Question

Let \( g(x) = f(x) + f(2 - x) \) for all \( x \in [0, 2] \), where \( f : [0, 2] \to \mathbb{R} \) is continuous on \( [0, 2] \) and twice differentiable on \( (0, 2) \). If \( g' \) denotes the first derivative of \( g \) and \( f'' \) denotes the second derivative of \( f \), then which of the following statements is NOT true?

• A. There exists \( c \in (0, 2) \) such that \( g'(c) = 0 \)
• B. If \( f''>0 \) on \( (0, 2) \), then \( g \) is strictly decreasing on \( (0, 1) \)
• C. If \( f''<0 \) on \( (0, 2) \), then \( g \) is strictly increasing on \( (1, 2) \)
• D. If \( f'' = 0 \) on \( (0, 2) \), then \( g \) is a constant function

Hint:

- The Mean Value Theorem guarantees a point where the derivative equals zero if the function is continuous and differentiable.
- If the second derivative of a function is positive, the function is concave up, and if it's negative, the function is concave down.

Answer 1

The Correct Option is C

1) Understanding the problem:
We are given that \( g(x) = f(x) + f(2 - x) \). Taking the first and second derivatives of \( g(x) \): \[ g'(x) = f'(x) - f'(2 - x) \] \[ g''(x) = f''(x) + f''(2 - x) \] 2) Analyzing each option:
- (A) There exists \( c \in (0, 2) \) such that \( g'(c) = 0 \): This is true by the Mean Value Theorem. Since \( g(x) \) is continuous on [0, 2] and differentiable on (0, 2), there exists a point where the derivative equals zero.

- (B) If \( f''>0 \) on \( (0, 2) \), then \( g \) is strictly decreasing on \( (0, 1) \): This is true. If \( f''>0 \), then \( f \) is concave up, and since \( g''(x) = f''(x) + f''(2 - x) \), \( g \) will be decreasing on \( (0, 1) \).

- (C) If \( f''<0 \) on \( (0, 2) \), then \( g \) is strictly increasing on \( (1, 2) \): This is NOT true. If \( f''<0 \), \( f \) is concave down, and we cannot conclude that \( g \) is increasing on \( (1, 2) \). The behavior of \( g \) depends on both \( f(x) \) and \( f(2 - x) \), and this does not guarantee that \( g \) is strictly increasing.

- (D) If \( f'' = 0 \) on \( (0, 2) \), then \( g \) is a constant function: This is true. If \( f'' = 0 \), \( f'(x) \) is constant, and \( g'(x) \) will also be constant. Hence, \( g(x) \) will be a linear function, and if \( g'(x) = 0 \), it will be a constant function.

Thus, the correct answer is (C).