Key Concepts and Formulas

• Determinant of a 2x2 Matrix: For a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $\det(A) = ad - bc$.
• Quotient Rule of Differentiation: The derivative of a quotient of two functions $u(x)$ and $v(x)$ is given by $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
• Logarithmic Differentiation: This technique can be useful when dealing with products or quotients of functions, or when exponents are involved. It involves taking the natural logarithm of both sides of an equation and then differentiating implicitly.

Step-by-Step Solution

Step 1: Interpret the given determinant equation. We are given that for all $x \in \mathbb{R}$, \left| {\matrix{ {f(x)} & {f'(x)} \cr {f'(x)} & {f''(x)} \cr } } \right| = 0 Using the formula for the determinant of a 2x2 matrix, this expands to: $f(x)f''(x) - (f'(x))^2 = 0$ This is a second-order ordinary differential equation.

Step 2: Rearrange the differential equation to identify a pattern. We can rewrite the equation as: $f(x)f''(x) = (f'(x))^2$ Since $f'(x) \ne 0$ for all $x \in \mathbb{R}$, we can divide both sides by $(f'(x))^2$ and by $f(x)$ (we need to confirm $f(x) \ne 0$). Let's rearrange it as: $\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$ This form is suggestive of derivatives of logarithmic functions.

Step 3: Integrate the rearranged differential equation. Consider the left side of the equation: $\frac{f''(x)}{f'(x)}$. We recognize that this is the derivative of $\ln|f'(x)|$ with respect to $x$, because $\frac{d}{dx}(\ln|f'(x)|) = \frac{1}{f'(x)} \cdot f''(x)$. Similarly, consider the right side of the equation: $\frac{f'(x)}{f(x)}$. We recognize that this is the derivative of $\ln|f(x)|$ with respect to $x$, because $\frac{d}{dx}(\ln|f(x)|) = \frac{1}{f(x)} \cdot f'(x)$. Therefore, the differential equation can be written as: $\frac{d}{dx}(\ln|f'(x)|) = \frac{d}{dx}(\ln|f(x)|)$ Integrating both sides with respect to $x$, we get: $\ln|f'(x)| = \ln|f(x)| + C$ where $C$ is the constant of integration.

Step 4: Solve for the relationship between $f(x)$ and $f'(x)$. Exponentiating both sides with base $e$: $e^{\ln|f'(x)|} = e^{\ln|f(x)| + C}$ $|f'(x)| = e^{\ln|f(x)|} \cdot e^C$ $|f'(x)| = |f(x)| \cdot K$ where $K = e^C$ is a positive constant. This implies: $f'(x) = \pm K f(x)$ Let $A = \pm K$. Then we have: $f'(x) = A f(x)$ This is a first-order linear homogeneous differential equation.

Step 5: Solve the first-order differential equation. The solution to $f'(x) = A f(x)$ is of the form $f(x) = B e^{Ax}$, where $B$ is another constant. We are given the initial conditions: $f(0) = 1$ and $f'(0) = 2$. Let's use these conditions to find the constants $A$ and $B$.

From $f(x) = B e^{Ax}$: $f(0) = B e^{A \cdot 0} = B e^0 = B \cdot 1 = B$. Since $f(0) = 1$, we have $B = 1$. So, $f(x) = e^{Ax}$.

Now, let's find $f'(x)$: $f'(x) = \frac{d}{dx}(e^{Ax}) = A e^{Ax}$.

Using the second initial condition, $f'(0) = 2$: $f'(0) = A e^{A \cdot 0} = A e^0 = A \cdot 1 = A$. Since $f'(0) = 2$, we have $A = 2$.

Therefore, the function is $f(x) = e^{2x}$. Let's verify if this function satisfies the original determinant equation and the initial conditions. $f(x) = e^{2x}$, $f'(x) = 2e^{2x}$, $f''(x) = 4e^{2x}$. $f(0) = e^0 = 1$ (Correct). $f'(0) = 2e^0 = 2$ (Correct). $f'(x) = 2e^{2x} \ne 0$ for all $x \in \mathbb{R}$ (Correct).

Determinant: $f(x)f''(x) - (f'(x))^2 = (e^{2x})(4e^{2x}) - (2e^{2x})^2 = 4e^{4x} - 4e^{4x} = 0$ (Correct).

Step 6: Calculate the value of f(1). Now that we have found the function $f(x) = e^{2x}$, we can find the value of $f(1)$: $f(1) = e^{2 \cdot 1} = e^2$.

Step 7: Determine the interval containing $f(1) = e^2$. We need to approximate the value of $e^2$. We know that $e \approx 2.718$. So, $e^2 \approx (2.718)^2$. Let's estimate: $2.7^2 = 7.29$ $2.71^2 \approx 7.3441$ $2.718^2 \approx 7.387524$

The value $e^2$ is approximately 7.389. Now let's look at the given options: (A) (0, 3) (B) (9, 12) (C) (3, 6) (D) (6, 9)

Our calculated value $e^2 \approx 7.389$ falls within the interval (6, 9).

Common Mistakes & Tips

• Forgetting the absolute value: When integrating $\frac{1}{u}$, the result is $\ln|u|$. While in this specific problem, $f(x)$ and $f'(x)$ are positive for $x \ge 0$ (since $f(0)=1$ and $f'(0)=2$, and they don't become zero), it's good practice to remember the absolute value.
• Incorrectly applying the quotient rule: Ensure the correct order of terms in the numerator of the quotient rule ($u'v - uv'$).
• Algebraic errors when solving the differential equation: Be careful with signs and exponents.

Summary

The problem requires solving a differential equation derived from a determinant. By recognizing the structure of the equation $f(x)f''(x) - (f'(x))^2 = 0$, we can rewrite it as $\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$. This form allows us to integrate both sides, leading to $\ln|f'(x)| = \ln|f(x)| + C$, which simplifies to $f'(x) = A f(x)$. Solving this first-order linear ODE with the given initial conditions $f(0)=1$ and $f'(0)=2$ yields the function $f(x) = e^{2x}$. Evaluating $f(1)$ gives $e^2$, which is approximately 7.389. This value lies in the interval (6, 9).

The final answer is \boxed{A}.