$f (x) = \frac{1}{\sqrt{[x] - x}}$ , where [ ] denotes the greatest integral function less than or equals to x. Then, number of elements in the domain of f(x) is

Open in App

Solution

$f (x) = \frac{1}{\sqrt{[x] - x}}$ exists, if
$[x] - x > 0$ i.e, $[x] > x$
But from definition of greatest integral function, we know,
$[x] \leq x$ [as $x = [x] + {x}]$
Thus, it is not possible that $[x] > x .$
Hence, domain $f (x) = ϕ$

Suggest Corrections

Similar questions

f (x) = \frac{1}{\sqrt{([x]^{2} - 7 [x] + 10)}}

(- \infty, a) \cup [b, \infty),

a + b

[x]

x

f (x) = \frac{{sec}^{- 1} x}{\sqrt{x - [x]}}

[x]

x

x

f (x) = \sqrt{x^{2} - {[x]}^{2}}

[x] =

x

Let $f (x) = x {(- 1)}^{[\frac{1}{x}]} . x \neq 0$ , where [x] denotes the greatest integer less than or equal to x. then $lim x \to 0 f (x)$

f (x) = [x]

[x]

x

Join BYJU'S Learning Program