If `f(x) = x+|x|+ cos ([ pi^(2) ]x)` and `g(x) =sin x,` where [.] denotes the greatest integer function, then

Options

A. f(x) + g(x) is continuous everywhere ,
B. f(x) + g(x) is differentiable everywhere
C. `f(x) xxg(x) ` is differentiable everywhere
D. `f(x) xx g(x) ` is continuous but not differentiable at x=0

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

\[ f(x) = x + |x| + \cos(\lfloor \pi^2 x \rfloor) \]
\[ g(x) = \sin x \]

where \( \lfloor x \rfloor \) denotes the greatest integer function.

### Step 1: Analyze \( f(x) \)

1. **Break down \( f(x) \)**:
- For \( x \geq 0 \):
\[ |x| = x \implies f(x) = x + x + \cos(\lfloor \pi^2 x \rfloor) = 2x + \cos(\lfloor \pi^2 x \rfloor) \]
- For \( x < 0 \):
\[ |x| = -x \implies f(x) = x - x + \cos(\lfloor \pi^2 x \rfloor) = \cos(\lfloor \pi^2 x \rfloor) \]

2. **Continuity of \( f(x) \)**:
- The function \( f(x) \) is continuous everywhere since both \( 2x \) and \( \cos(\lfloor \pi^2 x \rfloor) \) are continuous functions. The greatest integer function \( \lfloor x \rfloor \) is also continuous except at integer points, but since \( \cos \) is continuous, \( f(x) \) remains continuous.

### Step 2: Analyze \( g(x) \)

- The function \( g(x) = \sin x \) is continuous and differentiable everywhere.

### Step 3: Analyze \( f(x) + g(x) \)

- Since both \( f(x) \) and \( g(x) \) are continuous everywhere, their sum \( f(x) + g(x) \) is also continuous everywhere.

### Step 4: Analyze differentiability of \( f(x) + g(x) \)

- To check if \( f(x) + g(x) \) is differentiable everywhere, we note that \( f(x) \) is not differentiable at \( x = 0 \) because of the term \( |x| \).
- The derivative of \( |x| \) does not exist at \( x = 0 \) because:
\[ f'(0^+) = 1 \quad \text{and} \quad f'(0^-) = -1 \]
- Thus, \( f(x) + g(x) \) is not differentiable at \( x = 0 \).

### Step 5: Analyze \( f(x) \cdot g(x) \)

1. **Product of \( f(x) \) and \( g(x) \)**:
\[ f(x) \cdot g(x) = (x + |x| + \cos(\lfloor \pi^2 x \rfloor)) \cdot \sin x \]

2. **Continuity of \( f(x) \cdot g(x) \)**:
- Since both \( f(x) \) and \( g(x) \) are continuous everywhere, their product \( f(x) \cdot g(x) \) is also continuous everywhere.

3. **Differentiability of \( f(x) \cdot g(x) \)**:
- At \( x = 0 \):
- \( f(0) = 1 \) and \( g(0) = 0 \), so \( f(0) \cdot g(0) = 0 \).
- The left-hand derivative and right-hand derivative of \( f(x) \cdot g(x) \) at \( x = 0 \) will show that it is not differentiable at this point due to the non-differentiability of \( f(x) \).

### Conclusion

1. \( f(x) + g(x) \) is continuous everywhere but not differentiable at \( x = 0 \).
2. \( f(x) \cdot g(x) \) is continuous but not differentiable at \( x = 0 \).

### Final Answers:
- **Option 1**: \( f(x) + g(x) \) is continuous everywhere: **True**
- **Option 2**: \( f(x) + g(x) \) is differentiable everywhere: **False**
- **Option 3**: \( f(x) \cdot g(x) \) is continuous but not differentiable at \( x = 0 \): **True**