For each value of x, f(x) assumes the value of the greatest integer, less than or equal to x.

It is also called the Floor function and step function

Example

[2.56] =2

[.5]=0

[-6.45]=-7

[-.65]=-1

This can be generalized as

$[x] = –1 , –1 \leq x < 0$

$[x] = 0 , 0 \leq x < 1$

$[x] = 1 , 1 \leq x < 2$

$[x] = 2 , 2 \leq x < 3 $

For $f : R \rightarrow R , y = f(x) = [x]$ for each $x \in R$

Domain = R

Range = Z (integer set) as it only attains integer values

The graph of the modulus function is shown below. It coincide with the graph of identity function y=x for $x \geq 0$ and for x< 0, it is a graph of linear function y=-x

[1.56] =1

[.25]=0

[-5.45]=-5

[-.15]=-1

[-5]=-5

[1]=1

This can be written as

$ [x] \geq x$

Fractional part of x is defined as the difference between [x] and x. It is denoted as {x}

So,

{x}=x -[x]

Example

x=4.65 ,then [x]=4 ,{x}=.65

x= -.65, then [x]=-1 ,{x}= .35 ( This is very important thing to remember for negative number)

x =-1.45 ,then [x]=-2 ,{x}= .55 ( This is very important thing to remember for negative number)

x=3 ,then [x]=3 ,{x}=0

We know that

$ x-1 < [x] \leq x$

$ -x \leq -[x] < 1-x$

Adding x

$ 0 \leq x -[x] < 1$

$ 0 \leq {x} < 1 $

So Fractional part {x} is always non-negative and lies between (0,1).

So we can define Fractional part function as

y=f(x) = {x} =x -[x]

This is good for $x \in R$

Range of the fractional part function is [0,1)

Graph of the fractional part function is given below

1. If x is an integer

[x]=x

{x} =0

2. For $x \in R$ [[x]]= [x]

[{x}]=0

{[x]}=0

3. for $k \in Z$

a. $[x] \geq k \Rightarrow x \geq n , x \in [n,\infty)$

b. $[x] > k \Rightarrow x \geq n +1 , x \in [n+1,\infty)$

c. $[x] \leq k \Rightarrow x \leq n+1 , x \in (-\infty,n+1)$

d. $[x] < k \Rightarrow x < n , x \in (-\infty,n)$

e. [x+ k]=[x] +k

4.

a. $y=[x]+ [-x]=\begin{cases} 0 & \text{ if } x \in Z \\ -1 & \text{ if } x \notin Z \end{cases}$

b. $y=[x]- [-x]=\begin{cases} 2[x]+1 & \text{ if } x \notin Z \\ 2[x]] & \text{ if } x \in Z \end{cases}$

c. $y=\left\{ x \right\}- \left\{-x\right\}=\begin{cases} 0 & \text{ if } x \in Z \\ 1 & \text{ if } x \notin Z \end{cases}$

3. For real number x and y

a. [x +y] = [x] + [y+x - [x] ]

b. $[x+y]=\begin{cases} [x] + [y] & \text{ if } \left\{x \right\} + \left\{y \right\} < 1 \\ [x]+[y]+1 & \text{ if } \left\{x \right\} + \left\{y \right\} \geq 1 \end{cases}$

a. $y =\sqrt {[x] -1|}$

b. $y = \frac {1}{\sqrt { 2- [x]}}$

a. $y =\sqrt {[x] -1|}$

Clearly This is defined for $[x] -1 \geq 0$

or

$[x] \geq 1$

$ x \in [1,\infty)$

So Domain is $[1,\infty)$

b. $y = \frac {1}{\sqrt { 2- [x]}}$

Clearly This is defined for $2- [x] > 0$

or

2 > [x]

or [x] < 2

$ x \in (-\infty,2)$

So Domain is $(-\infty,2)$

2. Find the domain and range of function

$y= \frac {1}{\sqrt {x- [x]}}$

We have ,

$y= \frac {1}{\sqrt {x- [x]}}$

We know that

$0 \leq x -[x] < 1 , x \in R$

for $x \in Z ,x -[x]=0$

So Domain is R - Z

$0 < x- [x] < 1 , x \in R -Z$

$ 0 < \sqrt {x -[x]} < 1$

$ 1 < \frac {1}{\sqrt {x- [x]}} < \infty$

So Range is = $(1,\infty)$

3. Let $f : R \rightarrow R$ , $f(x) = x^2 + 2[x] -1$ for each $x \in R$

Find the values of f(x) at x= 1.2 ,-.5 ,-2.1

$f(x) = x^2 + 2[x] -1$

$f(1.2) = (1.2)^2 + 2[1.2] -1 =1.44 +2-1=2.44$

$f(-.5) = (-.5)^2 + 2[-.5] -1 =.25 -2-1=-2.75$

$f(-2.1) = (-2.1)^2 + 2[-2.1] -1 =4.41 -6-1=-2.59$

- Cartesian Products
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- What is relations?
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- What is Function
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- Domain of Function
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- Range of Function
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- Identity Function
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- Constant Function
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- Linear Function
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- Modules Function
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- Greatest Integer Function
- |
- Polynomial Function
- |
- Algebra of Real Function

Class 11 Maths Class 11 Physics