`f(x) : [0, 5] → R, F(x) = int_(0)^(x) x^2 g(x)`, f(1) = 3
g(x) = `int_(1)^(x) f(t) dt ` then correct choice is

Options

A. `F(x)` has local minimum at `x = 1`
B. `F(x)` has local maximum at `x = 1`
C. `F(x)` has point of inflection at `x = 1`
D. `F(x)` has no critical point

Answer

To solve the problem step by step, we need to analyze the given functions and their relationships.

### Step 1: Understand the functions
We are given:
- \( f(x) = \int_0^x x^2 g(x) \, dx \)
- \( g(x) = \int_1^x f(t) \, dt \)
- \( f(1) = 3 \)

### Step 2: Differentiate \( f(x) \)
To find the critical points, we first differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \int_0^x x^2 g(x) \, dx \right) \]
Using the Leibniz rule for differentiation under the integral sign, we get:
\[ f'(x) = x^2 g(x) \]

### Step 3: Evaluate \( f'(1) \)
Now, we substitute \( x = 1 \):
\[ f'(1) = 1^2 g(1) = g(1) \]
Next, we need to find \( g(1) \).

### Step 4: Evaluate \( g(1) \)
From the definition of \( g(x) \):
\[ g(1) = \int_1^1 f(t) \, dt = 0 \]
Thus, we find:
\[ f'(1) = g(1) = 0 \]

### Step 5: Differentiate \( f'(x) \) to find \( f''(x) \)
Now we differentiate \( f'(x) \) to find \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(x^2 g(x)) = 2x g(x) + x^2 g'(x) \]
Since \( g'(x) = f(x) \), we can rewrite this as:
\[ f''(x) = 2x g(x) + x^2 f(x) \]

### Step 6: Evaluate \( f''(1) \)
Substituting \( x = 1 \):
\[ f''(1) = 2 \cdot 1 \cdot g(1) + 1^2 \cdot f(1) = 2 \cdot 0 + 3 = 3 \]

### Step 7: Determine the nature of the critical point
Since \( f''(1) = 3 > 0 \), this indicates that \( f(x) \) has a local minimum at \( x = 1 \).

### Conclusion
The correct choice is that \( f(x) \) has a local minimum at \( x = 1 \).