The period of tan(ky) + sin(ky),where k = 1 + | vedclass

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

Question

(B) The value of $k$ is the sum of the squares of the first $20$ natural numbers:$k = \sum_{n=1}^{20} n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=20$.$k = \f

Vedclass · Accepted Answer

$\frac{2\pi}{1435}$

Vedclass · Answer

(B) The value of $k$ is the sum of the squares of the first $20$ natural numbers:$k = \sum_{n=1}^{20} n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=20$.$k = \frac{20 	imes 21 	imes 41}{6} = 70 	imes 41 = 2870$.We need the period of $f(y) = 	an(2870y) + \sin(2870y)$.The period of $	an(ky)$ is $T_1 = \frac{\pi}{k} = \frac{\pi}{2870}$.The period of $\sin(ky)$ is $T_2 = \frac{2\pi}{k} = \frac{2\pi}{2870} = \frac{\pi}{1435}$.The period of the sum of two functions is the Least Common Multiple $(LCM)$ of their individual periods.$LCM\left(\frac{\pi}{2870}, \frac{2\pi}{2870}ight) = \frac{LCM(\pi, 2\pi)}{GCD(2870, 2870)} = \frac{2\pi}{2870} = \frac{\pi}{1435}$.

The period of $\tan(ky) + \sin(ky)$,where $k = 1 + 4 + 9 + \ldots$ ($20$ terms),is

Similar Questions

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

Let $f$ and $g$ be periodic functions with the periods $T_{1}$ and $T_{2}$ respectively. Then $f+g$ is

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

The period of the function $f(x) = \frac{2 \sin \left(\frac{\pi x}{3}\right) \cos \left(\frac{2 \pi x}{5}\right)}{3 \tan \left(\frac{7 \pi x}{2}\right) - 5 \sec \left(\frac{5 \pi x}{3}\right)}$ is

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module