If the function y = m log x + `nx^(2) +x` has its critcal points at x = -1 and x =2 then the values of m and n respectively are

Options

A. `2 , (1)/(2)`
B. `- 2, (1)/(2)`
C. `2, - (1)/(2)`
D. `-2 , -(1)/(2)`

Answer

To solve the problem, we need to find the values of \( m \) and \( n \) for the function \( y = m \log x + nx^2 + x \) given that it has critical points at \( x = -1 \) and \( x = 2 \).

### Step-by-Step Solution:

1. **Understand Critical Points**:
Critical points occur where the derivative of the function is zero. Therefore, we first need to find the derivative of the function \( y \).

2. **Find the Derivative**:
The derivative of the function \( y \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{m}{x} + 2nx + 1 \]

3. **Set Derivative to Zero at Critical Points**:
We set the derivative equal to zero at the critical points \( x = -1 \) and \( x = 2 \).

- For \( x = -1 \):
\[ \frac{m}{-1} + 2n(-1) + 1 = 0 \]
Simplifying this gives:
\[ -m - 2n + 1 = 0 \quad \Rightarrow \quad m + 2n = 1 \quad \text{(Equation 1)} \]

- For \( x = 2 \):
\[ \frac{m}{2} + 2n(2) + 1 = 0 \]
Simplifying this gives:
\[ \frac{m}{2} + 4n + 1 = 0 \quad \Rightarrow \quad m + 8n = -2 \quad \text{(Equation 2)} \]

4. **Solve the System of Equations**:
Now we have a system of two equations:
\[ \begin{align*} m + 2n &= 1 \quad \text{(1)} \\ m + 8n &= -2 \quad \text{(2)} \end{align*} \]

Subtract Equation (1) from Equation (2):
\[ (m + 8n) - (m + 2n) = -2 - 1 \]
This simplifies to:
\[ 6n = -3 \quad \Rightarrow \quad n = -\frac{1}{2} \]

5. **Substitute \( n \) back to find \( m \)**:
Substitute \( n = -\frac{1}{2} \) back into Equation (1):
\[ m + 2(-\frac{1}{2}) = 1 \]
Simplifying gives:
\[ m - 1 = 1 \quad \Rightarrow \quad m = 2 \]

6. **Final Values**:
Therefore, the values of \( m \) and \( n \) are:
\[ m = 2, \quad n = -\frac{1}{2} \]

### Conclusion:
The values of \( m \) and \( n \) respectively are \( 2 \) and \( -\frac{1}{2} \).