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Statement-1 : f(x) = [{x}] is a periodic...

Statement-1 : `f(x) = [{x}]` is a periodic function with no fundamental period.
and
Statement-2 : f(g(x)) is periodic if g(x) is periodic.

A

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

B

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

C

Statement -1 is False, Statement -2 is False

D

Statement -1 is False, Statement -2 is True

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the given statements, we will evaluate each statement step by step. ### Statement 1: `f(x) = [{x}]` is a periodic function with no fundamental period. 1. **Understanding the Function**: - The function \( f(x) = \{x\} \) represents the fractional part of \( x \), which is defined as \( \{x\} = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). - The range of \( f(x) \) is \( [0, 1) \). 2. **Checking Periodicity**: - A function is periodic if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \). - For the function \( f(x) = \{x\} \), we can observe that: \[ f(x + 1) = \{x + 1\} = \{x\} \] - This shows that \( f(x) \) is periodic with a period of 1. 3. **No Fundamental Period**: - A function has no fundamental period if it is periodic but does not have the smallest positive period. - Since \( f(x) \) is periodic with period 1, but also periodic with any integer \( n \) (i.e., \( f(x + n) = f(x) \) for any integer \( n \)), it does not have a unique fundamental period. 4. **Conclusion for Statement 1**: - Therefore, Statement 1 is **correct**: \( f(x) = \{x\} \) is a periodic function with no fundamental period. ### Statement 2: \( f(g(x)) \) is periodic if \( g(x) \) is periodic. 1. **Understanding Periodicity of Composite Functions**: - Let \( g(x) \) be a periodic function with period \( T \), meaning \( g(x + T) = g(x) \) for all \( x \). - We need to determine if \( f(g(x)) \) is periodic. 2. **Applying the Definition of Periodicity**: - If \( f(g(x)) \) is periodic, there must exist a period \( P \) such that: \[ f(g(x + P)) = f(g(x)) \] - Since \( g(x) \) is periodic with period \( T \), we have: \[ g(x + T) = g(x) \] - Therefore: \[ f(g(x + T)) = f(g(x)) \] - This shows that \( f(g(x)) \) is periodic with period \( T \). 3. **Conclusion for Statement 2**: - Thus, Statement 2 is also **correct**: \( f(g(x)) \) is periodic if \( g(x) \) is periodic. ### Final Conclusion: - Both statements are correct, but Statement 2 does not provide an explanation for Statement 1. Therefore, we conclude: - Statement 1 is true. - Statement 2 is true. - Statement 2 is not a correct explanation for Statement 1.
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Each question has four choices, a,b,c and d,out of which only one is correct. Each question contains STATEMENT 1 and STATEMENT 2. if both the statements are true and statement 2 is the correct explanation of statement 1. If both the statements are true but statement 2 is not the correct explanation of statement 1. If statement is True and statement2 is false. If statement1 is false and statement2 is true. Consider the function f(x)=sin(k x)+{x}, where (x) represents the fractional part function. Statement 1 : f(x) is periodic for k=mpi, where m is a rational number Statement 2 : The sum of two periodic functions is always periodic.