Question

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

• A. many-one and into
• B. one-one and onto
• C. onto but not one-one
• D. one-one but not onto

Answer 1

The correct option is A many-one and into

$\mathrm{For}{x}_1,x_2\in R,\mathrm{Consider}f\left(x_1\right)=f\left(x_2\right)\Rightarrow \frac{x_1}{{x_1}^2+1}=\frac{x_2}{{x_2}^2+1}\Rightarrow x_1{x_2}^2+x_1=x_2{x_1}^2+x_2\Rightarrow x_1{x_2}^2+x_1-x_2{x_1}^2-x_2=0\Rightarrow x_1{x}_2\left(x_2-x_1\right)+\left(x_1-x_2\right)=0\Rightarrow \left(x_1-x_2\right)\left(x_1{x}_2-1\right)=0\Rightarrow x_1=x_2\mathrm{or}{x}_1{x}_2=1\Rightarrow x_1=x_2\mathrm{or}{x}_1=\frac{1}{x_2}\therefore f\mathrm{is}\mathrm{not}\mathrm{one}-\mathrm{one}.$

$\mathrm{Let}y=\frac{x}{x^2+1}\Rightarrow {\mathrm{yx}}^2+y=x\Rightarrow {\mathrm{yx}}^2-x+y=0......\left[1\right]\mathrm{Since}x\mathrm{can}\mathrm{take}\mathrm{all}\mathrm{real}\mathrm{values},\mathrm{the}\mathrm{discriminant}D\mathrm{for}\left[1\right]\mathrm{satisfies}D\ge 0\Rightarrow 1-4y^2\ge 0\Rightarrow 4y^2-1\le 0\Rightarrow y\in \left(\frac{-1}{2},\frac{1}{2}\right]\therefore \mathrm{Range}\left(f\right)=\left(\frac{-1}{2},\frac{1}{2}\right]\mathrm{Since}\mathrm{Range}\ne \mathrm{Co}-\mathrm{domain},f\mathrm{is}\mathrm{into}.$