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

Options

A. has finitely many discontinuities
B. is discontinuous at some x = c
C. is continuous on R
D. is a constant function

Answer

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

### Step 1: Understand the function \( g(x) \)

The function \( g(x) = x - [x] \) gives the fractional part of \( x \). This means:
- For any integer \( n \), \( g(n) = 0 \).
- For any non-integer \( x \), \( g(x) \) will be a value in the interval \( (0, 1) \).

### Step 2: Analyze the continuity of \( h(x) \)

We want to show that \( h(x) = f(g(x)) \) is continuous everywhere on \( \mathbb{R} \).

### Step 3: Consider the points where \( x \) is an integer

Let \( c \) be an integer. We will find the left-hand limit (LHL) and right-hand limit (RHL) of \( h(x) \) as \( x \) approaches \( c \).

- **Right-hand limit (RHL)**:
\[ \text{RHL} = \lim_{x \to c^+} h(x) = \lim_{x \to c^+} f(g(x)) = \lim_{x \to c^+} f(x - [x]) = \lim_{x \to c^+} f(x - c) = f(0) \]

- **Left-hand limit (LHL)**:
\[ \text{LHL} = \lim_{x \to c^-} h(x) = \lim_{x \to c^-} f(g(x)) = \lim_{x \to c^-} f(x - [x]) = \lim_{x \to c^-} f(x - c + 1) = f(1) \]

Since \( f(0) = f(1) \), we have:
\[ \text{RHL} = \text{LHL} = f(0) = f(1) \]

Thus, \( h(x) \) is continuous at integer points.

### Step 4: Consider the points where \( x \) is not an integer

Let \( c \) be a non-integer. For example, if \( c = 2.5 \):
- **Right-hand limit (RHL)**:
\[ \text{RHL} = \lim_{x \to 2.5^+} h(x) = \lim_{x \to 2.5^+} f(g(x)) = \lim_{x \to 2.5^+} f(x - [x]) = \lim_{x \to 2.5^+} f(x - 2) = f(0.5) \]

- **Left-hand limit (LHL)**:
\[ \text{LHL} = \lim_{x \to 2.5^-} h(x) = \lim_{x \to 2.5^-} f(g(x)) = \lim_{x \to 2.5^-} f(x - [x]) = \lim_{x \to 2.5^-} f(x - 2) = f(0.5) \]

Again, since both limits are equal, we find that \( h(x) \) is continuous at non-integer points as well.

### Conclusion

Since \( h(x) \) is continuous at both integer and non-integer points, we conclude that \( h(x) \) is continuous on \( \mathbb{R} \).

### Final Answer

The function \( h(x) = f(g(x)) \) is continuous on \( \mathbb{R} \).

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