Question

4. The function f : $f:R⟶\{x\in R:-1<x<1\}$ defined by $f\left(x\right)=$ $\frac{x}{1+\left|x\right|}$ is

• A. one-one and into
• B. one-one and onto
• C. Many-one and into
• D. Many-one and onto

Answer 1

The correct option is B one-one and onto

$f:R⟶\{x\in R:-1<x<1\}definedbyf\left(x\right)=\frac{x}{1+\left|x\right|}forx>0\left|x\right|=xforx<0\left|x\right|=-xasx⟶\infty ,f\left(x\right)=f\left(x\right)=\frac{x}{1+\left|x\right|}=\frac{1}{\frac{1}{x}+1}=1x-1-\infty ,f\left(x\right)=\frac{x}{1-\left|x\right|}=\frac{1}{\frac{1}{x}-1}=-1foreverydifferentx,f\left(x\right)hasdifferentvalue.f\left(x_1\right)=f\left(x_2\right)\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}{x}_1+x_1{x}_2=x_2+x_2{x}_1{x}_1=x_2$

so function is one-one and range is equal to co-domain so function is onto.