Let $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{1+x^{2}}, x \in R$. Then the range of $f$ is :
Correct Option: 1
Related Questions
Let $R_{1}$ and $R_{2}$ be two relations defined as follows :
$R_{1}=\left\{(a, b) \in \mathbf{R}^{2}: a^{2}+b^{2} \in Q\right\}$ and
$R_{2}=\left\{(a, b) \in \mathbf{R}^{2}: a^{2}+b^{2} \notin Q\right\}$, where $Q$ is the set of all rational numbers. Then :
The domain of the function $f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^{2}+1}\right)$ is $(-\infty,-a] \cup[a, \infty]$. Then $a$ is equal to :
If $R=\left\{(x, y): x, y \in \mathbf{Z}, x^{2}+3 y^{2} \leq 8\right\}$ is a relation on the set of integers $\mathbf{Z}$, then the domain of $R^{-1}$ is :
If $R=\left\{(x, y): x, y \in \mathbf{Z}, x^{2}+3 y^{2} \leq 8\right\}$ is a relation on the set of integers $\mathbf{Z}$, then the domain of $R^{-1}$ is :
The domain of the definition of the function $f(x)=\frac{1}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)$ is:
The range of the function $f(x)=\frac{x}{1+|x|}, x \in R$, is is
The domain of the function $f(x)=\frac{1}{\sqrt{|x|-x}}$ is
Domain of definition of the function $f(x)=\frac{3}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)$, is
Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x,[x]^{2}+2[x+2]-7=0$ has :
Let $f(x)$ be a quadratic polynomial such that $f(-1)+f(2)=$ 0 . If one of the roots of $f(x)=0$ is 3 , then its other root lies in: