Question

Determine the range $y=\frac{1}{\sqrt{x-\left[x\right]}}$.

• A. $[1,\infty ]$
• B. $[1,0]$
• C. $[-1,\infty ]$
• D. None of the above

Answer 1

Answer: (A)

$[1,\infty ]$

The function is defined for all $x\in R$ except when $x-\left[x\right]=0\Rightarrow x$ is an integer i.e. $x\in I$. Thus, Domain $=R-I,$ where $I$ is the set of integers.

Also $x-\left[x\right]>0$
$\therefore$ $x$ must be positive and $x-\left[x\right]$ is a positive fraction $<1$ ........(1)

Thus, $0<x-\left[x\right]<1$

Given, $y=\frac{1}{\sqrt{x-\left[x\right]}}$

$\Rightarrow \frac{1}{y^2}=x-\left[x\right]$

By (1), we have

$0<\frac{1}{y^2}<1$
$\therefore y^2>1\Rightarrow y^2-1>0$ $\Rightarrow (y+1)(y-1)>0$
$y<-1$ or $y>1$ but $y$ is positive

$\therefore y>1$

Hence range $=[1,\infty ]$