Let a, b R be such that the function f given by `f(x)""=""ln""|x|""+""b x^2+""a x ,""x!=0` has extreme values at `x""=""1` and `x""=""2` . Statement 1: f has local maximum at `x""=""1` and at `x""=""2` . Statement 2: `a""=1/2"and"b=(-1)/4` (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false

Options

A. statement 1 is false statement 2 is true
B. statement 1 is true statement 2 is true , statement 2 is a correct explanation for statement 1
C. statement 1 is true statement 2 is true , statement 2 is not a correct explanation for statement 1
D. statement 1 is true statement 2 is false

Explanation

To solve the problem, we need to analyze the function \( f(x) = \ln |x| + b x^2 + a x \) and determine the values of \( a \) and \( b \) such that the function has extreme values at \( x = 1 \) and \( x = 2 \). We will also check the nature of these extreme values (whether they are maxima or minima).

### Step 1: Find the first derivative \( f'(x) \)
The first derivative of the function is given by:
\[ f'(x) = \frac{1}{x} + 2bx + a \]

### Step 2: Set the first derivative to zero at the extreme points
Since \( f \) has extreme values at \( x = 1 \) and \( x = 2 \), we set \( f'(1) = 0 \) and \( f'(2) = 0 \).

1. For \( x = 1 \):
\[ f'(1) = 1 + 2b + a = 0 \quad \text{(Equation 1)} \]

2. For \( x = 2 \):
\[ f'(2) = \frac{1}{2} + 4b + a = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
From Equation 1:
\[ a = -1 - 2b \quad \text{(Substituting in Equation 2)} \]

Substituting \( a \) into Equation 2:
\[ \frac{1}{2} + 4b - 1 - 2b = 0 \]
\[ \frac{1}{2} - 1 + 2b = 0 \]
\[ 2b = \frac{1}{2} \]
\[ b = \frac{1}{4} \]

Now substituting \( b \) back into Equation 1 to find \( a \):
\[ a = -1 - 2 \left(\frac{1}{4}\right) = -1 - \frac{1}{2} = -\frac{3}{2} \]

### Step 4: Verify the second derivative to determine the nature of the extreme values
Now we find the second derivative:
\[ f''(x) = -\frac{1}{x^2} + 2b \]
Substituting \( b = \frac{1}{4} \):
\[ f''(x) = -\frac{1}{x^2} + \frac{1}{2} \]

1. For \( x = 1 \):
\[ f''(1) = -1 + \frac{1}{2} = -\frac{1}{2} < 0 \quad \text{(Local maximum)} \]

2. For \( x = 2 \):
\[ f''(2) = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4} > 0 \quad \text{(Local minimum)} \]

### Conclusion
- Statement 1: \( f \) has a local maximum at \( x = 1 \) (True) and a local minimum at \( x = 2 \) (False).
- Statement 2: \( a = -\frac{3}{2} \) and \( b = \frac{1}{4} \) (False).

Thus, the correct option is:
(4) Statement 1 is true, statement 2 is false.