, a) \cup [b, c) \cup [u, \infty)$ , a < b < c, then the value of a + b + c is

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answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\frac{| [x] | - 2}{| [x] | - 3}} \\ \frac{| [x] | - 2}{| [x] | - 3} \geq 0 \end{array}$

Let $| [x] | = t$

$| [x] | = 3 \Rightarrow x \in [- 3, - 2) \cup [3, 4)$

$Domain of x = [- \infty, - 3) \cup [- 2, 3) \cup [4, \infty)$

$\begin{array}{l} a = - 3 \\ b = - 2 \\ c = 3 \end{array}$

$∴ a + b + c = - 3 + (- 2) + 3 = - 2$

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