`f : R -> R` is one-one, onto and differentiable and graph of y = f (x) is symmetrical about the point (4, 0), then

Options

A. `f^(-1)(2010) + f^(-1) (-2010) = 8`
B. `int_(-2010)^(2018) f(x) dx = 0`
C. if `f'(-100) gt 0`, then roots of `x^(2) - f'(10) x - f'(10) = 0` may be non-real
D. if `f'(10) = 20`, then f'(-2) = 20

Answer

To solve the problem, we need to analyze the properties of the function \( f \) given that it is one-one, onto, differentiable, and symmetric about the point \( (4, 0) \).

### Step-by-Step Solution:

1. **Understanding Symmetry**:
The graph of \( y = f(x) \) is symmetric about the point \( (4, 0) \). This means that for any point \( (x, f(x)) \) on the graph, there exists a corresponding point \( (8 - x, -f(x)) \) that is also on the graph. This gives us the relationship:
\[ f(8 - x) = -f(x) \]

2. **Finding Inverse Relationships**:
Since \( f \) is one-one and onto, it has an inverse \( f^{-1} \). From the symmetry, we can derive:
\[ f^{-1}(y) + f^{-1}(-y) = 8 \]
This implies that:
\[ f^{-1}(2010) + f^{-1}(-2010) = 8 \]

3. **Using the Inverse Relationship**:
Let \( a = f^{-1}(2010) \) and \( b = f^{-1}(-2010) \). From the previous step, we have:
\[ a + b = 8 \]

4. **Finding Specific Values**:
We can express \( b \) in terms of \( a \):
\[ b = 8 - a \]
Now, substituting \( a = f^{-1}(2010) \) and \( b = f^{-1}(-2010) \):
\[ f^{-1}(2010) + f^{-1}(-2010) = 8 \]

5. **Evaluating the Integral**:
We need to evaluate the integral:
\[ \int_{-2010}^{2018} f(x) \, dx \]
By the symmetry property, we can transform the limits:
\[ \int_{-2010}^{2018} f(x) \, dx = \int_{-2010}^{2010} f(x) \, dx + \int_{2010}^{2018} f(x) \, dx \]
The first integral can be evaluated using the symmetry property, which will yield zero because the area above the x-axis will cancel out the area below the x-axis.

6. **Analyzing the Derivative**:
For the third option, we need to analyze the roots of the equation:
\[ x^2 - f'(10)x - f'(10) = 0 \]
The discriminant \( D \) of this quadratic equation is given by:
\[ D = (f'(10))^2 - 4(-f'(10)) \]
Since \( f'(10) > 0 \), the discriminant is positive, indicating that the roots are real.

7. **Final Derivative Evaluation**:
For the last option, we have:
\[ f'(10) = 20 \]
Using the symmetry property again, we can conclude that:
\[ f'(-2) = 20 \]

### Conclusion:
The correct options based on the analysis are:
- \( f^{-1}(2010) + f^{-1}(-2010) = 8 \)
- The integral \( \int_{-2010}^{2018} f(x) \, dx = 0 \)
- The roots of the quadratic equation are real.
- \( f'(-2) = 20 \)