Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a differentiable function such that \( f'(x)>f(x) \) for all \( x \in \mathbb{R} \), and \( f(0) = 1 \). Then \( f(1) \) lies in the interval

• A. \( (0, e^{-1}) \)
• B. \( (e^{-1}, \sqrt{e}) \)
• C. \( (\sqrt{e}, e) \)
• D. \( (e, \infty) \)

Hint:

When a function's derivative is greater than the function itself, the function grows faster than exponential growth.

Answer 1

The Correct Option is D

Step 1: Use the differential inequality.
We are given that \( f'(x)>f(x) \). This suggests that the growth rate of \( f(x) \) exceeds its own value. Integrating this inequality gives: \[ f(x)>f(0) e^x \text{ for all } x \in \mathbb{R}. \] Since \( f(0) = 1 \), we have \( f(x)>e^x \). In particular, for \( x = 1 \), we get: \[ f(1)>e. \]
Step 2: Conclusion.
Thus, \( f(1) \) must be greater than \( \sqrt{e} \) and less than \( e \), so the correct answer is \( \boxed{(C)} \).