Consider `f'(x)=(x^(2))/(2)-kx+1` such that `f(0)=0` and `f(3)=15`.
The value of k is

Options

A. `(5)/(3)`
B. `(3)/(5)`
C. `-(5)/(3)`
D. `-(3)/(5)`

Answer

To find the value of \( k \) given the function \( f'(x) = \frac{x^2}{2} - kx + 1 \), and the conditions \( f(0) = 0 \) and \( f(3) = 15 \), we can follow these steps:

### Step 1: Integrate \( f'(x) \)
We start by integrating \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int f'(x) \, dx = \int \left( \frac{x^2}{2} - kx + 1 \right) \, dx \]

### Step 2: Perform the integration
Now we perform the integration term by term:
\[ f(x) = \frac{x^3}{6} - \frac{kx^2}{2} + x + C \]
where \( C \) is the constant of integration.

### Step 3: Use the condition \( f(0) = 0 \)
Substituting \( x = 0 \) into \( f(x) \):
\[ f(0) = \frac{0^3}{6} - \frac{k \cdot 0^2}{2} + 0 + C = C \]
Since \( f(0) = 0 \), we have:
\[ C = 0 \]
Thus, the function simplifies to:
\[ f(x) = \frac{x^3}{6} - \frac{kx^2}{2} + x \]

### Step 4: Use the condition \( f(3) = 15 \)
Now we substitute \( x = 3 \) into \( f(x) \):
\[ f(3) = \frac{3^3}{6} - \frac{k \cdot 3^2}{2} + 3 \]
Calculating \( f(3) \):
\[ f(3) = \frac{27}{6} - \frac{9k}{2} + 3 \]
This simplifies to:
\[ f(3) = 4.5 - \frac{9k}{2} + 3 = 7.5 - \frac{9k}{2} \]
Setting this equal to 15:
\[ 7.5 - \frac{9k}{2} = 15 \]

### Step 5: Solve for \( k \)
Rearranging the equation:
\[ -\frac{9k}{2} = 15 - 7.5 \]
\[ -\frac{9k}{2} = 7.5 \]
Multiplying both sides by -2:
\[ 9k = -15 \]
Dividing both sides by 9:
\[ k = -\frac{15}{9} = -\frac{5}{3} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{-\frac{5}{3}} \]