Period of $f\left( x \right)=nx+n-\left[ nx+n \right],$ $\left( n\in N \right)$ where $\left[ {} \right]$ denotes the greatest integer function is
A. 1
B. $\dfrac{1}{n}$
C. $n$
D. none of these
Answer
540.6k+ views
Hint: We solve this question by using the concept that any number can be written as the sum of its integer part and fractional part. This is represented as $x=\left[ x \right]+\left\{ x \right\},$ where $\left[ x \right]$ is the greatest integer function that represents the integer part of x and $\left\{ x \right\}$ is the fractional part of x. Using this to simplify the above expression, we represent it in terms of a fractional part only. Then we use the concept of periodicity to solve the sum.
Complete step by step answer:
In order to solve this question, let us first simplify the given function. The given function is,
$\Rightarrow f\left( x \right)=nx+n-\left[ nx+n \right]$
We know that any number can be represented by the sum of its integer part and fractional part. This is given by the expression as $x=\left[ x \right]+\left\{ x \right\},$ where $\left[ x \right]$ is the integer part of x and $\left\{ x \right\}$ is the fractional part of x. Looking at the above function and using the above expression, we can say that the difference of the number and its integer part is nothing but the fractional part. Using this,
$\Rightarrow f\left( x \right)=\left\{ nx+n \right\}$
We need to know a property of periodicity to solve this question. This property can be given as follows:
If a function $f\left( x \right)$ is periodic with a period $T,$ then the function $f\left( ax \right)$ is periodic with a period $\dfrac{T}{a}.$
Therefore, taking n out common from the above function,
$\Rightarrow f\left( x \right)=\left\{ n\left( x+1 \right) \right\}$
Using the above concept, we can say that this is periodic with a period of $\dfrac{1}{n}.$
So, the correct answer is “Option B”.
Note: We need to know the properties of periodic functions in order to solve such questions. It is important to note the concept that any number can be represented by the sum of its integer part which is nothing but the greatest integer function and its fractional part. This concept applies to negative numbers as well.
Complete step by step answer:
In order to solve this question, let us first simplify the given function. The given function is,
$\Rightarrow f\left( x \right)=nx+n-\left[ nx+n \right]$
We know that any number can be represented by the sum of its integer part and fractional part. This is given by the expression as $x=\left[ x \right]+\left\{ x \right\},$ where $\left[ x \right]$ is the integer part of x and $\left\{ x \right\}$ is the fractional part of x. Looking at the above function and using the above expression, we can say that the difference of the number and its integer part is nothing but the fractional part. Using this,
$\Rightarrow f\left( x \right)=\left\{ nx+n \right\}$
We need to know a property of periodicity to solve this question. This property can be given as follows:
If a function $f\left( x \right)$ is periodic with a period $T,$ then the function $f\left( ax \right)$ is periodic with a period $\dfrac{T}{a}.$
Therefore, taking n out common from the above function,
$\Rightarrow f\left( x \right)=\left\{ n\left( x+1 \right) \right\}$
Using the above concept, we can say that this is periodic with a period of $\dfrac{1}{n}.$
So, the correct answer is “Option B”.
Note: We need to know the properties of periodic functions in order to solve such questions. It is important to note the concept that any number can be represented by the sum of its integer part which is nothing but the greatest integer function and its fractional part. This concept applies to negative numbers as well.
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