If `f(x) = 2^(|x|)` where `[x]` denotes the fractional part of x. Then which of the following is true ?

Options

A. f is periodic
B. `underset(0)overset(1)int2^([x])dx = 1/(ln2)`
C. `underset(0)overset(1)int2^([x]) dx = log_(2)e`
D. `underset(0)overset(100)int2^([x]) dx = 100 log_(2) e`

Answer

To solve the problem, we need to analyze the function \( f(x) = 2^{\{x\}} \), where \( \{x\} \) denotes the fractional part of \( x \). The fractional part of \( x \) is defined as \( \{x\} = x - \lfloor x \rfloor \), which means it takes values in the interval \([0, 1)\).

### Step-by-Step Solution:

**Step 1: Determine the range of the function \( f(x) \)**

Since the fractional part \( \{x\} \) varies from \( 0 \) to \( 1 \), we can evaluate \( f(x) \):
\[ f(x) = 2^{\{x\}} \]
When \( \{x\} = 0 \), \( f(x) = 2^0 = 1 \).
When \( \{x\} \) approaches \( 1 \), \( f(x) \) approaches \( 2^1 = 2 \).
Thus, the range of \( f(x) \) is \( [1, 2) \).

**Step 2: Check if the function is periodic**

The function \( f(x) \) is periodic with a period of \( 1 \) because \( f(x + 1) = f(x) \) for any integer \( n \):
\[ f(x + n) = 2^{\{x + n\}} = 2^{\{x\}} = f(x) \]
This confirms that \( f(x) \) is periodic.

**Step 3: Evaluate the integral \( \int_0^1 f(x) \, dx \)**

We can express the integral as:
\[ \int_0^1 f(x) \, dx = \int_0^1 2^{\{x\}} \, dx \]
Since \( \{x\} = x \) in the interval \( [0, 1) \), we have:
\[ \int_0^1 2^{\{x\}} \, dx = \int_0^1 2^x \, dx \]
Calculating this integral:
\[ \int 2^x \, dx = \frac{2^x}{\ln(2)} \]
Evaluating from \( 0 \) to \( 1 \):
\[ \left[ \frac{2^x}{\ln(2)} \right]_0^1 = \frac{2^1}{\ln(2)} - \frac{2^0}{\ln(2)} = \frac{2}{\ln(2)} - \frac{1}{\ln(2)} = \frac{1}{\ln(2)} \]

**Step 4: Evaluate the integral \( \int_0^{100} f(x) \, dx \)**

We can break this integral into intervals of length \( 1 \):
\[ \int_0^{100} f(x) \, dx = \sum_{n=0}^{99} \int_n^{n+1} f(x) \, dx = \sum_{n=0}^{99} \int_0^1 f(x) \, dx \]
Since \( f(x) \) is periodic:
\[ = 100 \cdot \int_0^1 f(x) \, dx = 100 \cdot \frac{1}{\ln(2)} \]

### Conclusion:

Based on the evaluations, we can conclude that:
1. The function \( f(x) \) is periodic.
2. The integral \( \int_0^1 f(x) \, dx = \frac{1}{\ln(2)} \).
3. The integral \( \int_0^{100} f(x) \, dx = \frac{100}{\ln(2)} \).

Thus, all the options provided in the question are correct.