The period of f(x) = x - [x],if it is periodi | vedclass

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

Question

(C) Let $f(x)$ be a periodic function with period $T > 0$.Then,by definition,$f(x + T) = f(x)$ for all $x \in \mathbb{R}$.Substituting the given funct

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $f(x)$ be a periodic function with period $T > 0$.Then,by definition,$f(x + T) = f(x)$ for all $x \in \mathbb{R}$.Substituting the given function $f(x) = x - [x]$:$(x + T) - [x + T] = x - [x]$$x + T - [x + T] = x - [x]$$T = [x + T] - [x]$For this equation to hold for all $x \in \mathbb{R}$,$T$ must be an integer.The smallest positive value of $T$ that satisfies this condition is $T = 1$.Therefore,the period of the function $f(x) = x - [x]$ (which is the fractional part function ${x}$) is $1$.

The period of $f(x) = x - [x]$,if it is periodic,is

Similar Questions

If $f(x) = \sin^2\left(\frac{\pi}{8} + \frac{x}{2}\right) - \sin^2\left(\frac{\pi}{8} - \frac{x}{2}\right)$,then the period of $f$ is

Let $f(x) = \cos px + \sin x$ be a periodic function,then $p$ must be

The period of the function $f(x) = \cos^2(\sin x) + \sin^2(\cos x)$ is

The period of $\sin \frac{x}{2} - \cos \frac{x}{3}$ is (in $\pi$)

If the period of the function $f(x) = \sin \left( \frac{x}{n} \right)$ is $4\pi$,then $n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module