If the rate of increase of `(x^(2))/(2)-2x+5` is twice the rate of decrease of it then x is

Options

A. 2
B. 3
C. 4
D. 1

Answer

To solve the problem, we need to find the value of \( x \) such that the rate of increase of the function \( y = \frac{x^2}{2} - 2x + 5 \) is twice the rate of decrease of the same function.

### Step-by-Step Solution:

1. **Define the function**:
\[ y = \frac{x^2}{2} - 2x + 5 \]

2. **Find the derivative of \( y \)**:
To find the rate of change of \( y \) with respect to \( x \), we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{2}\right) - \frac{d}{dx}(2x) + \frac{d}{dx}(5) \]
\[ \frac{dy}{dx} = x - 2 \]

3. **Express the rate of increase and decrease**:
- The rate of increase of \( y \) is given by:
\[ \text{Rate of increase} = \frac{dy}{dt} = (x - 2) \frac{dx}{dt} \]
- The rate of decrease of \( y \) is:
\[ \text{Rate of decrease} = -\frac{dy}{dt} = -(x - 2) \frac{dx}{dt} \]

4. **Set up the equation based on the given condition**:
According to the problem, the rate of increase is twice the rate of decrease:
\[ (x - 2) \frac{dx}{dt} = 2 \left(- (x - 2) \frac{dx}{dt}\right) \]

5. **Simplify the equation**:
This simplifies to:
\[ (x - 2) \frac{dx}{dt} = -2(x - 2) \frac{dx}{dt} \]
If \( \frac{dx}{dt} \neq 0 \), we can divide both sides by \( \frac{dx}{dt} \):
\[ x - 2 = -2(x - 2) \]

6. **Solve for \( x \)**:
Expanding the right side:
\[ x - 2 = -2x + 4 \]
Rearranging gives:
\[ x + 2x = 4 + 2 \]
\[ 3x = 6 \]
\[ x = 2 \]

### Conclusion:
The value of \( x \) is \( 2 \).