Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be a differentiable function such that \( f(0) = 0 \) and \( f'(x) + 2f(x)>0 \) for all \( x \in \mathbb{R} \), where \( f' \) denotes the derivative of \( f \) and \( \mathbb{R} \) denotes the set of all real numbers. Then which one of the following statements is true?

• A. \( f(x)>0, \, \text{for all} \, x>0 \quad \text{and} \quad f(x)<0, \, \text{for all} \, x<0 \)
• B. \( f(x)<0, \, \text{for all} \, x \neq 0 \)
• C. \( f(x)>0, \, \text{for all} \, x \neq 0 \)
• D. \( f(x)<0, \, \text{for all} \, x>0 \quad \text{and} \quad f(x)>0, \, \text{for all} \, x<0 \)

Hint:

In solving differential inequalities, remember to consider the behavior of the solution as \( x \) moves away from zero to understand the sign changes.

Answer 1

The Correct Option is A

We are given the condition \( f'(x) + 2f(x)>0 \) for all \( x \in \mathbb{R} \). This inequality implies that \( f(x) \) behaves in a specific way. Let's solve this inequality to determine the behavior of \( f(x) \). 
Step 1: Solve the differential inequality. 
Rewrite the inequality as: \[ f'(x)>-2f(x). \] This is a first-order linear differential inequality. The solution to the corresponding equation \( f'(x) = -2f(x) \) is: \[ f(x) = Ce^{-2x}, \] where \( C \) is a constant determined by initial conditions. For the inequality \( f'(x) + 2f(x)>0 \), this solution shows that \( f(x) \) is positive for \( x>0 \) and negative for \( x<0 \), confirming that option (A) is correct. 
Final Answer: \[ \boxed{f(x)>0, \, \text{for all} \, x>0 \quad \text{and} \quad f(x)<0, \, \text{for all} \, x<0}. \]