Question

Let $f$ be a twice differentiable function defined on $R$ such that $f (0)=1, f'(0)=2$ and $f'(x)\neq 0$ for all $x \in R$. If $\left|\begin{array}{cc}f(x) & f'(x) \\ f'(x) & f''(x)\end{array}\right|=0,$ for all $x \in R,$ then the value of $f (1)$ lies in the interval:

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

Answer 1

The Correct Option is B

To solve the given problem, we first need to analyze the provided determinant condition and the properties of the function \( f(x) \). Given the determinant:

\[\left|\begin{array}{cc}f(x) & f'(x) \\ f'(x) & f''(x)\end{array}\right|=0\]

This determinant is zero if the following equality holds:

\(f(x) \cdot f''(x) - (f'(x))^2 = 0\)

Rearranging, we get:

\(f(x) \cdot f''(x) = (f'(x))^2\)

This implies that we can write:

\(\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}\)

Separating the variables and integrating both sides, we obtain:

\(\int \frac{d}{dx}(\ln|f'(x)|) \, dx = \int \frac{d}{dx}(\ln|f(x)|) \, dx\)

Thus, integrating gives:

\(\ln|f'(x)| = \ln|f(x)| + C\)

Exponentiating both sides leads to:

\(f'(x) = k f(x)\)

where \( k = e^C \). This suggests that \( f(x) \) is an exponential function. Solving the differential equation \( f'(x) = k f(x) \), we get:

\(f(x) = A e^{kx}\)

Using the conditions \( f(0)=1 \) and \( f'(0)=2 \), we determine \( A \) and \( k \):

• From \( f(0) = A e^{0} = 1 \Rightarrow A = 1 \).
• From \( f'(x) = k e^{kx} \) and \( f'(0)=2 \Rightarrow k \cdot 1 = 2 \Rightarrow k = 2 \).

Therefore, the function is:

\(f(x) = e^{2x}\)

Now, we calculate \( f(1) \):

\(f(1) = e^{2 \cdot 1} = e^2\)

Since \(\approx 2.718\), \( e^2 \approx 7.389\). This value lies within the interval (6, 9\)

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