Let $f : R \rightarrow R$ be a function such that $f(x)=\frac{x^2+2 x+1}{x^2+1}$. Then
$f(x)$ is many-one in $(-\infty,-1)$
$f(x)$ is many-one in $(1, \infty)$
$f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$
$f ( x )$ is one-one in $(-\infty, \infty)$
The range of $f(x) = \cos (x/3)$ is
Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$ $\forall $ $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$ $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.
If $f:R \to R$ satisfies $f(x + y) = f(x) + f(y)$, for all $x,\;y \in R$ and $f(1) = 7$, then $\sum\limits_{r = 1}^n {f(r)} $ is
Domain of function $f(x) = {\sin ^{ - 1}}5x$ is
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: