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
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
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When a function's derivative is greater than the function itself, the function grows faster than exponential growth.
Updated On: Dec 11, 2025
- \( (0, e^{-1}) \)
- \( (e^{-1}, \sqrt{e}) \)
- \( (\sqrt{e}, e) \)
- \( (e, \infty) \)
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The Correct Option is D
Solution and Explanation
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 \( \boxed{(C)} \).
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 \( \boxed{(C)} \).
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