Question

The period of function f(x) = \(e^{log(sinx)}+(tanx)^3 - cosec(3x - 5)\)is

• A. π
• B. π/2
• C. 2π
• D. 2π/3

Answer 1

The Correct Option is C

To solve the problem, we need to find the period of the function $f(x) = e^{\log(\sin x)} + (\tan x)^3 - \csc(3x-5)$ by analyzing each term separately.

1. Analyzing the First Term:
The first term is $e^{\log(\sin x)} = \sin x$.
The period of $\sin x$ is $2\pi$.

2. Analyzing the Second Term:
The second term is $(\tan x)^3$.
The period of $\tan x$ is $\pi$, so the period of $(\tan x)^3$ is also $\pi$.

3. Analyzing the Third Term:
The third term is $\csc(3x-5)$.
The period of $\csc x$ is $2\pi$, so the period of $\csc(3x-5)$ is $\frac{2\pi}{3}$.

4. Determining the Overall Period:
The period of $f(x)$ is the least common multiple of the periods of the individual terms ($2\pi$, $\pi$, and $\frac{2\pi}{3}$), which is $2\pi$.

5. Considering the Domain:
The function $f(x)$ is only defined where $\sin x > 0$, which occurs in intervals $(2n\pi, (2n+1)\pi)$ for integer $n$.
Additionally, $\tan x$ must be defined, so $x \neq (2n+1)\frac{\pi}{2}$.
Thus, the domain is $x \in (2n\pi, (2n+1)\pi) \setminus \{(2n+1)\frac{\pi}{2}\}$.

6. Final Period Determination:
Despite the restricted domain, the pattern repeats every $2\pi$ units.
The least common multiple of all periods is $2\pi$, which matches the periodicity of the defining condition ($\sin x > 0$).

Final Answer:
The period of $f(x)$ is ${2\pi}$.