The range of the function $f(x) = \frac{x}{{1 + \left| x \right|}},\,x \in R,$ is
The range of the function $f(x) = \frac{x}{{1 + \left| x \right|}},\,x \in R,$ is
- [AIEEE 2012]
- A
$R$
- B
$(-1,1)$
- C
$R-\{0\}$
- D
$[-1,1]$
Similar Questions
Domain of the function $f(x) = {\sin ^{ - 1}}(1 + 3x + 2{x^2})$ is
Domain of the function $f(x) = {\sin ^{ - 1}}(1 + 3x + 2{x^2})$ 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:
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:
- [IIT 2018]
Let $f : N \rightarrow R$ be a function such that $f(x+y)=2 f(x) f(y)$ for natural numbers $x$ and $y$. If $f(1)=2$, then the value of $\alpha$ for which
$\sum \limits_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)$ holds, is
Let $f : N \rightarrow R$ be a function such that $f(x+y)=2 f(x) f(y)$ for natural numbers $x$ and $y$. If $f(1)=2$, then the value of $\alpha$ for which
$\sum \limits_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)$ holds, is
- [JEE MAIN 2022]
If the domain and range of $f(x){ = ^{9 - x}}{C_{x - 1}}$ contains $m$ and $n$ elements respectively, then
If the domain and range of $f(x){ = ^{9 - x}}{C_{x - 1}}$ contains $m$ and $n$ elements respectively, then
Let for $a \ne {a_1} \ne 0,$ $f\left( x \right) = a{x^2} + bx + c\;,g\left( x \right) = {a_1}{x^2} + {b_1}x + {c_1},p\left( x \right) = f\left( x \right) - g\left( x \right),$ If $p\left( x \right) = 0$ only for $ x=-1 $ and $p\left( { - 2} \right) = 2$ then value of $p\left( 2 \right)$ is
Let for $a \ne {a_1} \ne 0,$ $f\left( x \right) = a{x^2} + bx + c\;,g\left( x \right) = {a_1}{x^2} + {b_1}x + {c_1},p\left( x \right) = f\left( x \right) - g\left( x \right),$ If $p\left( x \right) = 0$ only for $ x=-1 $ and $p\left( { - 2} \right) = 2$ then value of $p\left( 2 \right)$ is
- [AIEEE 2011]