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Step-by-Step Solution
Step 1: Write the determinant condition
We are given that
$$
\begin{vmatrix}
f(x) & f'(x) \\
f'(x) & f''(x)
\end{vmatrix}
= 0
\quad \text{for all } x \in \mathbb{R}.
$$
This expands to
$$
f(x) \, f''(x) \;-\; \bigl(f'(x)\bigr)^2 \;=\; 0.
$$
Step 2: Rearrange the expression
Rewriting the above equation by dividing both sides by $f(x)^2$ (assuming $f(x) \neq 0$), we get
$$
\frac{f(x)\,f''(x) - \bigl(f'(x)\bigr)^2}{\bigl(f(x)\bigr)^2} \;=\; 0
\;\;\;\Longrightarrow\;\;\;
\frac{d}{dx}\!\Bigl(\frac{f'(x)}{f(x)}\Bigr) \;=\; 0.
$$
The last step uses the fact that
$$
\frac{d}{dx}\!\Bigl(\frac{f'(x)}{f(x)}\Bigr)
=\frac{f(x) f''(x) - \bigl(f'(x)\bigr)^2}{\bigl(f(x)\bigr)^2}.
$$
Step 3: Integrate to find the relationship between f'(x) and f(x)
Since the derivative is zero, we have
$$
\frac{f'(x)}{f(x)} \;=\; c,
$$
where $c$ is a constant. This implies
$$
f'(x) \;=\; c\,f(x).
$$
Step 4: Use initial conditions to determine the constant
We know from the problem that $f(0)=1$ and $f'(0)=2$. Substituting $x=0$ in
$$
\frac{f'(x)}{f(x)} = c,
$$
gives
$$
c \;=\; \frac{f'(0)}{f(0)} \;=\; \frac{2}{1} \;=\; 2.
$$
Hence,
$$
f'(x) \;=\; 2\,f(x).
$$
Step 5: Solve the differential equation
The differential equation
$$
f'(x) = 2\,f(x)
$$
can be rewritten as
$$
\frac{f'(x)}{f(x)} = 2.
$$
Integrating both sides with respect to $x$ gives
$$
\int \frac{f'(x)}{f(x)} \, dx \;=\; \int 2 \, dx
\;\;\;\Longrightarrow\;\;\;
\ln \bigl|f(x)\bigr| \;=\; 2x + C',
$$
where $C'$ is the constant of integration.
Step 6: Apply the initial condition f(0)=1 to find C'
Substitute $x=0$ and $f(0)=1$:
$$
\ln|f(0)| \;=\; \ln(1) \;=\; 0
\quad\text{and}\quad
0 \;=\; 2\cdot0 + C'
\;\;\;\Longrightarrow\;\;\;
C' \;=\; 0.
$$
Therefore,
$$
\ln\bigl|f(x)\bigr| = 2x
\;\;\;\Longrightarrow\;\;\;
f(x) = e^{2x},
$$
since $f(x)$ is positive for $f(0)=1$.
Step 7: Evaluate f(1)
Using $f(x) = e^{2x}$, we find
$$
f(1) \;=\; e^{2}.
$$
We know that $e \approx 2.71828$, so $e^{2} \approx (2.71828)^2 \approx 7.389.
$$
This value lies between 6 and 9.
Conclusion
Therefore, $f(1)$ lies in the interval $(6, 9)$, matching the correct answer.
