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Statement-1: Function f(x)=sin(x+3 sin x...

Statement-1: Function `f(x)=sin(x+3 sin x)` is periodic .
Statement-2: If g(x) is periodic then f(g(x)) periodic

A

Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given problem, we need to analyze both statements step by step. ### Step 1: Analyze Statement 1 The first statement is about the function \( f(x) = \sin(x + 3 \sin x) \). We need to determine if this function is periodic. 1. **Understanding Periodicity**: A function \( f(x) \) is periodic if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \). 2. **Components of the Function**: The function consists of \( \sin(x) \) which is periodic with a period of \( 2\pi \). However, we also have \( 3 \sin x \) inside the argument of the sine function. 3. **Finding the Period**: The argument of the sine function is \( x + 3 \sin x \). The term \( 3 \sin x \) oscillates between -3 and 3, which means that the overall argument \( x + 3 \sin x \) does not have a fixed period. 4. **Conclusion for Statement 1**: The function \( f(x) = \sin(x + 3 \sin x) \) is not periodic because the oscillation introduced by \( 3 \sin x \) does not lead to a repeating pattern in \( f(x) \). ### Step 2: Analyze Statement 2 The second statement claims that if \( g(x) \) is periodic, then \( f(g(x)) \) is also periodic. 1. **Understanding the Implication**: For \( f(g(x)) \) to be periodic, \( f(x) \) must also be periodic. If \( g(x) \) is periodic, it means there exists a period \( T \) such that \( g(x + T) = g(x) \). 2. **Consider Non-Periodic \( f(x) \)**: We already established that \( f(x) = \sin(x + 3 \sin x) \) is not periodic. Therefore, even if \( g(x) \) is periodic, the composition \( f(g(x)) \) will not necessarily be periodic. 3. **Conclusion for Statement 2**: The statement is false because the periodicity of \( g(x) \) does not guarantee the periodicity of \( f(g(x)) \) if \( f(x) \) itself is not periodic. ### Final Conclusion - **Statement 1**: False (the function is not periodic). - **Statement 2**: False (the periodicity of \( g(x) \) does not imply the periodicity of \( f(g(x)) \)). ### Answer The correct option is: **4. Statement 1 is false, but statement 2 is true.**
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