The curve `y =f (x)` satisfies `(d^(2) y)/(dx ^(2))=6x-4 and f (x)` has a local minimum vlaue 5 when `x=1.` Then `f^(prime)(0)` is equal to :

Options

A. 1
B. 0
C. 5
D. None of these

Answer

To solve the problem, we need to find \( f'(0) \) given that the second derivative of the function \( y = f(x) \) is \( \frac{d^2y}{dx^2} = 6x - 4 \) and that \( f(x) \) has a local minimum value of 5 when \( x = 1 \).

### Step-by-Step Solution:

1. **Integrate the Second Derivative**:
We start with the second derivative:
\[ f''(x) = 6x - 4 \]
Now, we integrate \( f''(x) \) to find the first derivative \( f'(x) \):
\[ f'(x) = \int (6x - 4) \, dx = 3x^2 - 4x + C \]
where \( C \) is the constant of integration.

**Hint**: Remember that integrating a polynomial involves increasing the power by one and dividing by the new power.

2. **Use the Local Minimum Condition**:
We know that \( f(x) \) has a local minimum value of 5 when \( x = 1 \). This implies that:
\[ f'(1) = 0 \]
(since the derivative at a local minimum is zero) and \( f(1) = 5 \).

First, we will use \( f'(1) = 0 \):
\[ f'(1) = 3(1)^2 - 4(1) + C = 0 \]
Simplifying this gives:
\[ 3 - 4 + C = 0 \implies C = 1 \]

**Hint**: Set the derivative equal to zero at the point of the local minimum to find the constant.

3. **Substitute \( C \) Back into \( f'(x) \)**:
Now that we have \( C \), we can write:
\[ f'(x) = 3x^2 - 4x + 1 \]

**Hint**: Always substitute back the constant of integration to get the complete expression.

4. **Find \( f'(0) \)**:
Now we need to find \( f'(0) \):
\[ f'(0) = 3(0)^2 - 4(0) + 1 = 1 \]

**Hint**: Substitute \( x = 0 \) into the expression for \( f'(x) \) to find the derivative at that point.

### Final Answer:
Thus, \( f'(0) = 1 \).