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The function $f(x) = \frac{{{{\sec }^{ - 1}}x}}{{\sqrt {x - [x]} }},$ where $[.]$ denotes the greatest integer less than or equal to $x$ is defined for all $x$ belonging to
$R$
$R - \{ ( - 1,\;1) \cup (n|n \in Z)\} $
${R^ + } - (0,\;1)$
${R^ + } - \{ n|n \in N\} $
Solution
(b) The function ${\sec ^{ – 1}}x$ is defined for all $x \in R – ( – 1,\,\,1)$ and the function $\frac{1}{{\sqrt {x – [x]} }}$ is defined for all $x \in R – Z.$
So the given function is defined for all $x \in R – \{ ( – 1,\,\,1)\,\, \cup \,\,(n\,\,|\,\,n \in Z)\} .$
Similar Questions
Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.
(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )
Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$
$LIST I$ | $LIST II$ |
$P$ The range of $f$ is | $1$ $\left(-\infty, \frac{1}{1- e }\right] \cup\left[\frac{ e }{ e -1}, \infty\right)$ |
$Q$ The range of $g$ contains | $2$ $(0,1)$ |
$R$ The domain of $f$ contains | $3$ $\left[-\frac{1}{2}, \frac{1}{2}\right]$ |
$S$ The domain of $g$ is | $4$ $(-\infty, 0) \cup(0, \infty)$ |
$5$ $\left(-\infty, \frac{ e }{ e -1}\right]$ | |
$6$ $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{ e }{ e -1}\right]$ |
The correct option is: