Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and `f '(x) ne 0` for all `x in R`. If `|[f(x)" "f'(x)], [f'(x)" "f''(x)]|= 0`, for all `x in R`, then the value of f(1) lies in the interval:

Options

A. (9, 12)
B. (6, 9)
C. (0, 3)
D. (3,6)

Answer

To solve the problem, we need to analyze the given conditions and derive the function \( f(x) \).

### Step 1: Understand the determinant condition
We are given that:
\[ \left| \begin{array}{cc} f'(x) & f'(x) \\ f'(x) & f''(x) \end{array} \right| = 0 \]
This determinant simplifies to:
\[ f'(x) \cdot f''(x) - f'(x) \cdot f'(x) = 0 \]
This can be rewritten as:
\[ f'(x) \cdot (f''(x) - f'(x)) = 0 \]
Since \( f'(x) \neq 0 \) for all \( x \in \mathbb{R} \), we must have:
\[ f''(x) - f'(x) = 0 \]
Thus, we can conclude:
\[ f''(x) = f'(x) \]

### Step 2: Solve the differential equation
The equation \( f''(x) = f'(x) \) is a second-order linear differential equation. The general solution is:
\[ f'(x) = C_1 e^x + C_2 \]
where \( C_1 \) and \( C_2 \) are constants.

### Step 3: Integrate to find \( f(x) \)
Integrating \( f'(x) \) gives:
\[ f(x) = C_1 e^x + C_2 x + C_3 \]
where \( C_3 \) is another constant of integration.

### Step 4: Use initial conditions
We know:
1. \( f(0) = 1 \)
2. \( f'(0) = 2 \)

Using \( f(0) = 1 \):
\[ f(0) = C_1 e^0 + C_2 \cdot 0 + C_3 = C_1 + C_3 = 1 \quad \text{(1)} \]

Using \( f'(0) = 2 \):
\[ f'(0) = C_1 e^0 + C_2 = C_1 + C_2 = 2 \quad \text{(2)} \]

### Step 5: Solve the system of equations
From equations (1) and (2):
1. \( C_1 + C_3 = 1 \)
2. \( C_1 + C_2 = 2 \)

Subtracting (1) from (2):
\[ C_2 - C_3 = 1 \quad \text{(3)} \]

From (1), we can express \( C_3 \):
\[ C_3 = 1 - C_1 \]
Substituting into (3):
\[ C_2 - (1 - C_1) = 1 \implies C_2 + C_1 - 1 = 1 \implies C_2 + C_1 = 2 \]
This is consistent with (2), so we can choose \( C_1 = 1 \) and \( C_2 = 1 \).

### Step 6: Find \( C_3 \)
Substituting \( C_1 = 1 \) into (1):
\[ 1 + C_3 = 1 \implies C_3 = 0 \]

### Step 7: Write the function
Thus, we have:
\[ f(x) = e^x + x \]

### Step 8: Evaluate \( f(1) \)
Now, we compute \( f(1) \):
\[ f(1) = e^1 + 1 = e + 1 \]
Since \( e \approx 2.718 \), we find:
\[ f(1) \approx 2.718 + 1 = 3.718 \]

### Step 9: Determine the interval
The value \( 3.718 \) lies in the interval \( (3, 4) \).

### Final Answer
The value of \( f(1) \) lies in the interval \( (3, 4) \).