Let f be a twice differentiable defined on R such that `f(0)=1,f'(0)=2`. If `|[f(x),f'(x)],[f'(x),f''(x)]|=0 AA n in R`, then the value of `f(1)` lie in the interval

Options

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

Answer

To solve the problem, we need to analyze the given conditions and the determinant condition provided. Let's break it down step by step.

### Step 1: Understand the Determinant Condition
We are given that the determinant of the matrix formed by \( f(x) \), \( f'(x) \), and \( f''(x) \) is zero for all \( x \in \mathbb{R} \):
\[ \left| \begin{array}{cc} f(x) & f'(x) \\ f'(x) & f''(x) \end{array} \right| = 0 \]
This determinant can be expanded as:
\[ f(x) f''(x) - (f'(x))^2 = 0 \]
This implies:
\[ f(x) f''(x) = (f'(x))^2 \]

### Step 2: Rearranging the Equation
From the equation \( f(x) f''(x) = (f'(x))^2 \), we can rearrange it as:
\[ \frac{f''(x)}{(f'(x))^2} = \frac{1}{f(x)} \]

### Step 3: Integrating Both Sides
Now, we will integrate both sides with respect to \( x \):
\[ \int \frac{f''(x)}{(f'(x))^2} \, dx = \int \frac{1}{f(x)} \, dx \]
The left-hand side can be simplified using the substitution \( u = f'(x) \), leading to:
\[ \int \frac{1}{u} \, du = \ln |u| + C_1 = \ln |f'(x)| + C_1 \]
The right-hand side gives us:
\[ \ln |f(x)| + C_2 \]
Thus, we have:
\[ \ln |f'(x)| = \ln |f(x)| + C \]
where \( C = C_2 - C_1 \).

### Step 4: Exponentiating Both Sides
Exponentiating both sides gives us:
\[ |f'(x)| = K |f(x)| \]
where \( K = e^C \) is a constant.

### Step 5: Solving the Differential Equation
This leads us to the differential equation:
\[ \frac{f'(x)}{f(x)} = K \]
Integrating this gives:
\[ \ln |f(x)| = Kx + D \]
where \( D \) is another constant. Exponentiating again, we find:
\[ f(x) = Ae^{Kx} \]
where \( A = e^D \) is a constant.

### Step 6: Using Initial Conditions
We know:
1. \( f(0) = 1 \)
2. \( f'(0) = 2 \)

Using \( f(0) = 1 \):
\[ f(0) = Ae^{K \cdot 0} = A = 1 \]
Thus, \( f(x) = e^{Kx} \).

Now, using \( f'(x) = Ke^{Kx} \):
\[ f'(0) = K = 2 \]
Thus, \( K = 2 \).

### Step 7: Final Form of the Function
Now we have:
\[ f(x) = e^{2x} \]

### Step 8: Finding \( f(1) \)
Now we can find \( f(1) \):
\[ f(1) = e^{2 \cdot 1} = e^2 \]

### Step 9: Approximation of \( e^2 \)
We know that \( e \approx 2.718 \), hence:
\[ e^2 \approx 2.718^2 \approx 7.389 \]

### Conclusion
Thus, the value of \( f(1) \) lies in the interval \( (6, 9) \).