Question

The Domain of f(x) = √x - 2{x} (where {.} denotes fractional part of x) is:

1) R

2) [0,∞)

3) [2,∞)

4) [0,1)

Answer 1

We have to find the domain of,

$f(x) = \sqrt{x-2\{x\}}$

Since  $f(x)$  is directly given as square root of  $x-2\{x\},$  we can say,

$\begin{aligned}&f(x)\geq 0\end{aligned}$

because,

$\begin{aligned}&(f(x))^2=x-2\{x\}\\ \\ \Longrightarrow\ \ &|f(x)| =\sqrt{x-2\{x\}}\\ \\ \Longrightarrow\ \ &|f(x)|=f(x)\\ \\ \Longrightarrow\ \ &f(x)\geq 0\end{aligned}$

So,

$\begin{aligned}&f(x)\geq 0\\ \\ \Longrightarrow\ \ &\sqrt{x-2\{x\}}\geq 0\\ \\ \Longrightarrow\ \ &x-2\{x\}\geq 0\\ \\ \Longrightarrow\ \ &x-\{x\}-\{x\}\geq 0\\ \\ \Longrightarrow\ \ &\lfloor x \rfloor-\{x\}\geq 0\\ \\ \Longrightarrow\ \ &\lfloor x\rfloor\geq \{x\}\end{aligned}$

This always holds true if and only if,

$\Large x=0\ \wedge \ x\geq 1\ \ \ \ \ \Longrightarrow\ \ \ \ \ \Large x\in \{0\}\cup[1, +\infty )$

Hence we can say that the domain is  {0} ∪ [1, +∞), ignoring the sign of ∞ we get  {0} ∪ [1, ∞),  but it's not in the options!

→  All elements in R are not included in  {0} ∪ [1, ∞).  Hence  R ⊄ {0} ∪ [1, ∞). So option (1) can't be.

→  Since  (0, 1) ⊄ {0} ∪ [1, ∞),  [0, ∞)  and  [0, 1)  can't be.  Thus options (2) and (4) can't be.

→  But  [2, ∞) ⊂ [1, ∞) ⊂ {0} ∪ [1, ∞),  so [2, ∞) can be said as the domain.

Hence, option (3) is the answer.