If If `g(x)=int_0^x f(t)dt` where `f(x)` is a continuous functions on [0,3] such that `forall x in [0,1], f(x)` has ranges `[1/3,1] and forall x in (1,3] , f(x)` has ranges `[0,1/2]` Then the maximum range in which `g(x)` can lie is

Options

A. `[1/3,2]`
B. `(1,2)`
C. `[5/6,3]`
D. `[1,4/3]`

Answer

To solve the problem, we need to find the maximum range of the function \( g(x) = \int_0^x f(t) \, dt \) given the constraints on \( f(x) \).

### Step-by-step Solution:

1. **Understanding the Function \( g(x) \)**:
- The function \( g(x) \) is defined as the integral of \( f(t) \) from 0 to \( x \).
- Since \( f(x) \) is continuous on the interval \([0, 3]\), \( g(x) \) will also be continuous.

2. **Analyzing the Range of \( f(x) \)**:
- For \( x \in [0, 1] \), \( f(x) \) has a range of \([1/3, 1]\).
- For \( x \in (1, 3]\), \( f(x) \) has a range of \([0, 1/2]\).

3. **Calculating \( g(3) \)**:
- We need to compute \( g(3) = \int_0^3 f(t) \, dt \).
- We can break this integral into two parts:
\[ g(3) = \int_0^1 f(t) \, dt + \int_1^3 f(t) \, dt \]

4. **Finding the Maximum and Minimum of Each Integral**:
- **For \( \int_0^1 f(t) \, dt \)**:
- The minimum value occurs when \( f(t) = 1/3 \) for the entire interval:
\[ \int_0^1 f(t) \, dt \geq \int_0^1 \frac{1}{3} \, dt = \frac{1}{3} \]
- The maximum value occurs when \( f(t) = 1 \) for the entire interval:
\[ \int_0^1 f(t) \, dt \leq \int_0^1 1 \, dt = 1 \]

- **For \( \int_1^3 f(t) \, dt \)**:
- The minimum value occurs when \( f(t) = 0 \) for the entire interval:
\[ \int_1^3 f(t) \, dt \geq \int_1^3 0 \, dt = 0 \]
- The maximum value occurs when \( f(t) = 1/2 \) for the entire interval:
\[ \int_1^3 f(t) \, dt \leq \int_1^3 \frac{1}{2} \, dt = \frac{1}{2} \times (3 - 1) = 1 \]

5. **Combining the Results**:
- Therefore, we have:
\[ g(3) \text{ can range from } \left(\frac{1}{3} + 0\right) \text{ to } (1 + 1) = 2 \]
- Thus, the minimum value of \( g(3) \) is \( \frac{1}{3} \) and the maximum value is \( 2 \).

6. **Conclusion**:
- The maximum range in which \( g(x) \) can lie is:
\[ \left[\frac{1}{3}, 2\right] \]