Suppose f(x) is a function satisfying the folowing conditions:
(i)f(0)=2,f(1)=1
(ii) f(x) has a minimum value at `x=(5)/(2)`
(iii) for all x f'(x)= `|{:(2ax,2ax-1,2ax+b+1),(b,b+1,-1),(2(ax+b),2ax+2b+1,2ax+b):}|`
The value of f(2)+f(3) is

Options

A. 1
B. `(3)/(2)`
C. 2
D. `(5)/(2)`

Answer

To solve the problem step by step, we will follow the conditions given for the function \( f(x) \) and its derivative \( f'(x) \).

### Step 1: Analyze the given conditions
We know:
1. \( f(0) = 2 \)
2. \( f(1) = 1 \)
3. \( f(x) \) has a minimum at \( x = \frac{5}{2} \)
4. The derivative \( f'(x) \) is given by a determinant.

### Step 2: Simplify the determinant
The expression for \( f'(x) \) is given as:
\[ f'(x) = \left| \begin{array}{ccc} 2ax & 2ax - 1 & 2ax + b + 1 \\ b & b + 1 & -1 \\ 2(ax + b) & 2ax + 2b + 1 & 2ax + b \end{array} \right| \]

We will perform column operations on the determinant:
- Replace \( C_2 \) with \( C_2 - C_1 \)
- Replace \( C_3 \) with \( C_3 - C_1 \)

After performing these operations, we get:
\[ f'(x) = \left| \begin{array}{ccc} 2ax & -1 & b + 1 \\ b & 1 & -1 \\ 2(ax + b) & 2b + 1 & b \end{array} \right| \]

### Step 3: Further simplify the determinant
Next, we will perform row operations:
- Replace \( R_2 \) with \( R_2 + R_1 \)
- Replace \( R_3 \) with \( R_3 + R_1 \)

This leads us to:
\[ f'(x) = \left| \begin{array}{ccc} 2ax & -1 & b + 1 \\ b + 2ax & 0 & b \\ 2(ax + b) + 2ax & 2b + 1 & b \end{array} \right| \]

### Step 4: Expand the determinant
Now we can expand the determinant. The first column can be factored out:
\[ f'(x) = 2ax \cdot \left| \begin{array}{cc} 0 & b \\ 2b + 1 & b \end{array} \right| + \text{other terms} \]

Calculating the determinant gives us:
\[ f'(x) = 2a \cdot (2ax + b) - 1 \]

### Step 5: Set the derivative to zero at the minimum point
Since \( f'(x) \) has a minimum at \( x = \frac{5}{2} \), we set:
\[ f'\left(\frac{5}{2}\right) = 0 \]
Substituting \( x = \frac{5}{2} \):
\[ 5a + b = 0 \quad \text{(Equation 1)} \]

### Step 6: Integrate to find \( f(x) \)
Integrating \( f'(x) \):
\[ f(x) = ax^2 + bx + c \]
Using the conditions:
1. \( f(0) = 2 \) gives \( c = 2 \)
2. \( f(1) = 1 \) gives \( a + b + 2 = 1 \) or \( a + b = -1 \quad \text{(Equation 2)} \)

### Step 7: Solve the equations
From Equation 1:
\[ b = -5a \]
Substituting into Equation 2:
\[ a - 5a = -1 \implies -4a = -1 \implies a = \frac{1}{4} \]
Then,
\[ b = -5 \cdot \frac{1}{4} = -\frac{5}{4} \]

### Step 8: Write the function
Thus, we have:
\[ f(x) = \frac{1}{4}x^2 - \frac{5}{4}x + 2 \]

### Step 9: Calculate \( f(2) \) and \( f(3) \)
Now we calculate \( f(2) \) and \( f(3) \):
\[ f(2) = \frac{1}{4}(2^2) - \frac{5}{4}(2) + 2 = \frac{1}{4}(4) - \frac{10}{4} + 2 = 1 - 2.5 + 2 = 0.5 \]
\[ f(3) = \frac{1}{4}(3^2) - \frac{5}{4}(3) + 2 = \frac{1}{4}(9) - \frac{15}{4} + 2 = 2.25 - 3.75 + 2 = 0.5 \]

### Step 10: Find \( f(2) + f(3) \)
Thus,
\[ f(2) + f(3) = 0.5 + 0.5 = 1 \]

### Final Answer
The value of \( f(2) + f(3) \) is \( \boxed{1} \).