Function f(x) = frac{{{{sec }^{ - 1}}x}}{{√ {x - [x]} }}, where [.] denotes greatest integer less

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1.Relation and Function

medium

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 $x \in R – \{ ( – 1,\,\,1)\,\, \cup \,\,(n\,\,|\,\,n \in Z)\} .$

Standard 12

Mathematics

If $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne – 1$, for what value of $\alpha $ is $f(f(x)) = x$

normal

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:

medium

(IIT-2018)

Let $f(x) = {(x + 1)^2} – 1,\;\;(x \ge – 1)$. Then the set $S = \{ x:f(x) = {f^{ – 1}}(x)\} $ is

hard

(IIT-1995)

Let $X$ be a non-empty set and let $P(X)$ denote the collection of all subsets of $X$. Define $f: X \times P(X) \rightarrow R$ by $f(x, A)=\left\{\begin{array}{ll}1, & \text { if } x \in A \\ 0, & \text { if } x \notin A^*\end{array}\right.$ Then, $f(x, A \cup B)$ equals

normal

(KVPY-2011)

If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$ and $g\left( x \right) = {-x^2} – 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval