Question

The function $f\left(x\right)=\frac{x}{1+\left|x\right|}\mathrm{is}$
(a) strictly increasing
(b) strictly decreasing
(c) neither increasing nor decreasing
(d) none of these

Answer 1

(a) strictly increasing

$$\begin{gathered} f\left(x\right)=\frac{x}{1+\left|x\right|} \\ \text{Case 1: When}x>0,\left|x\right|=x \\ f\left(x\right)=\frac{x}{1+\left|x\right|} \\ =\frac{x}{1+x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(1+x\right)1-x\left(1\right)}{{\left(1+x\right)}^2} \\ =\frac{1}{{\left(1+x\right)}^2}>0,\forall x\in R \\ \text{So,}f\left(x\right)\text{is strictly increasing when}\mathit{\text{x}}\text{> 0.} \\ \text{Case 2: When}x<0,\left|x\right|=-x \\ f\left(x\right)=\frac{x}{1+\left|x\right|} \\ =\frac{x}{1-x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(1-x\right)1-x\left(-1\right)}{{\left(1-x\right)}^2} \\ =\frac{1}{{\left(1-x\right)}^2}>0,\forall x\in R \\ \text{So,}f\left(x\right)\text{is strictly increasing when}\mathit{\text{x <}}\text{0.} \\ \text{Thus,}f\left(x\right)\mathrm{is\; strictly\; increasing\; on}\mathit{}R. \end{gathered}$$