The period of the function f(x) = e^{x - [x] | vedclass

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

Question

(A) The function is given by $f(x) = e^{\{x\} + |\cos \pi x| + |\cos 2\pi x| + \dots + |\cos n\pi x|}$,where $\{x\} = x - [x]$ is the fractional part

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The function is given by $f(x) = e^{\{x\} + |\cos \pi x| + |\cos 2\pi x| + \dots + |\cos n\pi x|}$,where $\{x\} = x - [x]$ is the fractional part function.$1$. The period of the fractional part function $\{x\}$ is $T_0 = 1$.$2$. For each term of the form $|\cos(k\pi x)|$,the period $T_k$ is given by $\frac{\pi}{k\pi} = \frac{1}{k}$ for $k = 1, 2, \dots, n$.$3$. The period of the function $f(x)$ is the least common multiple $(LCM)$ of the periods of its individual components: $T = 	ext{LCM}(T_0, T_1, T_2, \dots, T_n) = 	ext{LCM}(1, 1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n})$.$4$. The $LCM$ of a set of fractions $\frac{a}{b}, \frac{c}{d}, \dots$ is given by $\frac{	ext{LCM}(a, c, \dots)}{	ext{GCD}(b, d, \dots)}$.$5$. Thus,$T = \frac{	ext{LCM}(1, 1, 1, \dots, 1)}{	ext{GCD}(1, 2, 3, \dots, n)} = \frac{1}{1} = 1$.Therefore,the period of the function is $1$.

The period of the function $f(x) = e^{x - [x] + |\cos \pi x| + |\cos 2\pi x| + \dots + |\cos n\pi x|}$ (where $[.]$ denotes the greatest integer function) is:

Similar Questions

The period of the function $f(x) = \frac{|\sin x| + |\cos x|}{|\sin x - \cos x|}$ is

The period of $f(x) = \cos \left(\frac{x}{3}\right) + \sin \left(\frac{x}{2}\right)$ is (in $\pi$)

If $n \in N$ and the period of $\frac{\cos nx}{\sin \left(\frac{x}{n}\right)}$ is $4\pi$,then $n$ is equal to

The period of $f(x) = \sin \left( \frac{\pi x}{n - 1} \right) + \cos \left( \frac{\pi x}{n} \right)$,where $n \in \mathbb{Z}$ and $n > 2$,is:

Which of the following functions has a period of $2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module