Question

Let \( f : [0,1] \to [0, \infty) \) be a continuous function such that \[ (f(t))^2<1 + 2 \int_0^t f(s)\,ds, \quad \text{for all } t \in [0,1]. \] Then

• A. \(f(t)<1 + t\) for all \(t \in [0,1].\)
• B. \(f(t)>1 + t\) for all \(t \in [0,1].\)
• C. \(f(t) = 1 + t\) for all \(t \in [0,1].\)
• D. \(f(t)<1 + \frac{t}{2}\) for all \(t \in [0,1].\)

Hint:

When an integral inequality resembles a differential equation, solve the equality case to establish an upper or lower bound using comparison arguments.

Answer 1

The Correct Option is A

Step 1: Compare with equality case.
Let \(g(t)\) satisfy the equality \[ (g(t))^2 = 1 + 2 \int_0^t g(s)\,ds. \] Differentiate both sides: \[ 2g(t)g'(t) = 2g(t) \Rightarrow g'(t) = 1, \ g(0) = 1 \Rightarrow g(t) = 1 + t. \]
Step 2: Use comparison principle.
Since \(f(t)\) satisfies a strict inequality \[ f(t)^2<1 + 2 \int_0^t f(s)\,ds, \] it must stay strictly below the corresponding equality solution \(g(t)\) for all \(t\).
Step 3: Conclusion.
Hence, (A) \(f(t)<1 + t\) for all \(t \in [0,1]\).