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Algebra Level 3

Find the domain of the real-valued function $f(x)=\dfrac{1}{\sqrt{x^2-\lfloor x \rfloor ^2}}$ , where $\lfloor \cdot \rfloor$ represents the greatest integer function .

$\begin{array}{ll} (a)~\mathbb R - \mathbb Z &&&&&(b)~\mathbb R^+ - \mathbb Z\\ (c)~\mathbb Z &&&&&(d)~\mathbb R^+ \end{array}$

d c b a

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Md Zuhair
Feb 12, 2017

Solution :
For $f(x)$ to be real-valued, $x^2-\lfloor x \rfloor ^2$ has to be positive:
$\begin{aligned} x^2-\lfloor x \rfloor ^2&>0 \\ (x-\lfloor x \rfloor )(x+\lfloor x \rfloor )&>0 \\ \{x\} (x+\lfloor x\rfloor )&>0 \\\ x+ \lfloor x\rfloor &>0 \qquad (\text{when } \{x\} \neq 0) \\ x &\in \mathbb R^+. \end{aligned}$ As $\{ x \} \geq 0$ , excluding the values at which $\{ x \}=0,$ i.e. $x \in \mathbb Z,$ the domain of the real-valued function $f(x)$ is $\mathbb R^+ - \mathbb Z. \ _\square$

JEE main

Find the domain of the real-valued function f ( x ) = 1 x 2 − ⌊ x ⌋ 2 f(x)=\dfrac{1}{\sqrt{x^2-\lfloor x \rfloor ^2}} f ( x ) = x 2 − ⌊ x ⌋ 2 ​ 1 ​ , where ⌊ ⋅ ⌋ \lfloor \cdot \rfloor ⌊ ⋅ ⌋ represents the greatest integer function .

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Find the domain of the real-valued function $f(x)=\dfrac{1}{\sqrt{x^2-\lfloor x \rfloor ^2}}$ , where $\lfloor \cdot \rfloor$ represents the greatest integer function .