Question

Let f be a twice differentiable function defined on ℝ such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x ∈ ℝ. If $| f(x) \ f'(x) | \ | f'(x) \ f''(x) | = 0$, then the value of f(1) lies in the interval :

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

Hint:

The expression $f(x)f''(x) - [f'(x)]^2$ is the numerator of the derivative of $f'(x)/f(x)$. Setting it to zero implies $f(x)$ is an exponential function.

Answer 1

The Correct Option is C

Step 1: The determinant condition gives $f(x)f''(x) - (f'(x))^2 = 0$.
Step 2: This can be rewritten as $\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$.
Step 3: Integrate: $\ln(f'(x)) = \ln(f(x)) + \ln(C) \Rightarrow f'(x) = C \cdot f(x)$.
Step 4: Use $f(0)=1, f'(0)=2$: $2 = C(1) \Rightarrow C = 2$.
Step 5: Solve $\frac{f'(x)}{f(x)} = 2$: $\ln(f(x)) = 2x + C_1$. Since $f(0)=1, C_1=0$.
Step 6: $f(x) = e^{2x}$.
Step 7: $f(1) = e^2 \approx (2.718)^2 \approx 7.39$. This lies in $(6, 9)$.