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 need to find the value of \(f(1)\) when the determinant of the following 2x2 matrix is zero: 

\(f(x)\)	\(f'(x)\)
\(f'(x)\)	\(f''(x)\)

The determinant condition implies:

\(\left| \begin{array}{cc} f(x) & f'(x) \\ f'(x) & f''(x) \end{array} \right| = f(x) \cdot f''(x) - (f'(x))^2 = 0\)

Rearranging gives us:

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

This implies that \(f(x)\) and \(f'(x)\) satisfy the differential equation:

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

By separating the variables and integrating both sides, we get:

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

This simplifies to:

\(\ln|f'(x)| = \ln|f(x)| + C\), where \(C\) is the constant of integration.

Exponentiating both sides, we obtain:

\(|f'(x)| = k|f(x)|\), where \(k = e^C\).

Based on the initial conditions provided: \(f(0) = 1\) and \(f'(0) = 2\), we determine:

\(2 = k \cdot 1 \Rightarrow k = 2\).

Thus, the functional relationship is:\(|f'(x)| = 2|f(x)|\).

Simplifying further, it becomes clear that:

\(f'(x) = 2f(x)\) (assuming \(f(x) > 0\) due to continuous differentiability and nonzero derivative constraint).

This is a standard separable differential equation, solved as follows:

\(\frac{df}{f} = 2 \, dx\)

Integrating both sides gives:

\(\ln |f(x)| = 2x + C_1\)

\(f(x) = C \cdot e^{2x}\), where \(C = e^{C_1}\).

Using initial condition \(f(0) = 1\), we find:

\(1 = C \cdot e^0 \Rightarrow C = 1\).

Therefore, the function is \(f(x) = e^{2x}\).

Substituting for \(x = 1\) gives:

\(f(1) = e^2 \approx 7.389\).

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