Let the sets A and B denote the domain and ra | vedclass

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(A) The function is given by $f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$.For the function to be defined,the expression inside the square root must b

Vedclass · Accepted Answer

only $(S1)$ is true

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(A) The function is given by $f(x) = \frac{1}{\sqrt{\lceil x ceil - x}}$.For the function to be defined,the expression inside the square root must be strictly positive: $\lceil x ceil - x > 0$,which implies $\lceil x ceil > x$.This inequality holds for all $x 
otin \mathbb{Z}$. If $x \in \mathbb{Z}$,then $\lceil x ceil = x$,so $\lceil x ceil - x = 0$,making the denominator zero.Thus,the domain $A = \mathbb{R} - \mathbb{Z}$.For $x 
otin \mathbb{Z}$,we know that $\lceil x ceil = \lfloor x floor + 1$. Therefore,$\lceil x ceil - x = \lfloor x floor + 1 - x = 1 - (x - \lfloor x floor) = 1 - \{x\}$,where $\{x\}$ is the fractional part of $x$.Since $x 
otin \mathbb{Z}$,$0 Then,$0  1$.Thus,the range $B = (1, \infty)$.Now,$A \cap B = (\mathbb{R} - \mathbb{Z}) \cap (1, \infty) = (1, \infty) - \mathbb{Z}$. Since the intersection with $(1, \infty)$ only includes positive integers,$(1, \infty) - \mathbb{Z} = (1, \infty) - \mathbb{N}$. Thus,$(S1)$ is true.$A \cup B = (\mathbb{R} - \mathbb{Z}) \cup (1, \infty) = \mathbb{R} - \{0, -1, -2, \dots\}$. This is not equal to $(1, \infty)$. Thus,$(S2)$ is false.Therefore,only $(S1)$ is true.

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