Suppose the function `g_n(x)=x^(2n+1)+a_n x+b_n(N in N)` satisfes the equation `int_-1^1 (px + q)g_n(x)dx=0` for all linear functions `(px+q)` then

Options

A. `a_(n)=b_(n)=0`
B. `b_(n)=0,a_(n)=-(3)/(2n+3)`
C. `a_(n)=0,b_(n)=-(3)/(2n+3)`
D. `a_(n)=(3)/(2n+3),b_(n)=-(3)/(2n+3)`

Answer

To solve the problem, we need to analyze the given function \( g_n(x) = x^{2n+1} + a_n x + b_n \) and the condition that the integral

\[ \int_{-1}^{1} (px + q) g_n(x) \, dx = 0 \]

holds for all linear functions \( (px + q) \).

### Step-by-Step Solution:

1. **Set Up the Integral**:
We start with the integral:
\[ I = \int_{-1}^{1} (px + q) g_n(x) \, dx = \int_{-1}^{1} (px + q) (x^{2n+1} + a_n x + b_n) \, dx \]

2. **Expand the Integral**:
Distributing \( (px + q) \) gives:
\[ I = \int_{-1}^{1} \left( px \cdot x^{2n+1} + q \cdot x^{2n+1} + p a_n x^2 + q a_n x + p b_n x + q b_n \right) \, dx \]
This can be rewritten as:
\[ I = p \int_{-1}^{1} x^{2n+2} \, dx + q \int_{-1}^{1} x^{2n+1} \, dx + p a_n \int_{-1}^{1} x^2 \, dx + q a_n \int_{-1}^{1} x \, dx + p b_n \int_{-1}^{1} x \, dx + q b_n \int_{-1}^{1} 1 \, dx \]

3. **Evaluate the Integrals**:
- The integral \( \int_{-1}^{1} x^{2n+1} \, dx = 0 \) (since it is an odd function).
- The integral \( \int_{-1}^{1} x^{2n+2} \, dx = \frac{2}{2n+3} \).
- The integral \( \int_{-1}^{1} x^2 \, dx = \frac{2}{3} \).
- The integral \( \int_{-1}^{1} x \, dx = 0 \) (since it is an odd function).
- The integral \( \int_{-1}^{1} 1 \, dx = 2 \).

Thus, we have:
\[ I = p \cdot \frac{2}{2n+3} + 0 + p a_n \cdot \frac{2}{3} + 0 + 0 + q b_n \cdot 2 \]
Simplifying gives:
\[ I = \frac{2p}{2n+3} + \frac{2p a_n}{3} + 2q b_n \]

4. **Set the Integral to Zero**:
Since \( I = 0 \) for all \( p \) and \( q \), we can separate the coefficients of \( p \) and \( q \):
\[ \frac{2}{2n+3} + \frac{2a_n}{3} = 0 \quad \text{(coefficient of } p\text{)} \]
\[ 2b_n = 0 \quad \text{(coefficient of } q\text{)} \]

5. **Solve for \( b_n \)**:
From \( 2b_n = 0 \):
\[ b_n = 0 \]

6. **Solve for \( a_n \)**:
From \( \frac{2}{2n+3} + \frac{2a_n}{3} = 0 \):
\[ \frac{2a_n}{3} = -\frac{2}{2n+3} \]
Multiplying both sides by \( 3 \):
\[ 2a_n = -\frac{6}{2n+3} \]
Thus:
\[ a_n = -\frac{3}{2n+3} \]

### Conclusion:
The values we have found are:
- \( b_n = 0 \)
- \( a_n = -\frac{3}{2n+3} \)

Thus, the correct option is the one that states \( b_n = 0 \) and \( a_n = -\frac{3}{2n+3} \).