If `f (x)=a ln |x| +bx^(2) +x` has extremas at `x=1and x=3` then:

Options

A. `a =3/4, b=-1/8`
B. `a=3/4, b=1/8`
C. `a=-3/4,b =-1/8`
D. `a=-3/4, b=1/8`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) for the function \( f(x) = a \ln |x| + bx^2 + x \) given that it has extremas at \( x = 1 \) and \( x = 3 \).

### Step-by-Step Solution:

1. **Differentiate the Function**:
We start by finding the derivative \( f'(x) \).
\[ f'(x) = \frac{a}{x} + 2bx + 1 \]

2. **Set the Derivative to Zero**:
To find the extremas, we set the derivative equal to zero:
\[ f'(x) = 0 \implies \frac{a}{x} + 2bx + 1 = 0 \]
Multiplying through by \( x \) (since \( x > 0 \)):
\[ a + 2bx^2 + x = 0 \]

3. **Use the Known Extremas**:
We know that \( x = 1 \) and \( x = 3 \) are roots of the equation \( a + 2bx^2 + x = 0 \). Therefore, we can substitute these values into the equation.

- For \( x = 1 \):
\[ a + 2b(1)^2 + 1 = 0 \implies a + 2b + 1 = 0 \quad \text{(1)} \]

- For \( x = 3 \):
\[ a + 2b(3)^2 + 3 = 0 \implies a + 18b + 3 = 0 \quad \text{(2)} \]

4. **Set Up the System of Equations**:
Now we have a system of two equations:
\[ \begin{align*} a + 2b + 1 &= 0 \quad \text{(1)} \\ a + 18b + 3 &= 0 \quad \text{(2)} \end{align*} \]

5. **Subtract the Equations**:
We can eliminate \( a \) by subtracting equation (1) from equation (2):
\[ (a + 18b + 3) - (a + 2b + 1) = 0 \\ 18b - 2b + 3 - 1 = 0 \\ 16b + 2 = 0 \]
Solving for \( b \):
\[ 16b = -2 \implies b = -\frac{1}{8} \]

6. **Substitute Back to Find \( a \)**:
Now substitute \( b = -\frac{1}{8} \) back into equation (1):
\[ a + 2\left(-\frac{1}{8}\right) + 1 = 0 \\ a - \frac{1}{4} + 1 = 0 \\ a + \frac{3}{4} = 0 \implies a = -\frac{3}{4} \]

### Final Result:
The values of \( a \) and \( b \) are:
\[ a = -\frac{3}{4}, \quad b = -\frac{1}{8} \]