Question

Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]

Hint:

For solving integrals involving functions, consider the fundamental theorem of calculus and utilize the properties of the function to simplify the given conditions.

Answer 1

Correct Answer: 4

Step 1: Analyze the given integral equation \( \int_0^a f(x) \, dx = f(a) \) and use it to deduce the form of \( f(x) \).
Step 2: Use the given values \( f(1) = 1 \) and \( f(16) = \frac{1}{8} \) to solve for \( f^{-1}\left( \frac{1}{16} \right) \).
Step 3: Substituting in the given equation, we find \( 16 - f^{-1}\left( \frac{1}{16} \right) = 4 \).