Question

Find the range of

$f(x) = \frac{x - [x]}{1 -[x] + x }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = \dfrac{x - [x]}{1 -[x] + x }$

can be rewritten as

$\rm \: f(x) = \dfrac{x - [x]}{1 + x-[x]}$

We know, By definition of Greatest Integer function and fractional part,

$\boxed{ \rm{ \:x - [x] = \: \{x \} \: }}$

So, using this definition, we have

$\rm \: f(x) = \dfrac{\{x \}}{1 + \{x \} }$

can be rewritten as

$\rm \: f(x) = \dfrac{1 + \{x \} - 1}{1 + \{x \} }$

$\rm \: f(x) =1 - \dfrac{1}{1 + \{x \} }$

As we know, from definition of fractional part,

$\rm \: 0 \leqslant \{x \} < 1$

On adding 1 in each term, we get

$\rm \: 0 + 1\leqslant \{x \} + 1 < 1 + 1$

$\rm \: 1\leqslant \{x \} + 1 < 2$

$\rm\implies \:\dfrac{1}{2} < \dfrac{1}{\{x \} + 1} \leqslant 1$

On multiply by - 1, we get

$\rm\implies \: - 1 \leqslant \dfrac{ - 1}{\{x \} + 1} < - \dfrac{1}{2}$

On adding 1 in each term, we get

$\rm\implies \: 1- 1 \leqslant 1 - \dfrac{ 1}{\{x \} + 1} < 1 - \dfrac{1}{2}$

$\rm\implies \: 0 \leqslant f(x) < \dfrac{1}{2}$

$\rm\implies \:Range \: of \: f(x) \: \in \: \bigg[0, \: \dfrac{1}{2}\bigg)$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ |x| < y\rm\implies \: - y < x < y}\\ \\ \bigstar \: \bf{ |x| \leqslant y\rm\implies \: - y \leqslant x \leqslant y}\\ \\ \bigstar \: \bf{ |x| > y\rm\implies \: x < - y \: or \: x > y} \: \\ \\ \bigstar \: \bf{ |x| \geqslant y\rm\implies \:x \leqslant - y \: or \: x \geqslant y}\\ \\ \bigstar \: \bf{ |x - a| < y\rm\implies \:a - y < x < a + y}\\ \\ \bigstar \: \bf{ |x - a| \leqslant y\rm\implies \:a - y \leqslant x \leqslant a + y}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$