Let the sets A and B denote the domain and range respectively of the function , where denotes the smallest integer greater than or equal to x. Then among the statements and [2023]

Question

Let the sets A and B denote the domain and range respectively of the function $f (x) = \frac{1}{\sqrt{⌈ x ⌉ - x}}$ , where $⌈ x ⌉$ denotes the smallest integer greater than or equal to x. Then among the statements

$(S 1) : A \cap B = (1, \infty) - N$ and

$(S 2) : A \cup B = (1, \infty)$ [2023]

$(S 1) : A \cap B = (1, \infty) - N$ and

$(S 2) : A \cup B = (1, \infty)$ [2023]

Smart Achievers · Accepted Answer

(1)

$f (x) = \frac{1}{\sqrt{⌈ x ⌉ - x}}$

If $x \in I$ , $⌈ x ⌉ = [x]$ (greatest integer function)

If $x \notin I$ , $⌈ x ⌉ = [x] + 1$

$\Rightarrow f (x) =$ ${\begin{matrix} \frac{1}{\sqrt{[x] - x}}, & x \in I \\ \frac{1}{\sqrt{[x] + 1 - x}}, & x \notin I \end{matrix}$

$\Rightarrow f (x) =$ ${\begin{matrix} \frac{1}{\sqrt{- {x}}}, & x \in I & (does not exist) \\ - \frac{1}{\sqrt{1 - {x}}}, & x \notin I \end{matrix}$

$\Rightarrow Domain of f (x) = R - I$

Now, $f (x) = \frac{1}{\sqrt{1 - {x}}}, x \notin I$

$\Rightarrow 0 < {x} < 1 \Rightarrow 0 < \sqrt{1 - {x}} < 1 \Rightarrow \frac{1}{\sqrt{1 - {x}}} > 1$

$\Rightarrow Range = (1, \infty) \Rightarrow A = R - I and B = (1, \infty)$

$So, A \cap B = (1, \infty) - N \Rightarrow A \cup B \neq (1, \infty)$

$\Rightarrow S 1 is true.$

Solution

1 only (S1) is true

2 both (S1) and (S2) are true

3 only (S2) is true

4 neither (S1) nor (S2) is true

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