Question

$f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1},\forall x\in R$ is

• A. one-one
• B. onto
• C. bijective
• D. neither one one nor onto

Answer 1

The correct option is D neither one one nor onto

$\begin{array}{cc}f\left(x\right)& =\frac{x}{1+x^2}\\ -f\left(4\right)=& \frac{4}{1+16}=\frac{4}{17}\\ f\left(\frac{1}{4}\right)& =\frac{\frac{1}{4}}{1+\frac{1}{16}}=\frac{4}{17}\end{array}$

Hence two different input has Same output $\cdot$ so not one-one

$f\left(x\right)=y$

$y=\frac{x}{1+x^2}=1y+y{x}^2-x=0$

$x=\frac{1\pm \sqrt{1-4y^2}}{2y}$

$1-4y^2\ge 0$

$(1+2y)(1-xy)\ge 0$

$\frac{1}{-3}\le y<\frac{1}{2}$

Range of $f\left(x\right)$ is $[\frac{-1}{2},\frac{1}{2}]$

Range of $f\left(x\right)\ne$ codomain $f\left(x\right)$ Hence f(x) is neither one-one nor onto option D is correct.