`f (x) = int _(0) ^(x) e ^(t ^(3)) (t ^(2) -1) (t+1) ^(2011) dt (x gt 0)` then :

Options

A. The number of point iof inflections is atleast 1
B. The number of point of inflectins is 0
C. The number of point of local maxima is 1
D. The number of point of local minima is 1

Answer

To solve the problem step by step, we will analyze the function given and find its first and second derivatives to determine points of inflection and local maxima or minima.

### Step 1: Define the function
Given:
\[ f(x) = \int_{0}^{x} e^{t^3} (t^2 - 1) (t + 1)^{2011} \, dt \]

### Step 2: Find the first derivative \( f'(x) \)
Using the Fundamental Theorem of Calculus, we find the first derivative:
\[ f'(x) = e^{x^3} (x^2 - 1) (x + 1)^{2011} \]

### Step 3: Set the first derivative to zero to find critical points
To find points of inflection, we set \( f'(x) = 0 \):
\[ e^{x^3} (x^2 - 1) (x + 1)^{2011} = 0 \]
Since \( e^{x^3} \) is never zero, we focus on the other factors:
\[ (x^2 - 1) (x + 1)^{2011} = 0 \]
This gives us:
1. \( x^2 - 1 = 0 \) which implies \( x = 1 \) or \( x = -1 \)
2. \( (x + 1)^{2011} = 0 \) which implies \( x = -1 \)

Thus, the critical points are \( x = -1 \) and \( x = 1 \).

### Step 4: Analyze the sign of \( f'(x) \)
To determine the nature of these critical points, we check the sign of \( f'(x) \) around these points.

- For \( x < -1 \): Choose \( x = -2 \)
\[ f'(-2) = e^{-8}((-2)^2 - 1)((-2) + 1)^{2011} = e^{-8}(4 - 1)(-1)^{2011} < 0 \]

- For \( -1 < x < 1 \): Choose \( x = 0 \)
\[ f'(0) = e^{0} (0^2 - 1)(0 + 1)^{2011} = -1 < 0 \]

- For \( x > 1 \): Choose \( x = 2 \)
\[ f'(2) = e^{8} (2^2 - 1)(2 + 1)^{2011} = e^{8}(4 - 1)(3) > 0 \]

Thus, \( f'(x) \) changes from negative to positive at \( x = 1 \), indicating a local minimum.

### Step 5: Find the second derivative \( f''(x) \)
To find the second derivative, we differentiate \( f'(x) \):
\[ f''(x) = \frac{d}{dx} \left( e^{x^3} (x^2 - 1) (x + 1)^{2011} \right) \]
Using the product rule:
\[ f''(x) = e^{x^3} \left( (x^2 - 1) \frac{d}{dx}((x + 1)^{2011}) + (x + 1)^{2011} \frac{d}{dx}(x^2 - 1) + (x^2 - 1)(x + 1)^{2011} \cdot 3x^2 \right) \]

### Step 6: Evaluate \( f''(0) \)
Substituting \( x = 0 \) into \( f''(x) \):
\[ f''(0) = e^{0} \left( (0^2 - 1) \cdot 2011 \cdot 1 + 0 + 0 \right) = -2011 < 0 \]
This indicates that \( x = 0 \) is a point of local maxima.

### Conclusion
- The function has one point of inflection at \( x = 0 \).
- There is one local maximum at \( x = 0 \).