Let [x] denote the integral part of `x in R` and `g(x)=x -[x]`. Let `f(x)` be any continous function with f(0) =f(1), then the function `h(x) = f(g(x))`

Options

A. has finitely many discontinous
B. is continous on R
C. is discontinous at some x=c
D. is a constant function.

Answer

To solve the problem, we need to analyze the function \( h(x) = f(g(x)) \), where \( g(x) = x - [x] \) (the fractional part of \( x \)), and \( f(x) \) is a continuous function with \( f(0) = f(1) \).

### Step-by-Step Solution:

1. **Understanding \( g(x) \)**:
- The function \( g(x) = x - [x] \) gives the fractional part of \( x \). This means \( g(x) \) will always yield a value in the interval \([0, 1)\) for any real number \( x \).
- For example, if \( x = 2.3 \), then \( g(2.3) = 2.3 - 2 = 0.3 \).

**Hint**: Remember that the fractional part of any number is always between 0 and 1.

2. **Behavior of \( h(x) \)**:
- The function \( h(x) = f(g(x)) \) means we are applying \( f \) to the fractional part of \( x \).
- Since \( g(x) \) maps every \( x \) to \([0, 1)\), \( h(x) \) will take values of \( f \) only in the interval \([0, 1)\).

**Hint**: \( h(x) \) will depend on how \( f \) behaves between 0 and 1.

3. **Continuity of \( f \)**:
- Given that \( f \) is continuous and \( f(0) = f(1) \), we can conclude that as \( g(x) \) approaches 1 (which happens as \( x \) approaches any integer), \( f(g(x)) \) will approach \( f(0) \).
- However, at every integer \( n \), \( g(n) = 0 \), hence \( h(n) = f(0) \).

**Hint**: The continuity of \( f \) at the endpoints of the interval \([0, 1)\) is crucial.

4. **Discontinuities of \( h(x) \)**:
- At every integer \( n \), \( h(n) = f(0) \), but just before reaching \( n \) (for values like \( n - 0.1 \)), \( h(n - 0.1) \) will be \( f(g(n - 0.1)) = f(0.9) \) (or some value close to \( f(1) \)).
- Since \( f(0) \) and \( f(1) \) are equal, but the values of \( h(x) \) just before \( n \) and at \( n \) can differ, \( h(x) \) is discontinuous at every integer.

**Hint**: Check the behavior of \( h(x) \) around integers to identify discontinuities.

5. **Conclusion**:
- Since \( h(x) \) is discontinuous at every integer and there are infinitely many integers, we conclude that \( h(x) \) has infinitely many points of discontinuity.

### Final Answer:
The function \( h(x) = f(g(x)) \) has infinitely many discontinuities.