Question

The range of the function f(x) defined by $f\left(x\right)=\frac{x}{1+x^2}$ is

• A. $R-\{-1,1\}$
• B. $\left(-\frac{1}{2},\frac{1}{2}\right]$
• C. $\left(-\frac{1}{2},\frac{1}{2}\right]-\left\{0\right\}$
• D. $\left(-\frac{1}{2},\frac{1}{2}\right)-\left\{0\right\}$

Answer 1

The correct option is B $\left(-\frac{1}{2},\frac{1}{2}\right]$

$\mathrm{We}\mathrm{have},f\left(x\right)=\frac{x}{1+x^2}\mathrm{Let}y=f\left(x\right),\mathrm{Then},y=f\left(x\right)\Rightarrow y=\frac{x}{1+x^2}\Rightarrow y+{\mathrm{yx}}^2-x=0\Rightarrow x=\frac{1\pm \sqrt{1-4y^2}}{2y}\mathrm{Clearly},x\mathrm{will}\mathrm{assume}\mathrm{real}\mathrm{values},\mathrm{if}1-4y^2\ge 0\mathrm{and}y\ne 0\Rightarrow 4y^2-1\le 0\mathrm{and}y\ne 0\Rightarrow y^2-\frac{1}{4}\le 0\mathrm{and}y\ne 0\Rightarrow \left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right)\le 0\mathrm{and}y\ne 0\Rightarrow -\frac{1}{2}\le y\le \frac{1}{2}\mathrm{and}y\ne 0\Rightarrow y\in \left(-\frac{1}{2},\frac{1}{2}\right]-\left\{0\right\}\mathrm{Also},\mathrm{if}y=0,\mathrm{then}x=0\therefore \mathrm{The}\mathrm{range}\mathrm{of}f=\left(-\frac{1}{2},\frac{1}{2}\right]$