`bb"Statement I"` The function f(x) =xsinx and f'(x)=xcosx+sinx are both non-periodic.
`bb"Statement II"` The derivative of differentiable functions (non-periodic) is non-periodic funciton.

Options

A. (A) Statement I is true,Statement II is also true
B. (B) Statement I is false,Statement II is also false
C. (C)Statement I is true,Statement II is false
D. (D) Statement I is false ,Statement II is true

Answer

To solve the problem, we need to analyze the two statements given regarding the functions \( f(x) = x \sin x \) and its derivative \( f'(x) = x \cos x + \sin x \).

### Step 1: Check if \( f(x) = x \sin x \) is periodic
A function \( f(x) \) is periodic if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \).

1. Calculate \( f(x + T) \):
\[ f(x + T) = (x + T) \sin(x + T) \]
Using the sine addition formula:
\[ \sin(x + T) = \sin x \cos T + \cos x \sin T \]
Thus,
\[ f(x + T) = (x + T)(\sin x \cos T + \cos x \sin T) \]

2. Expand this expression:
\[ f(x + T) = (x \sin x \cos T + x \cos x \sin T) + (T \sin x \cos T + T \cos x \sin T) \]

3. Compare \( f(x + T) \) with \( f(x) = x \sin x \):
The terms involving \( T \) will not cancel out with \( f(x) \), indicating that \( f(x + T) \neq f(x) \) for any positive \( T \). Therefore, \( f(x) \) is **not periodic**.

### Step 2: Check if \( f'(x) = x \cos x + \sin x \) is periodic
Now we will check if \( f'(x) \) is periodic.

1. Calculate \( f'(x + T) \):
\[ f'(x + T) = (x + T) \cos(x + T) + \sin(x + T) \]
Using the cosine addition formula:
\[ \cos(x + T) = \cos x \cos T - \sin x \sin T \]
Thus,
\[ f'(x + T) = (x + T)(\cos x \cos T - \sin x \sin T) + (\sin x \cos T + \cos x \sin T) \]

2. Expand this expression:
\[ f'(x + T) = (x \cos x \cos T - x \sin x \sin T) + (T \cos x \cos T - T \sin x \sin T) + (\sin x \cos T + \cos x \sin T) \]

3. Again, the terms involving \( T \) will not cancel out with \( f'(x) \), indicating that \( f'(x + T) \neq f'(x) \) for any positive \( T \). Therefore, \( f'(x) \) is also **not periodic**.

### Conclusion for Statement I
Since both \( f(x) \) and \( f'(x) \) are non-periodic, **Statement I is true**.

### Step 3: Analyze Statement II
Statement II claims that the derivative of a non-periodic differentiable function is also non-periodic.

1. Consider a non-periodic function \( g(x) = x + \sin x \):
- This function is non-periodic as shown by checking \( g(x + T) \) against \( g(x) \).
- Calculate its derivative:
\[ g'(x) = 1 + \cos x \]

2. Check if \( g'(x) \) is periodic:
- Since \( \cos x \) is periodic with period \( 2\pi \), \( g'(x) = 1 + \cos x \) is also periodic with period \( 2\pi \).

### Conclusion for Statement II
Thus, Statement II is **false** because we found a non-periodic function whose derivative is periodic.

### Final Answer
- **Statement I**: True
- **Statement II**: False