Question

Range of the function $f\left(x\right)=\frac{x}{1+x}$ where x is {x} which denotes the fractional part function is

• A. $[0,1)$
• B. $[0,\frac{1}{2}]$
• C. $[0,\frac{1}{2})$
• D. $(0,\frac{1}{2})$

Answer 1

The correct option is C $[0,\frac{1}{2})$

The maximum value of fractional part of x tends to 1 but is never 1 actually. Its lowest value is 0 though.

$f^{\prime }\left(x\right)=\frac{1}{(1+\{x\}{)}^2}$.

This is always positive. So $f\left(x\right)$ is an increasing function.

Hence minimum value of the function is at x=0. $f\left(x\right)=0$

Maximum value of f(x) is as $\left\{x\right\}$ tends to 1. The limiting value of the fuction is $\frac{1}{2}$ but it is never actually attained by the function