# Greatest Integer Function

For any real number x, the greatest integer function ⌊x⌋is equal to greatest integer less than or equal to x.

In general, if n is an integer and x is any number satisfying n $\leqslant$ x < n + 1, then ⌊x⌋ = 2. It is also known as integral part function.

e.g., If 2 $\leqslant$ < 3, then ⌊2⌋ = 2.

Domain = R

Range = I

Properties of Greatest Integer Function

If n is an integer and x is any real number between n and n + 1, then

• ⌊-n] = –⌊n]
• ⌊x + n] = –⌊x] + n
• ⌊-x] = –⌊x] - 1, x is not an integer.
• ⌊x + y] $\geqslant$ ⌊x] + ⌊y]
• ⌊x] > n ⇒ x $\geqslant$ n + 1
• ⌊x] < n ⇒ x < n
• ⌊x + y] = ⌊x]  + ⌊y + x -⌊x]], for all y $\in$ R
• ⌊x] + $\left [ x + \frac{1}{n} \right ]$ + $\left [ x + \frac{2}{n} \right ]$ + . . . + $\left [ x + \frac{n-1}{n} \right ]$ = ⌊nx], n $\in$ N
Signum Function

The function defined by

f(x) = $\frac{|x|}{x}$ = $\left\{\begin{matrix} -1&x<0 \\ 0& x = 0\\ 1& x>0\end{matrix}\right.$

is called the signum function.

Sample Problem 1

The domain of the function f(x) = $\frac{1}{[x]^2 - [x] - 6}$ is
(a) (-∞, -2) $\cup$ [4, ∞)
(b) (-∞, -2] $\cup$ [4, ∞)
(c) (-∞, -2) $\cup$ (4, ∞)
(d) None of these

f(x) is defined, if $[x]^2$ - [x] - 6 > 0

⇒ ([x] - 3)([x] + 2) > 0

⇒ [x] > 3 or [x] < -2

But, [x] < -2 ⇒ [x] = -3, -4, -5, ...

So, x < -2

Also, [x] > 3 ⇒ [x] = 4, 5, 6, ...

So, x $\geqslant$ 4

Domain of f = (-∞, -2] $\cup$ [4, ∞)

Sample Problem 2 [NCERT Exemplar]

If $[x]^2$ - 5[x] + 6 = 0, where [$\cdot$] denote the greatest integer function, then
(a) x $\in$ [3, 4]
(b) x $\in$ (2, 3]
(c) x $\in$ [2, 3]
(d) x $\in$ [2, 4)

Given, $[x]^2$ - 5[x] + 6 = 0

⇒ $[x]^2$ - 3[x] - 2[x] + 6 = 0

⇒ [x]([x] - 3) - 2([x] - 3) = 0

⇒ ([x] - 3)([x] - 2) = 0

⇒ [x] = 3 or [x] = 2

⇒ x $\in$ [3, 4) or x $\in$ [2, 3)

So, x $\in$ [2, 3)

Sample Problem 3

Find domain of function given by:

f(x) = $\frac{1}{\sqrt{\pi}-[x]}$

The denominator of function is positive. This means :

⇒ $\pi$ - [$\pi$] > 0

⇒ [x] < $\pi$

The value of π is 3.14. Here, [x] returns integral value. Clearly, it can assume a maximum value of 3. But, GIF returns integer value “n” for x < n+1. The inequality, therefore, has solution given by:

⇒ [x] < 4

Domain = (-∞,4)

Sample Problem 4

Find domain of the function: f(x) = $\frac{1}{[x-2]}$

Given function is in rational form having GIF as its denominator. The denominator should not evaluate to zero for real values of x. The domain of GIF is real number set R. But, we know that GIF evaluates to zero in an interval which is spread over unit value. In order to know this interval, we determine interval of x for which [x-2] is zero.

⇒ [x - 2] = 0

We can write this function as:

⇒ [x + (-2)] = 0

Using property [x+y] = [x] + [y], provided one of x and y is an integer. This is the case here,

⇒ [x + (-2)] = [x] + [-2] = [x] - 2 = 0

[x] = 2  ⇒ 2 $\geqslant$ x < 3

⇒ x $\in$ [2, 3)