The period of the function f(x) = e^{log(sin | vedclass

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

Question

(C) Given the function $f(x) = e^{\log(\sin x)} + (	an x)^3 - \operatorname{cosec}(3x - 5)$.First, simplify the expression: $f(x) = \sin x + (	an x)

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(C) Given the function $f(x) = e^{\log(\sin x)} + (	an x)^3 - \operatorname{cosec}(3x - 5)$.First, simplify the expression: $f(x) = \sin x + (	an x)^3 - \operatorname{cosec}(3x - 5)$.Let $f_1(x) = \sin x$, $f_2(x) = (	an x)^3$, and $f_3(x) = -\operatorname{cosec}(3x - 5)$.The period of $f_1(x) = \sin x$ is $T_1 = 2\pi$.The period of $f_2(x) = (	an x)^3$ is $T_2 = \pi$.The period of $f_3(x) = -\operatorname{cosec}(3x - 5)$ is $T_3 = \frac{2\pi}{|3|} = \frac{2\pi}{3}$.The period of $f(x)$ is the $	ext{LCM}$ of $(T_1, T_2, T_3) = 	ext{LCM}(2\pi, \pi, \frac{2\pi}{3})$.To find the $	ext{LCM}$ of fractions, we use $\frac{	ext{LCM of numerators}}{	ext{HCF of denominators}} = \frac{	ext{LCM}(2\pi, \pi, 2\pi)}{	ext{HCF}(1, 1, 3)} = \frac{2\pi}{1} = 2\pi$.

The period of the function $f(x) = e^{\log(\sin x)} + (\tan x)^3 - \operatorname{cosec}(3x - 5)$ is

Similar Questions

The period of $\cos(x + 8x + 27x + \ldots + n^3x)$ is

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 $\sin ^4 x + \cos ^4 x$ is

If the period of the function $f(x) = \frac{\tan 5x \cos 3x}{\sin 6x}$ is $\alpha$,then find the value of $f\left(\frac{\alpha}{8}\right)$.

The period of $\frac{\sin \theta + \sin 2\theta}{\cos \theta + \cos 2\theta}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module