Consider `f(x) =x^(2)+ax+3 and g(x) =x+band F(x) = lim_( n to oo) (f(x)+x^(2n)g(x))/(1+x^(2n))`
If F(x) is continuous at x=-1, then

Options

A. a+b=-2
B. a-b=3
C. a+b=5
D. none of these

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) and the limit that defines \( F(x) \). We will also use the condition of continuity at \( x = -1 \).

### Step-by-Step Solution:

1. **Define the Functions**:
We have:
\[ f(x) = x^2 + ax + 3 \]
\[ g(x) = x + b \]

2. **Define \( F(x) \)**:
The function \( F(x) \) is defined as:
\[ F(x) = \lim_{n \to \infty} \frac{f(x) + x^{2n} g(x)}{1 + x^{2n}} \]

3. **Analyze the Limit**:
We need to consider the behavior of the limit as \( n \) approaches infinity. The term \( x^{2n} \) will dominate depending on the value of \( x \):
- If \( |x| < 1 \), \( x^{2n} \to 0 \).
- If \( |x| = 1 \), \( x^{2n} \) remains 1.
- If \( |x| > 1 \), \( x^{2n} \to \infty \).

For \( |x| < 1 \):
\[ F(x) = \lim_{n \to \infty} \frac{f(x)}{1} = f(x) \]

For \( |x| > 1 \):
\[ F(x) = \lim_{n \to \infty} \frac{x^{2n} g(x)}{x^{2n}} = g(x) \]

4. **Evaluate at \( x = -1 \)**:
Since we are interested in continuity at \( x = -1 \), we need to evaluate \( F(-1) \):
\[ F(-1) = \lim_{n \to \infty} \frac{f(-1) + (-1)^{2n} g(-1)}{1 + (-1)^{2n}} \]
Here, \( (-1)^{2n} = 1 \) for all \( n \):
\[ F(-1) = \frac{f(-1) + g(-1)}{2} \]

5. **Calculate \( f(-1) \) and \( g(-1) \)**:
\[ f(-1) = (-1)^2 + a(-1) + 3 = 1 - a + 3 = 4 - a \]
\[ g(-1) = -1 + b \]

6. **Set Up the Equation for Continuity**:
For \( F(x) \) to be continuous at \( x = -1 \), we need:
\[ F(-1) = f(-1) = g(-1) \]
Thus,
\[ \frac{(4 - a) + (-1 + b)}{2} = 4 - a \]

7. **Simplify the Equation**:
Multiplying through by 2:
\[ (4 - a) + (-1 + b) = 2(4 - a) \]
Simplifying gives:
\[ 4 - a - 1 + b = 8 - 2a \]
Rearranging leads to:
\[ a + b = 5 \]

### Final Answer:
Thus, the condition for continuity at \( x = -1 \) gives us:
\[ \boxed{a + b = 5} \]