Question

Let \[f(x) = \dfrac{{[x - [x]]}}{{[1 + x - [x]]}}\]; \[x \in R\] and \[\left[ t \right]\]
 is the greatest integer function. What is the range of \[f(R)\]?
A. \[[0,1]\]
B.\[\left[ {0,\dfrac{1}{2}} \right)\]
C. \[(0,1)\]
D. None of these

Answer 1

Hint: The greatest integer function is always smaller than or equal to the argument of the function. First, we find the interval in which \[x - [x]\] lies then we will find the interval \[1 + x - [x]\] lies.
Then find the maximum values of \[x - [x]\] and \[1 + x - [x]\] to get the maximum value of the ratio. By using the maximum and minima of \[\dfrac{{[x - [x]]}}{{[1 + x - [x]]}}\], we will calculate the range of \[f(R)\].

Complete step-by-step Solution:
The value of the greatest integer function is given by,
\[[x] = \left\{ \begin{array}{l}x,\,\,\,\,\,{\rm{if }}x \in N\\x - h,\,\,\,\,{\rm{if }}x\neq{ \in }N\end{array} \right.\] where \[h \in (0,1)\] such that \[x - h \in N\]
\[ \Rightarrow x - [x] \in [0,1)\;\forall \;x \in R\]
And, \[1 + x - [x] \in [1,2)\;\forall \;x \in R\]
\[ \Rightarrow x - [x] < 1\]
Adding \[x - [x]\] to the both sides
\[ \Rightarrow 2(x - [x]) < 1 + x - [x]\]……….(1)
Dividing both sides of the above inequality (1) and we get
\[ \Rightarrow \dfrac{{x - [x]}}{{1 + x - [x]}} < \dfrac{1}{2}\]
Then,
\[\forall \;x \in R,\]
\[f(x) = \dfrac{{[x - [x]]}}{{[1 + x - [x]]}} \in \left[ {0,\dfrac{1}{2}} \right)\]
\[f(R) = \left[ {0,\dfrac{1}{2}} \right)\]
Hence the correct answer is option B.

Note: Many students make a common mistake to calculate the interval of \[x{\rm{ }} - {\rm{ }}\left[ x \right]\]. They take it as \[\left[ {0,1} \right]\] which is incorrect. Because \[x{\rm{ }} - {\rm{ }}\left[ x \right]\] given the fractional part of \[x\] and fractional part always less than 1.