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 xf'(x)= `|{:(2ax,2ax-1,2ax+b+1),(b,b+1,-1),(2(ax+b),2ax+2b+1,2ax+b):}|`
The number of solutions of the equation f(x) +1=0 is

Options

A. (A)0
B. (B)1
C. (C)2
D. (D) infinite

Answer

To solve the problem, we need to analyze the function \( f(x) \) given the conditions provided. Let's break it down step by step.

### Step 1: Understanding the conditions
We know:
1. \( f(0) = 2 \)
2. \( f(1) = 1 \)
3. \( f(x) \) has a minimum value at \( x = \frac{5}{2} \)
4. The derivative \( f'(x) = |{(2ax, 2ax-1, 2ax+b+1), (b, b+1, -1), (2(ax+b), 2ax+2b+1, 2ax+b):}| \)

### Step 2: Analyzing the derivative
The determinant given for \( f'(x) \) can be expanded. We will denote the determinant as \( D \):
\[ D = \begin{vmatrix} 2ax & 2ax-1 & 2ax+b+1 \\ b & b+1 & -1 \\ 2(ax+b) & 2ax+2b+1 & 2ax+b \end{vmatrix} \]

### Step 3: Simplifying the determinant
To find \( f'(x) \), we can perform row operations on the determinant. The goal is to simplify it to find a manageable expression for \( f'(x) \).

1. **Row operations**:
- We can subtract the second row from the first and third rows to simplify the determinant.

### Step 4: Finding the function \( f(x) \)
From the conditions \( f(0) = 2 \) and \( f(1) = 1 \), we can assume a quadratic function of the form:
\[ f(x) = ax^2 + bx + c \]

Using the conditions:
1. \( f(0) = c = 2 \)
2. \( f(1) = a + b + 2 = 1 \) → \( a + b = -1 \)

### Step 5: Finding the coefficients
Since \( f(x) \) has a minimum at \( x = \frac{5}{2} \), we know:
\[ f'\left(\frac{5}{2}\right) = 0 \]
This gives us another equation to solve for \( a \) and \( b \).

### Step 6: Solving for \( a \) and \( b \)
From the derivative \( f'(x) = 2ax + b \):
1. Setting \( f'\left(\frac{5}{2}\right) = 0 \):
\[ 2a\left(\frac{5}{2}\right) + b = 0 \implies 5a + b = 0 \implies b = -5a \]

Substituting \( b = -5a \) into \( a + b = -1 \):
\[ a - 5a = -1 \implies -4a = -1 \implies a = \frac{1}{4} \]
Then,
\[ b = -5\left(\frac{1}{4}\right) = -\frac{5}{4} \]

### Step 7: Forming the function
Thus, we have:
\[ f(x) = \frac{1}{4}x^2 - \frac{5}{4}x + 2 \]

### Step 8: Finding the number of solutions for \( f(x) + 1 = 0 \)
We need to solve:
\[ \frac{1}{4}x^2 - \frac{5}{4}x + 1 = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ x^2 - 5x + 4 = 0 \]
Factoring:
\[ (x - 4)(x - 1) = 0 \]
Thus, the solutions are \( x = 4 \) and \( x = 1 \).

### Conclusion
The number of solutions to the equation \( f(x) + 1 = 0 \) is **2**.