Question

Let \( f: [0, 1] \to \mathbb{R} \) be a function defined as
\[ f(t) = \begin{cases} t^3 \left( 1 + \frac{1}{5} \cos(\log(e^t)) \right), & \text{if } t \in (0,1] \\ 0, & \text{if } t = 0 \end{cases} \] Let \( F: [0, 1] \to \mathbb{R} \) be defined as
\[ F(x) = \int_0^x f(t) \, dt \] Then \( F''(0) \) equals

• A. 0
• B. \( \frac{3}{5} \)
• C. \( -\frac{5}{3} \)
• D. \( \frac{1}{5} \)

Hint:

When differentiating piecewise functions, carefully consider the behavior of the function at the boundaries.

Answer 1

The Correct Option is A

Step 1: Understanding the function.
The function \( f(t) \) is piecewise defined, with a nonzero value for \( t \in (0, 1] \) and 0 at \( t = 0 \). We need to compute the second derivative of \( F(x) \) at \( x = 0 \).
Step 2: Compute the first and second derivatives of \( F(x) \).
First, compute the first derivative \( F'(x) \) using the Fundamental Theorem of Calculus: \[ F'(x) = f(x) \] Then, compute the second derivative \( F''(x) \) by differentiating \( f(x) \). Since \( f(t) \) involves a cosine term that vanishes as \( t \to 0 \), we find that \( F''(0) = 0 \).