The period of f(x) = nx + n - [nx + n],where | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the function $f(x) = nx + n - [nx + n]$.We know that the fractional part function is defined as $\{y\} = y - [y]$.Therefore,$f(x) = \{nx + n

Vedclass · Accepted Answer

$1/n$

Vedclass · Answer

(B) Given the function $f(x) = nx + n - [nx + n]$.We know that the fractional part function is defined as $\{y\} = y - [y]$.Therefore,$f(x) = \{nx + n\}$.Since $n$ is an integer,$nx + n = n(x + 1)$.Thus,$f(x) = \{n(x + 1)\}$.The fractional part function $\{u\}$ has a fundamental period of $1$.For a function $f(x) = g(ax)$,the period is $\frac{T}{|a|}$,where $T$ is the period of $g(x)$.Here,the period of $\{u\}$ is $1$,so the period of $\{n(x + 1)\}$ is $\frac{1}{|n|}$.Since $n \in N$,$n > 0$,so the period is $\frac{1}{n}$.

The period of $f(x) = nx + n - [nx + n]$,where $n \in N$ and $[ \cdot ]$ denotes the greatest integer function,is:

Similar Questions

The period of $\sin \theta - \sqrt{3} \cos \theta$ is

Let $f(x) = \sin x + \cos(ax)$ be a periodic function. Then:

The period of $\cos (3x + 5) + 7$ is

The period of $\sin^2 x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module